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### PZ-plot-demo

Course: ECE 4078, Fall 2009
School: Georgia Tech
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Word Count: 690

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4078 EE HANDOUT: POLES and ZEROS EE 4078 DIGITAL SIGNAL PROCESSING Prof. J. H. McClellan Georgia Tech 1994 POLES DETERMINE THE SHAPE OF THE IMPULSE RESPONSE POLES and ZEROS DETERMINE THE SHAPE OF THE FREQUENCY RESPONSE SECOND-ORDER FILTERS HAVE RESONANT PEAKS IN FREQUENCY DOMAIN BANDWIDTH in FREQUENCY INVERSELY RELATED TO DECAY RATE IN TIME ALL PLOTS of |H(ej )| USE NORMALIZED FREQUENCY: /(2) PLOT is shown...

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4078 EE HANDOUT: POLES and ZEROS EE 4078 DIGITAL SIGNAL PROCESSING Prof. J. H. McClellan Georgia Tech 1994 POLES DETERMINE THE SHAPE OF THE IMPULSE RESPONSE POLES and ZEROS DETERMINE THE SHAPE OF THE FREQUENCY RESPONSE SECOND-ORDER FILTERS HAVE RESONANT PEAKS IN FREQUENCY DOMAIN BANDWIDTH in FREQUENCY INVERSELY RELATED TO DECAY RATE IN TIME ALL PLOTS of |H(ej )| USE NORMALIZED FREQUENCY: /(2) PLOT is shown over the RANGE: 0 because |H(ej )| is EVEN-SYMMETRIC with respect to = 0 1 EXAMPLE of TIME, FREQUENCY & z DOMAINS POLES and ZEROS THE FIRST-ORDER TRANSFER FUNCTION IS B(z) 1 H(z) = = A(z) 1 - 0.9 z -1 Imaginary Part 1 0.5 IMPULSE RESPONSE: OUTPUT when x[n] = [n] h[n] = (0.9)n 0 for n 0 for n < 0 0 THE FREQUENCY RESPONSE MAGNITUDE IS |H(ej )| = H(ej ) PEAKS AT = 0 1 1.81 - 1.8 cos POLE -0.5 -1 -1 -0.5 0 0.5 1 Real Part IMPULSE RESPONSE 10 1 9 DTFT: FREQUENCY RESPONSE 0.8 an 1 MAGNITUDE 8 7 6 5 4 3 2 1 0.6 0.4 NORMALIZED FREQ = /2 0.2 0 -5 0 5 10 15 20 25 30 35 INDEX (n) 2 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 NORMALIZED FREQUENCY 0-->0.5 FIRST-ORDER EXAMPLE: HIGH-PASS FILTER POLES and ZEROS THE FIRST-ORDER TRANSFER FUNCTION: H(z) = 1-z 1 + 0.707 z -1 -1 1 0.5 DIFFERENCE EQUATION: y[n] = -0.707y[n - 1] + x[n] - x[n - 1] THE FREQUENCY RESPONSE MAG SQUARED: 2 - 2 cos |H(e )| = 1.5 + 2 cos j 2 Imaginary Part 0 -0.5 IT PEAKS AT = -1 -1 -0.5 0 0.5 1 Real Part IMPULSE RESPONSE 7 1 DTFT: FREQUENCY RESPONSE 0.5 1 ENVELOPE is G 2 n 6 5 0 MAGNITUDE 4 -0.5 3 -1 h[n] = (1 + -1.5 -1 2) 2 20 n u[n] - 2 2 [n] 1 -5 0 5 10 15 25 30 35 INDEX (n) 3 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 NORMALIZED FREQUENCY 0-->0.5 SECOND-ORDER EXAMPLE: TIME, FREQUENCY & z DOMAINS POLES and ZEROS THE TRANSFER FUNCTION: 1 + z -1 H(z) = 1 - 1.4248 z -1 + 0.8649 z -2 IMPULSE RESPONSE: OUTPUT when x[n] = [n] h[n] = A rn cos( n + ) u[n] POLES at 0.93 ej2/9 THE FREQUENCY RESPONSE MAG SQUARED: |H(ej )|2 = 2 + 2 cos (1 - 2r cos( - ) + r2 )(1 - 2r cos( + ) + r2 ) -1 -1 -0.5 0 0.5 1 1 0.5 Imaginary Part 0 -0.5 IT PEAKS AT = = 2 = 2(0.111 . . .) 