4 Amplitude/Linear · PDF file4 Amplitude/Linear Modulation 4.1. ... De nition 4.2.Amplitude...

4 Amplitude/Linear Modulation

4.1. A sinusoidal carrier signal A cos(2πfct+φ) has three basic parameters:amplitude, frequency, and phase. Varying these parameters in proportionto the baseband signal results in amplitude modulation (AM), frequency16

modulation (FM), and phase modulation (PM), respectively. Collectively,these techniques are called continuous-wave (CW) modulation [13, p111][3, p 162].

Definition 4.2. Amplitude modulation is characterized by the fact thatthe amplitude A of the carrier A cos(2πfct + φ) is varied in proportion tothe baseband (message) signal m(t).

• Because the amplitude is time-varying, we may write the modulatedcarrier as

A(t) cos(2πfct+ φ)

• Because the amplitude is linearly related to the message signal, thistechnique is also called linear modulation.

4.3. Linear modulations:

(a) Double-sideband amplitude modulation

(i) Double-sideband-suppressed-carrier (DSB-SC or DSSC or simplyDSB) modulation

(ii) Standard amplitude modulation (AM)

(b) Suppressed-sideband amplitude modulation

(i) Single-sideband modulation (SSB)

(ii) Vestigial-sideband modulation (VSB)

16Technically, the variation of “frequency” is not as straightforward as the description here seems tosuggest. For a sinusoidal carrier, a general modulated carrier can be represented mathematically as

x(t) = A(t) cos (2πfct+ φ(t)) .

Frequency modulation, as we shall see later, is resulted from letting the time derivative of φ(t) be linearlyrelated to the modulating signal. [14, p 112]

35

4.1 Double-sideband suppressed carrier (DSB-SC) modulation

Definition 4.4. In double-sideband-suppressed-carrier (DSB-SC orDSSC or simply DSB) modulation, the modulated signal is

x(t) = Ac cos (2πfct)×m(t).

We have seen that the multiplication by a sinusoid gives two shifted andscaled replicas of the original signal spectrum:

X(f) =Ac

2M (f − fc) +

Ac

2M (f + fc) .

• When we set Ac =√

2, we get the “simple” modulator discussed inExample 3.12.

• We need fc > B to avoid spectral overlapping. In practice, fc B.

4.5. Synchronous/coherent detection by the product demodulator:The incoming modulated signal is first multiplied with a locally generatedsinusoid with the same phase and frequency (from a local oscillator (LO))and then lowpass-filtered, the filter bandwidth being the same as the mes-sage bandwidth B or somewhat larger.

4.6. A DSB-SC modem with no channel impairment is shown in Figure 12.

× ×Channel

2 cos 2 cf t

y

2 cos 2 cf t

vLPF

Modulator Demodulator

Message(modulating signal)

22

Figure 12: DSB-SC modem with no channel impairment

36

Once again, recall that

X (f) =√

2

(1

2(M (f − fc) +M (f + fc))

)=

1√2

(M (f − fc) +M (f + fc)) .

Similarly,

v (t) = y (t)×√

2 cos (2πfct) =√

2x (t) cos (2πfct)

V (f) =1√2

(X (f − fc) +X (f + fc))

Alternatively, we can work in the time domain and utilize the trig. iden-tity from Example 2.4:

v (t) =√

2x (t) cos (2πfct) =√

2(√

2m (t) cos (2πfct))

cos (2πfct)

= 2m (t) cos2 (2πfct) = m (t) (cos (2 (2πfct)) + 1)

= m (t) +m (t) cos (2π (2fc) t)

Key equation for DSB-SC modem:

LPF

(m (t)×

√2 cos (2πfct)

)︸︷︷︸

x(t)

×(√

2 cos (2πfct)) = m (t) . (31)

37

4.7. Implementation issues:

(a) Problem 1: Modulator construction

(b) Problem 2: Synchronization between the two (local) carriers/oscillators

(c) Problem 3: Spectral inefficiency

4.8. Spectral inefficiency/redundancy: When m(t) is real-valued, itsspectrum M(f) has conjugate symmetry. With such message, the corre-sponding modulated signal’s spectrum X(f) will also inherit the symmetrybut now centered at fc (instead of at 0). The portion that lies above fc isknown as the upper sideband (USB) and the portion that lies below fcis known as the lower sideband (LSB). Similarly, the spectrum centeredat −fc has upper and lower sidebands. Hence, this is a modulation schemewith double sidebands. Both sidebands contain the same (and complete)information about the message.

4.9. Synchronization: Observe that (31) requires that we can generatecos (ωct) both at the transmitter and receiver. This can be difficult in prac-tice. Suppose that the frequency at the receiver is off, say by ∆f , and thatthe phase is off by θ. The effect of these frequency and phase offsets can bequantified by calculating

LPF(m (t)

√2 cos (2πfct)

)√2 cos (2π (fc + ∆f) t+ θ)

,

which givesm (t) cos (2π (∆f) t+ θ) .

Of course, we want ∆ω = 0 and θ = 0; that is the receiver must generatea carrier in phase and frequency synchronism with the incoming carrier.

38

4.10. Effect of time delay:

Suppose the propagation time is τ , then we have

y (t) = x (t− τ) =√

2m (t− τ) cos (2πfc (t− τ))

=√

2m (t− τ) cos (2πfct− 2πfcτ)

=√

2m (t− τ) cos (2πfct− φτ) .

Consequently,

v (t) = y (t)×√

2 cos (2πfct)

=√

2m (t− τ) cos (2πfct− φτ)×√

2 cos (2πfct)

= m (t− τ) 2 cos (2πfct− φτ) cos (2πfct) .

Applying the product-to-sum formula, we then have

v (t) = m (t− τ) (cos (2π (2fc) t− φτ) + cos (φτ)) .

In conclusion, we have seen that the principle of the DSB-SC modem isbased on a simple key equation (31). However, as mentioned in 4.7, thereare several implementation issues that we need to address. Some solutionsare provided in the subsections to follow. However, the analysis will requiresome knowledge of Fourier series which is reviewed in the next subsection.

39

4.2 Fourier Series

Definition 4.11. Exponential Fourier series: Let the (real or complex)signal r (t) be a periodic signal with period T0. Suppose the following Dirich-let conditions are satisfied

(a) r (t) is absolutely integrable over its period; i.e.,T0∫0

|r (t)|dt <∞.

(b) The number of maxima and minima of r (t) in each period is finite.

(c) The number of discontinuities of r (t) in each period is finite.

Then r (t) can be expanded in terms of the complex exponential signals(ejnω0t

)∞n=−∞ as

r (t) =∞∑

n=−∞cne

jnω0t = c0 +∞∑k=1

(cke

jkω0t + c−ke−jkω0t

)(32)

where

ω0 = 2πf0 =2π

T0,

ck =1

T0

α+T0∫

α

r (t) e−jkω0tdt, (33)

for some arbitrary α. In which case,

r (t) =

r (t) , if r (t) is continuous at tr(t+)+r(t−)

2 , if r (t) is not continuous at t

We give some remarks here.

• The parameter α in the limits of the integration (33) is arbitrary. Itcan be chosen to simplify computation of the integral. Some references

40

simply write ck = 1T0

∫T0

r (t) e−jkω0tdt to emphasize that we only need

to integrate over one period of the signal; the starting point is notimportant.

• The coefficients ck = 1T0

∫T0

r (t) e−jkω0tdt are called the (kth) Fourier

(series) coefficients of (the signal) r (t). These are, in general, com-plex numbers.

• c0 = 1T0

∫T0

r (t) dt = average or DC value of r(t)

• The quantity f0 = 1T0

is called the fundamental frequency of thesignal r (t). The nth multiple of the fundamental frequency (for positiven’s) is called the nth harmonic.

• ckejkω0t + c−ke−jkω0t = the kth harmonic component of r (t).

k = 1 ⇒ fundamental component of r (t).

4.12. Getting the Fourier coefficients from the Fourier transform:Consider a restricted version rT0(t) of r(t) where we only consider r(t) for

one specific period. Suppose rT0(t)F−−−−F−1

RT0(f). Then,

ck =1

T0RT0(kf0).

So, the Fourier coefficients are simply scaled samples of the Fouriertransform.

Example 4.13. Find the Fourier series expansion for the train of impulses

δ(T0)(t) =∞∑

n=−∞δ (t− nT0) drawn in Figure 13.

0T 02T t

Figure 13: Train of impulses

41

4.14. The Fourier series in Example 4.13 gives an interesting Fourier trans-form pair:

A special case when T0 = 1 is quite easy to remember:

We can use the scaling property of the delta function to generalize the specialcase:

Example 4.15. Find the Fourier coefficients of the square pulse periodicsignal [5, p 57]shown in Figure 14. Note that multiplication by this signalis equivalent to a switching (ON-OFF) operation.

1

T0

4T0

4T0 2T0T02T0

1

t

Figure 14: Square pulse periodic signal

4.16. Parseval’s Identity: Pr = 1T0

∫T0

|r (t)|2 dt =∞∑

k=−∞|ck|2

42

4.17. Fourier series expansion for real valued function: Supposer (t) in the previous section is real-valued; that is r∗ = r. Then, we havec−k = c∗k and we provide here three alternative ways to represent the Fourierseries expansion:

r (t) =∞∑

n=−∞cne

jnω0t = c0 +∞∑k=1

(cke

jkω0t + c−ke−jkω0t

)(34)

= c0 +∞∑k=1

(ak cos (kω0t)) +∞∑k=1

(bk sin (kω0t)) (35)

= c0 + 2∞∑k=1

|ck| cos (kω0t+ ∠ck) (36)

where the corresponding coefficients are obtained from

ck =1

T0

α+T0∫

α

r (t) e−jkω0tdt =1

2(ak − jbk) (37)

ak = 2Re ck =2

T0

∫T0

r (t) cos (kω0t) dt (38)

bk = −2Im ck =2

T0

∫T0

r (t) sin (kω0t) dt (39)

2 |ck| =√a2k + b2

k (40)

∠ck = − arctan

(bkak

)(41)

c0 =a0

2(42)

The Parseval’s identity can then be expressed as

Pr =1

T0

∫T0

|r (t)|2 dt =∞∑

k=−∞

|ck|2 = c20 + 2

∞∑k=1

|ck|2

43

4.18. To go from (34) to (35) and (36), note that when we replace c−k byc∗k, we have

ckejkω0t + c−ke

−jkω0t = ckejkω0t + c∗ke

−jkω0t

= ckejkω0t +

(cke

jkω0t)∗

= 2 Recke

jkω0t.

• Expression (36) then follows directly from the phasor concept:

Recke

jkω0t

= |ck| cos (kω0t+ ∠ck) .

• To get (35), substitute ck by Re ck+ j Im ckand ejkω0t by cos (kω0t) + j sin (kω0t).

Example 4.19. For the train of impulses in Example 4.13,

δ(T0)(t) =∞∑

k=−∞

δ (t− kT0) =1

T0

∞∑k=−∞

ejkω0t =1

T0+

2

T0

∞∑k=1

cos kω0t (43)

Example 4.20. For the rectangular pulse train in Example 4.15,

1 [cosω0t ≥ 0] =1

2+

2

π

(cosω0t−

1

3cos 3ω0t+

1

5cos 5ω0t−

1

7cos 7ω0t+ . . .

)(44)

Example 4.21. Bipolar square pulse periodic signal [5, p 59]:

sgn(cosω0t) =4

π

(cosω0t−

1

3cos 3ω0t+

1

5cos 5ω0t−

1

7cos 7ω0t+ . . .

)

1

-1

0T 0T− t

1

0T 0T− t

Figure 15: Bipolar square pulse periodic signal

44

4.3 Classical DSB-SC Modulators

To produce the modulated signal Ac cos(2πfct)m(t), we may use the follow-ing methods which generate the modulated signal along with other signalswhich can be eliminated by a bandpass filter restricting frequency contentsto around fc.

4.22. Multiplier Modulators [5, p 184] or Product Modulator[3, p180]: Here modulation is achieved directly by multiplying m(t) by cos(2πfct)using an analog multiplier whose output is proportional to the product oftwo input signals.

• Such a multiplier may be obtained from

(a) a variable-gain amplifier in which the gain parameter (such as thethe β of a transistor) is controlled by one of the signals, say, m(t).When the signal cos(2πfct) is applied at the input of this amplifier,the output is then proportional to m(t) cos(2πfct).

(b) two logarithmic and an antilogarithmic amplifiers with outputsproportional to the log and antilog of their inputs, respectively.

Key equation:A×B = e(lnA+lnB).

