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IFT 372 Practice Mid Term Exam Fall B 2019 $8.49   Add to cart

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IFT 372 Practice Mid Term Exam Fall B 2019

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1. A continuous-spectrum baseband signal has a spectrum extending from 3 kHz to 8 kHz. It is applied as one input to an ideal balanced modulator. The carrier input is a sinusoid having a frequency of 100 kHz. (a) Determine the range of frequencies contained in the DSB output spectrum. (b) Determi...

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  • February 12, 2022
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Name: ______________________ IFT-372 Mid-Term Practice Exam
Fall 2019



1. A continuous-spectrum baseband signal has a spectrum extending from 3 kHz to 8 kHz. It is applied as one input to an ideal balanced
modulator. The carrier input is a sinusoid having a frequency of 100 kHz. (a) Determine the range of frequencies contained in the DSB
output spectrum. (b) Determine the required transmission bandwidth for an AM modulated system.

a. This spectrum ranges from 3-8kHz. fc = 100kHz

Lowest frequency = 100kHz – 8kHz = 92kHz

Highest frequency = 100kHz + 8kHz = 108kHz

Range of frequencies is 92kHz to 108kHz

b. w = 8kHz

transmission bandwidth BT = 2W

= 2*8kHz = 16kHz

2. What are the (a) noise figure, (b) noise factor, and (c) effective temperature (Te) for a system with a 26 dB input SNR and a 21
dB output SNR?

a. Noise Figure (NF)= 26dB – 21dB = 5dB

b. Noise Factor(F) = 10(5/10)= 3.16

c. Effective Temperature Te = T0(F-1)

Room temperature T0 = 290K

Te = 290K(3.16-1) = 626.4K




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, Name: ______________________ IFT-372 Mid-Term Practice Exam
Fall 2019




3. Given a three stage amplifier system with Stage 1 having a gain of 19 dB and a NF=4.5 dB, Stage 2 gain of 20 dB and a NF = 5.8 dB,
Stage 3 having a gain of 16 dB and a NF = 5.8 dB, calculate the system noise figure.

3.5 dB => 2.24; 16 dB => 40; 4.8 dB => 3.02; 19 dB => 80; 6.2 dB => 4.17; 20 dB => 100;

F = dB(Stage 1 NF) + ((dB(Stage 2 NF) – 1)/ dB(Stage 1 Gain) + ((dB(Stage 3 NF) – 1)/(dB(Stage 1 Gain)* dB(Stage 2 Gain)

F = 2.24 + ((3.02 - 1)/40) + ((4.17 - 1)/(40 * 80)) = 2.29 = 3.6 dB




4. A baseband signal has frequency components from 5 kHz to 15 kHz. Determine (a) the theoretical minimum sampling rate and (b) the
maximum time interval between successive samples.

a. Signal frequency components = 5kHz-15kHz

fs = sampling rate, fm = baseband highest frequency
fs = 2*fm
fs = 2*15kHz = 30ksps or 30kHz




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