9 IMPULSE RESPONSE 25 2.5 2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -5 0 5 10 15 20 25 30 35 Real Part DTFT: FREQUENCY RESPONSE DECAY RATE INVERSE to BW 20 RESONANT PEAK at = 2/9 MAGNITUDE 15 10 5 INDEX (n) 4 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 NORMALIZED FREQUENCY 0-->0.5 SECOND-ORDER EXAMPLE: CLOSER TO UNIT CIRCLE POLES and ZEROS THE TRANSFER FUNCTION: H(z) = 1 - z -2 1 - 1.8418 z -1 + 0.9604 z -2 1 DIFFERENCE EQUATION: y[n] = 1.8418y[n - 1] - 0.9604y[n - 2] + x[n] - x[n - 2] POLES at 0.98 ej2/18 THE FREQUENCY RESPONSE SQUARED: Imaginary MAG Part 0.5 0 -0.5 2 - 2 cos(2) |H(ej )|2 = (1 - 2r cos( - ) + r2 )(1 - 2r cos( + ) + r2 ) -1 IT PEAKS AT = = 2 18 = 2(0.0555 . . .) -1 -0.5 0 0.5 1 Real Part IMPULSE RESPONSE 2 50 DTFT: FREQUENCY RESPONSE LESS DECAY vs. TIME 1.5 1 45 40 35 NARROWER PEAK RESONANT PEAK at = 2/18 MAGNITUDE 0 5 10 15 20 25 30 35 0.5 30 25 20 15 0 -0.5 -1 10 5 0 0 -1.5 -5 INDEX (n) 5 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 NORMALIZED FREQUENCY 0-->0.5 SECOND-ORDER EXAMPLE: HIGHER FREQUENCY POLES and ZEROS THE TRANSFER FUNCTION: 1 + z -1 H(z) = 1 + 0.85 z -1 + 0.7225 z -2 IMPULSE RESPONSE: OUTPUT when x[n] = [n] h[n] = A rn cos( n + ) u[n] POLES at 0.85 ej2/3 THE FREQUENCY RESPONSE MAG SQUARED: |H(ej )|2 = 2 + 2 cos (1 - 2r cos( - ) + r2 )(1 - 2r cos( + ) + r2 ) -1 -1 -0.5 0 0.5 1 1 0.5 Imaginary Part 0 -0.5 IT PEAKS AT = = 2 = 2(0.333 . . .) 3 IMPULSE RESPONSE 4.5 1 4 0.8 0.6 0.4 0.2 0 -0.2 -0.4 1 -0.6 0.5 -0.8 -5 0 5 10 15 20 25 30 35 0 0 Real Part DTFT: FREQUENCY RESPONSE "PERIOD" is 3 MAGNITUDE 3.5 3 2.5 2 1.5 PEAK at = 2/3 ENVELOPE is G(0.85)n INDEX (n) 6 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 NORMALIZED FREQUENCY 0-->0.5 SECOND-ORDER EXAMPLE: ALL-PASS FILTER THE TRANSFER FUNCTION: H(z) = 0.7225 + 0.85 z -1 + z -2 1 + 0.85 z -1 + 0.7225 z -2 1 POLES and ZEROS IMPULSE RESPONSE: OUTPUT when x[n] = [n] h[n] = 0.7225 + A rn cos( n + ) u[n - 1] POLES at 0.85 ej2/3 THE MAGNITUDE RESPONSE IS ONE, PHASE IS Imaginary Part 0.5 0 -0.5 H(ej ) = -2 - 2 tan-1 r sin( - ) - ... 1 - r cos( - ) -1 PHASE CHANGES RAPIDLY AT ...

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