EECE 340 Exam 2, Spring 2017

EECE 340-Signals and Systems-Quiz II - Spring 2017 Page 1 AMERICAN UNIVERSITY OF BEIRUT FACULTY OF ENGINEERING AND ARCHITECTURE ELECTRICAL AND COMPUTER ENGINEERING DEPARTMENT EECE 340 - Signals & Systems Quiz II - Spring 2017- Solution ______________________________________________________________________________________ Problem 1 (5 pts) f(t) has a spectrum ( ) F  shown below y(t) has a spectrum Y(ω) shown below Y(ω) 2 10 rad/s a. Is y(t) a real-valued signal? Why? (2 pts) No. No symmetry about the imaginary axis b. Write y(t) as a function of x(t). (3 pts) x t e j t y t 5 2 ( ) 1 ( )  Wrong Coefficient -1 Wrong shifting -2 Wrong answer: -3 EECE 340-Signals and Systems-Quiz II - Spring 2017 Page 2 Problem 2 (5 pts) f(t) has a spectrum ( ) F  shown below y(t) has a spectrum Y(ω) shown below 2 -35 -30 -25 25 30 35 rad/s Write y(t) as a function of f(t) Answer    t  j t j t j j t e e t fj f t e e t fj f t y t 25 35 35 25 ( )] 4[ ( ) 1 ( )] 4[ ( ) 1 ( )         OR cos(30 ) ( )] 2[ ( ) 1 ( )] 2[ ( ) 1 ( ) 5 5 t t e fj f t t e fj f t y t j t t j        -4 for incorrect answer EECE 340-Signals and Systems-Quiz II - Spring 2017 Page 3 Problem 3 (5 pts) Consider the signal t j sin t cos ) t(f c c   3  2  a. Determine the pre-envelope of f(t). (2 pts) t j t c c c c c e t j t j t j t fj f t t f       3 j 2 e ) 2 cos3 (sin 2 sin3 cos )ˆ ( ) ) ( c          b. Determine the complex envelope of f(t). (3 pts) t j t j c c e t e f f t   2 2 1 ( ) ( )       Problem 4 (4 pts) The spectrum M() of a base-band signal m(t) is given by: M( ) ( ) ( ) . ( ) ( ) . ( )                     2 1 05 2 1 05 2 Determine the spectrum of the Hilbert transform of m(t). Answer  ( ) ( ) . ( ) ( ) . ( ) M j j j j                   1 05 2 1 05 2 Problem 5 (6 pts) The output (modulated) signal from an AM modulator is given by: 5cos(2200 ) 20cos(2000 ) 1800 ) 5cos( ( ) t t t s t       a. Determine the modulating signal m(t). (2 pts) s(t) can be re-written as:   5.0 cos(200 ) cos(2000 ) 201 ( ) t t s t     ) cos(200 5.0 ( ) t k t m a   b. Determine the carrier signal c(t). (1 pt) 20cos(2000 ) ( ) t c t   c. Determine the modulation index. (1 pt) 5.0   d. Can m(t) be recovered using an envelope detector? Justify your answer. (1 pt). EECE 340-Signals and Systems-Quiz II - Spring 2017 Page 4 Yes. Under-modulation e. Determine the ratio of the power of the sidebands to the power of the carrier. (1 pt) 200 Watts 2 (20) 1 P 25 Watts 2 (5) 1 2 (5) 1 2 c 2 2     s  P 12 5. % 8 1 200 25   Ratio  Problem 6 (8 pts) A frequency modulated signal with carrier frequency of 2 MHZ is described by the following equation  ) 5sin(400 10cos 2 ( ) t f t s t c     a. Find the power of the modulated signal. (1 pt) Watts A P c 50 2 2   b. Find the instantaneous frequency. (2 pts)   ) rad/s ,2 000 cos(400 2 ) 5sin(400 2 ( ) t f dt t f t d t c c i           Or ) Hz 1000cos(400 ( ) t f f t c i    c. Find the peak phase deviation in radians. (1 pt) 2000 rad/s 400 5 5                  m m d. Determine the modulation index. Is it a Narrow-band of Wide