4.23. Square Modulator: When it is easier to build a squarer than amultiplier, use

(m (t) + c cos (ωct))2 = m2 (t) + 2cm (t) cos (ωct) + c2 cos2 (ωct)

= m2 (t) + 2cm (t) cos (ωct) +c2

2+c2

2cos (2ωct) .

45

• Alternative, can use(m(t) + c cos

(ωc2 t))3

.

4.24. Multiply m(t) by “any” periodic and even signal r(t) whose periodis Tc = 2π

ωc. Because r(t) is an even function, we know that

r (t) = c0 +∞∑k=1

ak cos (kωct) for some c0, a1, a2, . . ..

Therefore,

m(t)r (t) = c0m(t) +∞∑k=1

akm(t) cos (kωct).

See also [4, p 157]. In general, for this scheme to work, we need

( )M ω ( )m r ω×F

cω 2 cω cω− 2 cω−

0c A 112

Aa 2

12

Aa A

2 Bπ 2 Bπ 2c Bω π−

× ( )m t

( )r t

BPF ( ) ( )cos cm t tω

BPF

Figure 16: Modulation of m(t) via even and periodic r(t)

• a1 6= 0; that is Tc is the “least” period of r;

• fc > 2B (to prevent overlapping).

Note that if r(t) is not even, then by (36), the outputted modulatedsignal is of the form a1m(t) cos(ωct+ φ1).

46

4.25. Switching modulator : Set r(t) to be the square pulse train givenby (44):

r (t) = 1 [cosω0t ≥ 0]

=1

2+

2

π

(cosω0t−

1

3cos 3ω0t+

1

5cos 5ω0t−

1

7cos 7ω0t+ . . .

).

Multiplying this r(t) to the signal m(t) is equivalent to switching m(t) onand off periodically.

It is equivalent to periodically turning the switch on (letting m(t) passthrough) for half a period Tc = 1

fc.

186 AMPLITUDE MODULATIONS AND DEMODULATIONS

Figure 4.4 Switching modulator for DSB-SC.

m(t )

~

w(l)

nnnnnnnnnnnnn

(a)

(b)

0 I

J~

The square pulse train w(t) in Fig. 4.4b is a periodic signal whose Fourier series was found earlier in Example 2.8 [Eq. (2.86)] as

ll' (l) = ~ + ~ (cos We t - ~ cos 3wct + ~ cos Seve! - · · · ) 2 J[ .) )

(4 .S )

The signalm (t)lr (t) is given by

m (t)w(t) = -m(t) + - m (t) cos eve! - -m(!) cos 3wct + -::111 (1) cos Seve! - · · · l 2 [ I l ] 2 J[ 3 )

(4.6)

The signal m(l) ll '(t) consists not onl y of the component 111(1) bu t a lso of an infinite number of modulated signals with angul ar frequencies r»c, 3wc, Seve, .. .. Therefore, the spectrum of m(1)1-v(t) consists of multiple copies of the message spectrum M (f), shifted to 0, ±fc, ±~fc , ±Sf~- , .. . (with decreas ing re lat ive weights), as shown in Fig. 4.4c.

For modulation , we are interested in extracting the modulated component m.(t) cos We i

on ly. To separate this component from the rest of the crowd, we pass the signal m(t)w(t) through a bandpass filter of band width 28 Hz (or 4Tr8 rad/s) , centered at the frequency ±fc· Provided the carrier angular frequency.fc ::::: 28 (or We ::::: 4Tr8) , thi s will suppress all the spectral components not centered at ±fc to yield the des ired modulated signal (2/ Tr)m(t) cos We t (Fig. 4.4d).

We now see the real payoff of this method. Multipl ication of a signal by a square pulse train is in reality a switching operation. It involves switching the signal m(t) on and off periodically and can be accomplished by simple switching elements controlled by w(t). Figure 4 .Sa shows one such electronic switch, the diode bridge modulator, driven by a sinusoid A cos We t to produce the switching action. Diodes D 1, D2 and D3, D4 are matched pairs. When the signal cos Wet is of a polarity that wi ll make terminal c positive with respect to d, all the diodes

Figure 17: Switching modulator for DSB-SC [4, Figure 4.4].

4.26. Switching Demodulator :

LPFm(t) cos(ωct)× 1[cos(ωct) ≥ 0] =1

πm(t) (45)

[4, p 162]. Note that this technique still requires the switching to be in syncwith the incoming cosine as in the basic DSB-SC.

47

4.4 Energy and Power

Definition 4.27. For a signal g(t), the instantaneous power p(t) dissipatedin the 1-Ω resister is pg(t) = |g(t)|2 regardless of whether g(t) represents avoltage or a current. To emphasize the fact that this power is based uponunity resistance, it is often referred to as the normalized power.

Definition 4.28. The total (normalized) energy of a signal g(t) is givenby

Eg =

∫ +∞

−∞pg(t)dt =

∫ +∞

−∞|g(t)|2 dt = lim

T→∞

∫ T

−T|g(t)|2 dt.

4.29. By the Parseval’s theorem discussed in 2.39, we have

Eg =

∫ ∞−∞|g(t)|2dt =

∫ ∞−∞|G(f)|2df.

Definition 4.30. The average (normalized) power of a signal g(t) is givenby

Pg = limT→∞

1

T

T/2∫−T/2

|g (t)|2dt = limT→∞

1

2T

∫ T

−T|g(t)|2 dt.

Definition 4.31. To simplify the notation, there are two operators thatused angle brackets to define two frequently-used integrals:

(a) The “time-average” operator:

〈g〉 ≡ 〈g (t)〉 ≡ limT→∞

1

T

∫ T/2

−T/2g (t)dt = lim

T→∞

1

2T

∫ T

−Tg (t)dt (46)

(b) The inner-product operator:

〈g1, g2〉 ≡ 〈g1 (t) , g2 (t)〉 =

∫ ∞−∞

g1(t)g∗2(t)dt (47)

4.32. Using the above definition, we may write

• Eg = 〈g, g〉 = 〈G,G〉 where G = F g

• Pg =⟨|g|2⟩

48

• Parseval’s theorem: 〈g1, g2〉 = 〈G1, G2〉where G1 = F g1 and G2 = F g2

4.33. Time-Averaging over Periodic Signal: For periodic signal g(t) withperiod T0, the time-average operation in (46) can be simplified to

〈g〉 =1

T0

∫T0

g (t)dt

where the integration is performed over a period of g.

Example 4.34. 〈cos (2πf0t+ θ)〉 =

Similarly, 〈sin (2πf0t+ θ)〉 =

Example 4.35.⟨cos2 (2πf0t+ θ)

⟩=

Example 4.36.⟨ej(2πf0t+θ)

⟩= 〈cos (2πf0t+ θ) + j sin (2πf0t+ θ)〉

Example 4.37. Suppose g(t) = cej2πf0t for some (possibly complex-valued)constant c and (real-valued) frequency f0. Find Pg.

4.38. When the signal g(t) can be expressed in the form g(t) =∑k

ckej2πfkt

and the fk are distinct, then its (average) power can be calculated from

Pg =∑k

|ck|2

49

Example 4.39. Suppose g(t) = 2ej6πt + 3ej8πt. Find Pg.

Example 4.40. Suppose g(t) = 2ej6πt + 3ej6πt. Find Pg.

Example 4.41. Suppose g(t) = cos (2πf0t+ θ). Find Pg.Here, there are several ways to calculate Pg. We can simply use Ex-

ample 4.35. Alternatively, we can first decompose the cosine into complexexponential functions using the Euler’s formula:

4.42. The (average) power of a sinusoidal signal g(t) = A cos(2πf0t+ θ) is

Pg =

12 |A|

2, f0 6= 0,

|A|2cos2θ, f0 = 0.

This property means any sinusoid with nonzero frequency can be written inthe form

g (t) =√

2Pg cos (2πf0t+ θ) .

4.43. Extension of 4.42: Consider sinusoids Ak cos (2πfkt+ θk) whose fre-quencies are positive and distinct. The (average) power of their sum

g(t) =∑k

Ak cos (2πfkt+ θk)

is

Pg =1

2

∑k

|Ak|2.

50

Example 4.44. Suppose g (t) = 2 cos(2π√

3t)

+ 4 cos(2π√

5t). Find Pg.

4.45. For periodic signal g(t) with period T0, there is also no need to carryout the limiting operation to find its (average) power Pg. We only need tofind an average carried out over a single period:

Pg =1

T0

∫T0

|g (t)|2dt.

(a) When the corresponding Fourier series expansion g(t) =∞∑

n=−∞cne

jnω0t

is known,

Pg =∞∑

k=−∞

|ck|2

(b) When the signal g(t) is real-valued and its (compact) trigonometric

Fourier series expansion g(t) = c0 + 2∞∑k=1

|ck| cos (kω0t+ ∠ck) is known,

Pg = c20 + 2

∞∑k=1

|ck|2

Definition 4.46. Based on Definitions 4.28 and 4.30, we can define threedistinct classes of signals:

(a) If Eg is finite and nonzero, g is referred to as an energy signal.

(b) If Pg is finite and nonzero, g is referred to as a power signal.

(c) Some signals17 are neither energy nor power signals.

• Note that the power signal has infinite energy and an energy signal haszero average power; thus the two categories are mutually exclusive.

Example 4.47. Rectangular pulse

17Consider g(t) = t−1/41[t0,∞)(t), with t0 > 0.

51

Example 4.48. Sinc pulse

Example 4.49. For α > 0, g(t) = Ae−αt1[0,∞)(t) is an energy signal withEg = |A|2/2α.

Example 4.50. The rotating phasor signal g(t) = Aej(2πf0t+θ) is a powersignal with Pg = |A|2.Example 4.51. The sinusoidal signal g(t) = A cos(2πf0t + θ) is a powersignal with Pg = |A|2/2.

4.52. Consider the transmitted signal

x(t) = m(t) cos(2πfct+ θ)

in DSB-SC modulation. Suppose M(f − fc) and M(f + fc) do not overlap(in the frequency domain).

(a) If m(t) is a power signal with power Pm, then the average transmittedpower is

Px =1

2Pm.

(b) If m(t) is an energy signal with energy Em, then the transmitted energyis

Ex =1

2Em.

• Q: Why is the power (or energy) reduced?

• Remark: When x(t) =√

2m(t) cos(2πfct + θ) (with no overlappingbetween M(f − fc) and M(f + fc)), we have Px = Pm.

52

4.5 Amplitude modulation: AM

4.53. DSB-SC amplitude modulation (which is summarized in Figure 18)is easy to understand and analyze in both time and frequency domains.However, analytical simplicity is not always accompanied by an equivalentsimplicity in practical implementation.

1

×

2 cos 2 cf t

Modulator

Message(modulating signal)

Figure 18: DSB-SC modulation.

Problem: The (coherent) demodulation of DSB-SC signal requires thereceiver to possess a carrier signal that is synchronized with the incomingcarrier. This requirement is not easy to achieve in practice because themodulated signal may have traveled hundreds of miles and could even sufferfrom some unknown frequency shift.

4.54. If a carrier component is transmitted along with the DSB signal,demodulation can be simplified.

(a) The received carrier component can be extracted using a narrowbandbandpass filter and can be used as the demodulation carrier. (There isno need to generate a carrier at the receiver.)

(b) If the carrier amplitude is sufficiently large, the need for generating ademodulation carrier can be completely avoided.

• This will be the focus of this section.

53

Definition 4.55. For AM, the transmitted signal is typically defined as

xAM (t) = (A+m (t)) cos (2πfct) = A cos (2πfct)︸︷︷︸carrier

+m (t) cos (2πfct)︸︷︷︸sidebands

4.56. Spectrum of xAM (t):

• Basically the same as that of DSB-SC signal except for the two addi-tional impulses (discrete spectral component) at the carrier frequency±fc.

This is why we say the DSB-SC system is a suppressed carriersystem.

Definition 4.57. Consider a signal A(t) cos(2πfct). If A(t) varies slowly incomparison with the sinusoidal carrier cos(2πfct), then the envelope E(t)of A(t) cos(2πfct) is |A(t)|.

4.58. Envelope of AM signal : For AM signal, A(t) = A+m(t) and

E(t) = |A+m(t)| .

See Figure 19.

(a) If ∀t, A(t) > 0, then E(t) = A(t) = A+m(t)

• The envelope has the same shape as m(t).

• We can detect the desired signal m(t) by detecting the envelope(envelope detection).

(b) If ∃t, A(t) < 0, then E(t) 6= A(t).

• The envelope shape differs from the shape of m(t) because thenegative part of A+m(t) is rectified.

This is referred to as phase reversal and envelope distortion.

