Challenge_30

# Challenge_30 - ϖ)= 1000-10000 10100 20j =(1000(1(100 j2000...

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Digital Signal Processing Dr. Fred J. Taylor, Professor Lesson Title: Introduction to Frequency Response Lesson Number: [30] (Chapter 10) Challenge 30 A filter has an impulse response given by: h(t) = 10 e -10t sin(100t) u(t) (see Table 1, Lesson 27). What is the filter’s frequency response measured at ϖ 0 =100 r/s ? Response Observe that h(t) = 10 e -10t sin(100t) u(t) H(s)=10*100/((s+10) 2 +100 2 ) = 1000/(s 2 +20s+10100) The frequency response of the filter is defined by: H(s)| s=j ϖ = H(j ϖ ) = 1000/((j ϖ ) 2 +20 j ϖ + 10100) = 100/((- ϖ 2 + 10100) + j20 ϖ ) At ϖ = ϖ 0 =100 r/s , H(j

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Unformatted text preview: ϖ )= 1000/(-10000+10100) +20j) = (1000) (1/(100+j2000)) = 0.499 ∠-tan-1 (2000/100) ° = 0.499 ∠-87 ° . >> [H,w]=freqs([100*10],[1,20,10100]); >> plot(w,abs(H)) >> plot(w,180*angle(H)/pi); Chapter 10 - 1 Digital Signal Processing Dr. Fred J. Taylor, Professor Observe that the magnitude frequency response will be maximized when (-ϖ 2 + 10100) + j20 ϖ = j20 ϖ or ϖ 2 =10100 or ϖ =100.5 r/s not 100 r/s. At ϖ =100.5 r/s the filter’s gain is 0.5 and the phase shift ∠-90 ° . Chapter 10 - 2...
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## This note was uploaded on 04/15/2010 for the course EEL 3135 taught by Professor ? during the Spring '08 term at University of Florida.

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Challenge_30 - ϖ)= 1000-10000 10100 20j =(1000(1(100 j2000...

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