Z transform

1. Z-Transforms Aakanksha Thakre Ayush Agrawal Kunal Agrawal Akshay Phadnis 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=0 ∞ f(n) z -n =F(z) Aakanksha_Kunal_Ayush_Akshay

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

4. a Z{ n } = Z / (z-a) Z{ -a n } = Z / ( z+a ) Z{n}= z / (z-1) 2 Z{1/n!}= e 1/z Z{sin n ф }= zSin ф / (z -2zcos ф +1 ) 2 Z{ Cosn ф }= z- zCos ф / (z -2zCos ф + 1) 2 2 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 ] i =0 - i K- i K Aakanksha_Kunal_Ayush_Akshay

7. Division by n property :- Z{f(n)/n}= ∫ Z ∞ F(z)/z dz Division by n+k property:- Z{f(n) /( n+k )}= Z k ∫ z ∞ F(z)/ (Z) K+1 dz 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 X(s) = ∫ x(t) e dt ∞ - st 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 :- X(t) = ∫ X(s) e ds a+j ∞ a-j∞ st 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 X(e ) = ∑ x(m) e s ∞ m =0 - sm 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 X(z) = ∑ x(m) z ∞ m = 0 -m The two-sided z-transform is defined as:- X(z) = ∑ x(m) z ∞ m = -∞ -m Aakanksha_Kunal_Ayush_Akshay

13. The Relationship Between the Laplace, the Fourier, and the z-Transforms :- The Laplace transform, the Fourier transform and the z-transform are closely related in that they all employ complex exponential as their basis function. For right-sided signals (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 a generalisation of the Fourier transform of a discrete-time signal. Aakanksha_Kunal_Ayush_Akshay