54


Figure 4.8 AM signal and its envelope.

A+ m(l) > 0

t A

t (b)

I I

I

m(1)

~np

far all 1

A + m(1) ':> 0 far all 1

L\-------1~ t (~ 1--

/

(d)

En velope

~A+ m(l)

(e)

Envelope

lA + m(t)l

and m(t) cannot be recovered from the envelope. Consequently, demodulation of rpAM (t) in Fig. 4.8d amounts to simple envelope detection. Thus, the condition for envelope detection of an AM signal is

A + m(t) ::: 0 for all t (4.9a)

If m(t) ::: 0 for all/ , then A = 0 already sati sfies condition ( 4 .9a) . In thi s case there is no need to add any carrier because the enve lope of the DSB-SC signal m(t) cos Wet is rn(t) , and such a DSB-SC signal can be detected by envelope detection . In the following discuss ion we assume that 111(t) t_ 0 for all t; that is, m(t) can be negative over some range oft.

iVIessage Signals m (t) with Ze1·o Offset: Let ±mp be the maximum and the minimum va lues of m(t) , res pecti vely (Fig . 4.8) . This means that m(t) ::: - mp. Hence, the condition of envelope detection (4.9a) is equi valent to

(4.9b)

Thus, the minimum carrier amplitude required for the viability of envelope detection is mp· This is quite clear from Fig. 4.8. We define the modulation index fJ- as

mp fJ- =

A (4.10a)

Figure 19: AM signal and its envelope [5, Fig 4.8]

Definition 4.59. The positive constant

µ ≡maxt

(envelope of the sidebands)

maxt

(envelope of the carrier)=

maxt|m (t)|

maxt|A|

=mp

A

is called the modulation index.

• mp ≡ maxt|m (t)|

By the way mp is defined, the message m(t) is between ±mp.

• The quantity µ× 100% is often referred to as the percent modulation.

55

Example 4.60. Consider a sinusoidal (pure-tone) messagem(t) = Am cos(2πfmt).Suppose A = 1. Then, µ = Am. Figure 20 shows the effect of changing thevalue modulation index on the modulated signal.

1

Time

50% Modulation

0

−1.5

1.5

−0.5

0.5

Time

100% Modulation

0

−2

2

EnvelopeModulated Signal

Time

150% Modulation

0

−2.5

2.5

−0.5

0.5

Figure 20: Modulated signal in standard AM with sinusoidal message

4.61. It should be noted that the ratio that defines the modulation indexcompares the maximum of the two envelopes. In other references, the nota-tion for the AM signal may be different but the idea (and the correspondingmotivation) that defines the modulation index remains the same.

• In [3, p 163], it is assumed that m(t) is already scaled or normalized tohave a magnitude not exceeding unity (|m(t)| ≤ 1) [3, p 163]. There,

xAM (t) = Ac (1 + µm (t)) cos (2πfct) = Ac cos (2πfct)︸︷︷︸carrier

+Acµm (t) cos (2πfct)︸︷︷︸sidebands

.

56

mp = 1

The modulation index is then

maxt


maxt


maxt|Acµm (t)|

maxt|Ac|

=|Acµ||Ac|

= µ.

• In [14, p 116],

xAM (t) = Ac

(1 + µ

m (t)

mp

)cos (2πfct) = Ac cos (2πfct)︸︷︷︸

carrier

+Acµm (t)

mpcos (2πfct)︸︷︷︸

sidebands

.

The modulation index is then

maxt


maxt


maxt

∣∣∣Acµm(t)mp

∣∣∣max

t|Ac|

=|Ac|µmp

mp

|Ac|= µ.

4.62. Power of the transmitted signals.

(a) In DSB-SC system, recall, from 4.52, that, when

x(t) = m(t) cos(2πfct)

with fc sufficiently large, we have

Px =1

2Pm.

Therefore, all transmitted power are in the sidebands which containmessage information.

(b) In AM system,

xAM (t) = A cos (2πfct)︸︷︷︸carrier

+m (t) cos (2πfct)︸︷︷︸sidebands

.

If we assume that the average of m(t) is 0 (no DC component), then thespectrum of the sidebandsm(t) cos(2πfct+θ) and the carrierA cos(2πfct+θ) are non-overlapping in the frequency domain. Hence, when fc is suf-ficiently large

Px =1

2A2 +

1

2Pm.

57

• Efficiency:

• For high power efficiency, we want smallm2p

µ2Pm.

By definition, |m(t)| ≤ mp. Therefore,m2p

Pm≥ 1.

Want µ to be large. However, when µ > 1, we have phasereversal. So, the largest value of µ is 1.

The best power efficiency we can achieved is then 50%.

• Conclusion: at least 50% (and often close to 2/3[3, p. 176]) ofthe total transmitted power resides in the carrier part which isindependent of m(t) and thus conveys no message information.

4.63. An AM signal can be demodulated using the same coherent demod-ulation technique that was used for DSB. However, the use of coherentdemodulation negates the advantage of AM

• Note that, conceptually, the received AM signal is the same as DSB-SC signal except that the m(t) in the DSB-SC signal is replaced byA(t) = A + m(t). We also assume that A is large enough so thatA(t) ≥ 0.

• Recall the key equation of switching demodulator (45):

LPFA(t) cos(2πfct)× 1[cos(2πfct) ≥ 0] =1

πA(t) (48)

We noted before that this technique requires the switching to be insync with the incoming cosine.

58

4.64. Demodulation of AM Signals via rectifier detector: The receiverwill first recover A+m(t) and then remove A.

• When ∀t, A(t) ≥ 0, we can replace the switching demodulator bythe rectifier demodulator/detector . In which case, we suppressthe negative part of A(t) cos(2πfct) using a diode (half-wave rectifier:HWR).

Here, we define a HWR to be a memoryless device whose input-output relationship is described by a function fHWR(·):

fHWR (x) =

x, x ≥ 0,0, x < 0.

• This is mathematically equivalent to a switching demodulator in (45)and (48).

• It is in effect synchronous detection performed without using a localcarrier [4, p 167].

• This method needs A(t) ≥ 0 so that the sign of A(t) cos(2πfct) will bethe same as the sign of cos(2πfct).

• The dc term Aπ may be blocked by a capacitor to give the desired output

m(t)/π.

59


Figure 4.10 Rectifier detector for AM.

[a+ m(t)] cos wet

'

' /

[A + m(l)] cos wet

VR(t) /[A + m(t)]

-_f I " rr [A + 111(1)]

Low-pass filter

-·

I -;-[A + m(1)]

~

signal is multiplied by w(t). Hence, the half-wave rectified output vR(t) is

VR(t) =[A+ m(t)] COS Wet) w(t) (4.12)

=[A+ m(t)] cos Wet [ ~ + ~ (cos (Vet- ~cos 3wet + ~cos Swet- · · ·)] (4.13)

l = -[A+ m(t)] +other terms of higher frequencies (4.14)

][

When vR(t) is applied to a low-pass filter of cutoff B Hz, the output is [A+ m(t)]jn, and all the other terms of frequencies higher than B Hz are suppressed. The de term Ajn may be blocked by a capac itor (Fig. 4.10) to give the desired output m(t) j n. The output can be doubled by using a full-wave rectifi er.

It is interesting to note that because of the multip lication with ll '(l), rectifier detection is in effect synchronous detection performed without using a local carrier. The high carrier content in AM ensures that its zero crossings are periodic and the information about frequency and phase of the carrier at the transmitter is built in to the AM signal itself.

Envelope Detector: fn an enve lope detector, the output of the detector follows the envelope of the modulated signal. The simple circuit show n in Fig. 4. lla functions as an envelope detector. On the positive cycle of the input signal, the input grows and may exceed the charged vo ltage on the capacity vc(t), turning on the diode and allowing the capacitor C to charge up to the peak voltage of the input signal cycle. As the input signal fall s below this peak value, it falls quickly below the capacitor voltage (which is very nearly the peak voltage), thus caus ing the diode to open. The capacitor now di scharges through the resi stor R at a slow rate (with a time constant RC). During the next positive cycle, the same drama repeats . As the input signal rises above the capacitor voltage, the diode conducts again. The capacitor again charges to the peak value of this (new) cycle. The capacitor discharges slowly during the cutoff period.

During each positive cycle, the capacitor charges up to the peak voltage of the input signal and then decays slowly until the next positive cycle, as shown in Fig. 4 . ll b. Thus, the output voltage vc(t), close ly follows the (rising) envelope of the input AM signal. Equally important, the slow capacity discharge via the resistor R a llows the capacity vo ltage to follow

Figure 21: Rectifier detector for AM [5, Fig. 4.10].

Figure 4.11 Envelope detector for AM.

4 .4 Bandwidth-Efficient Amplitude Modulations 197

AM signal c

(a)

Envelope detector output

RC too large \

····· K'f<K~--~. . Enve lop~.--· ... ·· · '( K I"' I"" ~-~ . , .. · < i"" !'--, ., ~ ··· ···~"

W'~

.... ... -·· '

.. ··· · ... ·· .. ..

(b) ······

a declining envelope. Capacitor d ischarge between positi ve peaks causes a ripple s ignal of freque ncy We in the output. Thi s rip ple can be reduced by choosing a larger time constant RC so that the capac itor discharges very littl e between the positive peaks (RC » I /eve) . If RC were made too large, however, it would be imposs ible for the capac itor voltage to follow a fast declining e nvelope (Fig. 4 .11 b). Because the max imum rate of AM envelope dec line is do minated by the bandw idth B of the message signal m (r ) , the des ign criterion of RC should be

I /eve « RC < I / (2Jr8) or I

2Jr8 < - « (t!c RC

The envelope detector output is vc(t ) = A+ m(r) with a ripple o f frequency W e . The de term A can be blocked oul by a capacitor or a s imple RC high-pass filte r. The ripple may be reduced further by another (low-pass) RC filter.

4.4 BANDWIDTH-EFFICIENT AMPLITUDE MODULATIONS

As seen from Fig. 4.12, the DSB spectrum (including suppressed carrier and AM) has two sidebands: the upper sideband (USB) and the lower sideband (LSB~both containing complete informatinn about the baseband signal m (r ). As a result , for a baseband signal m (t) with bandwidth B Hz, DSB modulations require twice the radio-frequency bandwidth to transmit.

Figure 22: Envelope detector for AM [5, Fig. 4.11].

60

4.65. Demodulation of AM signal via envelope detector :

• Design criterion of RC:

2πB 1

RC 2πfc.

• The envelope detector output is A+m(t) with a ripple of frequency fc.

• The dc term can be blocked out by a capacitor or a simple RC high-passfilter.

• The ripple may be reduced further by another (low-pass) RC filter.

4.66. AM Trade-offs:

(a) Disadvantages :

• Higher power and hence higher cost required at the transmitter

• The carrier component is wasted power as far as information trans-fer is concerned.

• Bad for power-limited applications.

(b) Advantages :

• Coherent reference is not needed for demodulation.

• Demodulator (receiver) becomes simple and inexpensive.

• For broadcast system such as commercial radio (with a huge num-ber of receivers for each transmitter),

any cost saving at the receiver is multiplied by the number ofreceiver units.

it is more economical to have one expensive high-power trans-mitter and simpler, less expensive receivers.

(c) Conclusion: Broadcasting systems tend to favor the trade-off by mi-grating cost from the (many) receivers to the (fewer) transmitters.

4.67. References: [3, p 198–199], [5, Section 4.3] and [13, Section 3.1.2].

61

4.6 Quadrature Amplitude Modulation (QAM)

4.68. We are now going to define a quantity called the “bandwidth” of asignal. Unfortunately, in practice, there isn’t just one definition of band-width.

Definition 4.69. The bandwidth (BW) of a signal is usually calculatedfrom the differences between two frequencies (called the bandwidth limits).Let’s consider the following definitions of bandwidth for real-valued signals[3, p 173]

(a) Absolute bandwidth: Use the highest frequency and the lowest fre-quency in the positive-f part of the signal’s nonzero magnitude spec-trum.

• This uses the frequency range where 100% of the energy is confined.

• We can speak of absolute bandwidth if we have ideal filters andunlimited time signals.

(b) 3-dB bandwidth (half-power bandwidth): Use the frequencieswhere the signal power starts to decrease by 3 dB (1/2).

• The magnitude is reduced by a factor of 1/√

2.

(c) Null-to-null bandwidth: Use the signal spectrum’s first set of zerocrossings.

(d) Occupied bandwidth: Consider the frequency range in which X%(for example, 99%) of the energy is contained in the signal’s bandwidth.

