Exam-1-w-08-sol

Exam-1-w-08-sol - California State Polytechnic University...

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Unformatted text preview: California State Polytechnic University, Pomona Electrical & Computer Eng. Dr. Z. Aliyazicioglu ! " # " ") * " " $ + , $ ' $ xa (t ) = 4 cos(2 1000t ) - 2sin(2 1500t ) " x ( n) % $ & '( - Fs = 800 Hz " ' x ( n) " $ y (t ) ' '( x ( n) " ' x( n) $ - & .( '( ' - x (t ) 1000 1500 n) - 2sin(2 n) 800 800 5 18 x(n) = 4sin(2 n) - 2sin(2 n) 4 8 1 1 x(n) = 4 cos 2 (1 + )n) - 2sin 2 (2 - )n 4 8 1 1 x(n) = 4 cos 2 n + (2 n) - 2sin 4 n - 2 4 8 x(n) = xa (nT ) = 4sin(2 x( n) = 4 cos(2 n) cos 2 1 1 1 1 n - 4 sin(2 n) sin 2 n - 2 sin(4 n) cos(2 ) n + 2 cos(4 n) sin(2 ) n 4 4 8 8 x ( n) = 4 cos 2 b. 1 1 n + 2 sin(2 ) n 4 8 y (t ) = x ( n) 1 1 Fs t + 2 sin 2 Fs t 4 8 = 4 cos(2 200t ) + sin(2 100t ) n = Fs t = 4 cos 2 c. F fold = Fs 800 = = 400 Hz 2 2 d. FN = 2 Fmax = 2(1500) = 3KHz x(n) = xa (nT ) = 4 cos 2 1000 1500 1 1 n - 2sin 2 n = 4 cos 2 n - 2sin 2 n 3000 3000 3 2 By: Z. Aliyazicioglu ECE 306-1-1 " # " " $ ( '$ / x(t ) = 5cos 2 1000t " 0 $ ' Fs = 8 KHz x(n) ( 1 % $ - / ' 0 t 2ms = 0.001 " 2 % $ ' ( 1 38$ ' $ ' $ $ , + 3 " ( 4 5 . 6 7 %$ 38$ $ ' 3 " 8$ $ ( 1 '( ' ' 3 a. x(n) = x(t ) t = nT = 5cos 2 1000 1 n = 5cos 2 n 8000 8 5 4 3 2 1 x(t) 0 -1 -2 -3 -4 -5 0 0.2 0.4 0.6 0.8 1 t 1.2 1.4 1.6 1.8 x 10 2 -3 b. X max - X min 5+5 0.001 = b b 2 -1 2 -1 b = log 2 (10001) = 14 bits = 2b - 1 = 10 = 10000 0.001 c. d. 64 dB = 1.76 + 6.02b b= 64 - 1.76 = 11 bits 6.02 = X max - X min 5+5 = 11 = 0.0049 V b 2 -1 2 -1 Fq = 16( Fs ) = 16(8000) = 128 Kb / s = X max - X min 5+5 = 16 = 1.5(10-4 ) V b 2 -1 2 -1 By: Z. Aliyazicioglu ECE 306-1-2 9: " : $ $ ' y ( n) & $ % - ' $ ' % x(n) = {1, 2, 4,3} h(n) = {2, 4, -1} x(n) y1(n) h (n) + y (n) y2(n) z -1 h (n) y1 (n) = k x(k) h(-k) h(-1-k) h(1-k) h(2-k) h(3-k) h(4-k) h(5-k) -5 -4 h( n - k ) x ( k ) -2 -1 4 -1 1 4 2 -1 0 2 2 4 -1 1 4 2 4 -1 2 3 3 4 5 6 7 y(0)=4+4=8 y(-1)=2 y(1)=-1+8+8=15 y(2)= -2+16+6=20 y(3)=-4+12=8 y(4)=-3 n =- -3 -1 2 4 -1 2 4 2 y1 (n) = {2,8,15, 20,8, -3} y2 ( n ) = k x(k) h(-k) h(-1-k) h(1-k) h(2-k) h(3-k) h(4-k) h(5-k) h(n - k ) x(k - 1) -5 -4 -3 -1 -2 -1 4 -1 4 2 -1 0 1 2 4 -1 1 2 2 4 -1 2 4 3 3 4 5 6 7 y(0)=2 y(-1)=0 y(14+4=8 y(2)=-1+8+8=15 y(3)= -2+16+6=20 y(4)=-4+12=8 y(5)=-3 n =- 2 4 -1 2 4 -1 2 4 2 y2 (n) = {2,8,15, 20,8, -3} y (n) = y1 (n) + y2 (n) = {2,8,15, 20,8, -3} + {2,8,15, 20,8, -3} = {2,10, 23,35, 28,5, -3} By: Z. Aliyazicioglu ECE 306-1-3 6 8 " ' ' $ ' ' / ( ( $ ' & $ ' x (n ) = u(n ) " % " y ( n ), n 0 x(n) y(n) + z-1 + 1.1 -0.3 + z-1 -0.3 Solution: y (n) = 1.1 y (n - 1) - 0.3 y (n - 2) + x(n) - 0.3 x(n - 1) 1. Homogenous solution y h ( n ) = nu ( n ) n - 1.1 n -1 + 0.3 n - 2 = 0 n - 2 ( 2 - 1.1 + 0.3) = 0 1 = 0.5, 2 = 0.6 2 - 1.1 + 0.3 = 0 ( - 0.5)( - 0.6) = 0 y h ( n ) = c1(0.5)n + c2 (0.6)n u ( n ) 2. Particular Solution for x ( n ) = u ( n ) y p (n) = Ku (n) Ku (n) = 1.1Ku (n - 1) - 0.3Ku (n - 2) + x(n) - 0.3 x(n - 1) For n=2; Ku (2) = 1.1Ku (1) - 0.3Ku (0) + x(2) - 0.3 x(1) K = 1.1K - 0.3K + 1 - 0.3 (1 - 1.1 + 0.3) K = 0.7 0.7 K= = 3.5 0.2 y p (n) = 3.5u (n) The total solution; y ( n ) = y h ( n ) = y p ( n ) = c1(0.5)n + c2 (0.6)n + 3.5 u ( n ) By: Z. Aliyazicioglu ECE 306-1-4 For n=0 From diff equ. To find c1 and c2 y (0) = 1.1 y (-1) - 0.3 y (-2) + x(0) - 0.3 x(-1) y (0) = 1 From total solution y (0) = c1 (0.5)0 + c2 (0.6)0 + 3.5 u (0) y (0) = c1 + c2 + 3.5 Equation 1 c1 + c2 = -2.5 For n=1 From diff equ. y (1) = 1.1 y (0) - 0.3 y (-1) + x(1) - 0.3 x(0) y (1) = 1.1(1) - 0 + 1 - 0.3 y (1) = 1.8 From total solution y (1) = c1 (0.5)1 + c2 (0.6)1 + 3.5 u (1) y (1) = 0.5c1 + 0.6c2 + 3.5 Equation 2 0.5c1 + 0.6c2 = -1.7 c1 c2 -2.5 -1.7 c1 c2 2 -4.5 1 1 0.5 0.6 = = The total solution is y ( n) = 2(0.5) n - 4.5n(0.6) n + 3.5 u ( n) By: Z. Aliyazicioglu ECE 306-1-5 ...
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This note was uploaded on 03/11/2008 for the course ECE 306 taught by Professor Aliyazicioglu during the Winter '08 term at Cal Poly Pomona.

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