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The Z-Transform Let xn , n = 0, 1, 2, . . . be a sequence of numbers, x 0 , x1 , x2 , . . . The Z-Transform of xn is defined to be X(z) = +∞ X xn z −n n=0 This amounts to exchanging a sequence for a function of z. It turns out that the process is generally reversible. In fact, the sequence of numbers xn are the coefficients of the Taylor series expansion of X(w−1 ) where w = z −1 . We say that X(z) is the Z-transform of xn , and that xn is the inverse Z-transform of X(z). We write X(z) = X {xn } and xn = Z −1 {X(z)}. We say that xn lives in the time domain and that X(z) lives in the frequency domain. Think of z as a complex variable. 1 Transform 1 Next, we compute the transform of xn = rn where r is a constant. Using the definition, X(z) = +∞ X xn z −n = n=0 +∞ X rn z −n n=0 Now, let S= N X rk z −k = 1 + rz −1 + r2 z −2 + . . . + rN z −N k=0 and so rz −1 S = rz −1 + r2 z −2 + . . . + rN +1 z −N −1 Subtraction then gives S − rz −1 S = 1 − rN +1 z −N −1 = 1 − rz −1 S = 1 − rN +1 z −N −1 1 − rN +1 z −N −1 ⇒ S= 1 − rz −1 −1 If |z| > |r|, then rz < 1, so that rN +1 z −N −1 → 0 as N → ∞, and then S→ 1 1 − rz −1 as N → ∞ Hence, X(z) = +∞ X rn z −n = n=0 1 1 − rz −1 This proves that Z {rn } = 2 1 1 − rz −1 Property 1 - Linearity It’s clear that if X(z) = Z{xn } and Y (z) = Z{yn } then Z{c1 xn + c2 yn } = c1 X(z) + c2 Y (z) where c1 and c2 are any constants. Property 2 - Right Shift Suppose that the sequence xn is shifted right by one step. Let the resulting sequence be denoted by yn , so that y0 = 0, y1 = x0 , y2 = x1 etc. Then Y (z) = +∞ X yn z −n n=0 = +∞ X xn−1 z −n n=1 =z −1 +∞ X xn−1 z −n+1 n=1 =z −1 +∞ X xk z −k k=0 −1 = z X(z) This shows that Z {xn−1 } = z −1 X(z) So a right shift in the time domain corresponds to multiplication by z −1 in the frequency domain. Repeated use of the above property shows that Z {xn−k } z −k X(z) 3 Property 3 - Left Shift Suppose that the sequence xn is shifted left by one step. Let the resulting sequence be denoted by yn , so that y0 = x1 , y1 = x2 , y2 = x3 etc, and x0 falls off the edge, so to speak. Then Y (z) = +∞ X yn z −n n=0 = +∞ X xn+1 z −n n=0 = +x0 z − x0 z + +∞ X xn+1 z −n n=0 = −x0 z + +∞ X xn+1 z −n n=−1 A change in variable k = n + 1 gives Y (z) = −x0 z + +∞ X xk z −k+1 k=0 = −x0 z + z +∞ X xk z −k k=0 = −x0 z + zX(z) This shows that Z {xn+1 } = zX(z) − zx0 So a left shift in the time domain corresponds to multiplication by z with an initial condition term in the frequency domain. 4 Property 4 - Differentiation By definition, X(z) = +∞ X xn z −n n=0 Differentiate with respect to z. +∞ X dX xn (−n)z −n−1 (z) = dz n=0 Then, +∞ X dX −1 (z) = −z xn nz −n dz n=1 +∞ X dX ⇒ −z (z) = xn nz −n dz n=1 Hence, ⇒ −z dX (z) = Z {yn } dz where yn = nxn 5 Difference Equations The above transform and properties are enough to solve a broad class of difference equations. Consider an example. Find xn when xn+1 + 3xn = 2−n , subjevtto x0 = 1 Take the z-transform, Z {xn+1 + 3xn } = Z 2−n By linearity, Z {xn+1 } + 3Z {xn } = Z 2−n By transform 1 and property 3 above, zX(z) − zx0 + 3X(z) = 1 −1 1 − z2 Next, solve for X(z), ⇒ (z + 3) X(z) = ⇒ X(z) = 1− 1 z −1 2 1 −1 + zx0 1 − z2 (z + 3) + zx0 1 (z + 3) Using the given initial condition and some algebra, z z ⇒ X(z) = + 1 z − 2 (z + 3) (z + 3) Finally, we get the inverse z-transform, for which we need a partial fraction expansion (PFE). Thus, ⇒ X(z) = 6 z 6 1 + + 1 (z + 3) 7 (z + 3) 7 z − 2 1 −n 1 = Z{2 }, and property 2 then gives −1 1−z 2 −1 z −1 1 z −n+1 = = = Z{2 }, and similarly z + 3 1 + 3z −1 1 − z −1 21 Z{(−3)−n+1 }. Hence, n−1 z 1 1 6 n−1 un−1 + xn = (−3) un−1 + 7 7 2 (z + 3) Now, where un = 0 when n < 0 and un = 1 when n ≥ 0. Notes The z-transform is named after a famous Russian control systems researcher Ya Tzypkin. In the German language, his surname begins with the letter z. You should know that the above notes deal with the one-sided z-transform, and there is also a two-sided ztransform. The latter deals with two-sided sequences, such as 1 1 1 . . . , −8, 4, −2, 1, − , , − , . . . 2 4 8 −n In this example, xn = 2 , and n is not restricted to be positive. The ideas above are sufficient to work out the properties of the sided z-transform. 7