(e) Relative power spectrum bandwidth: the level of power outsidethe bandwidth limits is reduced to some value relative to its maximumlevel.

• Usually specified in negative decibels (dB).

• For example, consider a 200-kHz-BW broadcast signal with a max-imum carrier power of 1000 watts and relative power spectrumbandwidth of -40 dB (i.e., 1/10,000). We would expect the sta-tion’s power emission to not exceed 0.1 W outside of fc± 100 kHz.

62

Example 4.70. Message bandwidth and the transmitted signal bandwidth

4.71. Rough Approximation: If g1(t) and g2(t) have bandwidths B1 andB2 Hz, respectively, the bandwidth of g1(t)g2(t) is B1 +B2 Hz.

This result follows from the application of the width property18 of con-volution19 to the convolution-in-frequency property.

Consequently, if the bandwidth of g(t) is B Hz, then the bandwidth ofg2(t) is 2B Hz, and the bandwidth of gn(t) is nB Hz. We mentioned thisproperty in 2.38.

4.72. BW Inefficiency in DSB-SC system: Recall that for real-valued base-band signal m(t), the conjugate symmetry property from 2.28 says that

M(−f) = (M(f))∗ .

The DSB spectrum has two sidebands: the upper sideband (USB) and thelower sideband (LSB), both containing complete information about the base-band signal m(t). As a result, DSB signals occupy twice the bandwidthrequired for the baseband.

4.73. To improve the spectral efficiency of amplitude modulation, thereexist two basic schemes to either utilize or remove the spectral redundancy:

(a) Single-sideband (SSB) modulation, which removes either the LSB orthe USB so that for one message signal m(t), there is only a bandwidthof B Hz.

(b) Quadrature amplitude modulation (QAM), which utilizes spectral re-dundancy by sending two messages over the same bandwidth of 2BHz.

We will only discussed QAM here. SSB discussion can be found in [3, Sec4.4], [13, Section 3.1.3] and [4, Section 4.5].

18This property states that the width of x ∗ y is the sum of the widths of x and y.19The width property of convolution does not hold in some pathological cases. See [4, p 98].

63

Definition 4.74. In quadrature amplitude modulation (QAM ) orquadrature multiplexing , two baseband real-valued signals m1(t) andm2(t) are transmitted simultaneously via the corresponding QAM signal:

xQAM (t) = m1 (t)√

2 cos (2πfct) +m2 (t)√

2 sin (2πfct) .

1m t

Transmitter (modulator) Receiver (demodulator)

1v t LPH f 1m t

2m t 2v t LPH f 2m t

2 cos 2 cf t

2 sin 2 cf t

2 h t y t QAMx t

Channel

2 cos 2 cf t

2 sin 2 cf t

2

Figure 23: QAM Scheme

• QAM operates by transmitting two DSB signals via carriers of the samefrequency but in phase quadrature.

• Both modulated signals simultaneously occupy the same frequencyband.

• The “cos” (upper) channel is also known as the in-phase (I ) channeland the “sin” (lower) channel is the quadrature (Q) channel.

4.75. Demodulation : The two baseband signals can be separated at thereceiver by synchronous detection:

LPFxQAM (t)

√2 cos (2πfct)

= m1 (t)

LPFxQAM (t)

√2 sin (2πfct)

= m2 (t)

64

• m1(t) and m2(t) can be separately demodulated.

4.76. Sinusoidal form (envelope-and-phase description [3, p. 165]):

xQAM (t) =√

2E(t) cos(2πfct+ θ(t)),

where

envelope: E(t) =√m2

1(t) +m22(t)

phase: θ(t) = − tan−1

(m2(t)

m1(t)

)• The envelope is defined as nonnegative. Negative “amplitudes” can be

absorbed in the phase by adding ±180.

4.77. Complex form:

xQAM (t) =√

2Re

(m(t)) ej2πfct

where20 m(t) = m1(t)− jm2(t).

• We refer to m(t) as the complex envelope (or complex basebandsignal) and the signals m1(t) and m2(t) are known as the in-phaseand quadrature(-phase) components of xQAM (t).

• The term “quadrature component” refers to the fact that it is in phasequadrature (π/2 out of phase) with respect to the in-phase component.

• Key equation:

LPF

(

Rem (t)×

√2ej2πfct

)︸︷︷︸

x(t)

×(√

2e−j2πfct) = m (t) .

20If we use − sin(2πfct) instead of sin(2πfct) for m2(t) to modulate,

xQAM (t) = m1 (t)√

2 cos (2πfct)−m2 (t)√

2 sin (2πfct)

=√

2 Rem (t) ej2πfct

where

m(t) = m1(t) + jm2(t).

65

4.78. Three equivalent ways of saying exactly the same thing:

(a) the complex-valued envelope m(t) complex-modulates the complex car-rier ej2πfct,

• So, now you can understand what we mean when we say that acomplex-valued signal is transmitted.

(b) the real-valued amplitude E(t) and phase θ(t) real-modulate the am-plitude and phase of the real carrier cos(2πfct),

(c) the in-phase signal m1(t) and quadrature signal m2(t) real-modulatethe real in-phase carrier cos(2πfct) and the real quadrature carriersin(2πfct).

4.79. References: [3, p 164–166, 302–303], [13, Sect. 2.9.4], [4, Sect. 4.4],and [8, Sect. 1.4.1]

4.80. Question: In engineering and applied science, measured signals arereal. Why should real measurable effects be represented by complex signals?

Answer: One complex signal (or channel) can carry information abouttwo real signals (or two real channels), and the algebra and geometry ofanalyzing these two real signals as if they were one complex signal bringseconomies and insights that would not otherwise emerge. [8, p. 3 ]

66

4.7 Suppressed-Sideband Amplitude Modulation

4.81. The upper and lower sidebands of DSB are uniquely related by sym-metry about the carrier frequency, so either one contains all the messageinformation. Hence, transmission bandwidth can be cut in half if one side-band is suppressed along with the carrier.

Definition 4.82. Conceptually, in single-sideband modulation (SSB),a sideband filter suppresses one sideband before transmission. [3, p 185–186]

(a) If the filter removes the lower sideband, the output spectrum consistsof the upper sideband alone.

(b) If the filter removes the upper sideband, the output spectrum consistsof the lower sideband alone.

Definition 4.83. In vestigial-sideband modulation (VSB), one side-band is passed almost completely while just a trace, or vestige, of the othersideband is included. [3, p 191–192]

4.84. In (analog) television video transmission, an AM wave is applied toa vestigial sideband filter. This modulation scheme is called VSB pluscarrier (VSB + C). [3, p 193]

• The unsuppressed carrier allows for envelope detection, as in AM

Distortionless envelope modulation actually requires symmetric side-bands, but VSB + C can deliver a fair approximation.

67

Sirindhorn International Institute of Technology

Thammasat University

School of Information, Computer and Communication Technology

ECS332 2015/1 Part II.3 Dr.Prapun

5 Angle Modulation: FM and PM

5.1. We mentioned in 4.1 that a sinusoidal carrier signal

A cos(2πfct+ φ)

has three basic parameters: amplitude, frequency, and phase. Varying theseparameters in proportion to the baseband signal results in amplitude mod-ulation (AM), frequency modulation (FM), and phase modulation (PM),respectively.

5.2. As usual, we will again assume that the baseband signal m(t) is band-limited to B; that is, |M(f)| = 0 for |f | > B.

As in the AM section, we will also assume that

|m(t)| ≤ mp.

In other words, m(t) is bounded between −mp and mp.

68

Definition 5.3. Phase modulation (PM ):

xPM (t) = A cos (2πfct+ φ+ kpm (t))

Definition 5.4. The main characteristic21 of frequency modulation (FM)is that the carrier frequency f(t) would be varied with time so that

f(t) = fc + km(t), (49)

where k is an arbitrary constant.

• The arbitrary constant k is sometimes denoted by kf to distinguish itfrom a similar constant in PM.

Example 5.5. With a sinusoidal message signal in Figure 24a, the frequencydeviation of the FM modulator output in Figure 24d is proportional tom(t). Thus, the (instantaneous) frequency of the FM modulator output ismaximum when m(t) is maximum and minimum when m(t) is minimum.

4.1 Phase and Frequency Modulation Defined 159

(a)

(b)

(c)

(d)

Figure 4.2Angle modulation with sinusoidal messsage signal. (a) Message signal. (b) Unmodulated carrier. (c)Output of phase modulator with !("). (d) Output of frequency modulator with !(").

quadrature with the carrier component, whereas for AM they are not. This will be illustrated inExample 4.1.

The generation of narrowband angle modulation is easily accomplished using the methodshown in Figure 4.3. The switch allows for the generation of either narrowband FM or narrow-

m(t)

(.)dt2 fd

kp

Ac

π

ω ω

FM

PM

(t) ×

−sin ct cos ct

Σxc(t)

Carrieroscillator

90° phaseshifter

Figure 4.3Generation of narrowband angle modulation.

Figure 24: Different modulations of sinu-soidal message signal. (a) Message signal. (b)Unmodulated carrier. (c) Output of phasemodulator (d) Output of frequency modula-tor [14, Fig 4.2 p 159 ]

The phase deviation of the PM output is proportional to m(t). However,because the phase is varied continuously, it is not straightforward (yet) to

21Treat this as a practical definition. The more rigorous definition will be provided in 5.15.

69

see how Figure 24c is related to m(t). In Example 5.18, we will come backto this example and re-analyze the PM output.

Example 5.6. Figure 25 illustrates the outputs of PM and FM modulatorswhen the message is a unit-step function.158 Chapter 4 ∙ Angle Modulation and Multiplexing

m(t)

1

t0t

t

t

t

(a)

t0(b)

t0(c)

t0(d)

Frequency = fc + fdFrequency = fc

Figure 4.1Comparison of PM and FM modulatoroutputs for a unit-step input.(a) Message signal. (b) Unmodulatedcarrier. (c) Phase modulator output(!" =

12#). (d) Frequency modulator

output.

where Re(⋅) implies that the real part of the argument is to be taken. Expanding $%&(') in apower series yields

()(') = Re*)

[1 + %&(') − &2(')

2! −⋯]$%2#+)'

(4.11)

If the peak phase deviation is small, so that the maximum value of |&(')| is much less thanunity, the modulated carrier can be approximated as

()(') ≅ Re[*)$%2#+)' + *)&(')%$%2#+)']

Taking the real part yields

()(') ≅ *) cos(2#+)') − *)&(') sin(2#+)') (4.12)

The form of (4.12) is reminiscent of AM. The modulator output contains a carrier com-ponent and a term in which a function of ,(') multiplies a 90 phase-shifted carrier. Thefirst term yields a carrier component. The second term generates a pair of sidebands. Thus,if &(') has a bandwidth - , the bandwidth of a narrowband angle modulator output is 2- .The important difference between AM and angle modulation is that the sidebands are pro-duced by multiplication of the message-bearing signal, & ('), with a carrier that is in phase

Figure 25: Comparison of PM and FM mod-ulator outputs for a unit-step input. (a) Mes-sage signal. (b) Unmodulated carrier. (c)Phase modulator output (d) Frequency mod-ulator output. [14, Fig 4.1 p 158]

• For the PM modulator output,

the (instantaneous) frequency is fc for both t < t0 and t > t0

the phase of the unmodulated carrier is advanced by kp = π2 radians

for t > t0 giving rise to a signal that is discontinuous at t = t0.

• For the FM modulator output,

the frequency is fx for t < t0, and the frequency is fc+fd for t > t0

the phase is, however, continuous at t = t0.

70

AM

FM

PM

Modulatingsignal

Figure 5.1–2 Illustrative AM, FM, and PM waveforms.

212 CHAPTER 5 • Angle CW Modulation

carrier amplitude, we modulate the frequency by swinging it over a range of, say,50 Hz, then the transmission bandwidth will be 100 Hz regardless of the messagebandwidth. As we’ll soon see, this argument has a serious flaw, for it ignores the dis-tinction between instantaneous and spectral frequency. Carson (1922) recognizedthe fallacy of the bandwidth-reduction notion and cleared the air on that score.Unfortunately, he and many others also felt that exponential modulation had noadvantages over linear modulation with respect to noise. It took some time to over-come this belief but, thanks to Armstrong (1936), the merits of exponential modula-tion were finally appreciated. Before we can understand them quantitatively, wemust address the problem of spectral analysis.

Suppose FM had been defined in direct analogy to AM by writing xc(t) Ac cos vc(t) twith vc(t) vc[1 mx(t)]. Demonstrate the physical impossibility of this definition byfinding f(t) when x(t) cos vmt.

