Digital Signal Processing

Digital Signal Processing The -Transform and Its Application to the Analysis of LTI Systems Moslem Amiri, Václav Přenosil Embedded Systems Laboratory Faculty of Informatics, Masaryk University Brno, Cech Republic amiri@mail.muni.c prenosil@fi.muni.c March, 2012

The Direct -Transform -transform of x(n) is defined as power series: X () where is a complex variable For convenience n= x(n) n X () Z{x(n)} x(n) X () Since -transform is an infinite power series, it exists only for those values of for which this series converges Region of convergence (ROC) of X () is set of all values of for which X () attains a finite value Any time we cite a -transform, we should also indicate its ROC ROC of a finite-duration signal is entire -plane except possibly points = 0 and/or = 2 / 63

The Direct -Transform Example Determine -transforms of following finite-duration signals x(n) = {1, 2, 5, 7, 0, 1} X () = 1 + 2 1 + 5 2 + 7 3 + 5 ROC: entire -plane except = 0 x(n) = {1, 2, 5, 7, 0, 1} X () = 2 + 2 + 5 + 7 1 + 3 ROC: entire -plane except = 0 and = x(n) = δ(n) X () = 1 [i.e., δ(n) 1], ROC: entire -plane x(n) = δ(n k), k > 0 X () = k [i.e., δ(n k) k ], ROC: entire -plane except = 0 x(n) = δ(n + k), k > 0 X () = k [i.e., δ(n + k) k ], ROC: entire -plane except = 3 / 63

The Direct -Transform Example Determine -transform of x(n) = ( 1 2 )n u(n) -transform of x(n) X () = 1 + 1 2 1 + ( 1 2 )2 2 + ( 1 2 )n n + = ( 1 2 )n n = n=0 ( 1 2 1 ) n n=0 For 1 2 1 < 1 or > 1 2, X () converges to X () = 1 1 1 2 1, ROC: > 1 2 4 / 63

The Direct -Transform Expressing complex variable in polar form = re jθ where r = and θ = X () = n= In ROC of X (), X () < X () = x(n)r n e jθn X () n= 1 n= n=0 x(n)r n e jθn n= x(n)r n e jθn = n=1 n= x(n)r n + x(n) r n x( n)r n + n=0 x(n)r n x(n) r n If X () converges in some region of complex plane, both sums must be finite in that region 5 / 63

The Direct -Transform Figure 1: Region of convergence for X () and its corresponding causal and anticausal components. 6 / 63

The Direct -Transform Example Determine -transform of We have x(n) = α n u(n) x(n) = α n u(n) = X () = α n n = n=0 X () = { α n, n 0 0, n < 0 (α 1 ) n n=0 1, ROC: > α 1 α 1 7 / 63

The Direct -Transform Example (continued) Figure 2: The exponential signal x(n) = α n u(n) (a), and the ROC of its -transform (b). 8 / 63

The Direct -Transform Example Determine -transform of { x(n) = α n 0, n 0 u( n 1) = α n, n 1 We have where l = n X () = 1 n= ( α n ) n = (α 1 ) l l=1 α 1 x(n) = α n u( n 1) X () = 1 α 1 = 1 1 α 1 ROC: < α 9 / 63

The Direct -Transform Example (continued) Figure 3: Anticausal signal x(n) = α n u( n 1) (a), and the ROC of its -transform (b). 10 / 63

The Direct -Transform From two preceding examples Z{α n u(n)} = Z{ α n u( n 1)} = 1 1 α 1 This implies that a closed-form expression for -transform does not uniquely specify the signal in time domain Ambiguity can be resolved if ROC is also specified A signal x(n) is uniquely determined by its -transform X () and region of convergence of X () ROC of a causal signal is exterior of a circle of some radius r 2 ROC of an anticausal signal is interior of a circle of some radius r 1 11 / 63

The Direct -Transform Example Determine -transform of We have X () = α n n + n=0 x(n) = α n u(n) + b n u( n 1) 1 n= b n n = (α 1 ) n + (b 1 ) l n=0 First sum converges if > α, second sum converges if < b If b < α, X () does not exist If b > α X () = ROC: α < < b l=1 1 1 α 1 1 1 b 1 = b α α + b αb 1 12 / 63

