ECE-S Introduction to Digital Signal Processing Lecture 4 Part A The Z-Transform and LTI Systems
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1 ECE-S 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) systems. Z-Transform applied to discrete-time signals and LTI systems plays the same role as the Laplace transform for continuous signals and LTI systems. The Direct Z-Transform X(z) Z[x(n)] z x(n) X(z) z-transform of signal z(n) bilateral relationship X(z) x(n)z -n n = - infinite power series in z where z is a complex variable Above equation transforms the time-domain signal x(n) into its complex representation X(z). The inverse process is called the inverse z-transform which takes X(z) to x(n). There are uniqueness issues. Since an infinite power series, the z-transform only exists for those values of z for which the series converges. The region of convergence (ROC) of X(z) defines the set of all z for which X(z) attains a finite value. Hence we must always specify a ROC with the z-transform ECE-S352 DSP of
2 Some simple examples of discrete time signals and their z-transforms a) x(n) = {,2,5,7, 0, } X(z) = z 0 + 2z - + 5z z z -4 +z -5 X(z) = + 2z - + 5z z -3 +z -5 note X(z) has a finite value for all values of z except z=0 ROC: entire z-plane except z=0 b) x2(n) = {,2,5,7, 0, } time origin (n=0) X2(z) = z 2 + 2z + 5z 0 + 7z - + 0z -2 +z -3 X2(z) = z 2 + 2z z - + z -3 note X2(z) has a finite value for all values of z except z=0 and z = ROC: entire z-plane except z=0 and z = Key Observation: Given the direct form of the z-transform (series representation) the coefficients of the powers of z are values of the discrete-time sequence at time n where n is the power of z. d) x5(n) = δ(n) e) x6(n) = δ(n-k) k>0 f) x7(n) = δ(n+k) k>0 X5(z) = z 0 = ROC: entire z-plane X6(z) = z -k ROC: entire z-plane except z = 0 X6(z) = z k ROC: entire z-plane except z = Hence we see the following transform pairs: z z z δ(n) δ(n-k) z -k δ(n+k) z k k>0 k>0 ROC: entire z-plane entire z-plane entire z-plane except z = 0 except z = ECE-S352 DSP 2 of
3 ROC of a finite-duration signals is the entire z-plane with the exception of possibly z=0 or z =. So what happens if the series is infinite? Consider: x(n) = (½) n u(n) = {, ½, (½) 2, (½) 3 (½) n } which is clearly an infinite power series by definition: X(z) = + ½z - +(½) 2 z -2 + (½) 3 z -3 + (½) n z -n + X(z) = (½) n z -n = (½z - ) n n = 0 n = 0 however we would like a closed form solution for X(z) so that we may manipulate X(z) in an easier fashion An infinite geometric series: + A + A 2 + A 3 + = provided that A < A Using this relationship: X(z) = z 2 and the ROC is found by : ½z - < power series convergence ½ z - < ½ z - < z - <2 z > ½ inversion flip inequality Note that this ROC is described by the entire z-plane with a circle of radius ½, centered at z = 0+j0 removed and not z=0 as one may incorrectly think/guess/extrapolate by examining the power series. ECE-S352 DSP 3 of
4 Definition: Absolutely Summable Reference: Exell, R. H. B. (998). Number Systems and Analysis, Real Numbers. World Wide Web, Series A series is an infinite list of numbers added together: x + x 2 + x A series is summable when the sequence of partial sums x, x + x 2, x + x 2 + x 3,... has a finite limit s. This limit is the sum of the series, and we write: x + x 2 + x = s. We also say the series converges to s. A series which is not summable is said to diverge. If the series x + x 2 + x is summable, then the sequence x, x 2, x 3,... converges to zero. But the converse is false. For example, lim(, /2, /3, /4,...) = 0, but the series + /2 + /3 + / diverges. A series x + x 2 + x is absolutely summable, or absolutely convergent, when the series x + x 2 + x is summable. A series x + x 2 + x converges to a unique sum independent of the arrangement of the terms if and only if it is absolutely summable. EXAMPLE The following series are all rearrangements of the same series: - /2 + /3 - /4 + /5 - /6 + /7 - /8... = /3 - /2 + /5 + /7 - / = /2 + /3 + /5 - /4 + /7 + /9 + / - / diverges. These series are not absolutely summable because + /2 + /3 + / diverges. End of Definition ECE-S352 DSP 4 of
