Need for transformation?

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Beryl Tate
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1 Z-TRANSFORM

2 In today s class Z-transform Unilateral Z-transform Bilateral Z-transform Region of Convergence Inverse Z-transform Power Series method Partial Fraction method Solution of difference equations

3 Need for transformation? Why do we need to transform our signal from one domain to another? Information available in one domain is not sufficient for complete analysis Looking at a sine wave in time-domain, we cannot really know the frequency content So we have to look into the frequency domain An alternate domain may express the information more comprehensively A pole-zero map easily tells whether a systems is stable or not

4 Z-transform Digital counterpart for the Laplace transform used for analog signals Mathematically defined as, X (z) x[n] z n n This equation is in general a power series, where z is a complex variable.

5 Derivation The continuous-time Fourier transform of x(t) is given as, Fxt xt e j 2ft dt And the discrete-time Fourier transform of x[nt] is given as, F xnt D n xnt e j 2fnT The Z-transform of x[n] is given as the Fourier transform of x[n] multiplied by r n nt D n x F r x nt D

6 Bi-lateral Z-transform D x nt F D r D x nt x nt D x nt D z re D j 2 ft x nt n x nt r n x nt e n n n n x nt re x nt z n j 2 fnt x nt r n j 2 fnt e j 2 ft n

7 Uni-lateral Z-transform xnt D n xnt z xnt xnt z n n0 X z xnt z n n0 where z -1 would show a delay by one sample time n

8 Example 1: Find the z-transform of the following finite-length sequence 4 y nt ynt ynt 2 n 1T 4 n 3T 3 n 4T 2 n 5T

9 ynt Y z ynt z n n0 Y z 0 n z n Y z 0z 0 2z 1 0z 2 4z 3 3z 4 2z 5 0z 6 0z 7 Y z 2z 1 4z 3 3z 4 2z 5 So, Y(z) would exist on the entire z-plane except the point z=0

10 Z-transform as Rational Function Often it is convenient to represent Z-transform X(z) as a rational function X (z) P(z) Q(z) Where P(z) and Q(z) are two polynomials The values of z at which X(z) becomes zero (X(z) = 0) are called zeros The values of z at which X(z) becomes infinite (X(z) = called poles ), are

11 Significance of Poles & Zeros Poles Roots of the denominator Q(z) The point where H(z) becomes infinite The point where H(e jw ) shows a peak value System may become unstable Zeros Roots of the numerator P(z) The point where H(z) becomes zero The point where H(e jw ) shows maximum attenuation

12 Convergence issues A power series may not necessarily converge The infinite sum may not always be finite The set of values of z for which the z-transform converges is called Region of Convergence (RoC) The convergence of X(z) depends only on z and it converges for n n x [ n ] z

13 X ( z ) x [ n ] z n n Replacing z re j 2 fn re jw n x [ n ] re jw n n x [ n ] r n e jwn This equation can be segmented into two parts, one for the right-sided (causal) signal and second for the left-sided (non-causal) signal 1 X (z) x[n]r n n x[n]r n n0

14 X (z) x[n]r n n1 x[n] n n0 r For X(z) to exist in a particular region (for certain values of z), both summations must be finite in that region For the first summation, r should be small enough x[-n]r n converges when summed to infinite terms so that For the second summation, r should be large enough so that x[n]/r n converges when summed to infinite terms

So, there are two circles with radius r L & r R for the sequence x[n] If it

becomes zero (by definition), and the radius r L should be small enough to

right-sided sequence (causal), then the first summation becomes zero (by

15 So, there are two circles with radius r L & r R for the sequence x[n] If it is defined as a left-sided sequence (non-causal), then the second summation becomes zero (by definition), and the radius r L should be small enough to make the first summation converge rr RR r L If x[n] is defined as a right-sided sequence (causal), then the first summation becomes zero (by definition), and the radius r R should be large enough to make the second summation converge

