chapter 6 - the z-transform - university of california,...

Chapter 6 - The Z-Transform

The Z-transform is the Discrete-Time counterpart of the Laplace Trans-form.

Laplace : G(s) =

∫ ∞−∞

g(t)e−stdt

Z : G(z) =∞∑

n=−∞

g[n]z−n

It is

• Used in Digital Signal Processing

• Used to Define Frequency Response of Discrete-Time System.

• Used to Solve Difference Equations – use algebraic methods as we didfor differential equations with Laplace Transforms; it is easier to solvethe transformed equations since they are algebraic.

We will see that

1. Lines on the s-plane map to circles on the z-plane.

2. Role of jω-axis is replaced by unit circle, so

(a) The DT Fourier Transform exists for a signal if the ROC includesthe unit circle.

(b) A stable system must have an ROC that contains the unit circle.

(c) A causal and stable system must have poles inside the unit circle.

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Aside: You can relate the Z transform and Laplace transform directlywhen you are dealing with sampled signals. First recall the definition ofLaplace transform:

Take a CT signal g(t) and sample it:

gs(t) = g(t)∞∑

n=−∞

δ(t− nT ) =∞∑

n=−∞

g(nT )δ(t− nT )

The Laplace transform of the sampled signal is

L[gs(t)] =

∫ ∞−∞

[∞∑

n=−∞

g(nT )δ(t− nT )

]e−stdt

=∞∑

n=−∞

∫ ∞−∞

g(nT )δ(t− nT )e−stdt

=∞∑

n=−∞

g(nT )e−snT .

Let g[n] = g(nT ) be the discrete representation and z = esT , then

G(z) =∞∑

n=−∞

g[n]z−n

G(z)|z=esT =∞∑

n=−∞

g[n]e−sTn

=∞∑

n=−∞

g(nT )e−snT

= L[gs(t)]

Thus, the Z transform with z = esT is the same as the Laplace transformof a sampled signal! Of course, if the signal is already discrete, the notion ofsampling is unnecessary for understanding and using the Z transform.

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Definitions of Z-Transforms

G(z) =∞∑

n=−∞

g[n]z−n

is the bilateral (2-sided) Z-transform. Its inverse Z-transform is defined as:

Z−1[H(z)] = h[n] =1

2πj

∮H(z)zn−1dz

which is a counterclockwise contour integral along a closed path in the z-plane. We will see how to take inverse Z-transforms using tables and partialfraction expansion.

Some textbooks work with the unilateral Z-transform:

Hu(z) =∞∑

n=0

h[n]z−n.

IMPORTANT: We do not use this at all, but make sure you know thedifference in case you come across it later. In this course, our focus is on theBilateral Z-Transform, to which we simply refer as Z-transorm:

G(z) =∞∑

n=−∞

g[n]z−n,

defined for 2-sided, anticausal, and causal signals (i.e. all signals).Note that, whereas for Laplace Transform we considered where the inte-

gral converges, here we consider where the sum converges.

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One must consider the Region of Convergence (ROC) of the Z-transform,because left-sided and right-sided time functions will have the same Z-transformand only the ROC will distinguish between the two possible time functions.

Remember:∞∑i=0

ai =1

1− a, |a| < 1

You’ll use this a lot!

Ex. Find the Z transforms of

x1[n] = anu[n] and x2[n] = −(an)u[−n− 1]

and plot the ROCs and pole/zero diagrams.

We see that we must specify the ROC for the bilateral Z- transform tobe unique.

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Definitions and Regions of Convergence

• x[n] is right-sided if x[n] = 0, n < n0

• x[n] is left-sided if x[n] = 0, n > n0

We can write

X(z) = · · ·+ x[−2]z2 + x[−1]z + x[0] + x[1]z−1 + x[2]z−2 + · · ·

1. We’ve seen that right-sided signals have an ROC of the form |z| >rmax, i.e., it converges outside the largest magnitude pole.

(Infinite Egg White)

Examine for right-sided x[n]

X(z) =∞∑

n=n0

x[n]z−n

X(z) =∞∑

n=n0

x[n]

(1

z

)n

As n→∞, need (1/z)n → 0 for sum to converge.

This will happen for values of z outside rather than inside the pole, i.e.|z| > rmax.

What about z =∞ ?

If x[n] is not causal but is still right-sided, e.g. x[n] = u[n+ 1], then

X(z) =∞∑

n=−1

z−n = z +∞∑

n=0

z−n

Will not converge at z =∞, and we won’t include it in the ROC.

Thus we can tell if a system is causal from the ROC of the Z-transformof its impulse response.

|z| > rmax ⇒ CAUSAL

∞ > |z| > rmax ⇒ right-sided but not causal

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2. Left-sided signals have ROC of form |z| < rmin, i.e., it convergesINSIDE circle |z| < rmin (EGG YOLK).

