Wpzfilt7 - ECE 421 - Sum 2010 Detailed Solutions - Set 7:...

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Unformatted text preview: ECE 421 - Sum 2010 Detailed Solutions - Set 7: Filter Design by Pole-Zero Placement 1 7-1. The poles and zeros of a digital filter are POLES: z = : 8 e j = 6 , ZEROS: z = ; z = 0. With G = 1, compute the simplest form of the difference equation which implements this filter. DETAILED SOLUTION: The transfer function H z in terms of the poles p 1 and p 2 and the zeros z 1 and z 2 is given by H z = G z z 1 z z 2 z p 1 z p 2 a Substitute into a the values of the poles and zeros from the problem statement, as well as G = 1, giving H z = z z z : 8 e j = 6 z : 8 e j = 6 b Multiply out the factors in c , and use Eulers formula to obtain H z = z 2 z 2 1 : 6 z cos = 6 : 64 = z 2 z 2 1 : 386 z : 64 c Substitute H z = Y z = X z in c and write the right side of c using negative powers of z : Y z X z = 1 1 1 : 386 z 1 : 64 z 2 d Cross-multiplying terms in d then gives Y z 1 : 386 z 1 Y z : 64 z 2 Y z = X z e Taking the inverse Z-transform term-by-term of e then provides y [ m 1 : 386 y [ m 1 : 64 y [ m 2 = x [ m f Finally, solving f for y [ m then gives the desired form for the difference equation: y [ m = 1 : 386 y [ m 1...
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This note was uploaded on 09/07/2010 for the course ECE 421 taught by Professor Hallen during the Summer '08 term at N.C. State.

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Wpzfilt7 - ECE 421 - Sum 2010 Detailed Solutions - Set 7:...

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