Chapter 10 FOURIER TRANSFORMS...Chapter 10 FOURIER TRANSFORMS...

Chapter 10

FOURIER TRANSFORMS

Fourier transforms express a givenaperiodicfunction as alinear combinationof complex exponential functions. Unlike the Laplace transform, the functionis not restricted to be zero for negative times.

The main advantages of the Fourier transform are similar to those of theFourier series, namely (a) analysis of the transform ismuch easierthan analysisof the original function, and, (b) the transform allows us to view the signal in thefrequency domain. Unlike the Fourier series, since the function is aperiodic,there is no fundamental frequency. The frequency spectrum of the signal willcontaincontinuousfrequencies, not just multiples of a fundamental frequency.

The Fourier transform is most useful in characterizing the system that pro-duces an output signal from an input signal. The system is usually defined by adifferential equation, and is never periodic. However, because of the approxi-mation properties of the Fourier series, the input signals can be represented bysums of periodic signals. The combination of Fourier transforms and Fourierseries is extremely powerful.

1. Introduction1.1 Definition of the transform and spectrum

Definition: Consider a signalv(t), wheret ∈ (−∞,∞). Letω be a real number.The Fourier transform,V (ω), of the signalv(t) is defined by the integral

V (ω) =∫ ∞

−∞v(t)e−jωtdt (10.1)

4This integral exists whenever

515

516 MATHEMATICS: THE LANGUAGE OF ECE

∫ ∞

−∞|v(t)|dt < ∞

Like Fourier series, evaluation of the Fourier transform in Equation 10.1can be done by direct integration or (in a much easier fashion) by using theproperties of the transform (see Section 3).

Given the complex-valued functionV (ω), the functionv(t) can be found viathe inverse Fourier transform:

v(t)4=

12π

∫ ∞

−∞V (ω)ejωtdω. (10.2)

Note that the variable of integration is real, unlike the case of the inverseLaplace transform. We have seen how to calculate such integrals already, inChapter 9.

The Fourier transform can be defined to directly represent frequency in Hertzby the transformation pair

V (F ) =∫ ∞

−∞v(t)e−j2πFtdt,

v(t) =∫ ∞

−∞V (F )ej2πFtdF.

We will tend to use radial frequencyω, for computations, since it is morecompact.

The Fourier transform,V (ω), of a signalv(t), is a complex-valued functionof the real variableω (see Chapter 4, Section 6.1). As with the Fourier seriescase, for any fixed value ofω, the complex numberV (ω) can be represented viaits real and imaginary parts, or, as it is more usual in practice, via its magnitudeand phase.

Definition: The real-valued functions|V (ω)| and 6 V (ω) are called themag-nitude spectrumandphase spectrumof the signalv(t), respectively. Together,these two functions are called thefrequency spectrum(or simply the spectrum)of v(t).

4Since the parameterω takes both positive and negative values, the plot of

the spectrum should cover both ranges. However, if the signalv(t) is real-valued, which is almost always the case in our applications, for a fixedω, thecomplex numbersV (ω) andV (−ω) areconjugatesof each other; the proof ofthis property is left as an exercise. As with the Fourier series case, we have that

|V (−ω)| = |V (ω)|, 6 V (−ω) = −6 V (ω)

Fourier transforms 517

i.e., the magnitude spectrum is anevenfunction ofω and the phase spectrumis anodd function ofω. In the sequel, when we plot the spectrum, most of thetime we will only show positive values ofω.

The power spectrum of an aperiodic signal is defined in a manner analogousto that of a periodic one in Chapter 9, Section 3.3.

Definition: The power spectrum,Ps(ω), of a signals(t) with Fourier transformS(ω), is defined as

Ps(ω) = |S(ω)|2, ω ∈ (−∞,∞) (10.3)

As with the Fourier series case (see Equation 9.72, page 478), the powerspectrum can also be expressed in decibels:

Ps(ω) = 20 log(|S(ω)|) (dB), ω ∈ (−∞,∞) (10.4)

4

1.2 Relationship to Laplace transform and Fourier seriesThe Fourier transform is related to both the Laplace transform and Fourier

series.

1.2.1 Relationship to Laplace transform

Note the similarity of definition 10.1 to the Laplace transform. Ifv(t) = 0for t < 0, the Laplace transformLv(s) is also defined1. In this case,

V (ω)4=

∫ ∞

−∞v(t)e−jωtdt =

∫ ∞

0v(t)e−jωt = Lv(jω)

Therefore,

V (ω) = Lv(s)|s=jω (10.5)

Equation 10.5 says that the Fourier transform can be found from the Laplacetransform by the substitutions = jω. Inversely, the Laplace transform can befound from the Fourier transform by the substitutionω = s/j.

Example 10.1.The Laplace transform of the functionv(t) = eatu(t) wasfound to be

1In Chapter 8, we denoted the Laplace transform ofv(t) asV (s). We change the notation here to avoidconfusion, since we useV (ω) to denote the Fourier transform ofv(t).


Lv(s) =1

s− a

(see Table 1, page 404 in Chapter 8). Therefore, its Fourier transform is

V (ω) =1

jω − a

————————————————————————————-4

1.2.2 Relationship to Fourier series

The Fourier series coefficients of a periodic signal,sp(t), with periodT , aredetermined by

Sp(k) =1T

∫ T/2

−T/2sp(t)e−j 2π

Tktdt

Consider an aperiodic signals(t) that is equal to zero outside an interval(−T/2, T/2] (and takes arbitrary values inside(−T/2, T/2]). LetS(ω) denotethe Fourier transform ofs(t). Define theperiodic extensionof s(t) as theperiodic signalsp(t), whose period is equal toT and

sp(t) = s(t),−T/2 < t ≤ T/2

The Fourier series coefficientsSp(k) can be found from the Fourier transformS(ω), as follows:

Sp(k) =1T

∫ T/2

−T/2sp(t)e−j 2π

Tktdt

=1T

∫ T/2

−T/2s(t)e−j 2π

Tktdt

=1T

∫ ∞

−∞s(t)e−j 2π

Tktdt

=1T

S

(2πk

T

)(10.6)

Therefore,

Sp(k) =1T

S

(2πk

T

)(10.7)

Equation 10.7 says that the Fourier seriesSp(k) can be found from the FouriertransformS(ω) by dividing it byT and substitutingω = 2πk

T . This is basically


sampling the Fourier transform at an interval ofω0 = 2π/T . As you will learnin later courses, it is possible to reconstruct a signal from samples only underspecial conditions. We cannot, in general, go from the Fourier series to theFourier transform by the inverse substitutionk = Tω/2π.