band FM signal? (1 pt)   .5 Wide band FM EECE 340-Signals and Systems-Quiz II - Spring 2017 Page 5 e. The signal passes through a a band-pass filter centered at the carrier frequency and of bandwidth 10 KHz. Determine the signal at the output of the filter. (2 pts) At the input of the filter          n n c n t n f t J s t   400 (5)cos 2 10 ( ) At the output of the filter          2 2 400 (5)cos 2 10 ) ( n n c n t n f t J s t   f. What is the power ratio at the output of the filter. .0 25127 2( .0 047) 2( .0 328) ( .0 178) 2 2 2    P  Watts Problem 7 (6 pts) A frequency modulated signal is described by the equation  ) sin(2000 cos(4000 ) 10cos 2 ( ) t b t a f t s t c       a. Find the average power of this FM modulated signal. (1 pt) P=50 Watts b. Let a=10 and b=5. Find the peak frequency deviation in KHz. (2 pts) ) ,5 000cos(2000 ) 20,000sin(4000 ) ( 2 1 ( ) t t f dt t d t f c i           25 KHz c. For a=10 and b=5, find the deviation ratio. (2 pts) 12 5. 2 25 Deviation Maximum Frequency    Bandwidth D d. For a=10 and b=5, find the bandwidth of this signal. (1 pt) KHz D W BT 54 )1 2*2*(12 5. ) 1( 2      EECE 340-Signals and Systems-Quiz II - Spring 2017 Page 6 Problem 8 (6 pts) An FM modulator is used to transmit a single tone message with amplitude 4 Volts and frequecy 20 Hz. The frequency deviation constant kf for the modulator is 25 Hz/V, and the carrier wave has amplitude 10 Volts and frequncy 2000 Hz. a. What is the approximate bandwidth of the FM signal using Carson’s Rule. (3 pts) 240 Hz 5) 2 20 1( ) 1( 2        m T f B b. What is the approximate bandwidth of this FM signal using 1% rule? (3 pts) From the table 320 Hz 16 20 16     m T f B Problem 9 (5 pts) Determine the range of permissible cut-off frequencies (Bandwidth) for the ideal low-pass filter used to reconstruct the signal x t t t ( ) cos( )cos ( ) 10 600 2 1600   which is sampled at 4000 samples per second. Solution       ) 2 cos(3200 1 2 ) 1 10cos(600 ( ) t t x t   5cos(600 )cos(3200 ) 5cos(600 ) ( ) t t t x t      cos(2600 ) 5.2 5.2 cos(3800 ) 5cos(600 ) ( ) t t t x t       From the above, the bandwidth of the signal x(t) is:  3800 rad/s or 1900Hz The sampling period Ts=4000 samples per second. This implies that sf  4000 Hz Accordingly, the bandwidth of the filter will be between 1900 Hz and 2100 Hz EECE 340-Signals and Systems-Quiz II - Spring 2017 Page 7 EXTRA PROBLEM (5 PTS) Let f(t) be a signal with Fourier Transform F(ω). Suppose we are given the following facts a. f(t) is real, b. f(t)=0 for t<0, and c. The inverse Fourier transform of the Real part of F(ω) is:   t e t F FT   ( )  1 Re  Determine f(t). Show your work. Answer:   2 ( ) ( ) ( ) Re       F F F Since f(t) is a real value, ) ( ( )      F F Then   2 ) ( ( ) ( ) Re       F F F   )) ( ( 2 1 ( ( )) 2 1 ( ) Re 1 1 1          F FT F FT F FT t e t t f f t F FT F FT          ) ( 2 1 ( ) 2 1 )) ( ( 2 1 ( ( )) 2 1 1 1   t e t t f f t     ) ( 2 1 ( ) 2 1 Now, if t>0, then –t < 0, and ( ) 2 1 ) ( 2 1 ( ) 2 1 f t te t e t f f t t t        Therefore           0 0 0 2 ) ( t t te t f t