1 Today's lecture −Cascade Systems −Frequency Response −Zeros of H(z) −Significance of zeros of H(z) −Poles of H(z) −Nulling Filters

2 Cascade Example : Combining Systems

3 Factoring z-Polynomials Example 7.7: Split H(z) into cascade H(z) = 1- 2z z -2 - z -3 Given that one root of H(z) is z=1, so H 1 (z)= (1-z -1 ), find H 2 (z) H 2 (z) = H(z)/ H 1 (z) H 2 (z) = (1- z -1 + z -2 ) H(z) = H 2 (z) H 1 (z) H(z) = (1-z -1 ) (1- z -1 + z -2 )

4 Deconvolution or Inverse Filtering Can the second filter in the cascade undo the effect of the first filter? Y(z) = H 1 (z) H 2 (z) X(z) Y(z) = H (z) X(z) Y(z) = X(z) if H (z) = 1 or H 1 (z) H 2 (z) = 1 or H 1 (z) = 1/ H 2 (z)

5 Z-Transform Definition

6 Convolution Property

7 Special Case: Complex Exponential Signals −What if x[n] = z n with z = e jώ ? −y[n] = ∑ b k x [n-k] −y[n] = ∑ b k z [n-k] −y[n] = ∑ b k z n z –k − y[n] =( ∑ b k z -k ) z n −y[n] = H(z) x[n] K=0 M M M M

8 Three Domains

9 Frequency Response

10 Another Analysis Tool

11 Zeros of H(z)

12 Zeros, Poles of the H(z) Each factor of the form (1- az -1 ) can be expressed as (1- az -1 ) = (z-a)/z representing a zero at z = a and a pole at z = 0

13 Zeros of H(z)

14 Plot Zeros in z-Domain

15 Significance of the Zeros of H(z) − Zeros of a polynomial system function are sufficient to determine H(z) except for a constant multiplier. − System function H(z) determines the difference equation of the filter − Difference equation is a direct link b/w an input x[n] and its corresponding output y[n] − There are some inputs where knowledge of the zero locations is sufficient to make a precise statement about the output without actually computing it using the difference equation.

16 Poles of H(z)

17 − Signals of the form x[n] = z n for all n give output y[n] = H(z) z n − H(z) is a complex constant, which through complex multiplication causes a magnitude and phase change of the input signal z n − If z 0 is one of the zeros of H(z), then H(z 0 ) = 0 so the output will be zero.

18 Example 7.10 Nulling signals with zeros H(z) = 1- 2z z -2 –z -3 z 1 = 1 z 2 = ½ + j (3 -1/2 )/2 = e jπ/3 z 3 = ½ - j (3 -1/2 )/2 = e -jπ/3 All zeros lie on the unit circle, so complex sinusoids with frequencies 0, π/3 and - π/3 will be set to zero by the system. Output resulting from the following three inputs will be zero x 1 [n] = (z 1 ) n = 1 x 2 [n] = (z 2 ) n = e jπn/3 x 3 [n] = (z 3 ) n = e -jπn/3

19 Nulling Property of H(z)

20 Plot Zeros in z-Domain