The Direct -Transform Example (continued) Figure 4: ROC for -transform in the example. 13 / 63

The Direct -Transform Table 1: Characteristic families of signals with their corresponding ROCs 14 / 63

Properties of the -Transform Combining several -transforms, ROC of overall transform is, at least, intersection of ROCs of individual transforms Linearity If x 1 (n) X 1 () and then x 2 (n) x(n) = a 1 x 1 (n) + a 2 x 2 (n) X 2 () X () = a 1 X 1 () + a 2 X 2 () for any constants a 1 and a 2 To find -transform of a signal, express it as a sum of elementary signals whose -transforms are already known 15 / 63

Properties of the -Transform Example Determine -transform and ROC of Defining signals x 1 (n) and x 2 (n) According to linearity property Recall that x(n) = [3(2 n ) 4(3 n )]u(n) x 1 (n) = 2 n u(n) and x 2 (n) = 3 n u(n) x(n) = 3x 1 (n) 4x 2 (n) α n u(n) X () = 3X 1 () 4X 2 () 1 1 α 1, ROC: > α Setting α = 2 and α = 3 X 1 () = 1, ROC: > 2 and X 1 2 1 2 () = 1, ROC: > 3 1 3 1 Intersecting ROCs, overall transform is X () = 3 4, ROC: > 3 1 2 1 1 3 1 16 / 63

Properties of the -Transform Time shifting If x(n) X () then x(n k) k X () ROC of k X () is same as that of X () except for = 0 if k > 0 and = if k < 0 17 / 63

Properties of the -Transform Scaling in -domain If x(n) X (), ROC: r 1 < < r 2 then a n x(n) X (a 1 ), ROC: a r 1 < < a r 2 for any constant a, real or complex Proof Z{a n x(n)} = n= a n x(n) n = n= Since ROC of X () is r 1 < < r 2, ROC of X (a 1 ) is r 1 < a 1 < r 2 or a r 1 < < a r 2 x(n)(a 1 ) n = X (a 1 ) 18 / 63

Properties of the -Transform Time reversal If then Proof x(n) x( n) Z{x( n)} = X (), ROC: r 1 < < r 2 X ( 1 ), ROC: 1 r 2 < < 1 r 1 n= x( n) n = l= x(l)( 1 ) l = X ( 1 ) where l = n ROC of X ( 1 ) r 1 < 1 < r 2 or 1 r 2 < < 1 r 1 19 / 63

Properties of the -Transform Differentiation in -domain If then Proof differentiating both sides dx () d = n= x(n) nx(n) X () = X () dx () d n= x(n) n x(n)( n) n 1 = 1 = 1 Z{nx(n)} Both transforms have same ROC n= [nx(n)] n 20 / 63

Properties of the -Transform Convolution of two sequences If x 1 (n) x 2 (n) then X 1 () X 2 () x(n) = x 1 (n) x 2 (n) X () = X 1 ()X 2 () ROC of X () is, at least, intersection of that for X 1 () and X 2 () Proof: convolution of x 1 (n) and x 2 (n) x(n) = x 1 (k)x 2 (n k) -transform of x(n) X () = x(n) n = = n= k= x 1 (k) = X 2 ()X 1 () [ n= k= n= [ k= x 1 (k)x 2 (n k) x 2 (n k) n ] = X 2 () ] k= n x 1 (k) k 21 / 63

Properties of the -Transform Example Compute convolution x(n) of signals x 1 (n) = {1, 2, 1} and x 2 (n) = -transforms of these signals X 1 () = 1 2 1 + 2 { 1, 0 n 5 0, elsewhere X 2 () = 1 + 1 + 2 + 3 + 4 + 5 Convolution of two signals is equal to multiplication of their transforms Hence X () = X 1 ()X 2 () = 1 1 6 + 7 x(n) = {1, 1, 0, 0, 0, 0, 1, 1} 22 / 63