Narrowband PM and FMOur spectral analysis of exponential modulation starts with the quadrature-carrierversion of Eq. (1), namely

(9)

where

(10)xci1t 2 Ac cos f1t 2 Ac c1 1

2! f21t 2 p d

xc1t 2 xci1t 2 cos vct xcq1t 2 sin vct

EXERCISE 5.1–1

car80407_ch05_207-256.qxd 12/8/08 10:49 PM Page 212

Confirming Pages

Figure 26: Illustrative AM, FM, and PM waveforms. [3, Fig 5.1-2 p 212]

Example 5.7. Figure 26 illustrates the outputs of AM, FM, and PM mod-ulators when the message is a triangular (ramp) pulse.

To understand more about FM, we will first need to know what it actuallymeans to vary the frequency of a sinusoid.

5.1 Instantaneous Frequency

Definition 5.8. The generalized sinusoidal signal is a signal of the form

x(t) = A cos (θ(t)) (50)

where θ(t) is called the generalized angle.

• The generalized angle for conventional sinusoid is 2πfct+ φ.

• In [3, p 208], θ(t) of the form 2πfct + φ(t) is called the total instan-taneous angle.

Definition 5.9. If θ(t) in (50) contains the message information m(t), wehave a process that may be termed angle modulation.

• The amplitude of an angle-modulated wave is constant.

• Another name for this process is exponential modulation.

71

The motivation for this name is clear when we write x(t) asAReejθ(t)

.

It also emphasizes the nonlinear relationship between x(t) andm(t).

• Since exponential modulation is a nonlinear process, the modulatedwave x(t) does not resemble the message waveform m(t).

5.10. Suppose we want the frequency fc of a carrier A cos(2πfct) to varywith time as in (49). It is tempting to consider the signal

A cos(2πg(t)t), (51)

where g(t) is the desired frequency at time t.

Example 5.11. Consider the generalized sinusoid signal of the form 51above with g(t) = t2. We want to find its frequency at t = 2.

(a) Suppose we guess that its frequency at time t should be g(t). Then,at time t = 2, its frequency should be t2 = 4. However, when com-pared with cos (2π(4)t) in Figure 27a, around t = 2, the “frequency”of cos(2π

(t2)t) is quite different from the 4-Hz cosine approximation.

Therefore, 4 Hz is too low to be the frequency of cos(2π(t2)t) around

t = 2.

1

(a) (b)

Figure 27: Approximating the frequency of cos(2π (t2) t) by (a) cos (2π(4)t) and (b)cos (2π(12)t).

72

(b) Alternatively, around t = 2, Figure 27b shows that cos (2π(12)t) seemsto provide a good approximation. So, 12 Hz would be a better answer.

Definition 5.12. For generalized sinusoid A cos(θ(t)), the instantaneousfrequency 22 at time t is given by

f(t) =1

2π

d

dtθ(t). (52)

Example 5.13. For the signal cos(2π(t2)t) in Example 5.11,

θ (t) = 2π(t2)t

and the instantaneous frequency is

f (t) =1

2π

d

dtθ (t) =

1

2π

d

dt

(2π(t2)t)

= 3t2.

In particular, f (2) = 3× 22 = 12.

5.14. The instantaneous frequency formula (52) implies

θ(t) = 2π

∫ t

−∞f(τ)dτ = θ(t0) + 2π

∫ t

t0

f(τ)dτ. (53)

5.2 FM and PM

Definition 5.15. Frequency modulation (FM ):

xFM (t) = A cos

2πfct+ φ+ 2πkf

t∫−∞

m (τ)dτ

. (54)

The instantaneous frequency is given by

f (t) = fc + kfm (t) .

22Although f(t) is measured in hertz, it should not be equated with spectral frequency. Spectral frequencyf is the independent variable of the frequency domain, whereas instantaneous frequency f(t) is a time-dependent property of waveforms with exponential modulation.

73

5.16. Phase modulation (PM ): The phase-modulated signal is definedin Definition 5.3 to be

xPM (t) = A cos (2πfct+ φ+ kpm (t))

Its instantaneous frequency is(55)

Therefore, the instantaneous frequency of the output of the PM modu-lator is

• maximum when the slope of m(t) is maximum and

• minimum when the slope of m(t) is minimum.

Example 5.17. Sketch FM and PM waves for the modulating signal m(t)shown in 28a.

Figure 28: FM and PM waveforms generated from the same message.

This “indirect” method of sketching xPM(t) (using m(t) to frequency-modulate a carrier) works as long as m(t) is a continuous signal. If m(t)is discontinuous, this indirect method fails at points of discontinuities. Insuch a case, a direct approach should be used to specify the sudden phasechanges.

74

4.1 Phase and Frequency Modulation Defined 159

(a)

(b)

(c)

(d)

Figure 4.2Angle modulation with sinusoidal messsage signal. (a) Message signal. (b) Unmodulated carrier. (c)Output of phase modulator with !("). (d) Output of frequency modulator with !(").

quadrature with the carrier component, whereas for AM they are not. This will be illustrated inExample 4.1.

The generation of narrowband angle modulation is easily accomplished using the methodshown in Figure 4.3. The switch allows for the generation of either narrowband FM or narrow-

m(t)

(.)dt2 fd

kp

Ac

π

ω ω

FM

PM

(t) ×

−sin ct cos ct

Σxc(t)

Carrieroscillator

90° phaseshifter

Figure 4.3Generation of narrowband angle modulation.

Figure 29: xPM(t) and the corre-sponding m(t).

Example 5.18. Consider xPM(t) in Example 5.5. It is copied here in Figure29 along with the corresponding message m(t) which generates it.

5.19. Relationship between FM and PM:

• Equation (54) implies that one can produce frequency-modulated signalfrom a phase modulator.

• Equation (55) implies that one can produce phase-modulated signalfrom a frequency modulator.

• The two observations above are summarized in Figure 30.

( )FMx t ( )m t

( )t

m dτ τ−∞∫

Phase Modulator ∫

Frequency modulator

( )PMx t ( )m t

( )m t′ Frequency Modulator

ddt

Phase modulator

Figure 30: With the helpof integrating and dif-ferentiating networks, aphase modulator can pro-duce frequency modula-tion and vice versa [4, Fig5.2 p 255].

• By looking at an angle-modulated signal x(t), there is no way of tellingwhether it is FM or PM.

Compare Figure 24c and 24d in Example 5.5.

In fact, it is meaning less to ask an angle-modulated wave whetherit is FM or PM. It is analogous to asking a married man withchildren whether he is a father or a son. [5, p 255]

75

5.20. Generalized angle modulation (or exponential modulation):

x(t) = A cos (2πfct+ θ0 + (m ∗ h)(t))

where h is causal.

(a) Frequency modulation (FM ): h(t) = 2πkf1[t ≥ 0]

(b) Phase modulation (PM ): h(t) = kpδ(t).

5.21. So far, we have spoken rather loosely of amplitude and phase modula-tion. If we modulate two real signals a(t) and φ(t) onto a cosine to producethe real signal x(t) = a(t) cos(ωct + φ(t)), then this language seems unam-biguous: we would say the respective signals amplitude- and phase-modulatethe cosine. But is it really unambiguous?

The following example suggests that the question deserves thought.

Example 5.22. [8, p 15] Let’s look at a “purely amplitude-modulated”signal

x1(t) = a(t) cos(ωct).

Assuming that a(t) is bounded such that 0 ≤ a(t) ≤ A, there is a well-defined function

θ(t) = cos−1

(1

Ax1(t)

)− ωct.

Observe that the signal

x2(t) = A cos (ωct+ θ(t))

is exactly the same as x1(t) but x2(t) looks like a “purely phase-modulated”signal.

5.23. Example 5.22 shows that, for a given real signal x(t), the factorizationx(t) = a(t) cos(ωct+φ(t)) is not unique. In fact, there is an infinite numberof ways for x(t) to be factored into “amplitude” and “phase”.

76

5.3 Bandwidth of FM Signals

5.24. FM: The “Holy Grail” Technique for BW Saving?In the 1920s, the idea of frequency modulation (FM) was naively proposed

very early as a method to conserve the radio spectrum. The argument waspresented as follows:

• If m(t) is bounded between −mp and mp, then the maximum and mini-mum values of the (instantaneous) carrier frequency would be fc+kmp

and fc − kmp, respectively. (Think of this as a delta function shiftingto various location between fc + kmp and fc − kmp in the frequencydomain.)

• Hence, the spectral components would remain within this band with abandwidth 2kmp centered at fc.

• Conclusion: By using an arbitrarily small k, we could make the infor-mation bandwidth arbitrarily small (much smaller than the bandwidthof m(t).

In 1922, Carson argued that this is an ill-considered plan. We will illustratehis reasoning later. In fact, experimental results shows that

As a result of his observation, FM temporarily fell out of favor.

5.25. Armstrong (1936) reawakened interest in FM when he realized ithad a much different property that was desirable. When the kf is large, theinverse mapping from the modulated waveform xFM(t) back to the signalm(t) is much less sensitive to additive noise in the received signal than isthe case for amplitude modulation. FM then came to be preferred to AMbecause of its higher fidelity. [1, p 5-6]

77

Finding the “bandwidth” of FM Signals turns out to be a difficult task.Here we present a few approximation techniques.

5.26. First, from 5.20, we see that both FM and PM can be viewed as

x(t) = A cos (2πfct+ θ0 + φ(t)) (56)

where φ(t) = (m ∗ h)(t) if h(t) is selected properly.The Fourier transform of φ(t) is Φ(f) = M(f) ∗ H(f). So, if M(f) is

band-limited to B, we know that Φ(f) is also band-limited to B as well.Now, let us rewrite (56) as

x(t) = AReej(2πfct+θ0+φ(t))

= ARe

ej(2πfct+θ0)ejφ(t)

Recall that if φ(t) is band-limited to B, then φn(t) is band-limited to nB.So, we can make a rough sketch of |X(f)| as follows

So,we conclude that the absolute bandwidth would be infinite.

5.27. When φ(t) is small,

78

• The “approximated” expression of x(t) is similar to AM.

The modulator output contains a carrier component and a term inwhich a function of m(t) multiplies a 90 phase-shifted carrier.

The first term yields a carrier component. The second term gen-erates a pair of sidebands. Thus, if φ(t) has a bandwidth B, thebandwidth of x(t) is 2B.

• The important difference between AM and angle modulation is thatthe sidebands are produced by multiplication of the message-bearingsignal, φ(t), with a carrier that is in phase quadrature with the carriercomponent, whereas for AM they are not.

• For larger values of |φ(t)| the terms φ2(t), φ3(t), . . . cannot be ignoredand will increase the bandwidth of x(t).

• Recall, from (29) that

g(t) cos(2πfct+ φ)F−−−−F−1

1

2

(ejφG(f − fc) + e−jφG(f + fc)

).

Therefore, when

x (t) ≈ A cos (2πfct+ θ0)− Aφ (t) cos (2πfct+ θ0 − 90) ,

we have

X (f) ≈ A

2

(ejθ0δ(f − fc) + e−jθ0δ(f + fc)− ej(θ0−90)Φ(f − fc)− e−j(θ0−90)Φ(f + fc)

)=A

2

(ejθ0δ(f − fc) + e−jθ0δ(f + fc) + jejθ0Φ(f − fc)− je−jθ0Φ(f + fc)

).

5.28. For potentially wideband m(t), here, we present a technique toroughly estimate the bandwidth of xFM(t).

To do this, we consider m(t) that is a piecewise constant function (alsoknown as step function or staircase function); this implies that the instan-taneous frequency f(t) = fc + kfm(t) of xFM(t) is also piecewise constant.

79

For example, we can consider the transmitted signal xFM(t) constructedfrom five different tones. Its instantaneous frequency is increased from f1

to f5.

Assume that each tone lasts Ts = 1Rs

[s] where Rs is called the “(symbol)rate” of the data transmission. Increasing the value of Rs reduces the timeto complete the transmission.

Recall that the Fourier transform of a cosine contains simply (two shiftedand scaled) delta functions at the (plus and minus) frequency of the cosine.However, recall also that when we consider the cosine pulse, which is time-limited, its Fourier transform contains (two) sinc functions. In particular,the cosine pulse

p (t) =

cos (2πf0t) , t1 ≤ t < t2,0, otherwise,

can be viewed as the pure cosine function cos (2πf0t) multiplied by a rect-angular pulse r (t) = 1 [t1 ≤ t < t2]. By (28), we know that multiplicationby cos (2πf0t) will shift the spectrum R(f) of the rectangular pulse to ±fcand scaled its values by a factor of 1

2 :

P (f) =1

2R (f − f0) +

1

2R (f + f0)

where the Fourier transform23 R(f) of the rectangular pulse is given by

R (f) = (t2 − t1) e−jπf(t1+t2) sinc (πf (t2 − t1)) .