2. Fourier transforms of common signalsLet’s see now how we can calculate the Fourier transform of some common

signals.

Example 10.2.Letv(t) = δ(t− t0)

wheret0 is a given real number. We have

V (ω) =∫ ∞

−∞δ(t− t0)e−jωtdt

= e−jωt0 (10.8)

Example 10.3. Let

V (ω) = πδ(ω − ω0) + πδ(ω + ω0). (10.9)

Then

v(t) =12π

∫ ∞

−∞V (ω)ejωtdω

=12

∫ ∞

−∞δ(ω − ω0)ejωtdω +

12

∫ ∞

−∞δ(ω + ω0)ejωtdω

=12ejω0t +

12e−jω0t = cos(ω0t)

From the uniqueness property (discussed in Section 3), we recognizeV (ω)as the Fourier transform of the (periodic)cos(ω0t) signal.

————————————————————————————-4

Example 10.4. Consider the centered unit pulse of durationτ defined by

pτ (t) =

1, −τ/2 ≤ t ≤ τ/2,0, otherwise.

(10.10)

We have


Pτ (ω) =∫ ∞

−∞pτ (t)e−jωtdt

=∫ τ/2

−τ/2e−jωtdt

=1

−jωe−jωt

∣∣∣τ/2

−τ/2

=1

−jω

(e−jωτ/2 − e−jω(−τ/2)

)

=1jω

(ejωτ/2 − e−jωτ/2

)

=2ω

sin(ωτ/2) = τsin(ωτ/2)

ωτ/2(10.11)

So, finally,

Pτ (ω) = τsin(ωτ/2)

ωτ/2(10.12)

A plot of the spectrum ofPτ (ω) in Equation 10.12, forτ = 2 is given inFigure 10.1. Note that the plot is versus frequencyF in Hertz, not angularfrequencyω. Only positive values off are shown in the spectrum. Note thatthe zeros in the magnitude spectrum appear at the frequenciesF for whichsin(ωτ/2) = 0. Therefore, the magnitude spectrum is equal to zero for thefrequencies

F =1τ,2τ,3τ, · · · ,

whereτ is the width of the pulse.————————————————————————————-4

Example 10.5.Consider the signal

s(t) =

t, t ∈ [−T/2, T/2)0, otherwise.

(10.13)

whereT > 0 is a known parameter. This signal is equal to one period of theperiodic signal in Example 9.11.

The Fourier transform ofs(t) can be found as follows:

S(ω) =∫ ∞

−∞s(t)e−jωtdt


0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

frequency (Hz)

Mag

nitu

de

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

3

3.5

frequency (Hz)

Pha

se

Figure 10.1. The spectrum of the signal in Equation 10.12.

=∫ T/2

−T/2te−jωtdt

=1

(−jω)2e−jωt ((−jω)t− 1)

∣∣∣T/2

−T/2

=1ω2

e−jωt (jωt + 1)∣∣∣T/2

−T/2

=1ω2

e−jω T2

(jω

T

2+ 1

)− 1

ω2ejω T

2

(−jω

T

2+ 1

)

=1ω2

(jω

T

2

) [e−jω T

2 + ejω T2

]+

1ω2

[e−jω T

2 − ejω T2

]

=jT cos(ωT/2)

ω− 2j sin(ωT/2)

ω2(10.14)

The plot of this spectrum forT = 2 is shown in Figure 10.2. Note again thatthe plot is versus frequencyF in Hertz, not angular frequencyω.


0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

frequency (Hz)

Mag

nitu

de

0 1 2 3 4 5 6 7 8 9 10−2

−1

0

1

2

frequency (Hz)

Pha

se

Figure 10.2. The spectrum of the signal in Equation 10.13.

————————————————————————————-4

*Example 10.6.The Fourier series of the periodic ramp, one period of whichis defined as

sp(t) = t, t ∈ [−T/2, T/2)

can be evaluated now using Equation 10.7. We have, from Equation 10.14:

Sp(k) =1T

[jT cos(ωT/2)

ω− 2j sin(ωT/2)

ω2

]∣∣∣∣ω= 2πk

T

=1T

jT cos(

2πkT

T2

)

2πkT

−2j sin

(2πkT

T2

)

(2πkT

)2

=jT cos(πk)

2πk


which agrees with Equation 9.57, page 466.If we attempt to recover the transform from the series, we would start with

Equation 9.57. The substitutionk = Tω/2π will give

Sp(k) =j

kω0cos(kπ).

We could derive Equation 10.14 as

S(ω) = TSp(k)|k=Tω2π

= T · j

kω0cos(kπ)

∣∣∣∣k=Tω

2π

= T · jTω2π

2πT

cos(

Tω

2ππ

)

=jT

ωcos

(ωT

2

)

and we see that we cannot recover the term with thesin(ωT/2) in Equation10.14.

————————————————————————————-4Table 10.2 summarizes the Fourier transforms of some common signals.

Table 10.1. Fourier transforms of common signals.

v(t) V (ω)

δ(t) 1

sin(ω0t)12j

δ(ω − ω0)− 12j

δ(ω + ω0)

cos(ω0t)12δ(ω − ω0) + 1

2δ(ω + ω0)

eatu(t) 1jω−a

eat sin(ω0t)u(t) ω0(jω−a)2+ω2

0

eat cos(ω0t)u(t) jω−a

(jω−a)2+ω20

3. PropertiesThe Fourier transform, like the Laplace transform and Fourier series, pos-

sesses similar properties.

3.1 UniquenessFor each given functionv(t), there exists one and only one functionV (ω),

defined via Equation 10.1. Inversely, given a functionV (ω), there exists one


and only one functionv(t), defined via Equation 10.2, that has the functionV (ω) as its Fourier transform.