Properties of the -Transform Convolution property is one of most powerful properties of -transform It converts convolution of two signals (time domain) to multiplication of their transforms Computation of convolution of two signals using -transform 1 Compute -transforms of signals to be convolved X 1 () = Z{x 1 (n)} X 2 () = Z{x 2 (n)} 2 Multiply the two -transforms X () = X 1 ()X 2 () 3 Find inverse -transform of X () x(n) = Z 1 {X ()} 23 / 63

Properties of the -Transform Correlation of two sequences If x 1 (n) x 2 (n) then r x1x 2 (l) = n= x 1 (n)x 2 (n l) X 1 () X 2 () R x1x 2 () = X 1 ()X 2 ( 1 ) Proof r x1x 2 (l) = x 1 (l) x 2 ( l) Using convolution and time-reversal properties R x1x 2 () = Z{x 1 (l)}z{x 2 ( l)} = X 1 ()X 2 ( 1 ) ROC of R x1x 2 () is at least intersection of that for X 1 () and X 2 ( 1 ) 24 / 63

Properties of the -Transform The initial value theorem If x(n) is causal (x(n) = 0 for n < 0), then Proof: since x(n) is causal X () = x(0) = lim X () x(n) n = x(0) + x(1) 1 + x(2) 2 + n=0 as, n 0 and hence X () = x(0) 25 / 63

Properties of the -Transform Table 2: Some common -transform pairs Signal, x(n) -Transform, X () ROC δ(n) 1 All 1 u(n) 1 1 > 1 a n 1 u(n) 1 a 1 > a na n u(n) a 1 (1 a 1 ) 2 a n 1 u( n 1) 1 a 1 na n a u( n 1) 1 (cos ω 0 n)u(n) (sin ω 0 n)u(n) (a n cos ω 0 n)u(n) (a n sin ω 0 n)u(n) (1 a 1 ) 2 1 1 cos ω 0 > a < a < a 1 2 1 cos ω 0 + 2 > 1 1 sin ω 0 1 2 1 cos ω 0 + 2 > 1 1 a 1 cos ω 0 1 2a 1 cos ω 0 +a 2 2 a 1 sin ω 0 1 2a 1 cos ω 0 +a 2 2 > a > a 26 / 63

Rational -Transforms An important family of -transforms are those for which X () is a rational function X () is a ratio of two polynomials in 1 (or ) Some important issues of rational -transforms are discussed here 27 / 63

Poles and Zeros Zeros of a -transform X () are values of for which X () = 0 Poles of a -transform are values of for which X () = If X () is a rational function (and if a 0 0 and b 0 0) X () = B() A() = b 0 + b 1 1 + + b M M a 0 + a 1 1 + + a N N = = b 0 M M + (b 1 /b 0 ) M 1 + + b M /b 0 a 0 N N + (a 1 /a 0 ) N 1 + + a N /a 0 = b 0 M+N ( 1)( 2 ) ( M ) a 0 ( p 1 )( p 2 ) ( p N ) = G N M M k=1 ( k) N k=1 ( p k) M k=0 b k k N k=0 a k k X () has M finite eros at = 1, 2,..., M N finite poles at = p 1, p 2,..., p N N M eros if N > M or poles if N < M at = 0 X () has exactly same number of poles as eros 28 / 63

Poles and Zeros Example Determine pole-ero plot for signal x(n) = a n u(n), a > 0 From Table 2 X () = 1 1 a 1 = a, ROC: > a X () has one ero at 1 = 0 and one pole at p 1 = a Figure 5: Pole-ero plot for the causal exponential signal x(n) = a n u(n). 29 / 63

Poles and Zeros Example Determine pole-ero plot for signal { a x(n) = n, 0 n M 1 0, elsewhere where a > 0 -transform of x(n) X () = M 1 n=0 M 1 a n n = n=0 Since a > 0, M = a M has M roots at (a 1 ) n = 1 (a 1 ) M 1 a 1 = M a M M 1 ( a) k = ae j2πk/m, k = 0, 1,..., M 1 30 / 63

