EECS 451

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): 1. x(n) causal→1-sided z-transform X + (z)=2-sided z-transform X(z) 2. Unique inverse only for causal signals: no info on {x(n), n < 0} Equal causal parts but unequal anticausal parts→same X + (z) 3. ROC: Always |z| > |pmax | (largest pole of X + (z)), so don’t bother D 4a. D > 0 → Z + {x(n − D)} = z −D (X + (z) + n=1 x(−n)z n ) D−1 4b. D > 0 → Z + {x(n + D)} = z +D (X + (z) − n=0 x(n)z −n ) i. Shift right→must add in terms previously anticausal ii. Shift left→must subtract off terms now anticausal 2 x iii. Compare to L{ d 2 } = s2 X(s) − sx(0) − dx (0) dt dt SOLVING DIFFERENCE EQNS WITH INITIAL CONDITIONS Solve: 2y(n) + 3y(n − 1) + y(n − 2) = u(n) + u(n − 1) − u(n − 2) (Chen p.264) Take Z + : 2Y + (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) 1 + z −1 − z −2 (3y(−1) + y(−2)) + z −1 y(−1) Solve: Y (z) = U (z) − 2 + 3z −1 + z −2 2 + 3z −1 + z −2
+ zero−state response zero−input response 1 1−z −1 ;

Plug in: u(n)=unit step→ U (z) = U + (z) = Y + (z) =
Partial: fraction 1+z −1 −z −2 (2+3z −1 +z −2 )(1−z −1 ) 4/3 1+ 1 z −1 2

y(−1) = 2, y(−2) = −1: =
−4+4z −1 +z −2 2(1+ 1 z −1 )(1+z −1 )(1−z −1 ) 2

−

5+2z −1 2+3z −1 +z −2

Y + (z) =

−

7/2 1+z −1

+

1/6 . 1−z −1

Note no constant term.

Z −1 : y(n) =

7 4 1 n 1 (− ) u(n) − (−1)n u(n) + u(n). 3 2 2 6 natural response forced

EECS 451

ZERO-INPUT RESPONSE AND STABILITY

Given: 2y(n) + 3y(n − 1) + y(n − 2) = u(n) + u(n − 1) − u(n − 2) (previous) [3y(−1) + y(−2)] + z −1 y(−1) 1 + z −1 − z −2 U (z) − Soln: Y (z) = 2 + 3z −1 + z −2 2 + 3z −1 + z −2 zero−state response(ZSR) zero−input response(ZIR)

Poles: Roots of 2 + 3z −1 + z −2 = 0 → −1, − 1 from denominator of ZSR. 2 1 −1 −2 Modes: Roots of 2 + 3z + z = 0 → −1, − 2 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). 1 1 EX #2: y(−1) = 1, y(−2) = −2 → ZIR(z) = − 2+z−1 → zir(n) = − 2 (− 1 )n u(n). 2 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}.
1+z Soln: Change 1 + z −1 − z −2 so 2+3z−1−z −2 has pole-zero cancellation. +z EX: Change ZIR numerator to 1 + 0z −1 − z −2 = (1 + z −1 )(1 − z −1 ).
−1 −2

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. 1. Stable AND Causal⇔Poles inside unit circle⇔ |pi | < 1, i = 1 . . . N . 2. Stable+Anticausal⇔Poles outside unit circle⇔ |pi | > 1, i = 1 . . . N . 3. Stable+2-sided⇔ {|z| = 1} ⊂ ROC = annulus. INITIAL AND FINAL VALUE THEOREMS AND CORRELATION
LIM IVT: x(n) causal→ z→∞ X(z) = x(0) since X(z) = x(0) + x(1)z −1 + . . . LIM LIM FVT: n→∞ x(n) = z→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) = rxy (−n).