See Figure 31 for an example.

23To get this, first consider the rectangular pulse of width t2 − t1 centered at t = 0. From (13), thecorresponding Fourier transform is 2

(t2−t1

2

)sinc

(2π(t2−t1

2

)f). Finally, by time-shifting the rectangular

pulse in the time domain by t2+t12 , we simply multiply the Fourier transform by e−2πf(

t2−t12 ) in the

frequency domain.

80

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1

-0.5

0

0.5

1

t [s]

x(t)

-200 -150 -100 -50 0 50 100 150 2000

0.01

0.02

0.03

0.04

0.05

f [Hz]

|X(f)

|

Cos Pulse

1

cos 2 100 , 0.5 0.6,0, otherwise.

Figure 31: Cosine pulseand its spectrum whichcontains two sinc func-tions at ± freqeuncy ofthe cosine (which is 100Hz in the figure). Whenthe pulse only lasts fora short time period, thesinc pulses in the fre-quency domain are wide.

When m(t) is piecewise constant, xFM(t) is a sum of cosine pulses. There-fore, its spectrum X(f) will be the sum of the sinc functions centered at thefrequencies of the pulses as shown in Figure 32.

1

cos 2 cos 2 cos 2cos 2cos 2300 Hz100 Hz 200 Hz 500 Hz400 Hz

0 0.05 0.1 0.15 0.2 0.25-1

-0.5

0

0.5

1

Seconds

-1000 -800 -600 -400 -200 0 200 400 600 800 10000

0.01

0.02

0.03

Frequency [Hz]

Mag

nitude

Figure 32: A digital version of FM: xFM(t) and the corresponding XFM(f).

• X(f) extends to ±∞. It is not band-limited.

• One may approximate its bandwidth by assuming that “most” of theenergy in the sinc function is contained in its main lobe which is at± 1Ts

= ±Rs from its peak. Therefore, the bandwidth of xFM(t) becomes

81

Sirindhorn International Institute of Technology

Thammasat University

School of Information, Computer and Communication Technology

ECS332 2015/1 Part III.1 Dr.Prapun

6 Sampling, Reconstruction, and Pulse Modulation

6.1 Sampling

Definition 6.1. Sampling is the process of taking a (sufficient) number ofdiscrete values of points on a waveform that will define the shape of waveform.

• The signal is sampled at a uniform rate, once every Ts seconds.

m[n] = m(nTs) = m(t)|t=nTs.

• We refer to Ts as the sampling period, and to its reciprocal fs = 1/Tsas the sampling rate.

• The reverse process is called “reconstruction”.

6.2. Sampling = loss of information? If not, how can we recover the originalwaveform back.

• The more samples you take, the more accurately you can define a wave-form.

• Obviously, if the sampling rate is too low, you may experience distortion(aliasing).

82

• The sampling theorem, to be discussed in the section, says that whenthe waveform is band-limited, if the sampling rate is fast enough, we canreconstruct the waveform back and hence there is no loss of information.

This allows us to replace a continuous time signal by a discretesequence of numbers.

Processing a continuous time signal is therefore equivalent to pro-cessing a discrete sequence of numbers.

In the field of communication, the transmission of a continuoustime message reduces to the transmission of a sequence of numbers.

Example 6.3. Mathematical functions are frequently displayed as contin-uous curves, even though a finite number of discrete points was used toconstruct the graphs. If these points, or samples, have sufficiently closespacing, a smooth curve drawn through them allows us to interpolate in-termediate values to any reasonable degree of accuracy. It can therefore besaid that the continuous curve is adequately described by the sample pointsalone.

Example 6.4. Plot y = x2.

Example 6.5. Plot g(t) = sin(100πt).(See slides.)

Theorem 6.6. Sampling Theorem: In order to (correctly and com-pletely) represent an analog signal, the sampling frequency, fs, must beat least twice the highest frequency component of the analog signal.

Example 6.7. In example 6.5, the frequency of the sine wave is 50 Hz.Therefore, we need the sampling frequency to be at least 100.

Example 6.8. Suppose the sampling frequency is 200 samples/sec. Theanalog signal should not have the frequency higher than 100 Hz.(See slides)

83

Definition 6.9.

(a) Given a sampling frequency, fs, the Nyquist frequency is defined asfs/2.

(b) Given the highest (positive-)frequency component fmax of an analogsignal,

(i) the Nyquist sampling rate is 2fmax and

(ii) the Nyquist sampling interval is 1/(2fmax).

6.10. Much more can be said about the result of performing the samplingprocess on a signal. Here we will use g(t) to denote the signal under consid-eration. You may replace g(t) below by m(t) if you want to think of it asan analog message to be transmitted by a communication system. We useg(t) here because the results provided here work in broader setting as well.

Definition 6.11. In ideal sampling, the (ideal instantaneous) sampledsignal is represented by a train of impulses whose area equal the instanta-neous sampled values of the signal

gδ (t) =∞∑

n=−∞g [n]δ (t− nTs) .

6.12. The Fourier transform Gδ(f) of gδ (t) can be found by first rewritinggδ (t) as

gδ (t) =∞∑

n=−∞g (nTs)δ (t− nTs) =

∞∑n=−∞

g (t)δ (t− nTs)

= g (t)∞∑

n=−∞δ (t− nTs).

Multiplication in the time domain corresponds to convolution in the fre-quency domain. Therefore,

Gδ (f) = F gδ (t) = G (f) ∗ F

∞∑n=−∞

δ (t− nTs)

.

84

For the last term, the Fourier transform can be found by applying what wefound in Example 4.1324:

∞∑n=−∞

δ (t− nTs)F−−−−F−1

fs

∞∑k=−∞

δ (f − kfs).

This gives

Gδ (f) = G (f) ∗ fs∞∑

k=−∞

δ (f − kfs) = fs

∞∑k=−∞

G (f) ∗ δ (f − kfs).

Hence, we conclude that

gδ (t) =∞∑

n=−∞g [n]δ (t− nTs)

F−−−−F−1

Gδ (f) = fs

∞∑k=−∞

G (f − kfs). (57)

6.13. As usual, we will assume that the signal g(t) is band-limited to B

Hz ((G(f) = 0 for |f | > B)). In which case, the Fourier transform of thesampled signal is given by

6.14. Remarks:

(a) Gδ (f) is “periodic” (in the frequency domain) with “period” fs.

• So, it is sufficient to look at Gδ (f) between ±fs2

(b) The MATLAB script plotspect that we have been using to visualizemagnitude spectrum also relies on sampled signal. Its frequency domainplot is between ±fs

2 .

(c) Although this sampling technique is “ideal” because it involves the useof the δ-function. We can extract many useful conclusions.

(d) One can also study the discrete-time Fourier transform (DTFT) to lookat the frequency representation of the sampled signal.

24We also considered an easy-to-remember pair and discuss how to extend it to the general case in 4.14.

85

6.2 Reconstruction

6.15. From (57), we see that when the sampling frequency fs is largeenough, the replicas of G(f) will not overlap in the frequency domain. Insuch case, the original G(f) is still intact and we can use a low-pass filterwith gain Ts to recover g(t) back from gδ (t).

6.16. To prevent aliasing (the corruption of the original signal because itsreplicas overlaps in the frequency domain), we need

Theorem 6.17. A low-pass signal g whose spectrum is band-limited toB Hz (G(f) = 0 for |f | > B) can be reconstructed (interpolated) exactly(without any error) from its sample taken uniformly at a rate (samplingfrequency/rate) fs > 2B Hz (samples per second).[5, p 302]

6.18. Ideal Reconstruction: Continue from 6.15. Assuming that fs >2B, the low-pass filter that we should use to extract g(t) from Gδ(t) shouldbe

HLP (f) =

|f | ≤ B,

B < |f | < fs −B,|f | ≥ fs −B,

In particular, for “brick-wall” LPF, the cutoff frequency fcutoff should bebetween B and fs −B.

6.19. Reconstruction Equation: Suppose we use fs2 as the cutoff fre-

quency for our “brick-wall” LPF in 6.18,

The impulse response of the LPF is hLP (t) = sinc(

2π(fs2

)t)

= sinc(πfst).

86

The output of the LPF is

gr(t) = gδ (t) ∗ hLP (t) =

( ∞∑n=−∞

g [n]δ (t− nTs)

)∗ hLP (t)

=∞∑

n=−∞g [n]hLP (t− nTs) =

∞∑n=−∞

g [n] sinc (πfs (t− nTs)) .

When fs > 2B, this output will be exactly the same as g(t):

g (t) =∞∑

n=−∞g [n] sinc (πfs (t− nTs)) (58)

• This formula allows perfect reconstruction the original continuous-timefunction from the samples.

• At the sampling instants t = nTs, all sinc functions are zero at thesetimes save one, and that one yields g(nTs) which is the correct values.

• Note that at time t between the sampling instants, g(t) is interpolatedby summing the contributions from all the sinc functions.

• The LPF is often called an interpolation filter, and its impulse responseis called the interpolation function.

Example 6.20.

1 2 3 4 5 6 7 80

2

4

6

8

n

g[n]

= g

(nT s)

1 2 3 4 5 6 7 8-2

0

2

4

6

8

t [Ts]

g r(t)

Figure 33: Application of the reconstructionequation

87

Theorem 6.21. Sampling theorem for uniform periodic sampling: Ifa signal g(t) contains no frequency components for |f | ≥ B, it is completelydescribed by instantaneous sample values uniformly spaced in time withsampling period Ts ≤ 1

2B . In which case, g(t) can be exactly reconstructedfrom its samples (. . . , g[−2], g[−1], g[0], g[1], g[2], . . .) by the reconstructionequation (58).

Example 6.22. We now return to the sampling of the cosine function (si-nusoid).

1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-1

0

1

2

t

g[n]g(t)

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-1

-0.5

0

0.5

1

t

g[n]g(t)

Figure 34: Reconstruction of thesignal g(t) = cos(2π(2)t) by itssamples g[n]. The upper plot usesTs = 0.4. The lower plot usesTs = 0.2.

6.23. Remarks:

• Need a lot of g[n] for the reconstruction.

• Practical signals are time-limited.

Filter the message as much as possible before sampling.

6.24. The possibility of fs = 2B:

• If the spectrum G(f) has no impulse (or its derivatives) at the highestfrequency B, then the overlap is still zero as long as the sampling rateis greater than or equal to the Nyquist rate, that is, fs ≥ 2B.

• If G(f) contains an impulse at the highest frequency ±B, then fs = 2Bwould cause overlap. In such case, the sampling rate fs must be greaterthan 2B Hz.

88

Example 6.25. Consider a sinusoid g(t) = sin (2π(B)t). This signal isbandlimited to B Hz, but all its samples are zero when uniformly taken ata rate fs = 2B, and g(t) cannot be recovered from its (Nyquist) samples.Thus, for sinusoids, the condition of fs > 2B must be satisfied.

Let’s check with our formula (57) for Gδ(f). First, recall that

sinx =ejx − e−jx

2j=

1

2jejx − 1

2je−jx.

Therefore,

g (t) = sin (2π (B) t) =1

2jej2π(B)t − 1

2je−j2π(B)t =

1

2jej2π(B)t − 1

2jej2π(−B)t

and

Note that G(f) is pure imaginary. So, it is more suitable to look at theplot of its imaginary part. (We do not look at its magnitude plot becausethe information about the sign is lost. We also do not consider the real partbecause we know that it is 0.)

89

6.26. A maximum of 2B independent pieces (samples/symbols) of infor-mation per second can be transmitted, errorfree, over a noiseless channel ofbandwidth B Hz [4, p 260].

• Start with 2B pieces of information per second. Denote the sequenceof such information by mn.

• Construct a signalm(t) whose (Nyquist) sample valuesm[n] = m(n 1

2B

)agrees with mn by the reconstruction equation (58).

6.27. A bandpass signal whose spectrum exists over a frequency bandfc − B

2 < |f | < fc + B2 has a bandwidth B Hz. Such a signal is also

uniquely determined by samples taken at above the Nyquist frequency 2B.The sampling theorem is generally more complex in such case. It uses twointerlaced sampling trains, each at a rate of fs > B samples per second(known as second-order sampling). [5, p 304]

90

6.3 Analog Pulse Modulation

In Section 6.1 we saw that continuous bandlimited signals can be representedby a sequence of discrete samples. Moreover, in Section 6.2, we saw thatthe continuous signal can be reconstructed if the sampling rate is sufficientlyhigh.