Poles and Zeros Example (continued) Zero 0 = a cancels pole at = a. Thus X () = ( 1)( 2 ) ( M 1 ) M 1 which has M 1 eros and M 1 poles Figure 6: Pole-ero pattern for the finite-duration signal x(n) = a n, 0 n M 1 (a > 0), for M = 8. 31 / 63

Poles and Zeros Example Determine -transform and signal corresponding to following pole-ero plot Figure 7: Pole-ero pattern. 32 / 63

Poles and Zeros Example (continued) We use X () = G N M M k=1 ( k) N k=1 ( p k) There are two eros (M = 2) at 1 = 0, 2 = r cos ω 0 There are two poles (N = 2) at p 1 = re jω 0, p 2 = re jω 0 X () = G ( 1)( 2 ) ( p 1 )( p 2 ) = G ( r cos ω 0 ) ( re jω 0 )( re jω 0) 1 r 1 cos ω 0 = G 1 2r 1 cos ω 0 + r 2, ROC: > r 2 From Table 2 we find that x(n) = G(r n cos ω 0 n)u(n) 33 / 63

Poles and Zeros -transform X () is a complex function of complex variable X () is a real and positive function of Since represents a point in complex plane, X () is a surface -transform 1 2 X () = 1 1.2732 1 + 0.81 2 has one ero at 1 = 1 and two poles at p 1, p 2 = 0.9e ±jπ/4 Figure 8: Graph of X () for the above -transform. 34 / 63

Pole Location & Time-Domain Behavior for Causal Signals Characteristic behavior of causal signals depends on whether poles of transform are contained in region < 1 or > 1 or on circle = 1 Circle = 1 is called unit circle If a real signal has a -transform with one pole, this pole has to be real The only such signal is the real exponential x(n) = a n u(n) 1 X () =, ROC: > a 1 a 1 having one ero at 1 = 0 and one pole at p 1 = a on real axis 35 / 63

Pole Location & Time-Domain Behavior for Causal Signals Figure 9: Time-domain behavior of a single-real-pole causal signal as a function of the location of the pole with respect to the unit circle. 36 / 63

Pole Location & Time-Domain Behavior for Causal Signals A causal real signal with a double real pole has the form x(n) = na n u(n) X () = a 1 (1 a 1, ROC: > a ) 2 In contrast to single-pole signal, a double real pole on unit circle results in an unbounded signal 37 / 63

Pole Location & Time-Domain Behavior for Causal Signals Figure 10: Time-domain behavior of causal signals corresponding to a double (m = 2) real pole, as a function of the pole location. 38 / 63

Pole Location & Time-Domain Behavior for Causal Signals Configuration of poles as a pair of complex-conjugates results in an exponentially weighted sinusoidal signal x(n) = (a n cos ω 0 n)u(n) x(n) = (a n sin ω 0 n)u(n) X () = ROC: > a X () = ROC: > a 1 a 1 cos ω 0 1 2a 1 cos ω 0 + a 2 2 a 1 sin ω 0 1 2a 1 cos ω 0 + a 2 2 Distance r of poles from origin determines envelope of sinusoidal signal Angle ω 0 with real positive axis determines relative frequency 39 / 63

Pole Location & Time-Domain Behavior for Causal Signals Figure 11: A pair of complex-conjugate poles corresponds to causal signals with oscillatory behavior. 40 / 63

Pole Location & Time-Domain Behavior for Causal Signals Figure 12: Causal signal corresponding to a double pair of complex-conjugate poles on the unit circle. In summary Causal real signals with simple real poles or simple complex-conjugate pairs of poles, inside or on unit circle, are always bounded in amplitude A signal with a pole, or a complex-conjugate pair of poles, near origin decays more rapidly than one near (but inside) unit circle Thus, time behavior of a signal depends strongly on location of its poles relative to unit circle Zeros also affect behavior of a signal but not as strongly as poles E.g., for sinusoidal signals, presence and location of eros affects only their phase 41 / 63

