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# Z transform

## on Aug 20, 2011

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## Z transformPresentation Transcript

• Z-Transforms
AakankshaThakre
AyushAgrawal
KunalAgrawal
Aakanksha_Kunal_Ayush_Akshay
• Introduction:
Just like Laplace transforms are used for evaluation of continuous functions, Z-transforms can be used for evaluating discrete functions.
Z-Transforms are highly expedient in discrete analysis ,
Which form the basis of communication technology.
Definition:
If a function f(n) is defined for discreet values ( n=0,+1 or -1 , +2 or -2,etc ) & f(n)=0 for n<0,then z-transform of the function is defined as
Z{f(n)}=

-n
f(n) z
=F(z)
n=0
Aakanksha_Kunal_Ayush_Akshay
• Some standard results & formulae:
-n

n=0
2
Aakanksha_Kunal_Ayush_Akshay
• Z{
}
n
=
a
Z / (z-a)
Z{
n
}
=
-a
Z / (z+a)
2
Z{n}= z /
(z-1)
Z{1/n!}=
e
1/z
Z{sin nф}= zSinф / (z -2zcosф +1 )
2
2
2
Z{Cosnф}= z- zCosф / (z -2zCosф + 1)
Aakanksha_Kunal_Ayush_Akshay
• Properties:
Linearity: -
Z{a f(n)+b g(n)}=a Z{f(n)}+b Z{g(n)}
Damping rule:-
Z{a f(n)} = F(z/a)
Multiplication by positive integer n :-
Z{n f(n)}= -z d/dz ( F(z) )
Aakanksha_Kunal_Ayush_Akshay
• Initial value theorem:-
f(0)= lim F(z)
Z∞
Final value theorem:-
f(∞)= lim f(n) = lim (z-1) F(z)
n∞
Z1
Shifting Theorem:-
Z{ f (n+k) }= z [ F(z) - ∑ f(i) z ]
K-i
-i
K
i=0
Aakanksha_Kunal_Ayush_Akshay
• Division by n property:-

Z{f(n)/n}=
F(z)/z
dz
Z
Division by n+k property:-

k
Z{f(n) /(n+k)}=
Z
K+1
F(z)/ (Z)
dz
z
Aakanksha_Kunal_Ayush_Akshay
• Applications of Z-Transforms
The field of signal processing is essentially a field of signal analysis in which they are reduced to their mathematical components and evaluated. One important concept in signal processing is that of the Z-Transform, which converts unwieldy sequences into forms that can be easily dealt with. Z-Transforms are used in many signal processing systems
Z-transforms can be used to solve differential equations with constant coefficients.
Aakanksha_Kunal_Ayush_Akshay
• Derivation of the z-Transform
The z-transform is the discrete-time counterpart of the Laplace transform. In this
section we derive the z-transform from the Laplace transform a discrete-time signal.
Aakanksha_Kunal_Ayush_Akshay
• The Laplace transform X(s), of a continuous-time signal x(t), is given by the integral
∞ -st
X(s) = ∫ x(t) e dt
0-
where the complex variable s=a +jω, and the lower limit of t=0− allows the possibility
that the signal x(t) may include an impulse. The inverse Laplace transform is defined
by:-
a+j∞
st
X(t) = ∫ X(s) e ds
a-j∞
Aakanksha_Kunal_Ayush_Akshay
• where a is selected so that X(s) is analytic (no singularities) for s>a. The ztransform
can be derived from Eq. by sampling the continuous-time input signal
x(t). For a sampled signal x(mTs), normally denoted as x(m) assuming the sampling
period Ts=1, the Laplace transform Eq. becomes

s
-sm
X(e ) = ∑ x(m) e
m=0
Aakanksha_Kunal_Ayush_Akshay
• Substituting the variable e to the power s
in Eq. with the
variable z we obtain the one-sided
ztransform

-m
X(z) = ∑ x(m) z
m = 0
The two-sided z-transform is defined as:-

-m
X(z) = ∑ x(m) z
m = -∞
Aakanksha_Kunal_Ayush_Akshay
• The Relationship Between the Laplace, the Fourier, andthe z-Transforms :-The Laplace transform, the Fourier transform and the z-transform are closely related inthat they all employ complex exponential as their basis function. For right-sidedsignals (zero-valued for negative time index) the Laplace transform is a generalisation of the Fourier transform of a continuous-time signal, and the z-transform is ageneralisation of the Fourier transform of a discrete-time signal.
Aakanksha_Kunal_Ayush_Akshay