5 Since z is a complex number it can be written as: z = re jθ and the z- transform becomes: X(z) = x(n)(re jθ ) -n = x(n)r -n e -jθn n = - n = - Now for the ROC X(z) < X(z) = x(n)r -n e -jθn n = - and noting that the absolute value of the sum of terms is always less than or equal to the sum of the absolute value of the terms X(z) <= x(n)r -n e -jθn n = - since the magnitude of e -jθn is then the term on the right becomes X(z) <= x(n)r -n n = - and X(z) is finite if the sequence x(n)r -n is absolutely summable. Thus the problem of finding the ROC is equivalent to determining the range of values of r for which the sequence x(n)r -n is absolutely summable. - X(z) <= x(n)r -n + x(n)r -n n = - n = 0 - X(z) <= x(n)r -n + x(n)/r n n = - n = 0 X(z) <= x(-n)r n + x(n)/r n n = n = 0 For X(z) to converge in some region of the z-plane, both summations must be finite in that region. ECE-S352 DSP 5 of
6 a) If the left sum converges there must exist values of r small enough such that the product sequence x(-n)r n where <=n< is absolutely summable. The ROC for the left sum consists of all points in a circle of some radius r where r <. b) If the second sum converges then there must exist values of r large enough such that the product sequence x(n)/r n 0<=n< is absolutely summable. the ROC of the right term consists of all points outside a circle of radius r >r2. c) Since convergence of X(z) requires both sums to be finite then the result is an annular region in the z-plane r2<r<r which is the common region where both sums are finite. d) Note that if r2>r there is no common region of convergence and X(z) does not exist. r r2 anti-causal signal left hand side causal signal right hand side r r2 annular region for X(z), both sides ECE-S352 DSP 6 of
7 Example: Causal signal (decaying exponential see page 56 for DT plot) x(n) = a n u(n) = a n n >= 0 0 n < 0 X(z) = a n z -n = (az - ) n n = 0 n = 0 Using: + A + A 2 + A 3 + = A provided that A < X(z) = az and the ROC is: az - < or z > a which is the exterior of a circle having radius a Example: Anti-causal signal (exponential see page 57 for DT plot) x(n) = -a n u(-n-) = 0 n >= 0 -a n n <= - - X(z) = (-a n )z -n = - (a - z) l n = - l = Using: A + A 2 + A 3 + = A A provided that A < za X(z) = - za az = understand manipulation and the ROC is: a - z < having radius a or z < a which is the interior of a circle ECE-S352 DSP 7 of
8 Note the two transform pairs x(n) = a n u(n) X(z) = az ROC: z > a x(n) = -a n u(-n-) X(z) = az ROC: or z < a Here we see that the causal and anti-causal signals have the same z-transform. So we must specify the ROC to remove any ambiguity. A discrete time signal, x(n), is uniquely determined by its z- transform and the region of convergence of X(z) Another transform pair: Consider the pair: x(n) = a n u(n) X(z) = az ROC: z > a if we set a =, then this is the transform pair for the unit step signal x(n) = u(n) X(z) = z ROC: z > This is a method that is used to get transform pairs without having to do all the intricate details, as we shall see one may also derive transform pairs using the properties of the z-transform. Let us consider a signal that has both anti-causal and causal components such as: x(n) = a n u(n) + b n u(-n-) ECE-S352 DSP 8 of
9 here the right term is multiplied by a folded shifted step to position it in the proper location in time - X(z) = a n z -n + b n z -n = (az - ) n + (b - z) l n = 0 n = - n = 0 l = left summation: ROC z > a right summation: ROC z < b Consider multiple cases b < a b > a a b b a there is no overlap and X(z) does not exist annulus (overlap) and X(z) is given as: X(z) = az bz - b a a + b z abz = ROC a < z < b In general for an infinite duration two sided signal the ROC is an annular region in the z-plane. There are possible combinations of Causal, Anti-causal and Two-sided signals with finite and infinite duration. The following table shows the ROCs ECE-S352 DSP 9 of
10 ECE-S352 DSP 0 of
11 Unilateral or One-sided Z-Transform There is a special case of the z-transform called the unilateral or one-sided z- transform (to distinguish it from the bilateral or two-sided z-transform). In this case: X + (z) x(n)z -n n = 0 when x(n) is causal (that is x(n) = 0 for n <0) then the unilateral and bilateral z-transforms are equivalent, in any other case they are different. Some care must be taken with properties and solution of difference equations depending on the type of z-transform. ECE-S352 DSP of
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