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Challenge_19-1

# Challenge_19-1 - -3 Y(z)=H(z)X(z = 0.25z-0...

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EEL 3135: Dr. Fred J. Taylor, Professor Lesson Title: Convolution Theorem Lesson Number: 19 [Sections 7-3 to 7-5 with part of 7-6.] Challenge Consider a 4 th order MA filter having an impulse response h[n]=(1/4)[1,1,1,1] and a 4-sample input signal x[n]=cos(2 π n/4) = {1,0,-1,0}. What is the z-transform of the output (i.e., Y(z)=?). Response The z-transform are as follows: x[n]=cos(2 π n/4) = {1,0,-1,0} X(z) = 1z -0 + 0z -1 -1z -2 + 0z -3 h[n]=(1/4) {1,1,1,1} X(z) = 1/4z -0 + 1/4z -1 +1/4z -2 + 1/4z
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Unformatted text preview: -3 Y(z)=H(z)X(z) = 0.25z-0 +0.25z-1-0.25z-4-0.25z-5 or y[n]={0.25, 0.25, 0, 0, -0.25 .0.25, 0} (why isn’t the output of the MA zeros (the 4-sample average value of x[n]?) Pseudo MATALB code x=[1,0,-1,0]; h=.25*[1,1,1,1] h\$x {convolution} 0.25 0.25 0 0 -0.25 -0.25 xx=x&x&x&x&x&x&x&x; len(xx) 32 h\$xx 0: 0.25 0.25 0 0 0 0 0 0 0: 0 0 0 0 0 0 0 0 0: 0 0 0 0 0 0 0 0 0: 0 0 0 0 0 -0.25 -0.25 0 { length 32+4-1=35} Should the above response been anticipated? 1...
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