EECE 340 Exam 2, Spring 2016

American University of Beirut · Signals and Systems · 7 pages · Prof. Karim Kabalan

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EECE 340-Signals and Systems-Spring 2016-Solution Page 1 AMERICAN UNIVERSITY OF BEIRUT FACULTY OF ENGINEERING AND ARCHITECTURE ELECTRICAL AND COMPUTER ENGINEERING DEPARTMENT ______________________________________________________________________________________ EECE 340 -Signals & Systems Quiz II - Spring 2016 Solution _____________________________________________________________________________________ Problem 1 (5 pts) Let X(ω) be the Fourier Transform of x(t). As a function of X(ω), determine the Fourier transform of y(t) given by a. y(t)=(1-t)x(t) (2 pts) ) ( ( ) ( ) ( ) ( ) ( )     d j dX X Y tx t x t y t      b. 2 1 ) ( 0 ) 3(          t t d x e y t    (3 pts)        j e X Y x t u t d x e t y j t t                    3 ( ) ) ( 2 1 ( ) e 2 1 ) ( 2 3t - 0 ) ( 3 EECE 340-Signals and Systems-Spring 2016-Solution Page 2 Problem 2 (6 pts) The Fourier Transform (spectrum) of the pre-envelope of a real-valued signal x(t) is shown below. Determine the signal x(t). Pre-envelope 4 2 2.2 3 Khz Solution         6000 2 ( 4400 ) ( 2 ( )     X                                       6000 ( 6000 ) ( 4400 ( 4400 ) ( 1 6000 ( 6000 ) ( 2 4400 ( 4400 ) ( ) (                 X ) 2 cos(6000 ) 1 cos(4400 ( ) t t x t       Problem 3 (4 pts) Consider the system shown below s(t) Hilbert y(t) Transformer The input s(t) to this system is: t 1 a )t(s    . Determine the spectrum of the output signal. Solution ( ) ( ) 2 ( )      jsign a S                 0 0 0 1 -sign ( ) ( ) ( )) ( ) (2 ) ( 2          jsign jsign a S EECE 340-Signals and Systems-Spring 2016-Solution Page 3 Problem 4 (5 pts) Consider the following modulation system m(t) ∑ X s(t) A cos(2πfct) Assume that a tone signal m(t)=2cos(20πt) is transmitted using this modulation system with fc=500Hz. Also assume that A=2 a. What type of modulation does this correspond to? (2 pts) 1000 ) cos(20 )]cos( 1[2 ( ))cos ( ( ) t t m t A s t c        DSB-LC modulation AM Modulation -1 b. What is the power efficiency of this system? (3 pts) 33.33% 3 100 1 100 2 1 2 2           Power efficency Problem 5 (5 pts) Consider the DSB modulated signal given by ΦDSB(t)={5+m(t)}cos(3000t). Assuming m(t)=2cos(200t)+2.8cos(400t) a. Determine the modulation index. ( 2 pts) .0 96 10 6.9 8.2 ) 2 (5 8.2 ) 2 5 ( 8.2 ) 2 (5 8.2 ) 2 5 ( min max min max                 A A A A  b. Justify whether an envelope detector can be used on this signal. (3 pts). Yes. Envelope detector can be used (under modulation) EECE 340-Signals and Systems-Spring 2016-Solution Page 4 Problem 6 (7 pts) The output (modulated signal) from an AM modulator is given by: s(t)=5 cos(1800πt)+20cos(2000π)+5cos(2200πt). a. Determine the modulating signal m(t)? (2 pts) ) cos(200 5.0 ) ( ) )]cos(2000 cos(200 5.0 20 1[ ) ( t k t m t t t s a       b. Determine the carrier signal c(t)? (1 pts) 20cos(2000 ) ( ) t s t   c. Determine the modulation index? (2 pts) 5.0   d. Determine the ration of the power of the sidebands to the power of the carrier. (2 pts) 8 1 400 50 2 400 2 25 2 25    Ratio  Problem 7 (6 pts) An FM modulator is used to transmit a tone message of amplitude 4 Volts and frequency 20 Hz. The frequency deviation constant for the modulator is 25 Hz/V, and the carrier was has amplitude of 10 Volts and a frequency of 2000 Hz. a. What is the average power of the FM modulated signal? (2 pts) Watts A P c avg 50 2 100 2 2    b. What is the approximate bandwidth of the FM signal using Carson’s rule? (1 pt) 240 Hz 5) 40 1( 2 20 1 ) 1( 2               m m f m T k a f B   EECE 340-Signals and Systems-Spring 2016-Solution Page 5 c. The output of the FM modulator is passed through an ideal bandpass filter center at 2000 Hz. What should be the bandwidth of the filter such that 90% of the power in the modulated signal passes through? (3 pts) Using Bessel Table for   5 , we obtain the following Power ratio = ... (5) 2 (5) 2 1 2 0   J J We keep adding until the sum becomes equal to 0.9. This ends for n=4. Therefore, the filter bandwidth is Hz f n m 160 (2)(4)(20) 2     Problem 8 (5 pts) Consider the system depicted below where T=1. Let x(t)=cos(3πt/2) and H(ω) be an ideal LPF with bandwidth equal to π rad/s. Determine the output of the system denoted by y(t). Solution Given,               2 ) 3 ( 2 ) 3 ( ( ) 2 cos 3 ( )          X t x t The periodic function p(t) is given by:               n n n n n t nT t p t ) ( ) ( ( )   P(t) can be written in a Fourier Series representation with    2 2 0  T  H(ω) x(t) y(t)      n nT t p t ) ( ( )  EECE 340-Signals and Systems-Spring 2016-Solution Page 6       n n nt j ne P p t 2 ( ) and 1 ) ( ( ) 1 5.0 5.0 2 2 / 2 / 2               n n nt i T T nt j n dt n e t dt p t e T P    The Fourier Transform of the output of the multiplier, s(t), is given by ..... 2 ) ( 2 ) ( ( ) ) 2 ( ( )                      X X X n P X S n n n ..... 2 ) 7 ( 2) ( 2) ( 2 ) 7 ( 2 ) 3 ( 2 ) 3 ( ) (                                                S The filter output is:         2) ( 2) ( ( )        Y       t y t cos 2 ( )  EECE 340-Signals and Systems-Spring 2016-Solution Page 7 Problem 9 (6 pts) A signal x(t) is sampled through a non-ideal sampler using the periodic signal s(t) of period T as shown in the figure below. The signal x(t) is assumed band-limited (also as shown in the figure). Let s(t)=s1(t)-1 and determine s1(t) first. Then obtain the Fourier transform spectrum of s(t) using that of s1(t) and as a result obtain the spectrum of the sampled signal, y(t). Determine the smallest possible value of the sampling rate (1/T) for which there is NO ALIASING in the sampled signal. USE =T/4 in all computations. Step 1 ( ) ) rest of the period , ( 0 2 1( ) f t T f t the for t s t               Step 2 Over one period                                                     2 0 0 2 sin 2 0 0 2 2 2 2 2 4 4 2        TSa n P n T n n n T n T n e n e j T dt e P n jn jn T T t T jn n ANSWER: 5000 HZ