16 Imaginary Part Example 2: Find the z-transform of the following finite-length sequence x[n] 2 [n 2] [n 1] 2 [n] [n 1] 2 [n 2] The z-transform of this sequence is given as, X ( z) 2 z 2 z 2 z 1 2 z 2 it is clear to see that the sequence does not have any poles (denominator is 1), it has 4 zeros It can be observed that X(z) becomes undetermined at z = 0 and z =, so the RoC is entire z-plane except at z = 0 and z = Real Part

Example 3: Find the z-transform of the following right-sided sequence x[n] a n u[n] X (z) a n u(n)z n n a n z n n0 (az 1 ) n n0 1 z X (z) (az 1 ) n 1 n0 1 az z a For convergence we require

17 Example 3: Find the z-transform of the following right-sided sequence x[n] a n u[n] X (z) a n u(n)z n n a n z n n0 (az 1 ) n n0 1 z X (z) (az 1 ) n 1 n0 1 az z a For convergence we require X (z) Now, X(z) will not exist for z=a & RoC is entire z-plane except z=a However, since the z-plane is a circle so we have to use the following condition (the sequence is right-sided) z a

Example 4: Find the z-transform of the following right-sided sequence X (z) a n u(n 1)z n n 1 a n z n n a n z n n1 x[n] a n u[n 1] 1 a n z n n0 X (z) 1 1 z (a 1 z) n 1 1 n0 1 a z z a For convergence

18 Example 4: Find the z-transform of the following right-sided sequence X (z) a n u(n 1)z n n 1 a n z n n a n z n n1 x[n] a n u[n 1] 1 a n z n n0 X (z) 1 1 z (a 1 z) n 1 1 n0 1 a z z a For convergence we require X (z) Now, X(z) will not exist for z=a & RoC is entire z-plane except z=a However, since the z-plane is a circle so we have to use the following condition (the sequence is left-sided) z a

19 Concepts From the two examples we observe that the closed form equations for the z-transform of causal & noncausal signals come out to be same This creates an ambiguity about the existence of their z-transform Therefore, we require complimentary information apart from the closed form equations, i.e. the RoC

20 Properties of RoC Property 1: The RoC is a ring or disk in the z-plane centred at the origin; i.e., 0 r R z r L Property 2: The RoC cannot contain any poles Property 3: If x[n] is a finite-duration sequence i.e. a sequence that is zero except in a finite interval N 1 n N 2, then the RoC is the entire z-plane except possibly z=0 and z=

21 Property 4: If x[n] is a right-sided sequence i.e. a sequence that is zero for, n N 1, the RoC extends outward from the outermost (i.e. largest magnitude) finite pole in X(z) to z= Property 5: If x[n] is a left-sided sequence i.e. a sequence that is zero for, n N 2, the RoC extends outward from the outermost (i.e. largest magnitude) finite pole in X(z) to z=0

22 Z-transform pairs Sequence z-transform RoC ( n ) 1 All z ( n m ) z m All z except 0 (if m>0) or (if m<0) a n u ( n ) 1 1 az 1 z a a n u(n 1) 1 1 az 1 z a na n u(n) (1 az 1 ) 2 az 1 z a

23 Sequence z-transform RoC 1[cos 0 ]z [cos 0 n]u(n) 1 z 1 1[2 cos ]z 1 z 2 0 [sin 0 n]u(n) [sin 0 ]z 1 z 1 1 [2 cos ]z 1 z 2 0 n [r cos 0 n]u(n) 1[r cos ]z 1 1[2r cos ]z 1 r 2 z z r n [r sin 0 n]u(n) [r sin 0 ]z z r 1[2r cos 0 ]z r z

24 Example 5: Find the RoC of x[n] (0.5) n u[n] (0.4) n u[n] Using the properties of z-transform we get X (z) z z 1 z z z(z 0.4) z(z 0.5) z 0.5 z 0.4 (z 0.5)(z 0.4) It is clear that the RoC is given by z 0.4 and z 0.5 So we can conclude that the RoC is z 0.5

