Lesson_20

# Lesson_20 - Lesson20 Challenge19 Lesson20FIRfilters...

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Lesson 20 Challenge 19 Lesson 20 – FIR filters Challenge 20

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Challenge 19 A filter is clocked at a rate f s =1k Sa/s and has the pole-zero distribution  shown and is known to have DC gain of 0dB.  What is the filter’s magnitude  frequency and phase response at f=0.25kHz? X      0.5+j0.5 X     0.5-j0.5 -0.5+j0.5  0 -0.5-j0.5    0
Challenge 19 The transfer function is given by: H(z) = k (z+0.5+j0.5) (z+0.5-j0.5)/ ((z- 0.5+j0.5) (z- 0.5-j0.5))  Determine k H(z=1)=1=k(2.5)/(0.5)=5k.   Therefore k=0.2. X 0.5+j0.5 X 0.5-j0.5 0 -0.5+j0.5 0 -0.5-j0.5 A B C D

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Challenge 19 Determine A,B,C, and D at  ω  =  π /2 (z=0+j1) A=0.707 135 ° B=1.58 108.5 ° C=1.58 71.5 ° D=0.707 45 ° At test frequency is at  ω  =  π /2: H(z=j)=0.2(CD)/(AB)              = 0.2 (1 -127 ° ) X 0.5+j0.5 X 0.5-j0.5 0 -0.5+j0.5 0 -0.5-j0.5 A B C D
Challenge 19 Verify » b=[.2, .2, .2*.5];   {0.2(z 2 +z+0.5)} » a=[1, -1, 0.5];       {(z 2 -z+0.5)} » x=[1,zeros(1,100)]; » y=filter(b,a,x); » yf=fft(y); » plot(abs(yf(1,1:100))) » plot(360*angle(yf(1,1:100))/(2*pi))

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Challenge 19 1.0 0.2
Challenge 19 0.0 ° -127 °

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Lesson 20    Lesson 20:  Filters and FIR Properties
Lesson 20 FIR (basic ideas) Impulse Response Direct Architecture { } 1 1 0 ,..., , ] [ - = N h h h k h - = = 1 0 ] - [ ] [ N m m m k x h k y

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Lesson 20 FIR (basic ideas) Transfer function Frequency Response - = - = 1 0 ) ( N k k k z h z H ( 29 k j N k k j e z e h e H z H j ω ω ω - - = = = = 1 0 ) (
Lesson 20 Special Cases N th  order MA (N+1) st  order comb ( 29 1 1 - 1 1 - 1 1 1 1 1 1 1 1 1 1 ) ( 1 1 - 1 - 1 - 0 - 1 0 - - = - = - - - = = - = = - - - - = = - = z z z N z z N z z N z N z N z N z N z H N N N N N m m m m N m m MA ( 29 N N N C z z z z H 1 1 ) ( ± = ± = -

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Lesson 20 Special Cases Moving Average Comb
Lesson 20 Polynomial zeros: ( 29 ( 29 ] 1 - , 0 [ for 0 1 ] 1 - , 0 [ for 0 1 / 2 / / 2 N i e z z N i e z z N

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