Transcript Signals and Systems

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
5
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]
M
K=0
M
− y[n] = ∑ bkz [n-k]
K=0
M
− y[n] = ∑ bkzn z –k
K=0
M
− y[n] =( ∑ bkz-k) z n
K=0
− y[n] = H(z) x[n]
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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
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