z-transform derived from Laplace transform
Lecture 15
Consider a discrete-time signal x(t) below sampled every T sec.
x(t ) = x0δ (t ) + x1δ (t − T ) + x2δ (t − 2T ) + x3δ (t − 3T ) + .....
Discrete-Time System Analysis
using z-Transform
δ (t)
⇔1
The Laplace transform of x(t) is therefore (Time-shift prop. L6S13):
X ( s) = x0 + x1e − sT + x2e − s 2T + x3e − s 3T + .....
(Lathi 5.1)
z = esT = e(σ + jω )T = eσ T cos ωT + jeσ T sin ωT
Now define
X [ z ] = x0 + x1 z −1 + x2 z −2 + x3 z −3 + .....
Peter Cheung
Department of Electrical & Electronic Engineering
Imperial College London
URL: www.ee.imperial.ac.uk/pcheung/teaching/ee2_signals
E-mail: p.cheung@imperial.ac.uk
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
L5.8 p560
Lecture 15 Slide 1
PYKC 3-Mar-11
z-1 the sample period delay operator
From Laplace time-shift property, we know that z = e sT is time advance
by T second (T is the sampling period).
−1
− sT
Therefore z = e
corresponds to UNIT SAMPLE PERIOD DELAY.
As a result, all sampled data (and discrete-time system) can be
expressed in terms of the variable z.
More formally, the unilateral z-transform of a causal sampled
sequence:
x[n] = x[0] + x[1] + x[2] + x[3] +
is given by:
X ( s) = ∫ x(t )e − st dt
Fourier
transform
X (ω ) = ∫ x(t )e − jωt dt
−∞
∞
−∞
ω0 = 2π / T
−n
n =−∞
E2.5 Signals & Linear Systems
Continuous-time system
Converts integraldifferential equations to & signal analysis; stable
or unstable
algebraic equations
Laplace
transform
z
transform
The bilateral z-transform for a general sampled sequence is:
PYKC 3-Mar-11
∞
Suitable for ..
Converts finite time
signal to frequency
domain
Continuous-time; stable
system, convergent
signals only; best for
steady-state
∑
n =0
∑ x[n]z
Purpose
N0 −1
= ∑ x[n]z − n
X [ z] =
Definition
Discrete X [nω ] =
x[n]e jnω0T Converts finite discrete- Discrete time, otherwise
0
time signal to discrete
Fourier
n =0
same as FT
transform N 0 samples,T = sample period frequency domain
∞
∞
Lecture 15 Slide 2
E2.5 Signals & Linear Systems
Laplace, Fourier and z-Tranforms
X [ z ] = x0 + x1 z −1 + x2 z −2 + x3 z −3 + .....
(L6S5)
δ (t − T) ⇔ e −sT (L6S13)
∞
X [ z] =
∑ x[n]z
n =−∞
−n
Converts difference
equations into
algebraic equations
Discrete-time system &
signal analysis; stable
or unstable
L5.8 p560
Lecture 15 Slide 3
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
Lecture 15 Slide 4
Example of z-transform (1)
Example of z-transform (2)
Find the z-transform for the signal γnu[n], where γ is a constant.
By definition
Since u[n] = 1 for all n ≥ 0 (step function),
Apply the geometric progression formula:
Therefore:
Observe that a simple equation in z-domain results in an infinite
sequence of samples.
Observe also that
exists only for
.
For
X[z] may go to infinity. We call the region of z-plane
where X[z] exists as Region-of-Convergence (ROC), and is shown below.
z-plane
L5.1 p496
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
Lecture 15 Slide 5
L5.1 p496
PYKC 3-Mar-11
z-transforms of δ[n] and u[n]
Also, for
Therefore
Since
From slide 5, we know
Hence
Therefore
Since
L5.1 p499
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
Lecture 15 Slide 6
z-transforms of cosβn u[n]
Remember that by definition:
E2.5 Signals & Linear Systems
Lecture 15 Slide 7
L5.1 p500
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
Lecture 15 Slide 8
z-transforms of 5 impulses
Find the z-tranform of:
By definition,
Now remember the equation for sum of a power series:
n
∑r
k =0
Let
k
=
z-transform Table (1)
r n+1 − 1
r −1
−1
r = z and n = 4
z −5 − 1
X [ z ] = −1
z −1
z
=
(1 − z −5 )
z −1
L5.1 p500
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
Lecture 15 Slide 9
L5.1 p498
PYKC 3-Mar-11
z-transform Table (2)
E2.5 Signals & Linear Systems
Inverse z-transform
As with other transforms, inverse z-transform is used to derive x[n] from
X[z], and is formally defined as:
Here the symbol
indicates an integration in counterclockwise direction
around a closed path in the complex z-plane (known as contour integral).
Such contour integral is difficult to evaluate (but could be done using
Cauchy’s residue theorem), therefore we often use other techniques to
obtain the inverse z-tranform.
One such technique is to use the z-transform pair table shown in the last
two slides with partial fraction.
L5.1 p498
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
Lecture 15 Slide 10
Lecture 15 Slide 11
L5.1 p494
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
Lecture 15 Slide 12
Find inverse z-transform – real unique poles
Find the inverse z-transform of:
Step 1: Divide both sides by z:
Step 2: Perform partial fraction:
Step 3: Multiply both sides by z:
Find inverse z-transform – repeat real poles (1)
Find the inverse z-transform of:
Divide both sides by z and expand:
Use covering method to find k and a0:
We get:
To find a2, multiply both sides by z and let z→∞:
Step 4: Obtain inverse z-transform of each term from table (#1 & #6):
L5.1-1 p501
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
Lecture 15 Slide 13
L5.1-1 p502
PYKC 3-Mar-11
Find inverse z-transform – repeat real poles (2)
E2.5 Signals & Linear Systems
Find inverse z-transform – complex poles (1)
To find a1, let z = 0:
Therefore, we find:
Use pairs #6 & #10
Find inverse z-tranform of:
Whenever we encounter complex pole, we need to use a special partial
fraction method (called quadratic factors):
Now multiply both sides by z, and let z→∞:
We get:
L5.1-1 p502
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
Lecture 15 Slide 15
Lecture 15 Slide 14
L5.1-1 p503
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
Lecture 15 Slide 16
Find inverse z-transform – complex poles (2)
Find inverse z-transform – long division
To find B, we let z=0:
Consider this example:
Now, we have X[z] in a convenient form:
Perform long division:
Use table pair #12c, we identify A = -2, B = 16,
Thus:
Therefore
and a = -3.
Therefore:
L5.1-1 p504
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
Lecture 15 Slide 17
L5.1-1 p505
PYKC 3-Mar-11
E2.5 Signals & Linear Systems
Lecture 15 Slide 18