The Inverse z -Transform

1. The Inversez-Transform

2. The inverse z transformation can be performed by using a few results from complex variable theory. Suppose that we integrate one (1) from z=0 to z=1:

3. The integrals become more involved when the path of integration is in the complex plane. For example, suppose that we wish to integrate around a unit circle in the (complex) z-plane. To do the integration we make a substitution of variables: z=ejq. We have dz = j ejq dq.

4. Im{z} q = p/2 Re{z} q q = 0 q = p radius = 1 q = 3p/2

5. Suppose, first, that we wish integrate one (1) around a semicircleC around the origin from 0 to p. Im{z} C q q = p q = 0 Re{z}

7. Suppose that we integrate one (1) around a closed unit circle Caround the origin (0 to 2p). Im{z} C q q = 0 Re{z} q = 2p

8. We have, (The areas under the sine and cosine sum to zero over a full period.)

9. Suppose that we integrate z around a closed unitcircle C around the origin (0 to 2p).

10. Similarly, if we integrate z2 ,z3 or zk(k  0) around the unit circle we get zero. What about integrating z-1?

11. We do not get zero when we integrate z-1 around the unit circle. What about integrating z-2?

12. Similarly, if we integrate z-3 ,z-4 or z-k(k>1) around the unit circle we get zero. So integrating z-1 is unique, and

13. So if we were to integrate a z-transform we get (Only z-1 does not integrate to zero.)

14. In other words the integration “picks-out” the sample x1. We can “pick-out” another sample by adjusting the power of z: (The integral is zero except when –n+1 = -1, or n=2.)

15. Similarly, and

16. Thus, we have found an expression for the inversez-transform: The path of the integral must be carried-out over a region of convergence of X(z). [CROC.]

17. Example: Find the inverse z-transform of X(z)=1using the complex inversion integral. Solution: Plugging-in X(z)=1, we have

18. We know that the integral is equal to zero except when n=0. When n=0, the integral is equal to 2pj. Thus, the final inverse transform is equal to when n=0. Hence we have x[n]=d[n].

19. Example: Find the inverse z-transform of X(z)=z-kusing the complex inversion integral. Solution: Plugging-in X(z)=z-k, we have

20. We know that the integral is equal to zero except when n=0. When n-k=0, or when n=k. Hence we have x[n]=d[n-k].

21. Example: Find the inverse z-transform of using the complex inversion integral. Solution: Let us expand 1/(1-z-1) using the geometric series:

22. Solution: Plugging this into the complex inversion integral we have For n=0, the only term in 1+z-1+z-2+… which “survives” a non-zero integration is 1. For n=1, the only term in 1+z-1+z-2 which “survives” is z-1.

23. We see that there is exactly one term that “survives” when n0. The resultant integral in each case is 2pj, and the corresponding value for x[n] in each case is one. Thus,

24. Example: Find the inverse z-transform of using the complex inversion integral. Solution: Using the expansion for 1/(1-z-1), we have

25. Solution: Plugging this into the complex inversion integral we have For n=0, the only term in 1+z-1+z-2+… which “survives” a non-zero integration is 1. For n=1, the only term in 1+z-1+z-2 which “survives” is 2z-1.

26. The resultant inverse z-transform is

27. Example: Find the inverse z-transform of using the complex inversion integral. Solution: Expanding 1/(1-az-1) as a geometric series:

28. Plugging this into the complex inversion integral we have For n=0, the only term in 1+az-1+a2z-2+… which “survives” a non-zero integration is 1. For n=1, the only term in 1+az-1+a2z-2 which “survives” is az-1.

29. The resultant integral in the first case is 2pj, the resultant integral in the second case is 2pja, etc. The inverse z-transform is

30. Example: Find the inverse z-transform of using the complex inversion integral. Solution: We expand both 1/(1-az-1) and 1/(1-bz-1):

31. The product is

32. Plugging this into the complex inversion integral, we wind-up with

33. This same problem can be solved in a slightly different manner:

34. where

35. and The inverse z-transform is It can be shown algebraically (exercise) that this expression is equivalent to the previous expression for x[n].

36. In general, to find the inverse z-transform of we perform a partial fraction expansion:

37. where etc.

38. The resultant inverse z-transform is

39. Example: Find the inverse z-transform of using the complex inversion integral. Solution: We take the MacLaurin series for ez-1

40. The inverse z-transform follows rather naturally: We could obtain this result through the complex inversion integral, or we could recognize that x[n]=1/n! corresponds to the coefficients of the z-transform.