3.2 LinearityThe following theorem is the analogue of theorems 8.1 and 9.1; it can be

proven in an entirely similar fashion:

Theorem 10.1 Consider given constantsci (real or complex valued) andfunctionsvi(t) with known Fourier transformsVi(ω). Let

v(t) =∑

i

civi(t) (10.15)

denote a linear combination of the functionsvi(t). Then

V (ω) =∑

i

ciVi(ω) (10.16)

4

3.3 Time-shiftingLett0 be a given real-valued constant, positive or negative (unlike the Laplace

transform). The analogue of theorems 8.2 and 9.2 is:

Theorem 10.2 Consider a signalv(t), with known Fourier transformV (ω).Let y(t) = v(t − t0), wheret0 is a given real-valued constant. The Fouriertransform of the time-shifted signal is given by

Y (ω) = e−jωt0V (ω) (10.17)

4

Proof: We can write

Y (ω)4=

∫ ∞

−∞y(t)e−jωtdt =

∫ ∞

−∞v(t− t0)e−jωtdt

=∫ ∞

−∞v(t− t0)e−jωt0ejωt0e−jωtdt

= e−jωt0

∫ ∞

−∞v(t− t0)e−jω(t−t0)dt

= e−jωt0

∫ ∞

−∞v(t)e−jωtdt = e−jωt0V (ω)


Example 10.7. Let

y(t) = cos(ω0t + θ) (10.18)

be a cosine signal with a phase shift. FindY (ω), the Fourier transform ofy(t).4

Let v(t) = cos(ω0t). In Example 10.3, we have already calculated theFourier transformV (ω). Using Theorem 10.2, we can calculateY (ω) as fol-lows. Rewrite Equation 10.18 as

y(t) = cos(ω0(t + θ/ω0))

The time-shift is then equal to

t0 = −θ/ω0

From Equation 10.17 and Equation 10.9, page 519, we can now write

Y (ω) = e−jωt0V (ω)

= ejωθ/ω0 [πδ(ω − ω0) + πδ(ω + ω0)]= ejθπδ(ω − ω0) + e−jθπδ(ω + ω0) (10.19)

In deriving Equation 10.19, we have used the property of theδ(t) functionin Equation 2.17, page 61.

————————————————————————————-4

3.4 Time-reversalConsider a signalv(t), with known Fourier transformV (ω).

Theorem 10.3 Lety(t) = v(−t), be the time-reversed version ofv(t). TheFourier transform of the time-reversed signal is given by

Y (ω) = V (−ω) (10.20)

4

Proof: We have by definition

Y (ω)4=

∫ ∞

−∞y(t)e−jωtdt

=∫ ∞

−∞v(−t)e−jωtdt


=∫ −∞

∞v(u)e−jω(−u)(−du) (10.21)

= −∫ −∞

∞v(u)e−j(−ω)udu

=∫ ∞

−∞v(u)e−j(−ω)udu = (10.22)

= V (−ω)

4

3.5 Time-scalingConsider a signalv(t), with known Fourier transformV (ω). Let y(t) =

v(at), wherea 6= 0 is a given constant. The Fourier transform can be calculatedas follows. Consider first the casea > 0. We have by definition

Y (ω)4=

∫ ∞

−∞y(t)e−jωtdt

=∫ ∞

−∞v(at)e−jωtdt

=∫ ∞

−∞v(u)e−jω u

adu

a=

1a

∫ ∞

−∞v(u)e−j ω

audu (10.23)

=1aV (ω/a) (10.24)

In Equation 10.23, we have made the substitution of variablesu = at.The case wherea < 0 can be derived from Equations 10.20 and 10.24, by

observing that we can writea = −|a|. Then the signal

y1(t) = v(|a|t)has Fourier transformY1(ω) = 1

|a|V (ω/|a|), and the signal

y(t) = y1(−t)

has Fourier transformY (ω) = 1|a|V (−ω/|a|) = 1

|a|V (ω/a). Therefore,

Theorem 10.4 Consider a signalv(t), with known Fourier transformV (ω).Let y(t) = v(at), wherea 6= 0 is a given constant. The Fourier transform isgiven by

Y (ω) =1|a|V (ω/a) (10.25)

4


3.6 DerivativesThe Fourier transform of then-th derivative of a signal can also be related

to the Fourier transform of the signal itself. The analogue of theorems 9.5 and8.4 is the following.

Theorem 10.5 Consider a signalv(t) with known Fourier transformV (ω).The Fourier transform ofyk(t) = dkv(t)/dtk, thek-th order derivative of thesignalv(t), is given by

Yk(ω) = (jω)kV (ω) (10.26)

4

3.7 Examples

*Example 10.8.Determine the Fourier transform of the signals(t) in Equation 10.27

s(t) =

0, t < −1,−1, −1 ≤ t < 0,2, 0 ≤ t ≤ 2,0, t > 2.

(10.27)

using (a) the direct integration approach and (b) the properties of the Fouriertransform. The signal is shown in Figure 10.3.

Note that when applying the properties, we can determine this transform:

1 using the linearity/time-shifting properties and the transform of (two) cen-tered pulsespτ (t) (of width τ ):

pτ (t) =

1, −τ/2 ≤ t ≤ τ/2,0, otherwise.

(10.28)

One pulse,p1(t), of width 1, would cover the region−1 ≤ t < 0; the other,p2(t), of width 2, would cover the region0 ≤ t ≤ 2.

2 using the linearity/time-scaling properties and the transform of one cen-tered pulsep(t) = p1(t) with unit width. The transform of the pulses wascalculated in Example 10.4.

Approach 1. From Example 10.4, we have

Pτ (ω) = τsin(ωτ/2)

ωτ/2(10.29)


−5 −4 −3 −2 −1 0 1 2 3 4 5−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

3The signal s(t)

s(t)

time t

Figure 10.3. The signals(t) in Example 10.8.

We can write the signal in Equation 10.27 using the two simpler signals inEquation 10.28 as follows:

s(t) = −p1(t + 0.5) + 2p2(t− 1) (10.30)

Equation 10.30 simply says that the signals(t) consists of two pulses: oneof width τ = 1, shifted to the left by 0.5 time units and another of widthτ = 2,shifted to the right by 1 time unit. A decomposition like this would avoid usinga time scaling property along with time shifting.

