DSP-Lec 05-z-Transform

Chapter p z-Transform Ha Hoang Kha, Ph.D.Click to edit Master subtitle style Ho Chi Minh City University of Technology @ Email: hhkha@hcmut.edu.vn ™ The z-transform is a tool for analysis, design and implementation of discrete time signals and LTI systems discrete-time systems ™ Convolution in time-domain ⇔ multiplication in the z-domain Ha H Kha z-Transforms Content z-transform t f Properties of the z-transform Causality and Stability Inverse z-transform Ha H Kha Discrete-Time Systems The z-transform ™ The z-transform of a discrete-time signal x(n) is defined as the power series: X ( z) = ∞ −n x ( n ) z = ∑ x(−2) z − + x(−1) z + x(0) + x(1) z −1 + x(2) z − + n = −∞ ™ The region of convergence (ROC) of X(z) is the set of all values of z for f which hi h X( X(z)) attains i a finite fi i value l ROC = {z ∈ C | X ( z ) = ∞ ∑ x ( n) z −n ≠ ∞} n = −∞ ™ The z-transform of impulse response h(n) is called the transform function of the filter: H ( z) = ∞ −n h ( n ) z ∑ n = −∞ ∞ Ha H Kha z-Transforms Example ™ Determine the z-transform of the following finite-duration signals a) x1(n)=[1, 2, 5, 7, 0, 1] b) x2(n)=x1(n-2) c) x3(n)=x1(n+2) d) x4(n)=δ(n) e) x5(n)=δ(n-k), (n)=δ(n k) k>0 f) x6(n)=δ(n+k), k>0 Ha H Kha z-Transforms Example ™ Determine the z-transform of the signal a)) x(n)=(0.5) ( )=(0 5)nu(n) ( ) b) x(n)=-(0.5)nu(-n-1) Ha H Kha z-Transforms z-transform and ROC ™ It is possible for two different signal x(n) to have the same ztransform Such signals g can be distinguished g in the z-domain byy their region of convergence ™ z-transforms: and their ROCs: ROC of a causal signal is the exterior t i off a circle i l Ha H Kha ROC of an anticausal signal is the interior of a circle circle z-Transforms Example ™ Determine the z-transform of the signal x(n) = a nu (n) + b n u (− n − 1) ™ The ROC of two-sided signal is a ring (annular region) Ha H Kha z-Transforms Properties of the z-transform ™ Linearity: if and then z x1 (n) ← ⎯→ X ( z ) with ROC1 z x2 ( n ) ← ⎯→ X ( z ) with ROC z x(n) = x1 (n) + x2 (n) ← ⎯→ X ( z ) = X ( z ) + X ( z ) with ROC = ROC1 ∩ ROC ™ Example: Determine the z-transform and ROC of the signals a) x(n)=[3(2)n-4(3)n]u(n) b) x(n)=cos(w0 t)u(n) c) x(n)=sin(w0 t)u(n) Ha H Kha z-Transforms Properties of the z-transform ™ Time shifting: if then z x ( n) ← ⎯→ X ( z) z x(n − D) ← ⎯→ z − D X ( z) ™ The ROC of z − D X (z ) is the same as that of X(z) except for z=0 if D>0 and z=∞ if D max | pi | i the ROC of causal signals g are outside of the circle ™ A anticausal signal of the form x(n) = − A1 p1n u (− n − 1) − A2 p2nu (− n − 1) + X ( z) = A1 A2 + + −1 −1 − p1 z − p2 z ROC | z |< | pi | i the ROC of causal signals g are inside of the circle Ha H Kha 13 z-Transforms Causality and stability ™ Mixed signals have ROCs that are the annular region between two circles ™ It can be shown that a necessary and sufficient condition for the stability of a signal x(n) is that its ROC contains the unit circle Ha H Kha 14 z-Transforms Inverse z-transform − transform x(n) ⎯z⎯ ⎯⎯→ X ( z ), ROC transform X ( z ), ROC ⎯inverse ⎯ ⎯z -⎯ ⎯ ⎯→ x(n) z x ( n) ← ⎯→ → X ( z ), ROC ™ In inverting a z-transform, it is convenient to break it into its partial fraction (PF) expression form, i.e., into a sum of individual pole terms whose inverse z transforms are known ™ Note that with X ( z ) = -1 we have - az ⎧ a nu (n) if ROC | z |>| a | (causal ( l signals) i l) x ( n) = ⎨ n ⎩− a u (−n − 1) if ROC | z | M Ha H Kha 19 z-Transforms Exampleod ™ Compute all possible inverse z-transform of Solution: - Find the poles: 1-0.25z-2 =0 Ỉ p1=0.5, p2=-0.5 - We have N N=22 and M=2, M 2, i.e., N = M Thus, we can write where Ha H Kha 20 z-Transforms Exampleod Ha H Kha 21 z-Transforms Exampleod ™ Determine the causal inverse z-transform of Solution: - We have h N=55 and d M=2, i.e., i N > M Thus, Th we have h to divide di id the h denominator into the numerator, giving Ha H Kha 22 z-Transforms Partial fraction expression method ™ Complex-valued poles: since D(z) have real-valued coefficients, the complex-valued poles of X(z) must come in complex-conjugate pairs C id i the Considering h causall case, we h have Writing A1 and p1 in their polar form, say, with B1 and R1 > 0,, and thus,, we have A a result, As lt the th signal i l in i time-domain ti d i is i Ha H Kha 23 z-Transforms Exampleod ™ Determine the causal inverse z-transform of Solution: Ha H Kha 24 z-Transforms Exampleod Ha H Kha 25 z-Transforms