EE 160 - Spring 2011 San Jose State University

Solution of Midterm Exam # 2

Problem 1 (30 points) The output of an AM modulator is fed to a spectrum analyzerand the resulting display is shown below:

SPAN 2.5 MHz CENTER 3 MHz

RES BW 10 KHz V BW 30 KHz

REF -10 dBm

5 dB/div

(a) What type of AM modulator is this? Determine its center frequency and bandwidth.

Solution: This is a DSB-SC AM modulator. fc = 3 MHz and B = 2W = 2 MHz.

(b) Determine the power Pu of the modulated signal in dBm.

Solution: The power of the upper sideband is Pu = 2× 10−25/10 = 6.325× 10−3 mW,or 6.325 µW. Alternatively, Pu = −25 + 3 = −22 dBm.

(c) Sketch the modulated signal u(t). Assume a peak-to-peak amplitude equal to 2A.

Solution: A rough sketch is shown below. There are three cycles of the sinusoidalcarrier signal (3 MHz) for every cycle of the sinusoidal message signal (1 MHz).

t (µs)

u(t) = m(t)c(t)

2

Problem 2 (30 points) A DSB-SC AM modulator is designed by first multiplying a messagesignal m(t), of bandwidth 20 kHz, by a periodic train of rectangular pulses p(t) of amplitudes-A and A, fundamental frequency 100 kHz and duty cycle 1

3. The output of the multiplier is

then fed to a bandpass filter centered at 200 kHz of bandwidth 40 kHz. The output of thefilter is the modulated signal u(t).

(a) Sketch carefully the spectrum of the modulated signal u(t) if m(t) = cos(40000πt).

Solution: The spectrum of the output of the multiplier is X(f) = M(f)?P (f), where

P (f) =∞∑

n=−∞n 6=0

2A

3sinc

(n

3

)δ (f − nf0) ,

where f0 = 100000 Hz, and M(f) = 12

[δ(f + fm) + δ(f − fm)], with fm = 20000 Hz.It follows that

X(f) =A

3

∞∑n=−∞

sinc(n

3

)[δ (f + fm − nf0) + δ (f − fm − nf0)] .

The spectrum of the modulated signal out of the bandpass filter is

U(f) = GA

3sinc

(2

3

)[δ(fm + 2f0) + δ(−fm + 2f0) + δ(fm − 2f0) + δ(−fm − 2f0)] ,

where G is the filter gain.

(b) With A = 1, find the gain of the bandpass filter such that the amplitude of themodulated signal is the same as the amplitude of the message signal.

Solution: From part (a), and with A = 1, the modulated signal is

u(t) = G4

3sinc

(2

3

)cos(2πfmt) cos(2πf0t).

Therefore, the gain of the bandpass filter should be set to

G =3

4 sinc(23

) =π√3.

(c) Sketch a block diagram of a demodulator that uses a multiplier by the same periodictrain of rectangular pulses used in the modulator.

Solution:

LPF

p(t)

u(t) m(t)

G=2π / 3

W=20 kHz

Problem 3 (40 points) An AM modulator uses a carrier of center frequency fc = 1 MHz,amplitude 1 V and modulation index a = 0.5. The message signal is m(t) = cos(2πfmt),with fm = 250 kHz.

(a) Sketch carefully the modulated signal u(t).

Solution: The modulated signal is given by

u(t) = [1 + 0.5 cos(2πfmt)] cos(2πfct),

with fm = 0.25 MHz and fc = 1 MHz. We have fcfm

= Tm

Tc= 4. That is, there are four

cycles of the carrier signal for every cycle of m(t):

u(t)

t/Tm

21

1.5

1.0

0.5

- 0.5

- 1.0

- 1.5

(b) Sketch the power spectral density |U(f)|2 of the modulated signal u(t).

Solution:

|U(f)|2

f (MHz)1 1.250.75-0.75-1.25 1.250.75-1

1/41/4

1/641/641/641/64

(c) The communication channel has attenuation -40 dB. At the receiver, the noise level ismeasured as -80 dBm/Hz. Determine the SNR in dB at the output of the demodulator.

Solution: The transmitted power is Pu = 916

= 562.5 mW or 27.5 dBm. The receivedpower is PR = 27.5 − 40 = −12.5 dBm or 5.625 × 10−5 W. The signal bandwidth isW = 250 kHz. Therefore, the noise power is PN = N0W = 2×10−80/10 ·10−3 ·250000 =5× 10−6 W. It follows that the output SNR is

(S

N

)o

= 10 log10

(5.625× 10−5

5× 10−6

)= 10.5 dB.