Therefore (applying the linearity and time-shifting properties of the Fouriertransform) and using Equation 10.29:

S(ω) = −ej0.5ωP1(ω) + 2e−jωP2(ω)


= −ej0.5ω1 · sin(ω · 1/2)ω · 1/2

+ 2e−jω2sin(ω · 2/2)

ω · 2/2

= −ej0.5ω sin(ω/2)ω/2

+ 4e−jω sin(ω)ω

(10.31)

Approach 2. A decomposition that uses time scaling along with time shiftingcan be obtained by decomposing the signal in terms ofp1(t) only. The timescaling

y(t) = p1(t/2)

stretches the unit pulse from−1 to +1. We can then write

s(t) = −p1(t + 0.5) + 2y(t− 1) = −p1(t + 0.5) + 2p1(1/2(t− 1)) (10.32)

Then

S(ω) = −ej0.5ωP1(ω) + 2e−jωY (ω) (10.33)

where

P1(ω) =sin(ω/2)

ω/2(10.34)

and from the time-scaling property

Y (ω) = 2P1(2ω) = 2sin(ω)

ω(10.35)

Combining Equations 10.33, 10.34 and 10.35 we get

S(ω) = −ej0.5ω sin(ω/2)ω/2

+ 2e−jω2sin(ω)

ω(10.36)

which (of course) agrees with the functionS(ω) we derived in Equation 10.31.Approach 3. From its definition, we have

S(ω) =∫ ∞

−∞s(t)e−jωtdt =

∫ 0

−1(−1)e−jωtdt +

∫ 2

0(2)e−jωtdt

= −∫ 0

−1e−jωtdt + 2

∫ 2

0e−jωtdt

= −e−jωt

−jω

∣∣∣∣∣0

−1

+ 2e−jωt

−jω

∣∣∣∣∣2

0


=1jω

e−jωt∣∣∣0

−1− 2

1jω

e−jωt∣∣∣2

0

=1jω

[1− ejω

]− 2

1jω

[e−j2ω − 1

]

=1jω

[1− ejω + 2− 2e−j2ω

]

=1jω

[3− ejω − 2e−j2ω

](10.37)

Equations 10.37 and 10.31 look different. However, using Euler’s identitiesin Equation 10.31, we get

S(ω) = −ej0.5ω12j [e

jω/2 − e−jω/2]

ω/2+ 4e−jω

12j [e

jω − e−jω]

ω

=1jω

[−ej0.5ω[ejω/2 − e−jω/2] + 2e−jω

(ejω − e−jω

)]

=1jω

[−ejω + 1 + 2− 2e−j2ω

]

=1jω

[3− ejω − 2e−j2ω

](10.38)

which is the same as Equation 10.37.Figure 10.4 depicts the spectrum of this signal. Note that both negative and

positive frequencies are shown. From the graph, the zeros of the magnitudespectrum occur at frequencies

F = 1, 2, 3, 4 Hz.

Can you prove that the frequencyF = k, wherek is an integer, will also yielda zero, by using Equation 10.37?

————————————————————————————-4

Example 10.9.The Fourier transform of a signals(t) is given in Equation10.39:

S(ω) = e−j2ω sin(ω)ω

(10.39)

Write a Matlab script to plot the spectrum of the signal for a frequency rangeof 0–10 Hz, with a resolution of 0.001 Hz.

The spectrum consists of two plots, one for the magnitude ofS(ω) and onefor the phase ofS(ω). A Matlab script to plot the spectrum is the following:


−4 −3 −2 −1 0 1 2 3 40

1

2

3

4Magnitude spectrum

Mag

nitu

de

Frequency (Hz)

−4 −3 −2 −1 0 1 2 3 4−4

−2

0

2

4Phase spectrum

Pha

se

Frequency (Hz)

Figure 10.4. The spectrum ofS(ω) = 1jω

[3− ejω − 2e−j2ω

].

Matlab script 10.1. ————————————————————-

% Define the desired frequency range;%resolution of the range is 10 Hzfreq_range = 0:0.001:10;omega_range = 2 * pi * freq_range;

% Calculate the Fourier transform S(omega)% Note the .* and ./ use for element-wise operationsS = exp( -j * 2* omega_range ) .* sin( omega_range ) ..../ omega_range ;

% Of course, Matlab will complain about% the sin(0)/0 division,% which occurs for omega_range(1).% From De L’Hospital’s rule, we have the following:


S(1) = 1 ;

% Calculate the magnitude and phasemagnitude = abs( S );phase = angle( S );

% Now plot the spectrum, using subplot and grid

subplot(2,1,1)plot( freq_range, magnitude )xlabel(’frequency (Hz)’)ylabel(’Magnitude’)grid on

subplot(2,1,2)plot( freq_range, phase )xlabel(’frequency (Hz)’)ylabel(’Phase’)grid on

——————————————————————————–The spectrum produced by this script is shown in Figure 10.5. Note that

the horizontal axis is frequency in Hz, not in radians. When comparing thisspectrum to the one shown in Figure 10.1, notice that the magnitude parts areequal. The phase parts differ by a factor of2ω, as you can easily explain byusing the time-shifting property.

————————————————————————————-4

Example 10.10.For the Fourier transform of the previous example, evaluatethe value ofS(ω) for ω = 0 andω = π/2. Express your results in exponentialform. 4

We have

S(0) = e−j2·0 sin(0)0

= 1

S(π/2) = e−j2π/2 sin(π/2)π/2

= 2e−jπ 1π

=2π

e−jπ

————————————————————————————-4

*Example 10.11. Find the Fourier transform of the signal,v(t), in Figure10.6, by using time scaled and shifted versions of the centered unit pulse signalp1(t) in Equation 10.10, page 519, Example 10.4. 4


0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

frequency (Hz)

Mag

nitu

de

0 1 2 3 4 5 6 7 8 9 10−4

−2

0

2

4

frequency (Hz)

Pha

se

Figure 10.5. The spectrum ofS(ω) = e−j2ω sin(ω)ω

.

Setτ = 1 in Example 10.4. The signalp1(t) is given by

p1(t) =

0, t < −1/21, −1/2 ≤ t < 1/20, 1/2 ≤ t

with Fourier transform given by (see Equation 10.12 in Example 10.4):

P1(ω) =sin(ω/2)

ω/2

Let’s define two new signals, that are time-scaled versions ofp1(t):

y1(t) = p1(t/2)y2(t) = p1(t/6)

y1(t) is a centered pulse of width 2 andy2(t) is a centered pulse of width 6.Then, from the time-scaling property of the Fourier transform we obtain


−5 −4 −3 −2 −1 0 1 2 3 4 5

−2

−1

0

1

The signal v(t)

t

v(t)

Figure 10.6. The signalv(t) in Example 10.11.

