Transcript Z Transform Tutorial - Embry–Riddle Aeronautical University

Z Transform Primer
Basic Concepts
• Consider a sequence of values: {xk : k = 0,1,2,... }
• These may be samples of a function x(t), sampled at
instants t = kT; thus xk = x(kT).
• The Z transform is simply a polynomial in z having the xk
as coefficients:

X ( z )  Z xk    xk z k
k 0
Fundamental Functions
• Define the impulse function: {dk} = {1, 0, 0, 0,....}
( z)  Z d k   1
• Define the unit step function: {uk} = {1, 1, 1, 1,....}

U z   Z uk    z
k 0
k
1
z


1
1 z
z 1
(Convergent for |z| < 1)
Delay/Shift Property
• Let y(t) = x(t-T) (delayed by T and truncated at t = T)
yk = y(kT) = x(kT-T) = x((k-1)T) = xk-1 ; y0 = 0

Y ( z )  Z yk    yk z
k
k 1

  xk 1 z k
k 1
• Let j = k-1 ; k = j + 1

Y ( z)   x j z
j 0
 j 1
z
1

j
1
x
z

z
X ( z)
 j
j 0
• The values in the sequence, the coefficients of the polynomial,
slide one position to the right, shifting in a zero.
The Laplace Connection
• Consider the Laplace Transforms of x(t) and y(t):
Y s   Lyt   Lxt  T   eTs X s 
• Equate the transform domain delay operators:
z
1
e
Ts
z e
Ts
• Examine s-plane to z-plane mapping . . .
S-Plane to Z-Plane Mapping
z e
Ts
Anything in the Alias/Overlay region in the S-Plane will be overlaid on the Z-Plane
along with the contents of the strip between +/- jp/T. In order to avoid aliasing, there
must be nothing in this region, i.e. there must be no signals present with radian
frequencies higher than w  p/T, or cyclic frequencies higher than f = 1/2T. Stated
another way, the sampling frequency must be at least twice the highest frequency
present (Nyquist rate).
Mapping Poles and Zeros
A point in the Z-plane rejq will map to a point in the Splane according to:
lnr 
Res 
T
Ims 
q
T
Conjugate roots will generate a real valued
polynomial in s of the form:
s 2  2wn s  wn2
 lnr  
wn 
1 

T
 q 
q
2
 lnr 
 
w nT
Example 1: Running Average Algorithm
xk  xk 1  xk 2  xk 3
yk 
4
(Non-Recursive)
1  z 1  z 2  z 3
z3  z2  z  1
Y z   X z 
 X z 
4
4z 4
Block Diagram
Z Transform
Transfer Function
Y
z3  z2  z  1
z  
X
4z 4
Note: Each [Z-1] block can be thought of as a
memory cell, storing the previously applied value.
Example 2: Trapezoidal Integrator
T
yk  yk 1  xk  xk 1 
2

(Recursive)

T
Y z   z Y z   X z   z X z 
2
1
1
Z Transform
1  z 1  T
 z  1 T
Y z   X z 
 X z 
1 
 2
1

z
2
z

1



Block Diagram
Transfer Function
Y z  T

X z  2
 z  1
 z  1
Ex. 2 (cont)
Block Diagram Manipulation
Intuitive Structure
Equivalent Structure
Explicit representation of
xk-1 and yk-1 has been lost,
but memory element
usage has been reduced
from two to one.
Ex. 2 (cont) More Block Diagram
Manipulation
Y z  T

X z  2
 z  1
 z  1
Note that the final form is equivalent to a rectangular
integrator with an additive forward path. In a PI
compensator, this path can be absorbed by the
proportional term, so there is no advantage to be gained
by implementing a trapezoidal integrator.