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					EE406-041 Discrete-Time Signal Processing                                            Applications of the z-Transform


                   Applications of the z-Transform
One of the major applications of the z-transform is used as an analysis tool for discrete-time
LTI systems. In particular, we will use the z-transform for finding the frequency response
and evaluating the stability of discrete-time LTI systems.

From the convolution property of z-transform, we have the relationship between the z-
transforms of input and output sequences of a discrete-time LTI system as

                                         Y (z ) = H (z )X (z )

where X ( z ) , Y ( z ) and H ( z ) are the z-transforms of the system input, output and impulse
response, respectively. H ( z ) is referred as the system function or transfer function of the
system.



System Function From the Difference Equation Representation
Consider a LTI system for which the input and output satisfy a linear constant-coefficient
difference equation of the form

                                  N                        M

                                 ∑a
                                 k =0
                                         k   y[n − k ] = ∑ bk x[n − k ]
                                                          k =0


By applying the z-transform to both sides and using the linearity property and the time-
shifting property, we obtain

                                 N                        M

                                ∑ a k z − k Y ( z ) = ∑ bk z − k X ( z )
                                k =0                      k =0
                                          N                             M
                                Y ( z ) ∑ a k z − k = X ( z ) ∑ bk z − k
                                         k =0                           k =0
so that
                                                                 M

                                                  Y ( z)         ∑b z     k
                                                                               −k


                                       H ( z) =          =       k =0
                                                                  N

                                                                 ∑a
                                                  X ( z)
                                                                          k   z −k
                                                                 k =0


The system function for a system satisfying a linear constant-coefficient difference equation
is always rational.

If we factor the numerator and denominator of the system function H(z), we can rewrite it as:

                                                  ( z − z1 )( z − z2 ) ⋅⋅⋅
                                      H ( z) =
                                                  ( z − p1 )( z − p2 ) ⋅⋅⋅

where the zi are the zeros and the pi are the poles.

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EE406-041 Discrete-Time Signal Processing                            Applications of the z-Transform

Because we are considering only real coefficients, if any of the roots of the numerator or
denominator are complex then they always occur as complex conjugate pairs (e.g. the roots of
( z 2 + 1) are ± j ). Thus complex poles and zeros will always occur in complex conjugate
pairs.

The poles and zeros completely specify the system function and are an alternative
representation of it. The pole and zero representation is very useful for checking stability and
a way to visualize the frequency response of the system.


Stability and Causality
                                    (              )
For z evaluated on the unit circle i.e., z = e jω , H ( z ) reduces to the frequency response of
the system provided that the unit circle is in the ROC. For LTI systems, the BIBO stability is
equivalent to the absolute summability of its impulse response with

                                            ∞

                                          ∑ h[n] < ∞ .
                                          n = −∞



Thus the DTFT of h[n] exists and consequently the ROC of H ( z ) must include the unit
circle. This result is called the z-domain stability theorem:

• A LTI system is stable if and only if the unit circle is in the ROC of H ( z ) .

For LTI causality, we require that samples h[n]=0 for n < 0 (i.e. a right-sided sequence). This
implies that the ROC of H ( z ) must be outside of a circle. This is not a sufficient condition
since any right-sided sequence has a similar ROC. However, when the system is stable, its
causality is easy to check. We have the z-domain causal LTI stability theorem:

• A causal LTI system is stable if and only if the system function H ( z ) has all its poles
   inside the unit circle.

Thus we can verify that a causal LTI system is stable by finding the poles of its system
function and checking that all of them have magnitudes less than 1.


Example 1

Consider a causal linear time-invariant system with system function

                                                1 − a −1z −1
                                        H (z) =
                                                 1 − az −1
where a is real.

(a) Write the difference equation that relates the input and output of this system.

(b) For what range of values of a is this system stable?


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EE406-041 Discrete-Time Signal Processing                                       Applications of the z-Transform

(c) For a = 0.5, plot the pole-zero diagram and shade the region of convergence in z-plane.

(d) Find the impulse response h(n) for this system.

(e) Show that this system is an allpass system, i.e. the magnitude of the frequency response is
    a constant. Also, specify the value of the constant.

