Challenge_14-2 - π ω 2 For ω = 2 π/16 T=16 and t d = 1...

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Lesson Title: FIR Frequency Response Lesson Number: 14 (Section 6-1 to 6-2) Challenge Compare the frequency responses of two L=6 order LTI FIRs shown below: h 1 ={1,1,1,1,1,0} h 2 ={0,1,1,1,1,1} Compare them in terms of magnitude and phase frequency responses. Determine whether they are the same or different? Response Computing the frequency response H 1 (e j ω ) = 1 + e -j ω + e -j2 ω + e -j3 ω + e -j4 ω = = e -j2 ω ( e j2 ω + e j ω + 1 + e -j ω + e -j2 ω 29 = e -j2 ω ( 1+2cos( ω29 +2cos(2 ω29;-π≤ω<π H 2 (e j ω ) = e -j ω + e -j2 ω + e -j3 ω + e -j4 ω + e -j5 ω = = e -j3 ω ( e j2 ω + e j ω + 1 + e -j ω + e -j2 ω 29 = e -j3 ω ( 1+2cos( ω29 +2cos(2 ω29;-π≤ω<π Comparing the magnitude frequency responses one determines that they are equal. The phase response of the first filter is H 1 (e j ω )=-2 ω and for the second H 2 (e j ω )=-3 ω . The difference in phase is ω which corresponds to a time-delay td through: T t d 2245
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Unformatted text preview: π ω 2 For ω = 2 π /16, T=16 and t d = 1 sample. For ω = 2 π /8, T=8 and again t d = 1 sample. This is supported by the simple (non-MATLAB) simulation: h1=[1,1,1,1,1,0,zeros(32)] {filter 1} h2=[0,1,1,1,1,1,zeros(32)] {filter 2} h1f=fft(h1) {FFT filter 1} h2f=fft(h2) {FFT filter 2} graph(mag(h1f),mag(h2f)) {graph magnitude frequency response – Figure 1.a} graph(phs(h1f),phs(h2f)) {graph p[hse response – Figure 1.b} x=mkcos(1,1/16,0,64) {create a cosine at f=f s /16} y1=h1$x; y2=h2$x {convolve cosine with filters} ograph(y1[0:50],y2[0:50]) {display convolved signals – Figure 2} Figure 1: (a) Magnitude responses (left) (b) Phase responses (right) Figure 2: Filtered cosine at ω =1/16. 1-sample 16-samples...
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Challenge_14-2 - π ω 2 For ω = 2 π/16 T=16 and t d = 1...

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