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# Lesson_30 - Lesson 30 Lesson 30 Challenge 29 Lesson 30 –...

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Unformatted text preview: Lesson 30 Lesson 30 Challenge 29 Lesson 30 – Introduction to Frequency Response Challenge 30 Lesson 30 Challenge 29 Assume you have a filter having 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 =100 r/s ? Lesson 30 Challenge 29 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 =100 r/s , H(j ϖ )= 1000/(-10000+10100) +20j) = (1000) (1/(100+j2000)) = 0.499 ∠-tan-1 (2000/100) ° = 0.499 ∠-87 ° . Lesson 30 Challenge 29 H(s)=1000/(s 2 +20s+10100) >> [H,w]=freqs([100*10],[1,20,10100]); >> plot(w,abs(H)) >> plot(w,180*angle(H)/pi); ~0.5 ~-90 ° Lesson 30 Challenge 29 Observe that the magnitude frequency response of H(j ϖ ) =100/((- ϖ 2 + 10100) + j20 ϖ ) is maximized when (- ϖ 2 + 10100) + j20 ϖ = j20 ϖ or ϖ 2 =10100 or ϖ =100.5 r/s and not 100 r/s. At ϖ =100.5 r/s the filter’s gain is 0.5 and the phase shift ∠-90 ° . Lesson 30...
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Lesson_30 - Lesson 30 Lesson 30 Challenge 29 Lesson 30 –...

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