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Complex Variables 5.8 The z Transformation
Taken samples of a function f (t ) at intervals of T : f (nT )
DEFINITION (z Transfo rm) The z transform of the function f (z ), that is, Z[f (t )], is given
by
Where T >0. So, we define a function: F (z ) = Z[f (t )].
The convergent in a dom ain ser ies Z[f (t )] is a Laurent series with no positive
exponent in it.
This is not a unique transform ! Z[sin( t )] = Z[sin2( t )] = 0 (T =1)
Substitute: ,which gives a Maclaurin series , that converges
inside a disk , so F (z ) is an analyt ic func t ion in and
0;
n nn
n nT f c z c
z w 1
0;
n nn
n nT f cwc
0 r w r z 0)0(lim c f z F z
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Complex Variables
EXAMPLE 2 Find the z transform of f (t ) = tu (t ) .
Solut ion. First f (nT ) = nT . Thus
Recall:
Now for
and
Linear i ty of the Transform ation
For any constant c :
And (for the same T )
More:
For instance:
11 z z w
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EXAMPLE 3 If find
Solut ion . F (z ) is two term series, so: and (from the definition):
EXAMPLE 4 If , find
Solut ion . F (z ) is a Laurent series ( ):
Now:
and:
An in verse transform is pos sible OLNLY for fun ct ion s presented by ser ies .
Transfo rm fo r translations .
Let So:
If
Thus: 4
Complex Variables
1 z
211)( z z z F
1
21
1
1)(
z z
z z F
3,0;1)2(;0)0( 210 ncT f T f cc f c n
11)( z z z F
...
111
2
11
2
1
22 z z z z
z
z
1;2)(;1)0(0 nnT f c f c n
k nkT nT f t t f ,0)(0,0)(
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Complex Variables
Or:
First Translat ion Formu la
For
Consider k =1:
Or:
That is
Consider k =2:
That is
Secon d Translat ion Formula
for k 0 .
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EXAMPLE 5 Use Z[e at u (t )] = F (z ) = z /(z - e aT ) for |z | > |e aT | find G (z )=Z[e a (t-T )u (t - T )] and
H (z )= Z[e a (t+T )u (t + T )].
Solut ion. For the “t - T ” translation and k = 1:
For the “t + T ” translation and k = 1 (f (0) = 1):
z Transform s of Products o f Funct ions
Let and
where . Thus:
For two analytic functions F (z ) and G (z ) in the same domain |z | > R we write Laurent
series:
Complex Variables
)(and)( nT g d nT f c nn
0 0
0
;)(
;)(
n n
nn
n
n
n
m
m
m
w R z z wd w z d w z G
Rwwcw F
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Complex Variables
Product is uniformly convergent( ):
For circle of the radius > R we consider for
the following Laurent expansion (uniformly convergent inside)
So, after integration term by term around
But, all integrals in the sum are zero, except one: n – m = 0,
which is 2i z -1
Thus:
Or
with F and G analytic in
w R z Rw ;
R z w ;
w
Rw
)(w F
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Complex Variables
EXAMPLE 6 Find Z[te at u (t )]
Solution. Let f (t ) = tu (t ) and g (t ) = e at
We know: Z[f (t )]=Tz /(1- z )2=F (z ) and Z[g (t )] = z /(z - e aT )=G (z ) , and they are analytic except
z =1 and z =e aT . Thus
We need > R > 1 and R >|e aT |.
Observe that a singular point w = 1 is in the range of integration, but w = ze -aT is not
because: |we aT | < |w |R < |z |. Thus, we can use the Extender Cauchy integral formula
at w = 1:
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Complex Variables
Inverse z Transform of a Product of Two Funct ions
What is Z-1[F (z )G(z )] for given f (t ) and g (t ) ?
But first:
DEFINITION (Convolut ion )
actually: 0 k n
Observe:
Let: , t = nT .
Now: 0 k n
If f (kT ) = a k ; g ( jT ) = b j ;
Or and:
the z transform of the convolut ion of two fun ct ionsis the product of the z transform s of each funct ion or Z-1[F (z )G (z )] = f (t )*g (t )
)(t h
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Complex Variables
EXAMPLE 7 Using the concept of convolution, find
Solution. Observe, that if
we can write
We also have demonstrated
and
So,
Now u ((n - k )T ) = 0 for k > n and u ((n- k )T ) = 1 for n k . Thus
and
l bl
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Complex Variables
Difference Equat ion s and the z Transfo rm
A prob lem involv in g a di f ference equat ion : Let f (nT ) be a function defined for n = 0 , 1
, 2 , . . . and assume T > 0. Obtain a closed-form expression fo r the solu t ion of the
equation
given that f (0) = 1
Step by step [ f ((n+1)T )=2 f (nT ) ] : f (T ) = 2, f (2T ) = 4,…, f (nT ) = 2n
One more method : perform a z transformation on both sides of the given equation
taking:
and the translation formula (f (0) = 1):
The equation transform gives
Or
Thus: f (nT ) = 2n
l bl
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Complex Variables
The general form of th e l inear di f ference equat ion:
EXAMPLE 8 (The Fibon acci sequence ) The sequence is 0, 1, 1, 2, 3, 5, 8, 13, 21, 34,
… : f (n + 2) = f (n + 1) + f (n ) [f (0)=0; f (1)=1; f (2)=1;.... ]
or : f (n + 2) - f (n + 1) - f (n ) = 0
For the general form: T =1, N =2, a 0=1, a 1=-1, a 2 =-1, a n >2= 0.
Find a closed form solution.
Solut ion. Use
For (T =1; f (0)=0; f (1)=1):
We obtain: or
Now
which give the expansion:
)(n f
