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EECS 451 TRANSFER OR SYSTEM FUNCTIONS
DEF: H(z) = N (z)/D(z) = Y (z)/X(z) = B(z)/A(z) as defined below.
Poles: roots of D(z) = 0. ARMA:
a(i)y(n − i) =
b(j)x(n − j).
Zeros: roots of N (z) = 0. Impulse response: δ(n) → |h(n)| → h(n).
Transfer functions associated with zero initial conditions (ZSR).
I/O: Input x(n) → |h(n)| →output y(n) ⇐⇒ ︸︷︷︸
Y (z)/X(z)
H(z) ⇐⇒ ︸︷︷︸
Z{h(n)}
h(n)
ARMA:
a(i)y(n − i)=
b(j)x(n − j) ⇐⇒ ︸︷︷︸
B(z)/A(z)
H(z) ⇐⇒ ︸︷︷︸
C
z−zi z−pi
P-Z plot
EX #1: x(n) = (−2)
n u(n) → y(n) =
2
3
n u(n) +
1
3
u(n). Find h(n).
Soln: X(z) =
z
z+
.Y (z) =
2 z/ 3
z+
z/ 3
z− 1
z
2
(z+2)(z−1)
.H(z) =
z
z− 1
.h(n) = u(n).
First term of y(n)=forced response. Second term=natural response.
Difference eqn.:
Y (z)
X(z)
= H(z) =
z
z− 1
1
1 −z − 1 →^ y(n)−y(n−1) =^ x(n).
EX #2: Find difference equation implementing H(z) =
z
3 +7z
2
z 3 −z 2 +2z− 3
Soln: Write H(z) =
Y (z)
X(z)
1+7z
− 1
1 −z − 1 +2z − 2 − 3 z − 3 and^ cross-multiply:
Y (z)(1 − z
− 1
− 2 − 3 z
− 3 ) = X(z)(1 + 7z
− 1 ) →difference equation:
Z
− 1 y(n) − y(n − 1) + 2y(n − 2) − 3 y(n − 3) = x(n) + 7x(n − 1).
EX #3: Find step response of system with zero at 1, pole at 3, and H(0) = 1.
Soln: H(z) = 3
z− 1
z− 3
. U (z) =
z
z− 1
→ Y (z) = 3
z
z− 3
→ y(n) = 3 · 3
n u(n).
Y (z)
X(z)
= H(z) = 3
z− 1
z− 3
1 −z
− 1
1 − 3 z − 1 →^ y(n)−^3 y(n−1) = 3x(n)−^3 x(n−1).
Modes 6 = y(n) − 3 y(n − 1) + 2y(n − 2) = x(n) − x(n − 1). Modes: 1,2.
Poles: H(z) =
1 −z
− 1
1 − 3 z − 1 +2z
− 2 =^
1
1 − 2 z − 1. Poles: 2.^ {poles} ⊂ {modes}.
ZIR: y(n) = C 12
n u(n) + C 21
n u(n), depending on initial conditions.
ZSR: y(n) = C 32
n u(n) (natural response)+forced response∼ u(n).
If no pole-zero cancellation, then {poles} = {modes}.
BIBO LTI system BIBO stable iff
∞
n=−∞
|h(n)| < ∞ (is finite).
Stable Same as: {unit circle |z| = 1} ⊂ ROC of H(z) = Z{h(n)}.
Causal Causal LTI BIBO stable iff poles inside unit circle.
Anticausal BIBO stable iff poles outside unit circle.
APPLICATIONS OF THE z-TRANSFORM
Given: x[n] = 3
n u[n] → |y[n] − 2 y[n − 1] = x[n − 1] − x[n − 2]| → y[n]
Goal: Compute the response y[n] of the system to this particular input.
Z: Y (z) − 2 z
− 1 Y (z) = z
− 1 X(z) − z
− 2 X(z) and X(z) =
z
z− 3
here
→ Y (z) =
z
− 1 −z
− 2
1 − 2 z − 1
z
z− 3
z− 1
(z−2)(z−3)
2
z− 3
1
z− 2
[2 =
3 − 1
3 − 2
2 − 1
2 − 3
]
→ y[n] = [2(3)
n− 1 − (2)
n− 1 ]u[n − 1] =
FORCED
RESPONSE
like
x[n]
NATURAL
RESPONSE
like
h[n]
Given: x[n] = (
1
2
n u[n] → |LTI|^ →^ y[n] =^ {^0 ,^0 ,^1 }^ =^ δ[n^ −^ 2]
Goal: Compute the response of this system to 2 cos(
π
3
n).
H(z):
TRANSFER
FUNCTION
= H(z) = Z{y[n]}/Z{x[n]} = z
− 2 /[z/(z−
1
2
)] = (z−
1
2
)/z
3 .
h[n]:
IMPULSE
RESPONSE
= h[n] = Z
− 1 {(z−
1
2
)/z
3 } = Z
− 1 {z
− 2 −
1
2
z
− 3 } = { 0 , 0 , 1 , −
1
2
H(w):
FREQUENCY
RESPONSE
= H(ω) = H(z)|z=ejω^ = (e
jω −
1
2
)/e
j 3 ω .
ω =
π
3
: H(
π
3
) = [e
jπ/ 3 −
1
2
]/e
j 3 π/ 3 = (
1
2
√ 3
2
1
2
)/(−1) = −j
√ 3
2
√ 3
2
e
−jπ/ 2 .
Sol’n: 2 cos(
π
3
n) → |LTI| →
3 cos(
π
3
n −
π
2
3 sin(
π
3
n).
Given: x[n] → |y[n] = x[n] −
3
4
x[n − 1] +
1
8
x[n − 2]| → y[n]
Huh? x[n]=cell phone signal. y[n]=multipath due to buildings.
Goal: Compute the inverse filter that recovers x[n] from y[n]:
Huh? x[n] → |h[n]| → y[n] → |g[n]| → x[n]. That is, g[n] undoes h[n].
Idea: Systems in cascade (series) ⇔ h[n] ∗ g[n] = δ[n] ⇔ H(z)G(z) = 1.
Here: h[n] = { 1 , −
3
4
1
8
} → H(z) = 1 −
3
4
z
− 1
1
8
z
− 2 = (z
2 −
3
4
z +
1
8
)/z
2 .
→ G(z) = 1/H(z) = z
2 /[z
2 −
3
4
z +
1
8
] = z
2 /[(z −
1
2
)(z −
1
4
)].
Z
− 1 :
G(z)
z
z
(z− 1 /2)(z− 1 /4)
2
z− 1 / 2
1
z− 1 / 4
→ G(z) = 2
z
z− 1 / 2
z
z− 1 / 4
using: residues 2 = (1/2)/[(1/2) − (1/4)] and −1 = (1/4)/[(1/4) − (1/2)].
g[n]: g[n] = 2(
1
2
n u[n] − (
1
4
n u[n]=inverse filter for original system.
Note: Stable since zeros of H(z)=poles of G(z) are inside unit circle.
Note: g[0] 6 = 0 since both
numerator &
denominator
of G(z) have the same degrees.
