hwset2solution

# hwset2solution - ECE 453 Homework Assignment 2 Solution...

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ECE 453: Homework Assignment 2 Solution 1-11 Solution a. s ( t ) = 10 cos[2 π 10 8 t 2 cos(2 π 10 4 t + π / 4)] . This signal is in the form A cos( ω c t + θ ( t )) , where θ ( t ) = 2 cos(2 π 10 4 t + π / 4) . Hence, the signal is purely angle modulated. The message signal consists of a single tone with frequency f m = 10 4 Hz = 10 kHz. The peak phase deviation is 2 radians. The instantaneous frequency deviation ω ( t ) = d dt θ ( t ) = 4 π 10 4 sin(2 π 10 4 t + π / 4) . The peak frequency deviation, f max , is 20 kHz, so using Carson’s rule, the bandwidth is BW = 2( f m + f max ) = 2(10 + 20) kHz = 60 kHz. b. s ( t ) = 50 cos[2 π 10 8 t ] cos[2 π 10 4 t + π / 4] . This signal is in the form Am ( t ) cos[2 π 10 8 t ] , where m ( t ) = cos(2 π 10 4 t + π / 4) . Hence, the signal is amplitude modulated using DSB-SC. As in part a., the message signal, m ( t ) , is a tone with frequency f m = 10 kHz. Hence, the bandwidth occupied by s ( t ) is BW = 2 f m = 20 kHz. 1-12 Solution s ( t ) = 50 cos( ω c t + 5 sin(2 π 1000 t )) . This signal is in the form A cos( ω c t + θ ( t )) , where θ ( t ) = 5 sin(2 π 1000 t ) . Hence, the signal is purely angle modulated. The message signal consists of a single tone with frequency f m = 1000 Hz = 1 kHz. a. The peak phase deviation is 5 radians. b. The instantaneous frequency deviation ω ( t ) = d dt θ ( t ) = 10 4 π cos(2 π 1000 t ) s 1 .

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