SIGNALS AND SYSTEMS: PAPER 3C1
HANDOUT 10.
Dr David Corrigan1 .
Electronic and Electrical Engineering Dept.
corrigad@tcd.ie www.mee.tcd.ie/∼corrigad
The Z Transform, System Transfer
Function, Poles and Stability
• For differential equations and analogue system analysis, the Laplace
Transform is an invaluable tool. It allows us to examine stability
through the simple exercise of pole-zero plots and also allows us to
understand frequency response of systems through factorisation.
• A similar tool exists for digital signals, it is called the Z Transform.
It helps us avoid alot of tedious difference equation manipulation.
• You probably have had enough of transforms in this course, so
rather than heap upon you a large amount of proofs, we will just
be stating what the Z-Transform is and then we’ll use it to do
stuff.
• Where possible, we will be using similar analysis for Z- Transforms
as we did for Laplace transforms and so the proofs and so on will
be much more brief.
1
This handout is based on the set of notes produced by Prof. Anil Kokaram
3C1 Signals and Systems
1
www.sigmedia.tv
1 THE Z-XFORM
1
The Z-Xform
• The unilateral Z-Transform of a digital sequence xn is given by
Z(xn) = X(z) =
∞
∑
xnz −n
n=0
• z is just a complex number in what is called the z−plane. Just
like s is a complex number in the s− plane.
• The Z-Transform maps a discrete sequence xn from the sample
domain [n] into the complex plane z.
• It is a power series in z
Z(xn) = X(z) =
∞
∑
xnz −n
n=0
• Recall the unilateral Laplace Transform
∫ ∞
L (x(t)) = X(s) =
x(t)e−stdt
0
3C1 Signals and Systems
2
www.sigmedia.tv
1 THE Z-XFORM
Z(xn) = X(z) =
∞
∑
xnz −n
n=0
• Calculating the Z-transform of a finite length signal is straightforward.
• An example. Say xn = 1, 2, −1, 0.5, 0.25, 0, 0 and 0 otherwise,
furthermore xn is causal. The Z Transform of xn is then as follows.
Z(xn) = X(z) =
∞
∑
xnz −n
n=0
= x0z 0 + x1z −1 + x2z −2 + x3z −3 . . .
= 1 + 2z −1 − 1z −2 + 0.5z −3 + 0.25z −4
And that’s it. The Z-transform just takes an input sequence and
multiplies it by increasing negative powers of z to create a power
series in z.
3C1 Signals and Systems
3
www.sigmedia.tv
1 THE Z-XFORM
• An important case. xn = rn where |r| < 1
X(z) =
=
=
∞
∑
n=0
∞
∑
xnz −n
rnz −n
n=0
∞ (
∑
rz −1
)n
n=0
Sum to infinity of a Geometric Progression a, ar, ar2, ar3, . . .
a
. Comparing this to our formula for X(z), it is a GP with
is 1−r
a = 1 and common ratio rz −1. So
X(z) =
rn ↔
1
1 − rz −1
1
1 − rz −1
• What about the inverse? Well, the fancy way of doing the inverse
is to do contour integration of X(z) in the z-plane to get back xn.
But we’re going to use two simpler methods. One is just plain
cheating and the other is to use tables2.
2
Also a form of cheating
3C1 Signals and Systems
4
www.sigmedia.tv
1.1 Inverse Z-Xform 1
1.1
1 THE Z-XFORM
The Inverse Z-Transform: The plain cheat
Z(xn) = X(z) =
∞
∑
xnz −n
n=0
Suppose we are given X(z) = −1 + z −1 + 0.1z −2 + 0.8z −3 + 1.2z −4.
What is xn?
Well, the Z-Transform of xn is a power series in z right?
X(z) =
∞
∑
xnz −n
n=0
= x0z 0 + x1z −1 + x2z −2 + x3z −3 . . .
Compare this with what we are given
X(z) = −1 + z −1 + 0.1z −2 + 0.8z −3 + 1.2z −4
You can see that the coefficients of the power series are the values of
xn themselves! So we can extract the inverse from the power series to
yield
xn = −1, 1, 0.1, 0.8, 1.2
Magic!
