Consider the case where all of the poles are real except for two complex poles

Consider the case where all of the poles are real

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Consider the case where all of the poles are real except for two complex poles that are complex conjugates. W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 124 / 170
Inverse Z Transform For Systems With Complex Poles Partial fraction expansion can be used, for this case, to expand the transfer function H ( z ) as H ( z ) = N 1 X m = 1 F m ( z ) + N 2 X k = 1 C k z z - p k (130) where all of the C k and p k are real and F m ( z ) = D 1 m z z - p 1 m + D 2 m z z - p 2 m (131) The constants D 1 m and D 2 m are complex conjugates and the poles p 1 m and p 2 m are complex conjugates. W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 125 / 170
Inverse Z Transform For Systems With Complex Poles The inverse Z Transform of F m ( z ) can be determined to obtain f m ( n ) = [ D 1 m ( p 1 m ) n + D 2 m ( p 2 m ) n ] u ( n ) (132) The two complex conjugate poles, p 1 m and p 2 m , can be written in complex exponential form as p 1 m = e α + j ω p 2 m = e α - j ω (133) Then, f m ( n ) can be written in the form f m ( n ) = h D 1 m e ( α + j ω ) n + D 2 m e ( α - j ω ) n i (134) W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 126 / 170
Inverse Z Transform For Systems With Complex Poles However, e j ω n = cos ( ω n ) + jsin ( ω n ) e - j ω n = cos ( ω n ) - jsin ( ω n ) (135) Thus, f m ( n ) = ( D 1 m + D 2 m ) e α n cos ( ω n ) u ( n ) + j ( D 1 m - D 2 m ) e α n sin ( ω n ) u ( n ) (136) Observe that D 1 m + D 2 m = 2 < e ( D 1 m ) D 1 m - D 2 m = 2 j = m ( D 1 m ) (137) W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 127 / 170
Inverse Z Transform For Systems With Complex Poles Thus f m ( n ) = 2 < e ( D 1 m ) e α n cos ( ω n ) u ( n ) - 2 = m ( D 1 m ) e α n sin ( ω n ) u ( n ) (138) f m ( n ) can be further simplified by using the trigonometric identity cos( ω n + φ ) = cos( φ ) cos( ω n ) - sin( φ ) sin( ω n ) (139) W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 128 / 170
Inverse Z Transform For Systems With Complex Poles It follows that f m ( n ) can be written in the form f m ( n ) = Ae α n cos( ω n + φ ) (140) where φ = tan - 1 = m ( D 1 m ) < e ( D 1 m ) A = 2 q = m ( D 1 m ) 2 + < e ( D 1 m ) 2 (141) Example 1.9 provides an example to illustrate this concept. W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 129 / 170
Complex Poles Inverse Z Transform Example Example (1.9) We wish to find the impulse response of the discrete time system with its transfer function given by H ( z ) = 0 . 5276 z 3 + 1 . 5829 z 2 + 1 . 5829 z + 0 . 5276 z 3 + 1 . 76 z 2 + 1 . 1829 z + 0 . 2781 (142) If x ( n ) is an impulse, then X ( z ) = X n = -∞ δ ( n ) z - n = 1 (143) Thus, Y ( z ) = H ( z ) X ( z ) = H ( z ) (144) W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 130 / 170
Complex Poles Inverse Z Transform Example Example (1.9) For convenience in obtaining the geometric series form, we expand H ( z ) z = 0 . 5276 z 3 + 1 . 5829 z 2 + 1 . 5829 z + 0 . 5276 z ( z 3 + 1 . 76 z 2 + 1 . 1829 z + 0 . 2781 ) = D 1 z - p 1 + D 2 z - p 2 + C 1 z - p 3 + C 2 z (145) W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 131 / 170
Complex Poles Inverse Z Transform Example Example (1.9) It follows that p 1 = - 0 . 6252 + 0 . 3935 j p 2 = - 0 . 6252 - 0 . 3935 j = p * 1 p 3 = - 0 . 5097 p 4 = 0 (146) and D 1 = - 0 . 3221 + 0 . 1501 j D 2 = - 0 . 3221 - 0 . 1501 j = D * 1 C 1 = - 0 . 7254 C 2 = 1 . 8972 (147) W. Alexander & C. Williams (NCSU) FUNDAMENTAL DSP CONCEPTS ECE 513, Fall 2019 132 / 170
Complex Poles Inverse Z Transform Example Example (1.9) Define F 1 ( z ) = D 1 z z - p 1 + D 2 z z - p 2 (148) Then, the above development can be used to find f 1 ( n ) , which is the part of the impulse response corresponding to the complex pair of poles, p

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