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EECS 216 1-SIDED Z-TRANSFORM
DEF: Z+{x(n)} = X+(z) =
n=0 x(n)z
−n, even if x(n) is noncausal. Compare to 2-sided z-transform: X(z) =
n=−∞ x(n)z
−n. WHY?: To solve difference equations with nonzero initial conditions.
PROPERTIES OF 1-SIDED Z-TRANSFORM X+(z):
- x(n) causal→1-sided z-transform X+(z)=2-sided z-transform X(z)
- Unique inverse only for causal signals: no info on {x(n), n < 0 } Equal causal parts but unequal anticausal parts→same X+(z)
- ROC: Always |z| > |pmax| (largest pole of X+(z)), so don’t bother 4a. D > 0 → Z+{x(n − D)} = z−D^ (X+(z) +
∑D
n=1 x(−n)z
n) 4b. D > 0 → Z+{x(n + D)} = z+D^ (X+(z) −
∑D− 1
n=0 x(n)z
−n) i. Shift right→must add in terms previously anticausal ii. Shift left→must subtract off terms now anticausal iii. Compare to L{ d
(^2) x dt^2 }^ =^ s
(^2) X(s) − sx(0) − dx dt (0) SOLVING DIFFERENCE EQNS WITH INITIAL CONDITIONS Solve: 2 y(n) + 3y(n − 1) + y(n − 2) = u(n) + u(n − 1) − u(n − 2) (Chen p.264)
Take Z+: 2 Y +(z) + 3z−^1 (Y +(z) + y(−1)z) + z−^2 (Y +(z) + y(−1)z + y(−2)z^2 ) = (2 + 3z−^1 +z−^2 )Y +(z) + [(3y(−1) + y(−2)) + z−^1 y(−1)] = (1 + z−^1 − z−^2 )U (z)
Solve: Y +(z) =
1 + z−^1 − z−^2 2 + 3z−^1 + z−^2
U (z) ︸ ︷︷ ︸ zero−state response
(3y(−1) + y(−2)) + z−^1 y(−1)
︸ 2 + 3z−︷︷^1 +^ z−^2 ︸ zero−input response
Plug in: u(n)=unit step→ U (z) = U +(z) = (^1) −^1 z− 1 ; y(−1) = 2, y(−2) = −1:
Y +(z) = 1+z
− (^1) −z− 2 (2+3z−^1 +z−^2 )(1−z−^1 ) −^
5+2z−^1 2+3z−^1 +z−^2 =^
−4+4z−^1 +z−^2 2(1+ 12 z−^1 )(1+z−^1 )(1−z−^1 ) Partial: fraction Y^
+(z) = 4 /^3 1+ 12 z−^1 −^
7 / 2 1+z−^1 +^
1 / 6 1 −z−^1. Note no constant term.
Z−^1 : y(n) =
)nu(n) −
(−1)nu(n) ︸ ︷︷ ︸ natural response
u(n). ︸ ︷︷ ︸ forced
EECS 216 ZERO-INPUT RESPONSE AND STABILITY
Given: 2 y(n) + 3y(n − 1) + y(n − 2) = u(n) + u(n − 1) − u(n − 2) (previous)
Soln: Y (z) =
1 + z−^1 − z−^2 2 + 3z−^1 + z−^2
U (z) ︸ ︷︷ ︸ zero−state response(ZSR)
[3y(−1) + y(−2)] + z−^1 y(−1)
︸ 2 + 3z−︷︷^1 +^ z−^2 ︸ zero−input response(ZIR) Poles: Roots of 2 + 3z−^1 + z−^2 = 0 → − 1 , − 12 from denominator of ZSR. Modes: Roots of 2 + 3z−^1 + z−^2 = 0 → − 1 , − 12 from denominator of ZIR. Huh? {poles} ⊆ {modes} since poles can be canceled by zeros of H(z). Note: Can also cancel poles by proper choice of input u(n).
EX #1: u(n) = δ(n) + δ(n − 1) → U (z) = 1 + z−^1 → ZSR(z) = 1+z
− (^1) −z− 2 2+z−^1. Note: Can also cancel modes by proper choice of initial conditions y(−1), y(−2).
EX #2: y(−1) = 1, y(−2) = − 2 → ZIR(z) = − (^) 2+^1 z− 1 → zir(n) = − 12 (− 12 )nu(n).
How? Choose initial conditions y(−1), y(−2) to cancel the mode at −1. Soln: Need 3y(−1) + y(−2) = y(−1) so ZIR numerator is y(−1)(1 + z−^1 ).
EX #3: Change one MA-part coefficient so {poles} ⊂ {modes}.
Soln: Change 1 + z−^1 − z−^2 so 1+z
− (^1) −z− 2 2+3z−^1 +z−^2 has pole-zero cancellation. EX: Change ZIR numerator to 1 + 0z−^1 − z−^2 = (1 + z−^1 )(1 − z−^1 ). CAUSALITY AND BIBO STABILITY Fact: BIBO stable⇔
|h(n)| < ∞ ⇔ {|z| = 1} ⊂ ROC. Fact: Causal⇔ ROC = {|z| > A} for some constant A. Also Stable if |A| < 1.
- Stable AND Causal⇔Poles inside unit circle⇔ |pi| < 1 , i = 1... N.
- Stable+Anticausal⇔Poles outside unit circle⇔ |pi| > 1 , i = 1... N.
- Stable+2-sided⇔ {|z| = 1} ⊂ ROC = annulus.
INITIAL AND FINAL VALUE THEOREMS AND CORRELATION IVT: x(n) causal→ (^) zLIM→∞ X(z) = x(0) since X(z) = x(0) + x(1)z−^1 +... FVT: (^) nLIM→∞ x(n) = (^) zLIM→ 1 (z − 1)X+(z) (1-sided z-xform)(step).
Auto: r(n) = x(n) ∗ x(−n) =
x(i)x(i ± n) = r(−n) (even function). Cross: rxy (n) = x(n) ∗ y(−n) =
x(i)y(i − n) = ryx(−n) 6 = rxy (−n).
