Ch.7 The z-Transform and Discrete-Time Systems

7.1 The z-Transform Definition: –Consider the DTFT: X(Ω) = Σ all n x[n]e -jΩn (7.1) –Now consider a real number ρ and the factor ρ n –Then X(z) = Σ all n x[n] ρ -n e -jΩn = Σ all n x[n]z -n –Where z is the complex number ρe -jΩ. –This is the two-sided z-transform. –The one-sided z-transform has n=0,…,∾ –In this text, the one-sided z-transform is assumed.

Example 7.1 z-transform of the Unit Pulse Let x[n] = δ(n) = {1 for n=0 and 0 otherwise}. Then X(z) = Σ n≥0 x[n]z -n =δ(0) z 0 = 1 (7.6) So: δ(0) ↔ 1 And the region of convergence is all z, where z is a complex number.

Example 7.2 z-transform of a Shifted Unit Pulse Let x[n] = δ(n-q) Then X(z) = Σ n≥0 x[n]z -n =δ(0) z -q = z -q So: δ(n-q) ↔ z -q =1/ z q And the region of convergence is all z except for z=0.

Example 7.3 Unit-Step Function Consider x[n] = u[n] = {1 n≥0; 0 otherwise} X(z) = U(z) = Σ n≥0 x[n]z -n = Σ n≥0 z -n So: U(z) = 1 + z -1 + z -2 … Now (z-1)U(z) = (z+1+z -1 + z -2 …) – (1+z -1 + z -2 …) And U(z) = z/(z-1) So: U(z) ↔ z/(z-1 )= 1/(1-z -1 ) And the region of convergence will be |z|>1.

Example 7.4 z-transform of a n u[n] Let x[n] = a n u[n] Then X(z) = Σ n≥0 a n z -n =(1 + az -1 +a 2 z -2 …) Now (z-a)X(z) =(z + a +a 2 z -1 …) - (a + a 2 z -1 +a 3 z -2 …) And X(z) = z/(z-a) So X(z) ↔ z/(z-a) = 1/(1-az -1 ) With the region of convergence |z|> |a| and a is any real or complex number.

7.1.1 Relationship between the DTFT and the z-transform X(Ω) = X(z) with z = e jΩ only when the region of convergence includes the unit circle, ie |z|=1.

Properties of the z-Transform Linearity—Example 7.5 –Let x[n] = u[n] and v[n] = a n u[n], where a≠1 –Now u(n) ⟷ z/(z-1) and a n u[n]⟷ z/(z-a) –So x[n] + v[n] ⟷ z/(z-1) + z/(z-a) = –(2z 2 – (1 + a)z/(z-1)(z-a)

Example 7.6 z-Transform of a Pulse Using the right shift property: – x[n-q]u[n-q] ⟷ z -q X(z) Let p[n] = u[n] – u[n-q] be a pulse of 1’s from n=0 to n= (q-1) and u[n-q]=1 n-q u[n-q] Now u(n)⟷z/(z-1) and u[n-q]⟷z -q (z/z-1) So p[n]⟷z/(z-1)-z -q (z/z-1)=(z q -1)/z q-1 (z-1)

Example 7.8 z-Transform of na n u[n] Let x[n] = na n u[n]; then a n u[n] ⟷ z/(z-a) Also, the property of “multiplication by n” – nx[n] ⟷ -z (d/dz X(z)) So taking the derivative – d/dz (z/z-a)=1/(z-a) –z/(z-a) 2 =-a/(z-a) 2 And using the properties: –n a n u[n] ⟷ -z (-a)/(z-a) 2 = az/(z-a) 2 –And for a=1 we have nu[n] ⟷ z /(z-1) 2

Example 7.11 z-Transform of Sinusoids Let v[n] = (cos Ωn) u[n] The property of multiplication by sinusoid –(cos Ωn)x(n) ⟷ (1/2) { X(e jΩ z) + X(e –jΩ z)} Using the property –After “manipulation” we have –(cos Ωn) u[n]⟷ z 2 – (cosΩ)z/(z 2 -(2cosΩ)z + 1) Similarly it can be shown: –(sin Ωn) u[n]⟷ (sinΩ)z/(z 2 -(2cosΩ)z + 1)

Other Properties Summation Convolution Initial-Value Theorem Final-Value Theorem

7.3 Computation of the Inverse z-Transform Complex integral, with integration taken along a counterclockwise close circular contour that is contained in the region of convergence. When X(z) is a rational function then we can be computed by expansion of X(z) into a power series in z -1.