Y1(ω) = 2P1(2ω) = 2sin(2ω/2)

2ω/2= 2

sin(ω)ω

(10.40)

Y2(ω) = 6P1(6ω) = 6sin(6ω/2)

6ω/2= 6

sin(3ω)3ω

(10.41)

Note that we keep the transform in the form of the sinc functionsin(x)/x.From Figure 10.6, we can write

v(t) = y2(t + 1)− 2y1(t− 3)

from which and the time-shifting property of the Fourier transform we get:

V (ω) = ejωY2(ω)− 2e−3jωY1(ω)

Now, from Equations 10.40 and 10.41 we have

V (ω) = ejω6sin(3ω)

3ω− 2e−3jω2

sin(ω)ω

So, finally,

V (ω) = 6ejω sin(3ω)3ω

− 4e−3jω sin(ω)ω

(10.42)

Figure 10.7 depicts the spectrum ofV (ω). (It looks almost artistic.) Notethat the magnitude spectrum is zero for angular frequenciesω = π, 2π, 3π, . . .,


or for frequenciesf = 0.5, 1, 1.5, 2, . . . Note also how these frequencies arerelated to the width (2) of the pulse: they are multiples of 1/pulse width. (Thisrelationship will be important in later classes.)

−1.5 −1 −0.5 0 0.5 1 1.50

2

4

6

8Magnitude spectrum

Mag

nitu

de

Frequency f (Hz)

−1.5 −1 −0.5 0 0.5 1 1.5−4

−2

0

2

4Phase spectrum

Pha

se

Frequency f (Hz)

Figure 10.7. The spectrum of the transform in Equation 10.42.

————————————————————————————-4

*Example 10.12. Find the Fourier transform of the signal,v(t), in Figure10.6, by using time scaled and shifted versions of the centered unit pulse signalp2(t) in Equation 10.10, example 10.4. 4

Inspecting Figure 10.6, we can write

v(t) = p2(t + 3) + p2(t + 1) + p2(t− 1)− 2p2(t− 3)

The Fourier transform of the centered unit pulse signalp2(t) is given by

P2(ω) = 2sin(ω)

ω.

Then, from the linearity and time-shifting properties of the Fourier transform,we can write

V (ω) =(ej2π3ω + ej2πω + e−j2πω − 2e−j2π3ω

)P2(ω)


=(ej2π3ω + ej2πω + e−j2πω − 2e−j2π3ω

)2sin(ω)

ω(10.43)

Equations 10.43 and 10.42 appear to be different. Can you reconcile them?————————————————————————————-4

4. Solving differential equations with Fourier transformsAs with the Fourier series, the most general differential equation that we will

consider in this section is Equation 9.103, page 494, in Chapter 9, i.e., a linear,k-th order differential equation with constant coefficients, of the type:

dkv(t)dtk

+ ak−1dk−1v(t)dtk−1

+ · · ·+ a1dv(t)dt

+ a0v(t) = vs(t) (10.44)

Note that the driving force does not have to be a periodic function. Weare interested again in findingvp(t), the particular solution of the equation.Therefore, throughout this section we will assume that the initial conditions are0.

As with the Laplace transform approach, the solution methodology is to:

calculate the Fourier transform of both sides of Equation 10.44,

use Theorem 10.5, to reduce Equation 10.44 into analgebraicequation thatinvolvesVp(ω), the Fourier transform of the particular solution.

determineVp(ω) from this algebraic equation

find vp(t) using Equation 10.2 or properties of the Fourier transform.

Taking the Fourier transform of both sides of Equation 10.44, and applyingTheorem 10.5, we get

(jω)kVp(ω) + ak−1(jω)k−1Vp(ω) + · · ·+ a1(jω)Vp(ω) + a0Vp(ω)= Vs(ω) (10.45)

from which the Fourier transformVp(ω) is:

Vp(ω) =1

(jω)k + ak−1(jω)k−1 + · · ·+ a1(jω) + a0· Vs(ω) (10.46)


The particular solution can now be found as the following theorem summa-rizes:

Theorem 10.6 Consider the differential equation 10.44. LetVs(ω) denotethe Fourier transform of the driving force. Then the particular solution is givenby

vp(t) =∫ ∞

−∞Vs(ω)ejωt

(jω)k + ak−1(jω)k−1 + · · ·+ a1(jω) + a0dt (10.47)

4

Note that we will rarely compute the inverse transform but will use theproperties of the transform and tables similar to the Laplace transform to obtaina complete description ofvp(t).

As with the Fourier series case,

Definition: The function

H(ω)4=

1(jω)k + ak−1(jω)k−1 + · · ·+ a1(jω) + a0

(10.48)

is called thetransfer functionof the system represented by differential equation10.44. Note that, in Equation 10.44, there are no derivatives of the drivingforce, hence the numerator in Equation 10.48 is equal to 1. Compare Equations10.48 and 8.58.

4

*Example 10.13.Consider the differential equation

d3v(t)dt3

+ 2d2v(t)dt2

+ 3dv(t)dt

+ 4v(t) = 60 cos(2π100t). (10.49)

Determine the particular solution,vp(t), using Fourier transforms. 4Note that this is the same differential equation in Equation 9.113, page 498.

We have solved the equation already using Fourier series, see Equation 9.115,page 499.

Let vs(t) = 60 cos(2π100t) denote the driving force in Equation 10.49.From Example 10.3, the Fourier transform,Vs(ω), of the driving force is

Vs(ω) = 60πδ(ω − 2π100) + 60πδ(ω + 2π100). (10.50)


Setk = 3 in Equation 10.46. From Equations 10.49 and 10.50, the Fouriertransform of the particular solution is given by:

Vp(ω) =Vs(ω)

(jω)3 + 2(jω)2 + 3(jω) + 4

=60πδ(ω − 2π100) + 60πδ(ω + 2π100)

−jω3 − 2ω2 + 3jω + 4

=60πδ(ω − 2π100) + 60πδ(ω + 2π100)

(4− 2ω2) + jω(3− ω2)

=60π

[4− 2(2π100)2] + j(2π100)[3− (2π100)2]δ(ω − 2π100)

+60π

[4− 2(−2π100)2] + j(−2π100)[3− (−2π100)2]δ(ω + 2π100)

(10.51)

=60π

[4− 2(2π100)2] + j[6π100− (2π100)3]δ(ω − 2π100)

+60π

[4− 2(2π100)2]− j[6π100− (2π100)3]δ(ω + 2π100)

=60π√

[4− 2(2π100)2]2 + [6π100− (2π100)3]2ejθδ(ω − 2π100)

+60π√

[4− 2(2π100)2]2 + [6π100− (2π100)3]2e−jθδ(ω + 2π100)

(10.52)

In deriving Equation 10.51, we have used the property of theδ(t) functionin Equation 2.17, page 61.

