1. Z-Transforms AakankshaThakre AyushAgrawal KunalAgrawal AkshayPhadnis Aakanksha_Kunal_Ayush_Akshay

2. 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

3. Some standard results & formulae: -n ∞ n=0 2 Aakanksha_Kunal_Ayush_Akshay

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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