#### Transcript Ch. 7 Z-Transforms

```Ch.7 The z-Transform and
Discrete-Time Systems
7.1 The z-Transform
• Definition:
–
–
–
–
–
–
–
Consider the DTFT: X(Ω) = Σall nx[n]e -jΩn (7.1)
Now consider a real number ρ and the factor ρn
Then X(z) = Σall nx[n] ρ-n e -jΩn = Σall nx[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) = Σn≥0x[n]z-n =δ(0) z0 = 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) = Σn≥0x[n]z-n =δ(0) z-q = z-q
So: δ(n-q) ↔ z-q =1/ zq
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) = Σn≥0x[n]z-n = Σn≥0z-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 anu[n]
• Let x[n] = anu[n]
• Then X(z) = Σn≥0 anz-n =(1 + az-1 +a2 z-2 …)
• Now (z-a)X(z) =(z + a +a2 z-1 …) - (a + a2z-1
+a3 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 = ejΩ 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] = anu[n], where a≠1
– Now u(n) ⟷ z/(z-1) and anu[n]⟷ z/(z-a)
– So x[n] + v[n] ⟷ z/(z-1) + z/(z-a) =
– (2z2 – (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]=1n-qu[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)=(zq-1)/zq-1(z-1)
Example 7.8 z-Transform of
nanu[n]
• Let x[n] = nanu[n]; then anu[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 anu[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(ejΩ z) + X(e–jΩ z)}
• Using the property
– After “manipulation” we have
– (cos Ωn) u[n]⟷ z2 – (cosΩ)z/(z2-(2cosΩ)z + 1)
• Similarly it can be shown:
– (sin Ωn) u[n]⟷ (sinΩ)z/(z2-(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 zTransform via Long Division
• Let X(z) = (z2 -1)/(z3 + 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 = c0/z + c1/(z-p1) + …+cN/(z-pN)
– Here c0 = {z X(z)/z} evaluated at z=0.
– Also ci = {(z-pi)X(z)/z) evaluated at z=pi
– Then X(z) = c0 + c1z/ (z-p1) + …+zcN/(z-pN)
– And taking the inverse z-transform using the
Table of Pairs
• x[n] = c0 δ(0)+ c1p1n + …+cNpNn
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 realvalued terms; in this case the poles are complex
conjugates.
• If p1=a +jb is a complex pole, and p2 = a –jb is its
complex conjugate then when added we have c1p1n +
c1*p1*n which can be expressed as a real term:
2|c1|σn cos(Ωn + ∟c1).
• Here c2 =c1*, σ = |p1|, and Ω=∟p1.
Example 7.17 Complex Poles
• Let X(z) = (z3 +1)/(z3 – 2 –z -2)
• Use the MATLAB command roots to find the poles p1 = .05 – j0.866, p2 = -.5 + j0.866, and p3=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.
• c0, c1, c2, and c3 are found using the procedure outlined
for “distinct poles”; also note c2=c1*.
• The final result is:
– x[n] = -0.5δ(n) + 0.874 cos(4/3 n + 10.89º) + 0.643(2)n
Repeated Poles
• Discussed on pages 374-375.
• The residues (ci’s) are computed in a
similar fashion as shown in the distinct
pole case.
• Example 7.18 (p.375-376) 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-1Y(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) ={ (bz2/(z-1)(z + a)} – {a y[-1]z}/ (z + a)
• Now, expand the first term:
– {bz2/(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) ={ (bz2/(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)
– anu[n] ↔ z/(z-a)= z/(z+(-a))
• So taking the inverse z-transform of Y(z):
– y[n] = {ab/(a+1)(-a)nu[n] }+b/(a+1)u[n] – ay[-1](-a)nu[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] +a1y[n-1] +a2y[n-2] = b0x[n] + b1x[n-1]
– Result for y[n-1]=y[n-2] =0:
• H(z) = (b0z2 + b1z)/(z2 + a1z + a2)
Example 7.20 Second-Order
System
• The general result for the second order
system is given by equation 7.87.
• For the case with the a1 = 1.5, a2 = .5,
b0=1 and b1=-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] +a1y[n-1] +…+aNy[n-N] =b0x[n]+…+bMx[n-M]
And H(z) = B(z)/A(z)
Where H(z) =(b0+…+bMz-M)/(1 + a1z-1+ …+aNz-N)
Now multiply by (zN/zN)
H(z) = (b0zN +…+bMzN-M)/ (zN + a1zN-1 + …+ aN)
So we have
– B(z) = (b0zN +…+bMzN-M)
– A(z) = (zN + a1zN-1 + …+ aN)
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 z2 – 1.449z)/(z2 - .866z + .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 384-386 gives the example of the step
response for the system from example
7.21.
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 7.23.
• From the figure:
• zQ1(z) = Q2(z) + X(z)
• zQ2(z) = Q1(z) – 3Y(z)
• Y(z) = 2Q1(z) + Q2(z)
Example 7.23 (cont.)
• Solving for Y(z) from these equations gives:
• Y(z) = {(2z-1 +1)/(z – z-1) } {z-1X(z) – 3Y(z)}
+ 2z-1X(z)
• Then it can be shown that:
• H(z) = (2z+1)/(z2 +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 Nthorder 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 (Ω0n), n = 0, ±1, ±2, …
• Then y[n] = C|H(Ω0)| cos(Ω0n + ∟H(Ω0) ),
n = 0, ±1, ±2, …
• Note: this is the sinusoidal steady state;
see page 392 for the “transient response”.
```