From Equation 10.19, page 525, the Fourier transformVp(ω) corresponds toa cosine time signal, with amplitude equal to

60√[4− 2(2π100)2]2 + [6π100− (2π100)3]2

and phase

θ = arctan

((2π100)[3− (2π100)2]

4− 2(2π100)2

)≈ −π/2

Therefore,

vp(t) =60√

[4− 2(2π100)2]2 + [6π100− (2π100)3]2cos(2π100t− π/2)

=60√

[4− 2(2π100)2]2 + [6π100− (2π100)3]2sin(2π100t)


Example 10.14. Consider the differential Equation 9.108, page 495, inExample 9.26. What methods are there to determine theparticular solutionvp(t)? 4

The methods we have to find the particular solution in general, are:

1 Guess a solution of the form

vp(t) = B cos(100t + θ1) + C sin(500t + θ2)

or

vp(t) = B cos(100t) + C sin(100t) + D cos(500t) + E sin(500t)

(see Chapter 7).

2 Apply the Laplace transform to both sides of Equation 9.108

3 Apply the Fourier transform to both sides of Equation 9.108.

4 Apply the Fourier series to both sides of Equation 9.108.

Option 1 will work. Option 2 is not applicable, since the driving forcevs(t) = 20 cos(100t) + 40 sin(200t) is not 0 fort < 0. Option 3 is applicable.Option 4 is applicable, since the driving forcevs(t) is periodic, with fundamentalfrequencyω0 = 100.

————————————————————————————-4

5. Matlab commands for calculating Fourier transformsThe command

fourier(f)

wheref is a symbolic variable returns the Fourier transform of the functionf(t).

Example 10.15.The script

Matlab script 10.2. ————————————————————-

syms tf = exp(-t^2);fourier(f)

——————————————————————————–returns the expression

√πe−1/4∗ω2

(try to compute this by hand!).————————————————————————————-4


The command

ifourier(F)

calculates the functionf(t) whose Fourier transform is equal toF (ω).

Example 10.16.The script

Matlab script 10.3. ————————————————————-

syms wF = 1/(1+j*w) ;ifourier(F)

——————————————————————————–returns the expressione−tu(t).

————————————————————————————-4

6. Summary of main pointsRepresentation of an aperiodic signal with the Fourier transform.

Properties of the Fourier transform.

The frequency spectrum of an aperiodic signal.

Finding the particular solution of a differential equation using Fourier trans-forms.

Skillset to be developedDetermine analytically the Fourier transform of a signal using the directintegration approach.

Determine analytically the Fourier transform of a signal using the propertiesof the Fourier transform and the Fourier transform of simpler signals.

Determine analytically the signalv(t) in the time domain, from knowledgeof its Fourier transformV (ω).

Write Matlab scripts to plot the frequency and power spectrum of a signal.

Determine analytically the particular solution of a differential equation usingFourier transforms.


Connection to other ECE coursesYou will use Fourier transforms ad nauseam in ECE301, ECE402, ECE403,

ECE420, ECE421, ECE422 and ECE451. In these courses, you will go deeperandfaster; in some of these courses, you will study Fast Fourier Transforms,the celebrated FFT.

As was the case with Laplace transforms, the concept of the transfer functionand applications in filter design will be the major topics in such courses. Inthe two communications-related courses, ECE402 and ECE420, the Fouriertransform will be indispensable in understanding the fundamental notion ofbandwidth.


ProblemsCalculation of the Fourier transform

10.1.Using direct integration, find the Fourier transform of the signal

s(t) = e−25t cos(2π100t + π/4)u(t)

You can check your result by using the Laplace transform table in Chapter8.

10.2.Using direct integration, find the Fourier transform of the signal

s(t) = 20e−tejπtu(t)

*10.3. Show that the Fourier transform of the triangular pulse signal

s(t) =

1− |t|T , −T ≤ t ≤ T ,

0, otherwise.

is given by

S(ω) = T

(sin(ωT/2)

ωT/2

)2

10.4. Consider a signals(t) with Fourier transformS(ω). Let ω0 be a realnumber. Define a new (complex-valued) signal

y(t) = ejω0ts(t).

Show that

Y (ω) = S(ω − ω0)

————————————————————————————–

Properties of the Fourier transform

*10.5. Considerp(t), the unit pulse signal in Equation 2.12, page 59. Lett0 = 10.

DetermineP (ω) using time shifting and scaling properties, and the resultsof Example 10.4.

*10.6. Repeat the previous problem for a generict0. (Hint: let τ = 1 inExample 10.4. First time-shiftp1(t) by 0.5 time units to the left, to obtain asignaly(t) = p1(t− 0.5). Then time scaley(t).)


10.7.Consider the signalh1(t) in Equation 2.39, page 80.(a) Calculate its Fourier transform,H1(ω), using direct integration.(b) Calculate its Fourier transform,H1(ω), using time shifting and scaling

properties, and the results of Example 10.4.

10.8.Consider the signalh3(t) shown in Figure 2.16, page 81. Calculate itsFourier transform.

10.9.Consider the signalh4(t), defined in Equation 2.41, page 82. Calculateits Fourier transform.

10.10.Consider the signalh5(t), defined in Equation 2.43, page 82. Calcu-late its Fourier transform.

10.11. Consider the centered unit pulsep5(t), in Example 10.4. Calculatethe Fourier transform of the signaly(t) = 3p5(t/2).

*10.12. Consider the centered unit pulsep5(t), in Example 10.4. Calculatethe Fourier transform of the signaly(t) = 3p5(t/2 + 3).

10.13.Consider the signalv(t), shown in Figure 8.3, page 428.(a) Calculate the Laplace transformV (s).(b) Calculate the Fourier transform,V (ω) from the Laplace transformV (s).

*10.14. The Fourier transform of the signals(t) is defined as

S(ω) = 8ejω sin(2ω)2ω

+ 2e−jω sin(ω)ω

Find the Fourier transform of the time-scaled and shifted signaly(t) = s(2t−3).

*10.15. Using the formula for the Fourier transform of the centered unit pulsein Equation 10.10, page 519, together with time shifting and scaling properties,find the Fourier transform of the signal defined by

s(t) =

0, t < 02, 0 ≤ t ≤ 1−1, 1 < t ≤ 30, 3 < t

————————————————————————————–


Fourier transform and Fourier series

*10.16. The Fourier series for the periodic signal,s1(t), whose single periodis defined by

s1(t) =

sin(2πt/T ), 0 ≤ t ≤ T/20, T/2 < t < T

whereT = 1/60 seconds, is denoted byS1(k).(a) Let a single period of the periodic signals2(t), be defined by

s2(t) =

sin(2πt/T ), 0 ≤ t ≤ T/20, T/2 < t < T

whereT = 1/120 seconds. Find the Fourier series,S2(k), in terms ofS1(k).Hint: Use the relationship between the Fourier series and Fourier transform toobtain both series and look for the similarities.