Examine for left-sided x[n]

X(z) =

n0∑n=−∞

x[n]z−n

As n→ −∞, need (1/z)n → 0 or z∞ → 0

This happens for values of z inside rather than outside the poles.

What about z = 0 ?

If x[n] is left-sided but not strictly anticausal

(x[n] = 0 for n > n0 > 0 but x[n0] 6= 0)

e.g. x[n] = u[−n+ 1], then

X(z) =1∑

n=−∞

z−n = z−1 +∞∑

n=0

zn

does not converge at z = 0 so don’t include z = 0 in the ROC.

3. 2-sided signals have ROC of the form

r1 < |z| < r2 (BAGEL OR DONUT)

4. Finite Duration x[n] has ROC of entire z-plane except possibly z = 0or z =∞

δ[n− 1]↔ z−1, |z| > 0

δ[n+ 1]↔ z, |z| <∞

FACT: An ROC must contain the unit circle for stability – this holds forcausal, anticausal, and two-sided signals.

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Ex. Find the Z-Transform of x[n] = a|n| for |a| < 1.

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Ex. Find the Z-Transform of

x[n] = 3nu[−n− 1] + 4nu[−n− 1].

Ex. Find the Z-transform of 12δ[n− 1] + 3δ[n+ 1].

What is its ROC?

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Ex. Find the Z-transform of

x[n] = (.5)nu[n− 1] + 3nu[−n− 1].

If h is the impulse response of a system, would the system be BIBO stable?

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Ex. Find the Z-transform of x[n] = rn sin(bn)u[n] using Euler’s rule.

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Insights from the Pole-Zero Plot and ROC

Things that you can tell about a signal from its pole-zero plot (and ROC):

• When the ROC includes the unit circle, then the signal is absolutelysummable. (If the signal is an impulse response⇒ the system is stable.)

• A pole on the positive real axis corresponds to a simple decaying orgrowing function (of form an for a pole at z = a).

• Poles off the positive real axis correspond to an oscillating signal wherethe frequency of oscillation is the the angle from the positive real axis.(Poles on the negative real axis have an angle of π, so the frequency ofoscillation is π, as in (−1)n.) When the poles are . . .

– on the unit circle⇒ sinusoidal functions with constant amplitude

– not on the unit circle ⇒ sinusoidal functions with a decaying (orgrowing) envelope (rate of decay/growth depends on the distancefrom the pole to the origin).

• Poles and zeroes must come in complex conjugate pairs for the signal tobe real (consequence of the Z-transform property: x[n]∗ ←→ X∗(z∗)).

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Z-Transform Properties

The properties of Z-transform are inherited from properties of Laplace trans-form. The book has a longer list; here, we discuss the most important ones:

• Linearity:ax[n] + by[n]←→ aX(z) + bY (z)

where the new ROC R′ ⊃ Rx ∩Ry.

• Time shift:x[n− n0]←→ z−n0X(z)

where the new ROC is the same as Rx with the possible addition ordeletion of the origin or infinity.

• Convolution:y[n] = x[n] ∗ h[n]←→ X(z)H(z)

where the new ROC Ry ⊃ Rx ∩Rh.

Linearity and the time shift property will be useful for LCCDE systems,and the convolution property lets us avoid discrete-time convolutions. We’lluse these properties a lot.

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Convolution in Time

y[n] = x[n] ∗ h[n]↔∞∑

n=−∞

[∞∑

k=−∞

x[k]h[n− k]]z−n

because we have a Z-transform. Switching the order of the summations (OKexcept for pathological cases), we get:

=∞∑

k=−∞

x[k]∞∑

n=−∞

h[n− k]z−n

Now, let m = (n− k) and we get:

=∞∑

k=−∞

x[k][∞∑

m=−∞

h[m]z−(m+k)]

=∞∑

k=−∞

x[k]z−k

∞∑m=−∞

h[m]z−m = X(z)H(z)

The new ROC will depend on both the poles in X(z) and H(z), givingRx ∩Rh since the ROC cannot include poles. However, if one transform hasa zero that cancels a pole of the other then the ROC can be bigger, henceR′y ⊃ Rx ∩Rh.

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LTI System Applications

Transfer Functions The Z-transform properties are particularly usefulwhen you have an LTI system described by an LCCDE.

N∑k=0

aky[n− k] =M∑

k=0

bkx[n− k]

N∑k=0

akz−kY (z) =

M∑k=0

bkz−kX(z)

H(z) =Y (z)

X(z)=

∑Mk=0 bkz

−k∑Nk=0 akz−k

An alternative way to write this is:

H(z) =Y (z)

X(z)=p0

d0

∏Mk=0(z − ζk)∏Nk=0(z − λk)

We can use this to determine outputs of LTI systems by multiplying theZ-transform with the input with H(z) to get the Z-transform of the output.Then we can recover the time domain output using the Inverse Z-transform.

