Chapter 5: DiscreteTime Systems: ZTransform Analysis

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Chapter 5 Discrete-Time SystemsZ-Transform
Analysis
Discrete Time D-T
Continuous Time C-T
Impulse Response, h(t) Convolution Integral
Impulse Response, hk Convolution Sum
Time Domain T-D
Laplace Transform Fourier Series Fourier
Transform
Z-Transform D-T Fourier Transform
Frequency Domain F-D
Signal Analysis
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The Big Picture (Again)

• We still want to develop our understanding LTI
  systems. This has been done thus far by
  decomposing the input signal f(t), fk, into a
  set of simpler functions. The Z transform is
  just another method for accomplishing this
  decomposition

Time Domain
Frequency Domain
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5.1 The Z-Transform

• The Unilateral Z-Transform is defined through the
  pair of equations
• Notation Use the Z operator to denote the
  Z-transform
• Notation Use the arrow operator to denote a
  Z-transform pair

Z-Transform
Unilateral assumes causal system and signals
Inverse Z-Transform
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5.1 The Z-Transform

• Linearity of the Z-Transform

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5.1 The Z-Transform

• Region of Convergence (ROC) of Fz
• The Z-transform
  may not converge for all values of z.
• That is, as k??, Fz may go to infinity for some
  z.
• The ROC is the set of z for which Fz converges
  (or exists).
• See example 5.1

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5.1 The Z-Transform

• Existence of the Z-Transform
• The Z-transform exists when Fzlt ?. This
  occurs when
• for some z
• Therefore, any signal fz that does not grow
  faster than an exponential r0k has a Z-transform.
  This can be expressed as

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5.1-1 Finding the Inverse Transform

• Explicit solution of the inverse Z-transform
  requires solution of
• Complex integration is beyond the scope of this
  course.
• Therefore we restrict our scope to those
  functions whose inverse Z-transform is in table
  5.1
• The trick is to get Fz into the proper form.
  This will require use of partial fraction
  expansion (See B.8)

Im
ROC
Re
which requires integration across the complex
plane
Possible path of integration
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5.1-1 Finding the Inverse Transform

• Use partial fraction expansion as before to
  rewrite Fz into a form found in table 5.1.
• Slight modification find partial fraction
  expansion for then multiply through by
  z.
• This is done to get solution in the form of uk
  instead of uk-1
• This is from the following transform pairs

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5.2 Some Properties of the Z-Transform
Operation
f(t)
F(s)

• Addition
• Scalar Multiplication
• Right Shift (Delay)
• Left Shift (Advance)
• Convolution
• Multiplication by ?k
• Multiplication by k

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5.3 Z-Transform Solution of Linear Difference
Equations

• Use left and right shift properties of the
  Z-Transform to determine solution.
• See example 5.5

Right Shift
Left Shift
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5.3 Z-Transform Solution of Linear Difference
Equations

• Given the generic system equation
• Which can be written in the advanced operator
  form as
• Can be written in the delayed operator form by
  subtracting n
• A solution to the system can be determined by
  using this form of the system and the Z-Transform
  pair

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5.3-1 Zero-State Response of LTID Systems The
Transfer Function

• Recall the generic system equation for a LTID
  system
• Which can be written in the delay operator form
  as
• Recall that for the zero-state solution, all ICs
  are zeroed. Thus
• Also, we are assuming causal input. Therefore
• Using this information the Z-transform pair of a
  delay operator becomes
• Thus we find the Z-transform of the zero-state
  system to be

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5.3-1 Zero-State Response of LTID Systems The
Transfer Function

• Solving for the output Yz
• Now define Hz, as the system Transfer Function
  to be
• then

Hz
Fz
YzHzFz
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5.3 Z-Transform Solution of Linear Difference
Equations

• Note that the denominator of Hz is Qz, and
  recall that this polynomial decides the
  asymptotic stability of the system. Therefore,
  an identical method for determining stability of
  the system is through examination of the location
  of the roots of the denominator of Hz.
• An LTID system is asymptotic stable IFF all the
  poles of Hz are inside the unit circle.
• An LTID system is asymptotic unstable IFF any of
  the following is true
• any pole of Hz is outside the unit circle
• there are repeated poles of Hz on the unit
  circle
• An LTID system is marginally asymptotic stable
  IFF all the poles of Hz are inside the unit
  circle and there are some unrepeated poles on the
  unit circle.
• The roots of the denominator are called the Poles
  of the system since it is at these values of z,
  Hz goes to infinity.

