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EECS 216 LECTURE NOTES
FORWARD OR DIRECT Z-TRANSFORM
DEF: Z{x[n]} = X(z) =
n=0 x[n]z
−n (compare to L{x(t)} =
0 x(t)e
−st dt). Huh? Finite-duration signal x[n] →polynomial X(z) with coefficients=x[n]. Or: Infinite-duration x[n]=coefficients of Laurent (power) series X(z).
Finite x[n] = { 3 , 1 , 4 , 2 , 5 }. Note x[0] = 3 and finite duration=5→ length X(z) = 3 + 1z − 1
- 4z − 2
- 2z − 3
- 5z − 4 = (3z 4
- z 3
- 4z 2
- 2z + 5)/z 4 . signal Z{δ[n − D]} = z−D^ for any integer delay D ≥ 0. Z{δ[n]} = 1.
Expon- Z{anu[n]} =
n=0 a
nz−n (^) =
n=0(az
− (^1) )n (^) = 1/(1−az− (^1) ) = z/(z−a).
ential EX: Z{( 12 )nu[n]} = 1 1 − 12 z−^1
. EX: Z{(− 13 )nu[n]} = 1 1+ 13 z−^1
Sinu- Z{cos(ω 0 n)u[n]} = 1 2 Z{e
jω 0 nu[n]} + 1 2 Z{e
−jω 0 nu[n]} = (since linear)
soids 12 1 1 −ejω^0 z−^1
1 −e−jω^0 z−^1
1 −z−^1 cos(ω 0 ) 1 − 2 z−^1 cos(ω 0 )+z−^2 =^
z^2 −z cos(ω 0 ) z^2 − 2 z cos(ω 0 )+.
Linear: Z{{ 3 , 1 , 4 } + u[n]} = 3 z
(^2) +1z+ z^2 +^
z z− 1 =^
4 z^3 − 2 z^2 +3z− 4 z^3 −z^2 =^
RATIONAL FUNCTION.
Delay: If Z{x[n]} = X(z) and D ≥ 0, then Z{x[n − D]} = z−DX(z).
EX: Z{u[n] − u[n − 2]} = z z− 1 −^ z
− 2 z z− 1 =^
z z^2
z^2 − 1 z− 1 = 1 +^ z
Note: Note that u[n] − u[n − 2] = δ[n] + δ[n − 1], so these are consistent.
Also: Z{ 1 n u[n^ −^ 1]}^ =^
n=
z−n n =^ −^ log(1^ −^ z
− (^1) ) if you recognize this.
Note: Convolution↔polynomial multiplication: Z{x[n] ∗ y[n]} = X(z)Y (z).
Eigen- zn^ → |h[n]| → znH(z): zn^ in→ scaled zn^ out. H(z) =
n=0 h[n]z
−n.
funcs h[n]∗zn^ =
h[i]zn−i^ = zn^
h[i]z−i^ = znH(z). H(ejω^ ) = H(z)|z=ejω.
of LTI Note zn^ plays the same role that est^ plays in continuous time.
Note: H(z)=transfer function. H(ejω^ ) = H(z)|z=ejω =frequency response.
EX: Compute step response (to u[n]) of LTI system with h[n] = { 2 , − 3 , 1 }.
Soln: We need to compute y[n] = { 2 , − 3 , 1 } ∗ u[n]. Do this two ways:
#1: y[n] = (2δ[n] − 3 δ[n − 1] + 1δ[n − 2]) ∗ u[n]. Using u[n] ∗ δ[n] = u[n],
#1: y[n] = 2u[n] − 3 u[n − 1] + u[n − 2] = { 2 , − 1 } (try it!) has duration=2.
#2: Y (z) = H(z)U (z) = (2− 3 z−^1 +z−^2 ) 1 1 −z−^1 = 2−z
− (^1) → y[n] = { 2 , − 1 }.
Note: This system has wiped out the step input! This seldom happens.
EECS 216 LECTURE NOTES
APPLICATIONS OF THE z-TRANSFORM
Given: x[n] = 3nu[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 ;^ −1 =^
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] = ( 12 )nu[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
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 +^ j^
√ 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 −^1
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( 12 )nu[n] − ( 14 )nu[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.