(b) Let a single period of the periodic signals3(t), be defined by

s3(t) =

cos(2πt/T ), −T/4 ≤ t ≤ T/40, −T/2 < t < −T/40, T/4 < t < T/2

whereT = 1/120 seconds. Find the Fourier series,S3(k), in terms ofS1(k).

*10.17. The rectified sinusoids are very important in many ECE applications,particularly in communications and power conversion. It is important that youunderstand the relationships among the various forms, half-wave and full-waverectified signals, in both the time and frequency domains.

(a) Compute, by direct integration, the Fourier transform of the function,s1(t), defined by

s1(t) =

0, t ≤ −T/4cos(2πt/T ), −T/4 ≤ t ≤ T/40, T/4 < t

(b) AssumeT = 1/60; plot the magnitude and phase spectra from zero to10KHz. Plot the power spectrum in decibels as a function of frequency in Hertz.Recall that the power spectrum in dB is defined byPdB(ω) = 20 log10(|S(ω)|).

(c) The periodic extension ofs1(t) with a period of T is written

s1p(t) =

0, −T/2 < t ≤ −T/4cos(2πt/T ), −T/4 ≤ t ≤ T/40, T/4 < t ≤ T/2

Use the relationship of the Fourier transform and Fourier series to obtain theFourier series coefficients ofs1p(t).


(d) Use the information found above to obtain the Fourier series coefficientsof s2p(t) = | cos(2πt/T )|. This is the full-wave rectified sinusoid. Pay partic-ular attention to the fundamental frequencies ofs1p(t) ands2p(t).

(e) Use the information found above and Fourier properties (time scaling andshifting) to obtain the Fourier series coefficients ofs3p(t) = | sin(2π10t/T )|.

————————————————————————————–

Frequency and power spectrum

*10.18. Let pτ (t) be the unit centered pulse of widthτ , in Equation 10.10of Example 10.4.

(a) Find the power spectrum of a clock generator signal in Equation 2.21,page 63. For this problem, you will periodically extend the single pulse witha periodT = 1

100 seconds, and setτ = T/2. The power spectrum does notdepend on the phase of the signal, so you may use the symmetric pulse to makecomputation easy.

(b) Plot the power spectrum of the clock generator signal in part (a) in deci-bels.

*10.19. Let s3(t) be defined as

s3(t) = e−αt cos(2πft + θ)u(t)

(a) Compute the Fourier transform ofs3(t). Hint: Use Euler’s identity.(b) For the case whereα = 50, f = 200, andθ = π/4, plot s3(t) from 0

to 125 ms. Using subplot, plot the magnitude spectrum in decibels from 0 to 2KHz.

(c) At what frequency does the peak of the spectrum occur? Is this what youwould expect? What parameter(s) would you change to make the peak moredistinct (sharper)? less distinct (broader)?

*10.20. Two time-domain signals,s1(t) ands2(t), and their Fourier trans-forms,SA(F ) andSB(F ), are shown in Figures 10.8 and 10.9. Using yourknowledge of the location of zeros of the spectra of rectangular pulses, matchthe spectra to the corresponding time domain signal. The zeros in the spectracan be identified as the deep negative dips. Using standard computation, it isunlikely that an exact zero will be produced.

————————————————————————————–


0 1 2 3 4 5 6 7 8−1

−0.5

0

0.5

1

1.5

2

time domains1

(t)

0 1 2 3 4 5 6 7 8−1

−0.5

0

0.5

1

1.5

2

s2(t

)

time in seconds

Figure 10.8. The signals in the time domain.

Solution of differential equations

10.21.A fourth order filter (circuit) is described by the differential equation

d4v(t)dt4

+ 26d3v(t)dt3

+ 341d2v(t)dt2

+ 2613dv(t)dt

+ 10000v(t) = 10000s(t)(10.53)

(a) Compute the transfer function of the system represented by the differentialequation.

(b) Compute the output (solution) of the system when the input is

s(t) = cos(5t)

(c) Compute the output (solution) of the system when the input is

s(t) = cos(20t)

(d) Plot the frequency response of the system from0 ≤ ω ≤ 50.


0 5 10 15 20 25−40

−20

0

20

40

60Frequency domain

SA

(F)

0 5 10 15 20 25−50

0

50

SB

(F)

frequency in hertz

Figure 10.9. The Fourier transforms.

10.22.Given a system defined by the differential equation

d2v(t)dt2

+ 30dv(t)dt

+ 900v(t) = 1800vs(t),

use the Fourier transform to find the particular solutionvp(t), if the input,vs(t)is defined by

vs(t) = cos(30t)

10.23.Consider the differential equation

d2v(t)dt2

+R

L

dv(t)dt

+1

LCv(t) =

1LC

vs(t) (10.54)

whereL = 100µH, C = 10µF andvs(t) = 5 cos(2π5000t). Use Fouriertransforms to:

(a) find the particular solutionvp(t) whenR = 0.6Ω.(b) find the particular solutionvp(t) whenR = 12Ω.


10.24.Consider the differential equation obtained from an RLC circuit

d2v(t)dt2

+R

L

dv(t)dt

+1

LCv(t) =

1LC

vs(t)

whereR = 0.5Ω, L = 10µH, C = 40µF , vs(t) = u(t)− u(t− 1), v(0) = 0andv(0) = 0.

Use Fourier transforms to find the particular solutionvp(t).

————————————————————————————–

Problems you will see in other coursesThis set of problems is intended to highlight the connection of the material

in this chapter to other ECE courses.

*10.25. (ECE301)Filter design.A Butterworth filter of orderK is definedby the transfer function

|H(ω)|2 =1

1 +(

ωωc

)2K(10.55)

In Equation 10.55,K is a positive integer andωc is called the cutoff frequencyof the filter.

1 Plot the transfer function forK = 1, 2, . . . , 10 andωc = 1 KHz. Vary ωbetween -10 and 10 KHz. Observe how the transfer function approximatesa rectangular function. Such a function would represent an ideal low-passfilter with cutoff frequencyωc = 1 KHz.

