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Today's lecture

This lecture explores cascade systems, frequency response, and the significance of zeros and poles in H(z). Examples and analysis tools are provided.

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Today's lecture

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  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-1 + 2z-2 - z-3 Given that one root of H(z) is z=1, so H1(z)= (1-z-1), find H2(z) H2(z) = H(z)/ H1(z) H2(z) = (1- z-1 + z-2) H(z) = H2(z) H1(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) = H1(z) H2(z) X(z) Y(z) = H(z) X(z) Y(z) = X(z) if H(z) = 1 or H1(z) H2(z) = 1 or H1(z) = 1/ H2(z)

  5. Z-Transform Definition

  6. Convolution Property

  7. Special Case: Complex Exponential Signals • What if x[n] = zn with z= e jώ? • y[n] = ∑ bkx [n-k] • y[n] = ∑ bkz [n-k] • y[n] = ∑ bkzn z –k • y[n] =( ∑ bkz-k)z n • y[n] = H(z)x[n] M K=0 M K=0 M K=0 M K=0

  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] = znfor all n give output y[n] = H(z)zn • H(z) is a complex constant, which through complex multiplication causes a magnitude and phase change of the input signal zn • Ifz0is one of the zeros of H(z), then H(z0) = 0 so the output will be zero.

  18. Example 7.10 Nulling signals with zeros H(z) = 1- 2z-1 + 2z-2 –z-3 z1 = 1 z2 = ½ + j (3 -1/2)/2 = e jπ/3 z3 = 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 x1[n] = (z1)n = 1 x2[n] = (z2)n = e jπn/3 x3[n] = (z3)n = e -jπn/3

  19. Nulling Property of H(z)

  20. Plot Zeros in z-Domain

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