Example 7.15 Inverse z- Transform via Long Division Let X(z) = (z 2 -1)/(z 3 + 2z +4) Then by long division we have: –X(z) = z -1 -3z -3 – 4 z -4 … And by the definition of the z-transform –x[0] = 0; x[1] = 1; x[2]= 0; x[3] =-3; x[4] =-4; …

Inversion via Partial Fraction Expansion Suppose X(z) = B(z)/A(z) is a rational function. If the degree of B(z) is equal to A(z) then dividing A(z) into B(z) yields X(z) = x[0] + R(z)/A(z). But also, X(z)/z = B(z)/zA(z), which can be expanded into partial fractions.

Distinct Poles Suppose the poles of X(z) are distinct: –X(z)/z = c 0 /z + c 1 /(z-p 1 ) + …+c N /(z-p N ) –Here c 0 = {z X(z)/z} evaluated at z=0. –Also c i = {(z-p i )X(z)/z) evaluated at z=p i –Then X(z) = c 0 + c 1 z/ (z-p 1 ) + …+zc N /(z-p N ) –And taking the inverse z-transform using the Table of Pairs x[n] = c 0 δ(0)+ c 1 p 1 n + …+c N p N n

Complex Poles If all the poles of X(z) are real, the terms comprising the signal are all real-valued. If two or more poles are complex, the corresponding terms will be complex-valued. These complex terms can be combined to yield real- valued terms; in this case the poles are complex conjugates. If p 1 =a +jb is a complex pole, and p 2 = a –jb is its complex conjugate then when added we have c 1 p 1 n + c 1 * p 1 *n which can be expressed as a real term: 2|c 1 |σ n cos(Ωn + ∟c 1 ). Here c 2 =c 1 *, σ = |p 1 |, and Ω=∟p 1.

Example 7.17 Complex Poles Let X(z) = (z 3 +1)/(z 3 – 2 –z -2) Use the MATLAB command roots to find the poles p 1 = -.05 – j0.866, p 2 = j0.866, and p 3 =2; or you could recognize that (z-2) is a factor of the denominator and then factor the denominator accordingly. Since the order of the numerator and denominator are equal, X(z)/z is expanded. c 0, c 1, c 2, and c 3 are found using the procedure outlined for “distinct poles”; also note c 2 =c 1 *. The final result is: –x[n] = -0.5δ(n) cos(4  /3 n º) (2) n

Repeated Poles Discussed on pages The residues (c i ’s) are computed in a similar fashion as shown in the distinct pole case. Example 7.18 (p ) illustrates the concepts.

Pole Location and the Form of the Signal Relationships between signal poles and terms is similar to that in the Laplace Transform Theory. In particular, x[n] converges to 0 as n→ ∾ if and only if all of the poles have magnitudes that are strictly less than 1 (lies inside the unit circle.)

7.4 Transfer Function Representation First – Order Case –Let y[n] + ay[n-1] = bx[n] –Then take the z-transform to get: Y(z) + a{ z -1 Y(z) + y[-1] }= b X(z) –Simplifying: Y(z) (1 + az -1 ) = b X(z)-a y[-1] Y(z) = (b X(z)-a y[-1])/ (1 + az -1 ) Y(z) ={ b X(z)/(1 + az -1 )} – {a y[-1]}/ (1 + az -1 ) –If y[-1] =0 then: Y(z) = b X(z)/(1 + az -1 )= z b X(z)/(z + a) –And we have the transfer function H(z): –Y(z) = H(z) X(z) so H(z) = b z /(z + a)

Example 7.19 Step Response of a First Order System Let x≠1 and x[n] = u[n]; Then X(z) = z/(z-1) Let Y(z)={bX(z)/(1+az -1 )} – {ay[-1]}/(1+az -1 ) from general result of a first-order system. Then : –Y(z) ={ (z/z-1)b/(1 + az -1 )} – {a y[-1]}/ (1 + az -1 ) –Y(z) ={ (z/z-1)bz/(z + a)} – {a y[-1]z}/ (z + a) –Y(z) ={ (bz 2 /(z-1)(z + a)} – {a y[-1]z}/ (z + a) Now, expand the first term: – {bz 2 /(z-1)(z + a)}(1/z) = {ab/(a+1)(z+a)}+{b/(a+1)(z-1)}

Example 7.19 (cont.) So we have: –Y(z) ={ (bz 2 /(z-1)(z + a)} – {a y[-1]z}/ (z + a) –Y(z) ={abz/(a+1)(z+a)}+{bz/(a+1)(z-1)}–{ay[-1]z}/(z+a) Now we have the following pairs: – u[n] ↔ z/(z-1) –a n u[n] ↔ z/(z-a)= z/(z+(-a)) So taking the inverse z-transform of Y(z): –y[n] = {ab/(a+1)(-a) n u[n] }+b/(a+1)u[n] – ay[-1](-a) n u[n] –y[n] = {b/(a+1)}{-(-a) n+1 +1} – ay[-1](-a) n n=0,1,2,… And if y[-1] =0 then –{b/(a+1)}{-(-a) n+1 +1} n=0,1,2…. (text is different)