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5.3-2 The Relationship Between hk and Hz

• From the idea of the transfer function we have
• Recall that the impulse response, hk was the
  result of letting the input fk be ?k.
• Also recall that the Z-transform of an impulse is
  1. Therefore
• Using the definitions of the Z-transform and its
  inverse, we also have

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5.3-3 Frequency-Domain Interpretation of the
Transfer Function

• Let the input be an everlasting exponential zk
• Conclusion Exciting an LTID system with a
  discrete time exponential (fkzk) produces an
  output signal that also contains a discrete time
  exponential ykHzzk.

LTID
ykhk?fkhk?zk
fkzk
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5.3-4 Physical Interpretation of the Z-Transform
Im
ROC
Re
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5.4 System Realization
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5.4 System Realization
Let n 3
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5.4 System Realization
For n 3
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What is the response of a LTID to fk zk, an
everlasting DT exponential ?
If the input is a complex exponential, i.e.,
ejk?, then the output is also a complex
exponential, yk ejk?Hejk? where Hejk?
is a complex quantity, having both magnitude and
phase. Just as with CT systems, the transfer
function provides an indication how a DT system
will respond to various frequencies. Furthermore,
Fz, provides a frequency spectrum for the DT
signal fk.
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5.5 Frequency Response of a DT System
From the previous discussion
Expressing Hej? in polar form
The response to a sinusoidal input yk is also
sinusoidal. The magnitude and phase response of
the system at the applied input frequency
determine how the amplitude and phase of the
input signal is altered.
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5.5 Frequency Response of a DT System
The steady-state response response of a
asymptotically stable LTID system to a sinusoidal
input fk Acos(?k ?) is yk AHej
?k cos(?k ? ? Hej ?k )
These results are very similar to those obtained
for CT systems H(s) ? H(j?), where H(j ?) and
? H(j ?) determine the magnitude and phase
response. An important difference for DT
systems is that Hz ? Hej?k ? Hj ?k. Since
ej?k is periodic, ej?k ej(?k2?), Hej?k is
also periodic. The example 5.9 illustrates.
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The Periodic Nature of the Frequency Responseand
Aliasing
Therefore, is periodic (as seen in
previous example).
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Aliasing
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The DT transfer function HZ
The DT transfer function Hz Yz/Fz
provides an indication of the response of a
system at various values of complex frequency z.
For the ac steady-state response to a DT
sinusoid, z ej?, where ? may represent the
sampled CT frequency ? ?TS ? may take on any
value, but only the range 0 to ? produces a
unique set of values for H(ej ?). (Note ? 0
corresponds to dc.) Hz can be expressed as the
ratio of two polynomials, and each polynomial can
be factored as shown

• The roots Pz, z1, z2, ,zm, are the zeros of
  Hz (the values which cause Hz 0)
• The roots Qz, p1, p2, ,pm, are the poles of
  Hz (the values which cause Hz ?)

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The DT transfer function HZ

• Hz is a complex-valued function, having both
  magnitude and phase, and both the magnitude and
  phase will vary as z varies.
• The magnitude of Hz is found by multiplying the
  magnitudes of the first-order terms in the
  numerator and dividing by the magnitudes of the
  first-order terms in the denominator.
• The phase of Hz is found by adding the phase
  angles of the first-order terms in the numerator
  and subtracting the phase angles of the
  first-order terms in the denominator

The behavior of the magnitude and phase of Hz
is dictated by the behavior of the first-order
terms.
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The DT transfer function HZ

• Let us examine the magnitude and phase of the
  first-order term (z - a), where
• a is any complex constant
• z is a complex variable for ac steady-state, z
  ej?. As ? varies, the magnitude of z is constant
  (1), but the phase angle, ?, varies. ej? 1??.

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The DT transfer function HZ

• If a is a zero, (z-a) is in the numerator, Hz
  will dip as z approaches a (Hz will be zero
  if a is on the unit circle.)
• If a is a pole, (z-a) is in the denominator,
  Hz will peak as z approaches a
• The angle of (z-a) will approach ?a as z
  approaches a .

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z - a, a 0.5?1rad
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? (z - a), a 0.5?1rad
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DT HPF wT 0pi/100pi z exp(jwT) p1
-0.8 p2 0.9exp(j0.9pi) p3
0.9exp(-j0.9pi) H z.3./((z-p1).(z-p2).(z
-p3)) magH abs(H) angH angle(H) hold off
plot (wT,magH,'b') ylabel('Magnitude of
Hz') hold on plot (wT,angH,'r')
ylabel('Phase of z/(z-0.8)') xlabel('DT HPF,
poles _at_ -0.8, 0.9lt(/-0.9pi)') grid on
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