Note that finding inverse needs paying special attention to ROC.

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Ex. Given a difference equation,

y[n]− .3y[n− 1] = x[n]

find the Z-transform of the equation and then find the response Y (z) of thesystem to an input x[n] = (.6)nu[n].

What if you wanted to find the response in the time domain?⇒ We can use Partial Fraction Expansion to invert the Z-transform.

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As we saw for Laplace Transforms,

Y (z) =N(z)

D(z)=

N∑k=1

rkz

z − pk

pk = pole rk = residue

where

rk = [Y (z)

z(z − pk)]|z=pk

Then use tables to invert the Z-transform, e.g.

anu[n]↔ z

z − a

Back to our previous example . . .

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Ex. Find Inverse Z-Transform of

X(z) =2z2 − 5z

(z − 2)(z − 3), |z| > 3

Expand:X(z)

z=

2z − 5

(z − 2)(z − 3)

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Ex. Given h[n] = anu[n] (|a| < 1) and x[n] = u[n], find y[n] = x[n]∗h[n].

What if x[n] = u[n− 2]?

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Ex. Find the output y[n] to an input x[n] = u[n] and an LTI systemwith impulse response

h[n] = −3nu[−n− 1].

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Another method to invert Z-transforms is the Power Series Expansion.Using

δ[n− k]←→ z−k

X(z) =∑∞

k=0 x[k]z−k = x[0] + x[1]z−1 + x[2]z−2 + · · ·x[n] =

∑∞k=0 x[k]δ[n− k] = x[0]δ[n] + x[1]δ[n− 1] + x[2]δ[n− 2] + · · ·

So if you can expand X(z) like this as a series in z−1, you can pick offx[n] as the coefficients of the series.

Ex. Find the Inverse Z-Transform of

H(z) = 1 + 2z−1 + 3z−2

Note: If H(z) has only finitely many zm, m ∈ Z terms, we can concludethat h[n] has only a finite set of non-zero elements. Such a scenario, then,corresponds to a FIR.

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Ex. Find the inverse Z-transform of

X(z) =8z − 19

z2 − 5z + 6

|z| > 3

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Ex. Find the inverse Z-transform of

H(z) =2z2 − 5

2z

z2 − 52z + 1

,

1

2< |z| < 2.

Would a system having this Z-transform be BIBO stable?

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Ex. Find the inverse Z-transform of

W (z) =z−4

z2 − 2z − 3, |z| > 3

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Frequency Response

If we evaluate H(z) at z = ejω, i.e., on the unit circle (as long as unit circleis in the ROC), then we get the Fourier transform H(ejω) (or as we will seethe DTFT H(Ω)) which we call the frequency response. H(ejω) is periodicwith period 2π. You will discuss the DTFT Next.

Ex. Find the magnitude of H(ejω)

|H(ejω)| =√H(ejω)H∗(ejω) =

√H(ejω)H(e−jω)

Find the phase of H(ejω)

∠H(ejω) =1

2jln

[H(z)

H(z−1)

]z=ejω

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Now we can calculate the frequency response for a stable rathional trans-fer function with real-coefficents (page 336 of the text):

|H(ejω)2 =

∣∣∣∣p0

d0

∣∣∣∣2 ∏Mk=1 |(ejω − ζk)(ejω − ζ∗k)|∏Nk=1 |(ejω − λk)(ejω − λ∗k)|

and

∠H(ejω) = ∠p0

d0

+ ω(N −M) +M∑

k=1

∠(ejω − ζk) +M∑

k=1

∠(ejω − λk)

Geometric Interpretation:

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StabilityAs we saw earlier, for BIBO stability of a causal LTI system, all roots of

the system characteristic equation lie within the unit circle in the z-plane.

This is equivalent to stating that all poles of the transfer function H(z)must lie within the unit circle on the z-plane. We point out that H(z) doesnot converge at its poles.

Because causal systems have Regions of Convergence that lie outside thelargest magnitude pole, an equivalent condition for BIBO stability is thatthe ROC must contain the unit circle.

Ex. Find the Z-Transform of the unit step u[n]. Would an LTI systemwith u[n] as its system function be BIBO stable?

Ex. Find the Z-transform of x[n] = (.9)nu[n]. Would an LTI system withx[n] as its system function be BIBO stable?

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Invertibility

h[n] ∗ hi[n] = δ[n]⇒ H(z)Hi(z) = 1

Ex. Find the inverse system hi[n] of h[n] = anu[n].Check your results by taking the convolution of h[n] with hi[n].

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Ex. Find the inverse system of h[n] where

H(z) =z − az − b

.

For BIBO stability of both systems (assuming they are both causal),where must all poles and zeros of H(z) lie?

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