Z-Transform Concepts and Applications
Explore the concepts and applications of Z-Transform, relating it to Fourier transform and Laplace transform. Understand the significance of Z-Transform in transforming discrete-time series and its relation to operational calculus.
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Engineering Analysis Assit.Lec. Shaimaa Shukri
Z -Transform Concepts And Applications
Primary Exam. 1- What is the z-transform? 2- The one-sided z-transform of a function x(n) is ----? 3- What is the Fourier transform of x(n)?
The z-transform is the most general concept for the transformation of discrete-time series. The Laplace transform is the more general concept for the transformation of continuous time processes. For example, the Laplace transform allows you to transform a differential equation, and its corresponding initial and boundary value problems, into a space in which the equation can be solved by ordinary algebra. The switching of spaces to transform calculus problems into algebraic operations on transforms is called operational calculus. The Laplace and z transforms are the most important methods for this purpose.
The Laplace transform of a function f(t): st = F ( s ) f ( t ) e dt 0 The one-sided z-transform of a function x(n): = 0 n The two-sided z-transform of a function x(n): = n n = X ( z ) x ( n ) z n = X ( z ) x ( n ) z
Note that expressing the complex variable z in polar form reveals the relationship to the Fourier transform: = n i i n = X ( re ) x ( n )( re ) , or = n i n i n = = X ( re ) x ( n ) r e , and if r , 1 = n i i n = ( = X ( e ) X ) x ( n ) e which is the Fourier transform of x(n).
The z-transform of x(n) can be viewed as the Fourier transform of x(n) multiplied by an exponential sequence r-n, and the z-transform may converge even when the Fourier transform does not. By redefining convergence, it is possible that the Fourier transform may converge when the z-transform does not. For the Fourier transform to converge, the sequence must have finite energy, or: = n n x ( n ) r
The power series for the z-transform is called a Laurent series: = n n = X ( z ) x ( n ) z The Laurent series, and therefore the z-transform, represents an analytic function at every point inside the region of convergence, and therefore the z-transform and all its derivatives must be continuous functions of z inside the region of convergence. In general, the Laurent series will converge in an annular region of the z-plane.
Some Special Functions First we introduce the Dirac delta function (or unit sample function): = 0 , 1 n This allows an arbitrary sequence x(n) or continuous- time function f(t) to be expressed as: = k , 0 t 0 , 0 n 0 or = ( t ) = ( n ) = , 1 t 0 = x ( n ) x ( k ) ( n k ) f = f ( t ) ( x ) ( x t dt )
These are referred to as discrete-time or continuous- time convolution, and are denoted by: * ) ( ) ( t f t f = = x n x n ( n ) ( ) ( * ) ( t ) We also introduce the unit step function: = , 0 n Note also: n u ) ( , 1 n 0 , 1 t 0 = u ( n ) or u ( t ) 0 = k , 0 t 0 = ( k )
When X(z) is a rational function, i.e., a ration of polynomials in z, then: 1. The roots of the numerator polynomial are referred to as the zeros of X(z), and 2. The roots of the denominator polynomial are referred to as the poles of X(z). Note that no poles of X(z) can occur within the region of convergence since the z-transform does not converge at a pole. Furthermore, the region of convergence is bounded by poles.
Region of convergence n = x ( n ) a u ( n ) a The z-transform is given by: = n = n n n 1) n = = X ( z ) a u ( n ) z ( az 0 Which converges to: 1 z = = X ( z ) for z a 1 z a 1 az Clearly, X(z) has a zero at z = 0 and a pole at z = a.
Suppose that only a finite number of sequence values are nonzero, so that: = 1 n n n 2 n = X ( z ) x ( n ) z Where n1 and n2 are finite integers. Convergence requires x ( n ) for n n n . 1 2 So that finite-length sequences have a region of convergence that is at least 0 < |z| < , and may include either z = 0 or z = .
Z-Transform Time Domain ik u(k)= e 1 U(z) = i 1 - - 1 e z -ik u(k)= e 1 U(z) = i - 1 - - 1 e z ik ik e e + u(k) cos(k ) = = 1 1 U(z) ( )/2 = + 2 i 1 - i 1 - 1 e z 1 e z 1 1 ( )/2 = + 1 1 1 1 1 cos z isin z 1 cos z + isin z ik ik e e 1 1 cos z u(k) sin(k ) = = = 2i 1 2 1 2 (1 cos z ) (sin z ) + 1 1 cos z = 1 2 1 2cos z z + -1 sin z U(z) = 1 - 2 - 1 2cos z z +
1 0.8 0.6 u(k)=cos(k*pi/6)*0.9k 0.4 0.2 0 i k i k i k i k (ae ) + (ae ) (ae ) (ae ) k k u (k) a cos(k ) u (k) a sin(k ) = -0.4 = = = -0.2 expcos expsin 2 2i -0.6 -0.8 -1 0 2 4 6 8 10 12 14 16 18 -1 a sin z -1 a sin z U(z) = U(z) = 1 - 2 2 1 - 2 2 - 1 2a cos z a z + - 1 2a cos z a z + A damped oscillating signal a typical output of a second order system