3C1 Signals and Systems
5
www.sigmedia.tv
1.2 Inverse Z-Xform 2
1.2
1 THE Z-XFORM
The Inverse Z-Transform: The tables cheat
Suppose we are given X(z) as follows. What is xn?
3 − 56 z −1
X(z) =
(1 − 14 z −1)(1 − 31 z −1)
(1)
Use partial fractions
3 − 56 z −1
A
B
=
+
(1 − 14 z −1)(1 − 13 z −1) 1 − 14 z −1 1 − 13 z −1
Use cover up rule for A, B
Put z
−1
Put z −1
3 − 20
6
= 4 and cover up A =
1 − 34
= −1
3 − 15
6
= 3 and cover up B =
1 − 43
=2
Hence:
3 − 56 z −1
−1
2
+
=
(1 − 14 z −1)(1 − 13 z −1) 1 − 14 z −1 1 − 13 z −1
( )n
( )n
1
1
Z −1(X(z)) = xn = −1
+2
4
3
3C1 Signals and Systems
6
www.sigmedia.tv
2 SOME Z XFORMS
2
Some Z Xforms
• Given xn = δn what is Z(xn)?
X(z) =
=
∞
∑
n=0
∞
∑
xnz −n
δnz −n
n=0
But δn is a sequence that is unity only where n = 0, otherwise
its 0. Hence
= 1 × z −0 + 0 × z −1 + 0 × z −2 + . . .
=1
• Given xn = un what is Z(xn)? (The Z-xform of a step function)
X(z) =
=
∞
∑
n=0
∞
∑
xnz −n
unz −n
n=0
But un is a sequence that is unity for all n ≥ 0. Hence
=
∞
∑
z −n
n=0
Remember sum to ∞ of a GP is a/(1 − r), and here common
ratio is z −1
=
3C1 Signals and Systems
7
1
1 − z −1
www.sigmedia.tv
3 SOME Z-XFORM RELATIONS
3
Some Z-Xform Relations
• Convolution. Assuming xn and yn are causal what is the Z-Xform
of the convolution of xn and yn?
Let wn = xn ∗ yn
∞
∑
wn = xn ∗ yn =
xk yn−k
⇒ Z{wn} =
=
k=−∞
∞
∑
wnz −n
n=0
∞ (
∑
n=0
Let m = n − k
=
∞
∑
)
xk yn−k z −n
k=−∞
∞ ( ∑
∞
∑
m=−k
)
xk ym z −(m+k)
k=−∞
Collect together the summations of like terms . . .
(∑
)( ∑
)
∞
∞
=
xk z −k
ymz −m
k=−∞
∞
∑
m=−k
(
=
xk z −k
k = −∞
= X(z)Y(z)
)(
∞
∑
ymz −m
)
m = −k
This is THE SAME AS the relationship between TIME DOMAIN CONVOLUTION AND THE LAPLACE OR FOURIER
XFORMS.
3C1 Signals and Systems
8
www.sigmedia.tv
3 SOME Z-XFORM RELATIONS
• Time shift (very important in digital systems theory). If xn ↔
X(z) what is the Z-Xform of xn−1?
Let yn = xn−1
∞
∑
Z(yn) =
ynz −n
=
n=0
∞
∑
xn−1z −n
n=0
Substitute m = n − 1
=
=
∞
∑
m=−1
∞
∑
xmz −(m+1)
xmz −mz −1
m=−1
∞
∑
−1
=z
xmz −m + z −1x−1z
m=0
= z −1X(z) + x−1
But we are dealing with CAUSAL signals so ...
Z(yn) = z −1X(z)
So z −1 represents a shift in time of ONE SAMPLE. Hence Z{xn} =
z −1X(z); Z{xn−2} = z −2X(z); Z{xn−3} = z −3X(z); and so
on.
• Note that we are assuming that the signal is causal. See tables
for exact Z-Xforms of non-causal signals.