2 Is |H(ω)| an odd or even function?

3 Determine at what frequency,ω0, the power is equal to 3 dB. Recall that forthis frequency,|H(ω0)|2 = 0.5.

4 LetK = 3 andωc = 1. Determine the poles of the denominator in Equation10.55.

5 Consider the transfer function

|H(ω)| = 1(jω)6 + 3.9(jω)5 + 7.5(jω)4 + 9.1(jω)3 + 7.5(jω)2 + 3.9(jω) + 1

(10.56)

Does this function represent a Butterworth filter? Hint: you must show thatEquation 10.56 can be cast in the form of Equation 10.55. To do that, findthe conjugate,H∗(ω), of the given transfer function and compute


|H(ω)|2 = |H(ω)| · |H∗(ω)|

6 Consider the circuit shown in Figure 10.10. The input is the current source;the output is the voltage across the resistor. Write the transfer function.

i(t)1 F

v(t)1 Ω

Figure 10.10. Implementation of a first-order Butterworth filter.

7 Show that this circuit implements a Butterworth filter of order 1. Show thatthe cutoff frequency isωc = 1 Hz.

8 Consider the circuit shown in Figure 10.11. The input is the current source;the output is the voltage across the capacitor. Write the transfer function.


In ECE301, this is called apassive filter. An improved implementation thatuses operational amplifiers from ECE200 would be called an active filter.You will study even betterdigital filter implementations in ECE421.

i(t)1 F

v(t)1 Ω1 F

1 H

Figure 10.11. Implementation of a third-order Butterworth filter.

9 Show that this circuit implements a Butterworth filter of order 3. Show thatthe cutoff frequency isωc = 1 Hz. Show that the transfer function for thisfilter is given by

|H(ω)| = 1(jω)3 + 2(jω)2 + 2(jω) + 1


*10.26. (ECE402)Line codes.Consider Problem 2.46.

1 Determine the magnitude spectrum of the Unipolar Non-Return-to-Zero(UNRZ) code signal. Plot the spectrum.

2 Determine the magnitude spectrum of the Polar Non-Return-to-Zero (PNRZ)code signal. Plot the spectrum.

3 Determine the magnitude spectrum of the Bipolar Return-to-Zero (BRZ)code signal. Plot the spectrum.

4 Suppose that any signal power lower than -30 dB is practically zero. Whichsignal consumes more bandwidth?

*10.27. (ECE402)On-Off Keying (OOK).Consider Problem 2.47.

1 Determine the magnitude spectrum of the OOK signal in equation 2.60,page 106. Plot the spectrum.

2 Suppose that any signal power lower than -30 dB is practically zero. Is theOOK code more expensive, in terms of bandwidth, than the signals in theprevious problem?

*10.28. (ECE402)Binary Phase Shift Keying (BPSK).Consider Problem2.48.

1 Determine the magnitude spectra of the BPSK signals in Equations 2.62 and2.63, page 107. Plot the two spectra.

2 Suppose that any signal power lower than -30 dB is practically zero. Is theBPSK code more expensive, in terms of bandwidth, than the signals in theprevious two problems?

*10.29. (ECE402)Frequency Shift Keying (FSK).Consider Problem 2.49.

1 Determine the magnitude spectra of the FSK signals in equations 2.64 and2.65, page 109. Plot the two spectra.

2 Suppose that any signal power lower than -30 dB is practically zero. Is theFSK code more expensive, in terms of bandwidth, than the signals in theprevious three problems?


*10.30. (ECE402)Binary Phase Shift Keying (BPSK) using raised cosinesignals.Consider the Binary Phase Shift Keying (BPSK) code in problem 2.48.A variation of this code uses the signal

c1(t) =

Ac2 [1 + cos(2πfct)], |t| < 0.5fc,

0, |t| ≥ 0.5fc.(10.57)

to transmit bit 1, while bit 0 is represented by the signal

c0(t) = −c1(t). (10.58)

The duration of the signalsc1(t) andc0(t) is T seconds, resulting in a rateR = 1/T bps. Figure 10.12 shows how the ASCII letter V (with ASCII code1010110) would be transmitted using this code, over a 1 bps channel. Note thatT = 1 sec for such a channel. Compare the signal waveforms to that of theBPSK code, in Figure 2.26, page 108.

−1 0 1 2 3 4 5 6 7 8−1

−0.5

0

0.5

1

1.5

2

1 0 1 0 1 1 0

T 2T 3T 4T 5T 6T 7T

The letter V in ASCII code

m(t

)

−1 0 1 2 3 4 5 6 7 8 9−2

−1

0

1

2

1 0 1 0 1 1 0

The signal for letter V in raised cosine BPSK

s(t)

time (sec)

Figure 10.12. The raised cosine Binary Phase Shift Keying (BPSK) code.

The reason for even considering use of a “complicated” signal like that inEquation 10.57 is the bandwidth savings it provides, compared to other BPSKsignals (see question 6).

1 Let fc = 1/(2T ). Sketch the waveform transmitted over a 50 kbps channel,for the letter B. Be sure to mark your abscissa accurately.


2 Calculate and plot the magnitude spectrum of the signal used to send bit 1.

3 Calculate and plot the magnitude spectrum of the signal used to send bit 0.Are they the same? Why?

4 Calculate and plot the magnitude spectrum of the signal used to send bit 1,for the signal used in the code of Problem 2.48.

5 Calculate and plot the magnitude spectrum of the signal used to send bit 0,for the signal used in the code of Problem 2.48. Are they the same? Why?

6 Is this BPSK code more or less expensive, in terms of bandwidth, than thecode in Problem 2.48? Suppose that any signal power lower than -30 dB ispractically zero.

7 In yet another BPSK variation, bit 1 is represented by the signal

c1(t) =

Ac, |t| < 0.5fc,0, |t| ≥ 0.5fc.

(10.59)

while bit 0 is represented by the signal

c0(t) = −c1(t). (10.60)

Calculate and plot the magnitude spectrum of the signalsc1(t) andc0(t).

8 Suppose that any signal power lower than -30 dB is practically zero. Usingthe spectrum plots or the formulas you have calculated, show that this BPSKcode is 5 times more expensive, in terms of bandwidth, than the raised cosineBPSK code.

————————————————————————————–

Chapter 10 FOURIER TRANSFORMS...Chapter 10 FOURIER TRANSFORMS...

Documents

Transcript of Chapter 10 FOURIER TRANSFORMS...Chapter 10 FOURIER TRANSFORMS...