Inversion of the -Transform There are three methods for evaluation of inverse -transform 1 Direct evaluation by contour integration 2 Expansion into a series of terms, in variables and 1 3 Partial-fraction expansion and table lookup Inverse -transform by contour integration x(n) = 1 X () n 1 d 2πj C Integral is a contour integral over a closed path C C encloses origin and lies within ROC of X () Figure 13: Contour C. 42 / 63

The Inverse -Transform by Power Series Expansion Given X () with its ROC, expand it into a power series of form X () = c n n n= By uniqueness of -transform, x(n) = c n for all n When X () is rational, expansion can be performed by long division Long division method becomes tedious when n is large Although this method provides a direct evaluation of x(n), a closed-form solution is not possible Hence this method is used only for determining values of first few samples of signal 43 / 63

The Inverse -Transform by Power Series Expansion Example Determine inverse -transform of X () = 1 1 1.5 1 + 0.5 2 1 When ROC: > 1 2 When ROC: < 0.5 ROC: > 1 Since ROC is exterior of a circle, x(n) is a causal signal. Thus we seek negative powers of by dividing numerator of X () by its denominator X () = 1 1 3 2 1 + 1 = 1 + 3 2 2 2 1 + 7 4 2 + 15 8 3 + 31 16 4 + x(n) = {1, 3 2, 7 4, 15 8, 31 16,...} 44 / 63

The Inverse -Transform by Power Series Expansion Example (continued) ROC: < 0.5 Since ROC is interior of a circle, x(n) is anticausal. To obtain positive powers of, write the two polynomials in reverse order and then divide X () = 1 1 3 2 1 + 1 2 2 = 22 + 6 3 + 14 4 + 30 5 + 62 6 + x(n) = { 62, 30, 14, 6, 2, 0, 0 } 45 / 63

The Inverse -Transform by Partial-Fraction Expansion In table lookup method, express X () as a linear combination X () = α 1 X 1 () + α 2 X 2 () + + α K X K () X 1 (),..., X K () are expressions with inverse transforms x 1 (n),..., x K (n) available in a table of -transform pairs Using linearity property x(n) = α 1 x 1 (n) + α 2 x 2 (n) + + α K x K (n) If X () is a rational function X () = B() A() = b 0 + b 1 1 + + b M M a 0 + a 1 1 + + a N N dividing both numerator and denominator by a 0 X () = B() A() = b 0 + b 1 1 + + b M M 1 + a 1 1 + + a N N This form of rational function is called proper if a N 0 and M < N 46 / 63

The Inverse -Transform by Partial-Fraction Expansion An improper rational function (M N) can always be written as sum of a polynomial and a proper rational function Example Express improper rational function X () = 1 + 3 1 + 11 6 2 + 1 3 3 1 + 5 6 1 + 1 6 2 in terms of a polynomial and a proper function Terms 2 and 3 should be eliminated from numerator Do long division with the two polynomials written in reverse order Stop division when order of remainder becomes 1 X () = 1 + 2 1 + 1 6 1 1 + 5 6 1 + 1 6 2 47 / 63

The Inverse -Transform by Partial-Fraction Expansion Any improper rational function (M N) can be expressed as X () = B() A() = c 0 + c 1 1 + + c M N (M N) + B 1() A() Inverse -transform of the polynomial can easily be found by inspection We focus our attention on inversion of proper rational transforms 1 Perform a partial fraction expansion of proper rational function 2 Invert each of the terms 48 / 63

The Inverse -Transform by Partial-Fraction Expansion Let X () be a proper rational function (a N 0 and M < N) X () = B() A() = b 0 + b 1 1 + + b M M 1 + a 1 1 + + a N N Eliminating negative powers of Since N > M, the function X () = b 0 N + b 1 N 1 + + b M N M N + a 1 N 1 + + a N X () = b 0 N 1 + b 1 N 2 + + b M N M 1 N + a 1 N 1 + + a N is also proper To perform a partial-fraction expansion, this function should be expressed as a sum of simple fractions First factor denominator polynomial into factors that contain poles p 1, p 2,..., p N of X () 49 / 63