3C1 Signals and Systems
9
www.sigmedia.tv
4 SYSTEM XFER FUNCTIONS
4
System Xfer Functions
• Now we are in a position to make it easier to manipulate difference equations using the Z-Xform, in the same way that we can use the Laplace
Xform to help us solve differential equations.
• Lets look at our simple IIR difference equation example again
yn − 0.9yn−1 = xn
(2)
Let’s try to find an expression for the output Y(z) in terms of X(z). Take
Z-Xforms of both sides remembering
yn ↔ Y(z)
yn−1 ↔ z −1 Y(z)
xn ↔ X(z)
So taking Z-Xforms we have
Y(z) − 0.9Y(z)z −1 = X(z)
⇒ Y(z)[1 − 0.9z −1 ] = X(z)
1
1 − 0.9z −1
Y(z)
1
⇒
=
X(z) 1 − 0.9z −1
⇒ Y(z) = X(z)
Y(z)
The function X(z)
is the SYSTEM TRANSFER FUNCTION for the difference equation. And given xn we can use the above to work out the
output given ANY input using the Z-Xform.
• ALL THE SAME METHODOLOGY AS FOR ANALOGUE SYSTEM
TRANSFER FUNCTIONS ALSO APPLIES INCLUDING THE BLOCK
DIAGRAM ALGEBRA.
• Thus if two systems G1 (z) and G2 (z) are in cascade the NET transfer
function is G1 (z)G2 (z).
3C1 Signals and Systems
10
www.sigmedia.tv
4.1 BLOCK DIAGRAMS
4.1
4 SYSTEM XFER FUNCTIONS
BLOCK DIAGRAMS
yn + 0.2yn−1 − 0.48yn−2 = xn + 0.5xn−1
⇒ yn = 0.48yn−2 − 0.2yn−1 + xn + 0.5xn−1
Y(z) = Y(z)[0.48z −2 − 0.2z −1 ] + X(z)[1 + 0.5z −1 ]
x[n]
y[n]
T
X(z)
-1
0.5
0.48
-0.2
T
T
z-1 X(z)
Y(z)
0.5
z
-0.2
0.48
z-2 Y(z)
z-1
3C1 Signals and Systems
z-1 Y(z)
11
z-1
www.sigmedia.tv
4.2 Example 1
4.2
4 SYSTEM XFER FUNCTIONS
Example 1
Given xn = δn input into the system G(z) = 1/(1 − 0.9z −1), what
is the output hn? (This is another way of asking you to calculate the
impulse response of the system G(z)). Let hn be the output sequence
(the impulse response). BTW: G(z) is IIR.
H(z) = G(z)X(z)
X(z) = Z(δn) = 1
⇒ H(z) = G(z)
⇒ hn = Z −1(G(z))
1
= Z −1(
)
1 − 0.9z −1
From tables
hn = 0.9n
THE SYSTEM IMPULSE RESPONSE IS THE INVERSE Z XFORM
OF THE SYSTEM TRANSFER FUNCTION.
in exactly the same way that the time domain impulse response
of an analogue system is the inverse Laplace Xform of its system
transfer function.
3C1 Signals and Systems
12
www.sigmedia.tv
4.3 Example 2
4.3
4 SYSTEM XFER FUNCTIONS
Example 2
Given xn = un (the step function) input into the system G(z) =
1/(1 − 0.9z −1), what is the output yn? (This is another way of asking
you to calculate the step response of the system G(z)). BTW: G(z)
is IIR.
yn is the output sequence (the step response in this case).
Y(z) = G(z)X(z)
1
1 − z −1
1
⇒ Y(z) = G(z)
−1
(1 − z )
G(z)
⇒ yn = Z −1
1 − z −1
(
)
1
= Z −1
(1 − 0.9z −1)(1 − z −1)
1
−9
10
=
+
(1 − 0.9z −1)(1 − z −1) 1 − 0.9z −1 1 − z −1
yn = −9(.9)n + 10
X(z) = Z(un) =
THE SYSTEM STEP RESPONSE IS THE INVERSE Z XFORM
OF [ THE SYSTEM TRANSFER FUNCTION MULTIPLIED BY
1
].