25 Example 6: Find the RoC of x[n] (0.5) n u[n] (0.9) n u[n 1] Using the properties of z-transform we get X (z) z z 1 z z z(z 0.9) z(z 0.5) z 0.5 z 0.9 (z 0.5)(z 0.9) The RoC due to the first part is z 0.5 since it is a right-sided sequence however, the second part is a left-hand sequence, therefore its RoC is z 0.9 So we can conclude that the RoC for X(z) is 0.5 z 0.9

26 Inverse Z-transform Power Series method Simple Tedious for large n Not accurate Partial Fraction method Complicated More accurate

27 IZT: Power Series method In this method we divide the numerator of a rational Z-transform by its denominator The basic idea is Given a Z-transform X(z) with its corresponding RoC, we can expand X(z) into a power series of the form X (z) c n z n which converges in the given RoC n

28 Example 7: Find the Inverse Z-transform of X(z) 1 X(z) 11.5z 1 0.5z 2 RoC z >1 Since RoC is the exterior of the circle, so we expect a right-sided sequence, so we seek an expansion in the negative powers of z By dividing the numerator of X(z) by its denominator, we obtain the power series z 1 1 z z 1 7 z 2 15 z 3 31 z x[n] = [1, 3/2, 7/4, 15/8, 31/16,. ]

29 Example 8: Find the Inverse Z-transform of X(z) 1 X(z) 11.5z 1 0.5z 2 RoC z <1 Since RoC is the interior of the circle, so we expect a left-sided sequence, so we seek an expansion in the positive powers of z By dividing the numerator of X(z) by its denominator, we obtain the power series z 1 1 z z 2 6 z 3 14 z 4 30 z 5... x[n] = [., 30, 14, 6, 2, 0, 0]

30 IZT: Partial Fraction method Steps to follow Eliminate the negative powers of z for the z-transform function X(z) Determine the rational function X(z)/z (assuming it is proper), and apply the partial fraction expansion to the determined rational function X(z)/z using formulae in table (next slide)

31 Partial fraction(s) and formulas for constant(s) Partial fraction with the first-order real pole: A z p Partial fraction with the first-order complex poles: A (z p) X (z) z p z Az A* z A (z p) X (z) z p z p * z p z, A A* Partial fraction with mth-order real poles: A k A A k 1 1 A z p (z p) 2 (z p) k k k 1 1 d (z p) k (k 1)! dz k 1 X (z) z z p

32 An example for Simple Real Poles

33 An example for Multiple Real Poles f n =[9(0.3) n 8(0.2) n +2n(0.2) n ]u(n)

34 Pulse Transfer Function Pulse transfer function H(z) is defined as the ratio of the Z-transform of the input x[n] to the Z-transform of the output y[n] H z Y z X z

35 Derivation l y n b i xn i a i y n i i 0 i 1 k Applying Z-transform and moving the terms of y to one side Y z k i1 Y za z i i l i i0 b X zz i Y z Y z Y z1 k l i a i z X z i i1 i0 k i i1 a z i X z l i i0 b z b z i i H z Y z X z 1 l b i z i0 k i1 i a i z i

36 Example 8: Find the Pulse Transfer function of the difference equation yn 0.1yn yn 2 2xn xn 1 a b 0 2 H z 1 a b b i z i0 2 i i1 i a z i H z Y z X z l i b i z i0 k i a z i 1 i1 H z 2z 1z z z z 1 0.1z z 2

ECE-S Introduction to Digital Signal Processing Lecture 4 Part A The Z-Transform and LTI Systems

ECE-S Introduction to Digital Signal Processing Lecture 4 Part A The Z-Transform and LTI Systems ECE-S352-70 Introduction to Digital Signal Processing Lecture 4 Part A The Z-Transform and LTI Systems Transform techniques are an important tool in the analysis of signals and linear time invariant (LTI)