[Solution]

          1 − a −1z −1 Y ( z )
H (z) =               =        ,
           1 − az −1    X (z)

As the system is causal, the ROC is |z| > |a|

(a) Y(z) – a z-1Y(z) = X(z) – a-1z-1X(z)

      Taking the inverse transform, we have

          y[n] − ay[n − 1] = x[n] − a −1 x[n − 1]


(b) For stability, the ROC must include the unit circle. So, the system is stable for |a| < 1.

(c) a = 0.5
                                                          1 − 2 z −1
                                                H ( z) =
                                                         1 − 0.5 z −1


                                                          Ζ




                                                                            Ζ
                                                          1/2           2




                         1      a −1 z −1
(d)       H ( z) =            −           ,     z > a
                     1 − az −1 1 − az −1

                              1
          h[n] = a n u[n] −     (a ) n−1 u[n − 1]
                              a



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EE406-041 Discrete-Time Signal Processing                                                           Applications of the z-Transform

                                         1 − a −1e − jω
(e)        H (ω ) = H ( z )|z = e jω =
                                          1 − ae − jω

                   2                1 − a −1e − jω 1 − a −1e jω
           H (ω ) = H (ω ) H (ω ) = *
                                                   ⋅
                                     1 − ae − jω 1 − ae jω

                                                          1/ 2                               1/ 2
                     1 + a −2 − 2a −1 cos ω                     1  a 2 + 1 − 2a cos ω               1
           H (ω ) = 
                     1 + a 2 − 2a cos ω                       =                                =
                                                                a  1 + a 2 − 2a cos ω 
                                                                                                      a

           As the magnitude response of the system is equal to a constant (1/a), the system is an all-
           pass system.


Example 2

A causal LTI system has impulse response h[n], for which the z-transform is

                                                                      1 + z −1
                                              H ( z) =
                                                            (1 − 0.5z −1 )(1 + 0.25z −1 )

(a)     What is the region of convergence of H(z)?

(b)    Is the system stable? Explain.

(c)     Find the z-transform X(z) of an input x[n] that will produce the output

                                  y[n] = (-1/3) (-0.25)n u[n] - (4/3) (2)n u[-n –1]

(d) Find the impulse response h[n] of the system.


[Solution]

                    1 + z −1             2           1
H ( z) =                        −1 =        −1 −
                    −1
           (1 − 0.5z )(1 + 0.25z ) (1 − 0.5z ) (1 + 0.25z −1 )

(a) The impulse response h[n] is causal => The ROC is |z| > 0.5

(b) ROC includes the unit circle |z|=1 => The system is stable.

            1 1           4
(c) y[n] = − (− ) n u[n] − (2) n u[− n − 1]
            3 4           3

                   − 1/ 3         4/3                1 + z −1
      Y ( z) =                 +         =                                ,                 1/ 4 < z < 2
               1 + (1 / 4) z −1 1 − 2z −1 (1 + (1 / 4) z −1 )(1 − 2z −1 )




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EE406-041 Discrete-Time Signal Processing                                  Applications of the z-Transform

Since
             Y ( z ) (1 − 0.5z −1 )
     X (z) =        =               ,      z <2
             H ( z ) (1 − 2z −1 )

    ∴ x[n] = −(2) n u[−n − 1] + 0.5(2) n −1 u[− n]

                      2            1
(d) H ( z ) =                −
                          −1
                (1 − 0.5 z ) (1 + 0.25 z −1 )

    ∴ h[n] = 2(0.5) n u[n] − (−0.25) n u[n]


Summary
Three different frequency domain representations of an aperiodic discrete-time sequence have
been introduced and their properties reviewed. Two of these representations, the discrete-time
Fourier transform (DTFT) and the z-transform, are applicable to any arbitrary sequences,
whereas the third one, the discrete Fourier transform (DFT), can be applied only to finite-length
sequences.


Relationships between Different Transforms



          write H ( z ) in z −1
                                                                            inverse
                                                           H (z )
                                                                            z − transform
          cross multiply and inverse
                                        take z − transform
                                                                    z − transform
                                        solve for Y X

                          Diff. Eqn.                                                  h(n )


            take z − transform
                                                      ( )
                                                  z = e jω
                                                                     Inverse
                                                                     DTFT
            solve for Y X                                                           Fourier
                                                             ( )
                                                        H e jω                      transform

                                     if system is stable




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