1−z −1
in similar fashion to analogue systems.
3C1 Signals and Systems
13
www.sigmedia.tv
4.3 Example 2
4 SYSTEM XFER FUNCTIONS
Impulse response of 1\(1−0.9z^{−1})
1
0.8
0.6
0.4
0.2
0
0
10
20
Sample (n)
30
40
Step response of 1/(1−0.9z−1)
10
8
6
4
2
0
0
3C1 Signals and Systems
10
20
Sample (n)
14
30
40
www.sigmedia.tv
5 POLES AND ZEROS
5
Poles and Zeros
• Same deal as for analogue systems. Poles are the values of z that
make the denominator zero, and zeros are the values of z that
make the numerator go to zero. NOT z −1, just z !!!
1
1 + 0.2z −1 − 0.48z −2
z2
= 2
z + 0.2z − 0.48
z2
=
(z + 0.8)(z − 0.6)
G(z) =
(3)
1
0.8
0.6
Imaginary Part
0.4
0.2
2
0
−0.2
−0.4
−0.6
−0.8
−1
−1
−0.5
0
Real Part
0.5
1
You must always mark out the UNIT CIRCLE on the z-plane.
We’ll see why next . . .
3C1 Signals and Systems
15
www.sigmedia.tv
6 STABILITY
6
Stability
• Just like for analogue systems, a digital system is stable if its
impulse response is absolutely summable i.e.
∞
∑
|hn| = B < ∞
n=0
• A DIGITAL SYSTEM IS STABLE IF ALL ITS POLES LIE
WITHIN THE UNIT CIRCLE IN THE Z-PLANE. We will not
prove this in depth. Instead we will use the following example.
• Lets look at the system G(z) as below
1
1 − az −1
This system has a pole at z = a, and a zero at z = 0
G(z) =
• We know the impulse response of the system above is hn = an.
Hence
∞
∞
∑
∑
|hn| =
|an|
n=0
=
n=0
∞
∑
|a|n
n=0
• For this sequence to be stable it must have a finite sum. As
∑∞
n
n=0 |a| repesents the sum of a geometric progession, the sum
is finite if and only if |a| < 1
• That means that the POLE must be within the unit circle!
• In general, to test stability for digital systems, just check that
all the poles are within the unit circle. If any are not, then the
3C1 Signals and Systems
16
www.sigmedia.tv
6 STABILITY
system is unstable. We have not proved this exactly, but you
can get the picture. The proof is almost identical to what we did
using the Analogue System Transfer Functions.
• The location of zeros does not affect stability.
• THEREFORE FIR FILTERS ARE ALWAYS STABLE! They do not have poles!
• IIR Filters always have poles.
• If poles lie ON the unit circle then the system may be marginally
stable. We will not deal with that in this course. All you will
need to know is that z-plane poles must lie WITHIN the unit
circle for a digital system to be stable.
3C1 Signals and Systems
17
www.sigmedia.tv
6 STABILITY
A relationship between the s and z planes
Im
Im
Stable Systems
have poles
over here
The Z-Transform maps
the LHS of the S-Plane
into the interior of the
Unit Circle
Re
1
1
-1
THE UNIT CIRCLE
THE S-PLANE
Re
Stable Systems
have poles in here
-1
THE Z-PLANE
Z-LAND
LAPLACE-LAND
DIGITAL ANALYSIS
ANALOGUE ANALYSIS
Why does it help?
It turns out that if you are designing digital systems that will be
processing signals generated by sampling, and then the output will be
reconstructed to give an analogue signal, this connection helps.
An example of this would be an audio processing system where an
analogue signal from a microphone is digitized for processing before
being converted back into an analogue signal and sent to a loudspeaker.
But we will not be dealing with that in this course. That’s got
more to do with design. In this course we just want you to get a
handle on the tools, and the rough lay of the land.
3C1 Signals and Systems
18
www.sigmedia.tv