6.28. Because the sequence m[n] completely contains the information aboutm(t), in this section, instead of trying to send m(t), we consider transmittingthe message in the form of pulse modulation.

Definition 6.29. In analog pulse modulation, some attribute of a pulsevaries continuously in one-to-one correspondence with a sample value.

• Example of a pulse:

• Three attributes can be readily varied: amplitude, width, and position.

• These lead to pulse-amplitude modulation (PAM), pulse-width modu-lation (PWM), and pulse-position modulation (PPM) as illustrated inFigure 35.

Definition 6.30. Unmodulated pulse train:∞∑

n=−∞p (t− nTs)

Definition 6.31. In Pulse-Amplitude Modulation (PAM), the samplevalues modulate the amplitude of a pulse train:

xPAM (t) =∞∑

n=−∞m [n] p (t− nTs)

91

Example 6.32.

6.33. One advantage of using pulse modulation is that it permits the si-multaneous transmission of several signals on a time-sharing basis.

• When a pulse-modulated signal occupies only a part of the channeltime, we can transmit several pulse-modulated signals on the samechannel by interweaving them.

• One User: TDM (time division multiplexing).

Transmit/multiplex multiple streams of information simultaneously.

• Multiple Users: TDMA (time division multiple access).

6.34. Frequency-Domain Analysis of PAM:

xPAM (t) =∞∑

n=−∞m [n] p (t− nTs) =

∞∑n=−∞

m [n] p (t) ∗ δ (t− nTs)

= p (t) ∗

( ∞∑n=−∞

m [n] δ (t− nTs)

)= p (t) ∗mδ (t)

Therefore,XPAM (f) = P (f)Mδ (f) .

6.35. Figure 35 compares different types of analog pulse modulation.

Definition 6.36. Pulse-Width Modulation (PWM): A PWM waveformconsists of a sequence of pulses with the width of the nth pulse is propor-tional to the value of m[n].

92

;rJ0r

;equence of : error if the p ic of pulse )Ll[se varies 1hich some

PWM signal

PPM signal

0 T, 2T, 9T s

3.5 Analog Pulse Mo dul a ti o n 183

Figure 3.56 Illustration of PAM, PWM, and PPM.

attribu te of a pu lse can take on a certa in va lue from a set of allowable va lues. In thi s section we examine analog pulse mod ulation . In the fo ll owing secti on we exami ne a coup le of examples of dig ital pul se modulat ion.

As mentioned, analog pulse modularion results when some attr ibute of a pu lse varies conti nuou sly in one-to-one cotTespondence wit h a samp le value . T hree attributes can be readi ly varied: amplitude, width , and position . T hese lead to pu lse ampl itude mod ul at ion (PAM), pul se-width modul ati on (PWM), and pulse-positio n mod ul ation (PPM) as ill ustrated in Figure 3.56.

3.5.1 Pulse-Amplitude M odulation

A PAM waveform consists of a sequence of fl at-topped pul ses des ignat ing sample va lues. The ampli tude of each pulse corresponds to the value of the message s ig nal m( t ) at the leading edge of the pulse. The essential di fference between PAM and the sampling operati on discussed in the previous chapter is that in PAM we allow the sampling pulse to have fi nite width . The fin itewidth pulse can be generated from the impul se-train sampling functio n by passing the impulsetrain samples through a holding circuit as shown in Figure 3.57. The impulse response of the ideal hold ing circuit is given by

Figure 35: Illustration of PAM, PWM, andPPM.

• Seldom used in modern communications systems.

• Used extensively for DC motor control in which motor speed is propor-tional to the width of the pulses . Since the pulses have equal amplitude,the energy in a given pulse is proportional to the pulse width.

Definition 6.37. Pulse-Position Modulation (PPM): A PPM signalconsists of a sequence of pulses in which the pulse displacement from a spec-ified time reference is proportional to the sample values of the information-bearing signal.

• Have a number of applications in the area of ultra-wideband commu-nications.

6.38. Pulse-modulation scheme are really baseband coding schemes, andthey yield baseband signal.

93

7 Inter-symbol Interference and Pulse Shaping

7.1. Recall that, in Pulse-Amplitude Modulation (PAM), we start witha sequence of numbers

· · · ,m[−3],m[−2],m[−1],m[0],m[1],m[2],m[3], · · ·

as shown in Figure 36.

PAM: Pulse Amplitude Modulation

1

T 2T 3T 4T-T-2T-3T-4Tt

m[0]m[1]

m[2]

m[3]m[4]

m[-1]m[-2]

m[-3]

m[-4]Figure 36: Sequence ofNumbers for PAM

• These numbers may come from sampling a continuous-time signal m(t).Alternatively, it may directly represent (digital) information that in-trinsically available in discrete-time.

• Because the m[n] may not come from sampling, we call each m[n] asymbol.

We use these numbers to modify (modulate) the height (amplitude) of apulse train. A single pulse is denoted by p(t). This pulse occurs every T

seconds.PAM: Pulse Amplitude Modulation

1

p t

T 2T 3T 4T-T-2T-3T-4Tt

t-.5T

1

.5T

Pulse

m[0]m[1]

m[2]

m[3]m[-1]

m[-2]

m[-3]

m[-4]

Figure 37: PAM

The PAM signal is then

x (t) =∞∑

n=−∞m [n] p (t− nT )

94

This signal is transmitted via the communication channel which usuallycorrupts it. At the receiver, the received signal is y(t). A more advancedreceiver would try to first cancel the effect of the channel. However, forsimplicity, let’s assume that our receiver simply samples y(t) every T secondsto get

y[n] = y(t)|t=nTand we will take this to be the estimate m[n] of our m[n].

• If m[n] is the sampled version of m(t), then at the receiver, after werecover m[n], we can reconstruct m(t) by using the reconstructing equa-tion (58).

Because our assumed receiver is so simple, we are going to also assume25

that y(t) = x(t).

7.2. In this section, our goal is to design a “good” pulse p(t) that satisfiestwo important properties

(a) m[n] = m[n] for all n. Under our assumptions above, this means wewant x[n] ≡ x(nT ) = m[n] for all n.

(b) P (f) is band-limited and hence X(f) is band-limited.

We will first give examples of “poor” p(t).

Example 7.3. Let’s consider the rectangular pulse used in Figure 37:

p(t) = 1[|t| ≤ T/2].

(a) m[n] = m[n] for all n.

(b) The Fourier transform of the rectangular pulse is a sinc function. So,it is not band-limited.

Example 7.4 (Slide). Let’s try a wider rectangular pulse:

p(t) = 1[|t| ≤ 1.5T ].

Here, we face a problem called inter-symbol interference (ISI) in oursequence m[n] at the receiver. The pulses are too wide; they interfere withother pulses at the sampling time instants (decision making instants), mak-ing m[n] 6= m[n].

25Alternatively, we may assume that there is an earlier part of the receiver that (perfectly) eliminatesthe effect of the channel for us.

95

Example 7.5 (slide). p(t) = 1[|t| ≤ T/4].

• When the pulse p(t) is narrower than T , we know that the pulses inPAM signal will not overlap and therefore we won’t have any ISI prob-lem.

Example 7.6 (slides).

p t

t

1

34

34

p t

tT-T

1 p t

tT-T

1

2T-2T

Figure 38: Examples of pulses that do not cause ISI.

• Even when the pulses are wider than T , if they do not interfere withother pulses at the sampling time instants (decision making instants),we can still have no ISI.

7.7. We can now conclude that a “good” pulse satisfying condition (a)in 7.2 must not cause inter-symbol interference (ISI): at the receiver,the nth symbol m[n] should not be affected by the preceding or succeedingtransmitted symbol m[k], k 6= n. This requirement means that a “good”pulse should have the following property:

p (t)|t=nT =

1, n = 0,0, n 6= 0.

(59)

Combining this with condition (b) in 7.2, we then want “band-limited pulsesspecially shaped to avoid ISI (by satisfying (59))” [3, p 506].

7.8. An obvious choice for such p(t) would be the sinc function that weused in the reconstruction equation (58):

96

Recall Figure 33, repeated here (with modified labels) as Figure 39.

1 2 3 4 5 6 7 80

2

4

6

8

n

m[n

]

1 2 3 4 5 6 7 8-2

0

2

4

6

8

t [Ts]

x(t)

Figure 39: Using the sinc pulse inPAM

Practically, there are problems that force us to seek better pulse shape.

(a) Infinite duration

(b) Steep slope at each 0-intercept.

(c) maxtx (t) could be a lot larger than max

nm [n].

2 4 6 8 10 12 14 16-2

-1

0

1

2

n

m[n

]

2 4 6 8 10 12 14 16-2

-1

0

1

2

3

t [Ts]

x(t)

Figure 40: Using the sinc pulse inPAM can cause high peak.

97

7.9. Because the sinc function may not be a good choice, we now have toconsider other pulses that are band-limited and also satisfy (59). To checkthat a signal is band-limited, we need to look in the frequency domain.However, condition (59) is specified in the time domain. Therefore, we willtry to translate condition (59) into a requirement in the frequency domain.

7.10. Note that condition (59) considers p (t)|t=nT which can be thought ofas the samples p[n] of the pulse p(t) where the sampling period is Ts = T .Recall, from (57), that

gδ (t) =∞∑

n=−∞g [n]δ (t− nTs)

F−−−−F−1

Gδ (f) = fs

∞∑k=−∞

G (f − kfs).

Therefore,

pδ (t) =∞∑

n=−∞p [n]δ (t− nT )

F−−−−F−1

Pδ (f) =1

T

∞∑k=−∞

P

(f − k

T

). (60)

On the LHS, by condition (59), the only nonzero term in the sum is theone with n = 0. Therefore, condition (59) is equivalent to pδ (t) = δ(t).

However, recall that δ(t)F−−−−F−1

1. Therefore, we must have Pδ (f) ≡ 1.

Hence, to check condition (59), we can equivalently check that the RHS of(60) must be ≡ 1.

Note that Pδ(f) is “periodic” (in the freq. domain) with “period” 1T .

(Recall that Gδ(f) is “periodic” (in the freq. domain) with “period” fs.)Therefore, the checking does not need to be performed across all frequencyf . We only need to focus on one period: |f | ≤ 1

2T .This observation is formally stated as the “Nyquist’s criterion” below.

7.11. Nyquist’s (first) Criterion for Zero ISI: A pulse p(t) whoseFourier transform P (f) satisfies the criterion

∞∑k=−∞

P

(f − k

T

)≡ T, |f | ≤ 1

2T(61)

has sample values satisfying condition (59):

p[n] = p (t)|t=nT =

1, n = 0,0, n 6= 0.

98

• Using this pulse, there will be no ISI in the sample values of

y (t) =∞∑

n=−∞m [n] p (t− nT )

Definition 7.12. A pulse p(t) is a Nyquist pulse if its Fourier transformP (f) satisfies (61) above.

Example 7.13. We know that the sinc pulse we used in Example 7.8 works(causing no ISI). Let’s check it with the Nyquist’s criterion:

Example 7.14.

Example 7.15.

99

Example 7.16.

Example 7.17. An important family of Nyquist pulses is called the raisedcosine family. Its Fourier transform is given by

PRC (f ;α) =

T, 0 ≤ |f | ≤ 1−α

2TT2

(1 + cos

(πTα

(|f | − 1−α

2T

))), 1−α

2T ≤ |f | ≤1+α2T

0, |f | ≥ 1+α2T

with a parameter α called the roll-off factor.

1

f

T

2

12

12

Figure 41: Raised cosinepulse (in the frequencydomain)

1

f

T

2

12

12

00.51

34

1

Figure 42: Raised co-sine pulse (in the fre-quency domain) with dif-ferent values of the roll-off factor

100

1

t

00.5

1

1

22

Figure 43: Raised cosinepulse (in the time do-main) with different val-ues of the roll-off factor

1

n

x t m n p t nT

0 0.5 1 1.5 2 2.5 3 3.5-1.5

-1

-0.5

0

0.5

1

1.5

t

0 0.5 1 1.5 2 2.5 3 3.5-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

t

0 0.5 1 1.5 2 2.5 3 3.5-1.5

-1

-0.5

0

0.5

1

1.5

t

;0RCp t p t

;1RCp t p t

;0.5RCp t p t

a)

b)

c)

Figure 44: Using the raised cosine pulses in PAM

101

8 Introduction to Digital Communications

8.1 Digitization and PCM

8.1. Generally, analog signals are continuous in time and in range (ampli-tude); that is, they have values at every time instant, and their values canbe anything within the range. On the other hand, digital signals exist onlyat discrete points of time, and their amplitude can take on only finitely (orcountably) many values.

