MIT OpenCourseWare http://ocw.mit.edu MAS.160 / MAS.510 / MAS.511 Signals, Systems and Information for Media Technology Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Z-transforms : Part I MIT MAS 160/510 Additional Notes, Spring 2003 R. W. Picard 1 Relation to Discrete-Time Fourier Transform Consider the following discrete system, written three different ways: y[n] = b−1y[n + 1] + b1y[n − 1] + a−1x[n + 1] + a0x[n] + a2x[n − 2] Y (z) = b−1zY (z) + b1z−1Y (z) + a−1zX(z) + a0X(z) + a2z−2X(z) Y (z) a−1z + a0 + a2z−2 H(z) = = (1)X(z) −b−1z + 1 − b1z−1 Simple substitution finds the Z-transform for a discrete system represented by a linear constant coefficient difference equation (LCCDE). Simply replace y[n] with Y (z), x[n] with X(z), and shifts of n0 with multiplication by zn0 . That’s almost all there is to it. Set jωˆz = reand let r = 1 for the moment. Then a shift in time by n0 becomes a multiplication in the Z-domain by ejωn0 . This should look familiar given what you know about Fourier analysis. Now here’s the formula for the Z-transform shown next to the discrete-time Fourier transform of x[n]: ∞Z-transform : X(z) = � x[n]z−n n=−∞ j ˆ∞ωn DTFT: X(eω) = � x[n]e−j ˆ, n=−∞ where we have used the notation X(ejωˆ) instead of the equivalent X(ˆω), to emphasize similarity with the Z-transform . Substituting z = rejωˆin the Z-transform , ∞ωn X(z) = � (x[n]r−n)e−j ˆ, n=−∞ reveals that the Z-transform is just the DTFT of x[n]r−n . If you know what a Laplace transform is, X(s), then you will recognize a similarity between it and the Z-transform in that the Laplace transform is the Fourier transform of x(t)e−σt . Hence the Z-transform generalizes the DTFT 1� � �in the same way that the Laplace transform generalizes the Fourier transform. Whereas the Laplace transform is used widely for continuous systems, the Z-transform is used widely in design and analysis of discrete systems. You may have noticed that in class we’ve already sneaked in use of the “Z-plane” in talking about the “unit circle.” The “Z-plane” contains all values of z, whereas the unit circle contains only z = ejωˆ. The Z-transform might exist anywhere in the Z-plane; the DTFT can only exist on the unit circle. One loop around the unit circle is one period of the DTFT. (For those who know Laplace transforms, which are plotted in the S-plane, there is a nonlinear mapping between the S-plane and the Z-plane. The Im{s} axis maps onto the unit circle, the left half plane (σ < 0) maps inside the unit circle, and the right half plane (σ > 0) maps outside the unit circle. Poles and zeros in the Laplace transform are correspondingly mapped to poles and zeros in the Z-transform .) M� 2 Discrete transfer function Equation (1) gave the transfer function, H(z), for a particular system. In general the form for LCCDE systems is N�aly[n − l] bkx[n − k]= l=−N N= M� k=−M which transforms to alz−lY (z) bkz−kX(z) M� l=−N k=−M bkz−k H(z) = Y (z)= k=−M NX(z) l=−N Note that the limits M, N do not impose symmetry on the number of terms in either direction. We could have b−1 = b1, etc. One could also have N = 0 with M finite, yielding an FIR system, or N = 0 and M = ∞ yielding an IIR system, or an IIR system can be created with N > 0 and M = 0. There are many possibilities, but in each case the transfer function H(z) completely characterizes the linear shift invariant system. The following relationships hold: y[n] = x[n] ∗ h[n] Y (z) = X(z)H(z) We can learn to look at H(z) and infer properties of the system. Is it low-pass, single-pole, FIR, IIR, stable, etc.? In order to do this, we need to consider its poles, zeros, and (like the Laplace transform) regions of convergence. 2 alz−l0h[n]n3 Poles, Zeros, Regions of Convergence Example 1: Infinite-length right-sided Consider the causal sequence: h1[n] = a n u[n] This impulse response might be used to model signals which decay over time, like echos. Consider its Z-transform : ∞nH1(z) = � a u[n]z−n n=−∞ ∞n = � a z−n n=0 ∞= �(az−1)n . (2) n=0 Remember the following useful result from mathematics of series summations: ∞1� = , γ < 1. n=0 γn 1 − γ | | Notice this result is only true when γ < 1. Forgetting this constraint is dangerous. When| ||γ| ≥ 1 then the series does not converge, and we say “it does not exist” or “it blows up.” To proceed with (2) we must therefore have az−1< 1, or z > a . This completes the result: | | | | | |nZ 1 a u[n] , z > a . ←→ 1 − az−1 | | | |The region in the Z-plane where the transform exists is called the region of convergence (ROC), e.g., z > a for this example. Even though we can evaluate 1 for z a , it is 1−az−1| | | | | | ≤ | |usually inappropriate to do so since this expression for the Z-transform is only defined within its ROC. Let’s draw a picture of this. First we need to think about the poles and zeros of H1(z). Poles are the values of z for which the Z-transform is ∞ and zeros are the values of z for which 3z>aa� the Z-transform is 0. Every Z-transform has the same number of poles as it has of zeros. It is good practice to count them up when you think you’ve found them all. For the example above there is one pole and one zero, and the ROC is outside the pole: 1 pole: z = a H1(z) = , z > a . 1 − az−1 | | | |zero: z = 0 The zero at z = 0, outside the ROC, comes only from the expression 1 without considering 1−az−1 its ROC. If you are troubled that there is a finite value (namely zero) found outside the ROC then you are currently in good company – because the staff is still troubled by this as well: how can the function be said to be zero outside its ROC when it doesn’t even exist outside its ROC? Note we do not have trouble saying there is a pole outside the ROC, because a pole is where the function “blows up” which is just what you’d expect when you’re not in the ROC. Nonetheless, all the standard texts put that zero there, outside the ROC. We will keep you posted if we convince the authors of the standard texts not to do that, or if they convince us not to worry about it. Does the DTFT exist for h1[n]?