7.4.2 Second-Order Case Consider the following difference equation: –Y[n] +a 1 y[n-1] +a 2 y[n-2] = b 0 x[n] + b 1 x[n-1] –Result for y[n-1]=y[n-2] =0: H(z) = (b 0 z 2 + b 1 z)/(z 2 + a 1 z + a 2 )

Example 7.20 Second-Order System The general result for the second order system is given by equation For the case with the a 1 = 1.5, a 2 =.5, b 0 =1 and b 1 =-1 we have –y[n] = 0.5(-.05) n - 3(-1) n n=0,1,2,3,… Note that the first term dies off but the second term continues (see fig. 7.2).

Nth Order Case y[n] +a 1 y[n-1] +…+a N y[n-N] =b 0 x[n]+…+b M x[n-M] And H(z) = B(z)/A(z) Where H(z) =(b 0 +…+b M z -M )/(1 + a 1 z -1 + …+a N z -N ) Now multiply by (z N /z N ) H(z) = (b 0 z N +…+b M z N-M )/ (z N + a 1 z N-1 + …+ a N ) So we have –B(z) = (b 0 z N +…+b M z N-M ) –A(z) = (z N + a 1 z N-1 + …+ a N )

7.4.4 Transform of the Input/Output Convolution Sum Y(z) = H(z) X(z) where h[n] ↔ H(z). Let H(z) = B(z)/A(z) Then Y(z) = {B(z)/A(z)} X(z) A causal liner time-invariant discrete-time system is said to be finite dimensional if the transfer function H(z) is a rational function of z.

Example 7.21 Computation of aTransfer Function Let h[n] = 3(2 -n )cos(  n/6 +  /12) n=0,1,2,… Note that cos(a+b) = cos(a)cos(b) – sin(a)sin(b). So h[n] = 3(2 -n ){cos(  n/6) cos(  /12) - sin(  n/6)sin(  /12)} n=0,1,2,… And h[n] = 2.898(2 -n )cos(  /12) – 0.776(2 -n )sin(  n/6), n=0,1,2,… Taking the z-transform of h[n] gives: H(z) = (2.898 z 2 – 1.449z)/(z z +.25)

7.22 Computation of Step Response If x[n] = u[n] then X(z) = z/(z-1). Since Y(z) = H(z) X(z) the step response can be calculated as Y(z) = H(z) {z/(z-1)}. Page gives the example of the step response for the system from example Figure 7.3 illustrates the plot of the step response.

Transfer Function of Interconnections Unit Delay –y[n] = x[n-1] –Y(z) = z -1 X(z)

Example 7.23 Computation of the Transfer Function of Interconnections Consider a discrete time system given by Figure From the figure: zQ 1 (z) = Q 2 (z) + X(z) zQ 2 (z) = Q 1 (z) – 3Y(z) Y(z) = 2Q 1 (z) + Q 2 (z)

Example 7.23 (cont.) Solving for Y(z) from these equations gives: Y(z) = {(2z -1 +1)/(z – z -1 ) } {z -1 X(z) – 3Y(z)} + 2z -1 X(z) Then it can be shown that: H(z) = (2z+1)/(z 2 +3z +5)

7.5 System Analysis Using the Transfer Function Representation Consider the general linear time-invariant discrete-time system with transfer function H(z). Let H(Z) = B(z)/A(z) where B(z) is an Mth-order polynomial and A(z) is an Nth- order polynomial and M≤ N and the polynomials do not have any common factors.

7.5 (cont.) The time variation of h[n] is directly determined by the poles of the system. The system is considered stable if its unit-pulse response, h[n], converges to zero as n tends to infinity. To be stable, the poles must lie inside the “open” unit disk, in the z-plane. The system is marginally stable if the unit-pulse is bounded (non-repeated poles could lie on the unit circle. The system is unstable if the magnitude of the unit pulse function grows without bound (one or more poles located outside the unit circle or one or more repeated poles on the unit circle).

7.5.1 Response to a Sinusoidal Input Let x[n] = C cos (Ω 0 n), n = 0, ±1, ±2, … Then y[n] = C|H(Ω 0 )| cos(Ω 0 n + ∟H(Ω 0 ) ), n = 0, ±1, ±2, … Note: this is the sinusoidal steady state; see page 392 for the “transient response”.