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

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Cascade Example : Combining Systems

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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)

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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)

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Z-Transform Definition

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Convolution Property

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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]

K=0

K=0

K=0

K=0

M

M

M

M

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Three Domains

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Frequency Response

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Another Analysis Tool

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Zeros of H(z)

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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

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Zeros of H(z)

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Plot Zeros in z-Domain

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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.

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Poles of H(z)

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−Signals of the form x[n] = zn for 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

−If z0 is one of the zeros of H(z), then H(z0) = 0 so

the output will be zero.

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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

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Nulling Property of H(z)

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Plot Zeros in z-Domain