8.2. The analog-to-digital (A/D) converter or ADC enables digitalcommunication systems to convey analog source signals such as audio andvideo.

8.3. Suppose we want to convey an analog message m(t) from a source toour destination. We now have many options.

(a) Use m(t) to modulate a carrier A cos(2πfct) using AM, FM, or PMtechniques studied earlier.

(b) Sample the continuous-time message m(t) to get a discrete-time mes-sage m[n].

Note that m[n] is a sequence of numbers. (There are uncountably manypossibilities for these numbers).

(i) Send m[n] using analog pulse modulation techniques (PAM, PWM,PPM).

(ii) Quantize m[n] into mq[n] which has finitely (or countably) manylevels.

i. Send mq[n] using pulse modulation techniques (PAM, PWM,PPM).

ii. Pulse Code Modulation (PCM): Convert mq[n] into binarysequence. Then use two basic pulses to represent 1 and 0.

102

Digitization (analog to digital)

1

Time

Vertical lines are used for sampling

Horizontal lines are used for quantization

001

000

010

011

100

101

110

111111

100

100

111

011

100

101

010

000

001 001

100111111100001000010100011001

Figure 45: An overviewof digitization (sampling+ quantizing) and PCM

Definition 8.4. Through quantization, each sample is approximated, or“rounded off,” to the nearest quantized level [5, p 320] or quantum level[3, p 545] (permissible number).

• This process introduces permanent errors that appear at the receiveras quantization noise in the reconstructed signal.

Example 8.5. Simple quantizer: Suppose amplitudes of the message signalm(t) lie in the range (−mp,mp). A simple quantizer may partition thesignal range into L intervals. Each sample amplitude is approximated bythe midpoint of the interval in which the sample value falls. Each sample isnow represented by one of the L numbers.

• Such a signal is known as an L-ary digital signal.

8.6. From practical viewpoint, a binary digital signal (a signal that cantake on only two values) is very desirable because of its simplicity, economy,and ease of engineering. We can convert an L-ary signal into a binary signalby using pulse coding.

• A binary digit is called a bit.

• L = 2` levels cam be mapped into (represented by) ` bits.

103

Example 8.7. Suppose L = 8. The binary code can be formed by thebinary representation of the 8 decimal digits from 0 to 7.

Example 8.8. Telephone (speech) signal:

• The components above 3.4 kHz are eliminated by a low-pass filter.

For speech, subjective tests show that signal intelligibility is notaffected if all the components above 3.4 kHz are suppressed.

• The resulting signal is then sampled at a rate of 8,000 samples persecond (8 kHz).

This rate is intentionally kept higher than the Nyquist samplingrate of 6.8 kHz so that realizable filters can be applied for signalreconstruction.

• Each sample is finally quantized into 256 levels (L = 256), which re-quires eight bits to encode each sample (28 = 256).

[5, p 320]

Example 8.9. Compact disc (CD) audio signal:

• High-fidelity: Require the audio signal bandwidth to be 20 kHz.

• The sampling rate of 44.1 kHz is used.

• The signal is quantized into L = 65, 536 of quantization levels, each ofwhich is represented by 16 bits (16-bit two’s complement integer) toreduce the quantizing error.

[5, p 321]

104

8.2 Digital PAM Signals

8.10. Digital message representation (at baseband) commonly takes theform of an amplitude-modulated pulse train. We express such signals bywriting

x(t) =∑n

m[n]p(t− nTs)

where the modulating amplitude m[n] represents the nth symbol in themessage sequence.

• The amplitudes belong to a set A of M discrete values called the al-phabet.

8.11. Note that Ts does not necessarily equal the pulse duration but ratherthe pulse-to-pulse interval or the time allotted to one symbol. Thus, thesignaling rate (or symbol rate) is Rs = 1

Ts. measured in symbols per

second, or baud.

• In the special but important case of binary signaling (M = 2), we writeTs = Tb for the bit duration and the bit rate is Rb = 1

Tb.

8.12. Figure 46 depicts various PAM formats or line codes for the binarymessage 10110100, taking rectangular pulses for clarity.

(a) The simple on-off waveform in Figure 46a represents each 0 by an “off”pulse and each 1 by an “on” pulse.

(i) In the a return-to-zero (RZ) format, the pulse duration is smallerthan Tb after which the signal return to the zero level.

(ii) A nonreturn-to-zero (NRZ) format has “on” pulses for full bitduration Tb.

(b) The polar signal in Figure 46b has opposite polarity pulses

• Its DC component will be zero if the message contains 1s and 0sin equal proportion.

(c) Figure 46c, we have bipolar signal where successive 1s are representedby pulses with alternating polarity.

• Also known as pseudo-trinary or alternate mark inversion (AMI)

105

(d) The split-phase Manchester format in Figure 46d represents 1s witha positive half-interval pulse followed by a negative half-interval pulse,and vice versa for the representation of 0s.

• Also called twinned binary.

• Guarantee zero DC component regardless of the message sequence.

(e) Figure 46e shows a quaternary signal derived by grouping the mes-sage bits in blocks of two and using four amplitude levels to prepresentthe four possible combinations 00, 01, 10, and 11.

• Quaternary coding can be generalized to M-ary coding in whichblocks of n message bits are represented by an M-level waveformwith M = 2n.

(a)

(b)

(c)

(d)

(e)

Tb Tb

A/2

– A/2

t

t

t

t

t

1

0

0

A

– A

0

A/2

A/2

– A/2

– A/2

– 3A/2

0

0

0 1 1 0 1 0

RZ

0

1 0 1 1 0 1 0 0

1 0 1 1 0 1 0 0

A

Ts

NRZ

3A/2

Figure 11.1–1 Binary PAM formats with rectangular pulses: (a) unipolar RZ and NRZ; (b)polar RZ and NRZ; (c) bipolar NRZ; (d) split-phase Manchester; (e) polar quaternary NRZ.

11.1 Digital Signals and Systems 483

Finally, Fig. 11.1–1e shows a quaternary signal derived by grouping the mes-sage bits in blocks of two and using four amplitude levels to prepresent the four pos-sible combinations 00, 01, 10, and 11. Thus, D 2Tb and . Differentassignment rules or codes may relate the ak to the grouped message bits. Two suchcodes are listed in Table 11.1–1. The Gray code has advantages relative to noise-induced errors because only one bit changes going from level to level.

Quaternary coding generalizes to M-ary coding in which blocks of n messagebits are represented by an M-level waveform with

(4a)

Since each pulse now corresponds to n log2 M bits, the M-ary signaling rate hasbeen decreased to

M 2n

r rb>2

car80407_ch11_479-542.qxd 12/17/08 6:35 PM Page 483

Confirming Pages

Figure 46: Line codes withrectangular pulses: (a) unipo-lar RZ and NRZ; (b) po-lar RZ and NRZ; (c) bipolarNRZ; (d) split-phase Manch-ester; (e) polar quaternaryNRZ.

106

8.3 Digital PAM with Noise

8.13. In this section, the transmitted signal is still in the form of digitalPAM as in the previous subsection:

x(t) =∑n

m[n]p(t− nTs)

However, here, we also consider the effect of additive noise. Therefore, thereceived signal is

y(t) = x(t) +N(t)

where N(t) is a random noise process.

0 1 2 3 4 5 6 7 8 9-1

-0.5

0

0.5

1

t [T]

x(t)

0 1 2 3 4 5 6 7 8 9-5

0

5

t [T]

y(t)

PAM with Noise

1

m[0] = 1

m[1] = -1

m[2] = -1

Figure 47: Digital PAM with Noise

8.14. Note that

• The noise N(t) is random.

• The message m[n] should be random (at least from the perspective ofthe receiver; if the receiver had known in advance the value of m[n],there would have been no point of transmitting m[n]).

107

This makes x(t) and y(t) random.

To emphasize the randomness in the signals under consideration, wesometimes write M [n], X(t), and Y (t) using capital letters26.

8.15. Simple receiver: For simplicity, let’s assume that our receiver sim-ply samples y(t) every T seconds to get

y[n] = y(t)|t=nT .

When the alphabet A contains only two symbols of opposite sign (A =−a, a, where a > 0), the decoded value m[n] of our m[n] can be found by

m [n] =

a, y [n] ≥ 0,−a, y [n] < 0.

Here we use “0” as the threshold level/value. Turn out that this middlepoint is the optimal threshold to use when the two possible symbol valuesfrom the alphabet are equally likely.

Example 8.16. In Figure 47, A = −1, 1.

n 0 1 2 3 4 5 6 7 8 9

m[n] 1 -1 -1 1 -1 1 1 1 1 -1

y[n] 1.69 -0.27 -0.81 4.20 1.58 1.04 -0.34 0.35 2.19 -1.51

m[n] 1 -1 -1 1 1 1 -1 1 1 -1

Note that the value of the noise N(t) at t = 4T is too positive. Even when“-1” was transmitted, the received value is 1.58 which exceeds 0. Therefore,we get an error at n = 4.

Similarly that the value of the noise N(t) at t = 6T is too negative. Evenwhen “1” was transmitted, the received value is −0.34 which is lower than0. Therefore, we get an error at n = 6.

Among the ten symbols sent in this example, there are two symbol errors.Therefore, the symbol error rate (SER) or symbol error probability is2/10.

Because there are two symbols in the alphabet, each symbol transmissionconveys 1 bit. Hence, the bit error rate (BER) or bit error probability isalso 2/10.

26Caution: Here, capital letters represent random variables/processes. In earlier sections, we used capitalletter to represent Fourier transform. However, we won’t talk about Fourier transform here; so confusioncan be avoided.

108

8.17. Additive White Gaussian Noise: At each time instant t, thenoise N(t) is usually modeled by a Gaussian random variable with mean 0and standard deviation σN ,

fN(t)(n) =1√

2πσNe− 1

2

(nσN

)2.

Furthermore, the “white” part means that the noise values at different timeinstants are independent.

Definition 8.18. In general, a Gaussian (normal) random variable Xwith mean m and standard deviation σ is characterized by its probabilitydensity function (PDF):

fX(x) =1√2πσ

e−12(

x−mσ )

2

.

To talk about such X, we usually write X ∼ N (m,σ2). Probability involv-ing X can be evaluated by

P [X ∈ A] =

∫A

fX(x)dx.

In particular,

P [X ∈ [a, b]] =

∫ b

a

fX(x)dx = FX(b)− FX(a)

where FX(x) =∫ x−∞ fX(t)dt is called the cumulative distribution function

(CDF) of X.We usually express probability involving Gaussian random variable via

the Q function which is defined by

Q (z) =

∞∫z

1√2πe−

x2

2 dx.

Note that Q(z) is the same as P [Z > z] where Z ∼ N (0, 1); that is Q (z)is the probability of the “tail” of N (0, 1).

It can be shown that

• Q is a decreasing function

109

( )0,1N

z 0

( )Q z

-3 -2 -1 0 1 2 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

z

10,2

⎛ ⎞⎜ ⎟⎝ ⎠

N

z

( )erf z

0

( )2Q z

Figure 48: Q-function

• Q (0) = 12

• Q (−z) = 1−Q (z)

This is useful for converting the argument of the Q function topositive value.

• For X ∼ N (m,σ2),

P [X > c] = Q

(c−mσ

).

8.19. Three important noise probabilities for N ∼ N (0, σ2N):

P [N > c] = , P [N < c] = , P [a < N < b] =

Note that all strict inequalities above can also be replaced by the onesthat also include equalities because the noise is a continuous random variableand hence including one particular noise value does not change probability.

8.20. For the simple receiver in 8.15, suppose N(t) ∼ N (0, σ2).

(a) When a “−a” was transmitted, error occurs when N(t) > a

(b) When a “+a” was transmitted, error occurs when N(t) < −a

110

A formula that connects these events with the (combined) error proba-bility is called the Total Probability Theorem : If a (finite or infinitely)countable collection of events B1, B2, . . . is a partition of Ω, then

P (A) =∑i

P (A|Bi)P (Bi). (62)

In particular,

P (A) = P (A|B)P (B) + P (A|Bc)P (Bc).

Here, we replace event A by the error event E . Event B is defined to bethe event that the transmitted symbol is “a”. The error probability is then

Example 8.21. In a digital PAM system, equally-likely symbols are selectedfrom an alphabet set A = −4, 4. The pulse used in the transmitted signalis a Nyquist pulse. The additive noise at each particular time instant isGaussian with mean 0 and standard deviation 2.

(a) Find the probability that the received signal at a particular time is > 6.

(b) Find the symbol error probability when 0 is used as the threshold levelfor the decoding decision at the receiver.

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