The Inverse -Transform by Partial-Fraction Expansion Distinct poles Suppose poles p 1, p 2,..., p N are all different. We seek expansion X () = A 1 p 1 + A 2 p 2 + + A N p N To determine coefficients A 1, A 2,..., A N, multiply both sides by each of terms ( p k ), k = 1, 2,..., N, and evaluate resulting expressions at corresponding pole positions, p 1, p 2,..., p N ( p k )X () = ( p k)a 1 p 1 A k = ( p k)x (), =pk + + A k + + ( p k)a N p N k = 1, 2,..., N 50 / 63

The Inverse -Transform by Partial-Fraction Expansion Example Determine partial-fraction expansion of X () = 1 + 1 1 1 + 0.5 2 Eliminate negative powers by multiplying by 2 X () = + 1 2 + 0.5 p 1 = 1 2 + j 1 2 and p 2 = 1 2 j 1 2 p 1 p 2 X () A 1 = ( p 1)X () A 2 = ( p 2)X () = + 1 ( p 1 )( p 2 ) = A 1 + A 2 p 1 p 2 1 2 p 2 = + j 1 2 + 1 1 =p1 2 + j 1 2 1 2 + j 1 = 1 2 j 3 2 2 = + 1 =p1 = + 1 =p2 p 1 = =p2 1 2 j 1 2 + 1 1 2 j 1 2 1 2 j 1 2 = 1 2 + j 3 2 Complex-conjugate poles result in complex-conjugate coefficients 51 / 63

The Inverse -Transform by Partial-Fraction Expansion Multiple-order poles If X () has a pole of multiplicity m (there is factor ( p k ) m in denominator), partial-fraction expansion must contain the terms A 1k + A 2k p k ( p k ) 2 + + A mk ( p k ) m Coefficients {A ik } can be evaluated through differentiation 52 / 63

The Inverse -Transform by Partial-Fraction Expansion Example Determine partial-fraction expansion of X () = 1 (1 + 1 )(1 1 ) 2 Expressing in terms of positive powers of X () = 2 ( + 1)( 1) 2 X () has a simple pole at p 1 = 1 and a double pole at p 2 = p 3 = 1 X () = 2 ( + 1)( 1) 2 = A 1 + 1 + A 2 1 + A 3 ( 1) 2 A 1 = ( + 1)X () = 1 = 1 4 53 / 63

The Inverse -Transform by Partial-Fraction Expansion Example (continued) A 3 = ( 1)2 X () = 1 =1 2 To obtain A 2 ( 1) 2 X () = ( 1)2 + 1 A 1 + ( 1)A 2 + A 3 Differentiating both sides and evaluating at = 1, A 2 is obtained [ ( 1) 2 ] X () A 2 = d d =1 = 3 4 54 / 63

The Inverse -Transform by Partial-Fraction Expansion Having performed partial-fraction expansion, final step in inversion is as follows If poles are distinct X () = A 1 + A 2 + + p 1 p 2 p N 1 X () = A 1 1 p 1 1 + A 1 2 1 p 2 1 + + A 1 N 1 p N 1 x(n) = Z 1 {X ()} is obtained by inverting each term and taking the corresponding linear combination From table 2 { } { Z 1 1 (pk ) 1 p k 1 = n u(n), if ROC: > p k (causal) (p k ) n u( n 1), if ROC: < p k (anticausal) If x(n) is causal, ROC is > p max, where p max = max{ p 1, p 2,..., p N } In this case all terms in X () result in causal signal components A N x(n) = (A 1 p n 1 + A 2 p n 2 + + A N p n N)u(n) 55 / 63

The Inverse -Transform by Partial-Fraction Expansion If all poles are distinct but some of them are complex, and if signal x(n) is real, complex terms can be reduced into real components If p j is a pole, its complex conjugate pj is also a pole If x(n) is real, the polynomials in X () have real coefficients If a polynomial has real coefficients, its roots are either real or occur in complex-conjugate pairs Their corresponding coefficients in partial-fraction expansion are also complex-conjugates Contribution of two complex-conjugate poles is x k (n) = [A k (p k ) n + A k (p k )n )]u(n) Expressing A j and p j in polar form A k = A k e jα k and p k = r k e jβ k which gives x k (n) = A k rk n[ej(β k n+α k ) + e j(β k n+α k ) ]u(n) or x k (n) = 2 A k rk n cos(β kn + α k )u(n) Thus ( Z 1 A k 1 p k 1 + A ) k 1 pk = 2 A k r n 1 k cos(β k n + α k )u(n) if ROC is > p k = r k 56 / 63

The Inverse -Transform by Partial-Fraction Expansion In case of multiple poles, either real or complex, inverse transform of terms of the form A/( p k ) n is required In case of a double pole, from table 2 { Z 1 p 1 } (1 p 1 ) 2 = np n u(n) provided that ROC is > p In case of poles with higher multiplicity, multiple differentiation is used 57 / 63

The Inverse -Transform by Partial-Fraction Expansion Example Determine inverse -transform of X () = 1 If ROC: > 1 2 If ROC: < 0.5 3 If ROC: 0.5 < < 1 Partial-fraction expansion for X () 2 1 1 1.5 1 + 0.5 2 p 1=1 X () = p 2=0.5 X () = 2 1.5 + 0.5 ( 1)X () A 1 = = 2 =1 ( 0.5)X () A 2 = = 1 =0.5 ( 1)( 0.5) = A 1 1 + A 2 0.5 58 / 63

The Inverse -Transform by Partial-Fraction Expansion Example (continued) X () = 2 1 1 1 1 0.5 1 When ROC is > 1, x(n) is causal and both terms in X () are causal 1 1 p k 1 (p k ) n u(n) x(n) = 2(1) n u(n) (0.5) n u(n) = (2 0.5 n )u(n) When ROC is < 0.5, x(n) is anticausal and both terms in X () are anticausal 1 1 p k 1 (p k ) n u( n 1) x(n) = [ 2 + (0.5) n ]u( n 1) 59 / 63

The Inverse -Transform by Partial-Fraction Expansion Example (continued) When ROC is 0.5 < < 1 (ring), signal x(n) is two-sided One of the terms corresponds to a causal signal and the other to an anticausal signal Since the ROC is overlapping of > 0.5 and < 1, pole p 2 = 0.5 provides causal part and pole p 1 = 1 anticausal x(n) = 2(1) n u( n 1) (0.5) n u(n) 60 / 63

The Inverse -Transform by Partial-Fraction Expansion Example Determine causal signal x(n) whose -transform is X () = 1 + 1 1 1 + 0.5 2 We have already obtained partial-fraction expansion as X () = A 1 1 p 1 1 + A 2 1 p 2 1 A 1 = A 2 = 1 2 j 3 2 and p 1 = p 2 = 1 2 +j 1 2 For a pair of complex-conjugate poles (A k = A k e jα k and p k = r k e jβ k ) ( Z 1 A k 1 p k 1 + A ) k 1 pk = 2 A k r n 1 k cos(β k n + α k )u(n) A 1 = ( 10/2)e j71.565 and p 1 = (1/ 2)e jπ/4 x(n) = 10(1/ 2) n cos(πn/4 71.565 )u(n) 61 / 63

The Inverse -Transform by Partial-Fraction Expansion Example Determine causal signal x(n) having -transform X () = 1 (1 + 1 )(1 1 ) 2 We have already obtained partial-fraction expansion as For causal signals X () = 1 1 4 1 + 1 + 3 1 4 1 1 + 1 2 (1 1 ) 2 1 1 p 1 (p) n u(n) and 1 p 1 (1 p 1 ) 2 np n u(n) x(n) = 1 4 ( 1)n u(n) + 3 4 u(n) + 1 2 nu(n) = [ 1 4 ( 1)n + 3 4 + n 2 ] u(n) 62 / 63

References John G. Proakis, Dimitris G. Manolakis, Digital Signal Processing: Principles, Algorithms, and Applications, Prentice Hall, 2006. 63 / 63