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Discrete-Time LTI Systems and Analysis Excitation Response Discrete-Time Discrete-Time Signal Signal Dr Discrete-Time LTI Systems Discrete-time Systems Input-Output Description of Dst-Time Systems x(n) y(n) input/ Discrete-time output/ System Discrete-Time LTI Systems and Analysis excitation response Discrete-time Discrete-time signal signal Dr. Deepa Kundur University of Toronto I Input-output description (exact structure of system is unknown or ignored): y(n) = T [x(n)] I \black box" representation: x(n) −!T y(n) Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 1 / 61 Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 2 / 61 Discrete-Time LTI Systems Discrete-time Systems Discrete-Time LTI Systems Discrete-time Systems System Properties Terminology: Implication If\ A" then\ B" Shorthand: A =) B Why is this so important? Example 1: I mathematical techniques developed to analyze systems are often it is snowing= ) it is at or below freezing temperature contingent upon the general characteristics of the systems being Example 2: considered α ≥ 5:2= ) α is positive Note: For both examples above, B 6=) A I for a system to possess a given property, the property must hold for every possible input to the system I to disprove a property, need a single counter-example I to prove a property, need to prove for the general case Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 3 / 61 Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 4 / 61 Discrete-Time LTI Systems Discrete-time Systems Discrete-Time LTI Systems Discrete-time Systems Terminology: Equivalence Common Properties Time-invariant system: input-output characteristics do not If\ A" then\ B" Shorthand: A =) B I change with time and I a system is time-invariant iff If\ B" then\ A" Shorthand: B =) A T T x(n) −! y(n) =) x(n − n0) −! y(n − n0) can be rewritten as for every input x(n) and every time shift n0. \A" if and only if\ B" Shorthand: A () B We can also say: I A is EQUIVALENT to B = I A = B Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 5 / 61 Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 6 / 61 Discrete-Time LTI Systems Discrete-time Systems Discrete-Time LTI Systems Discrete-time Systems Common Properties Additivity: I Linear system: obeys superposition principle I a system is linear iff T [a1 x1(n) + a2 x2(n)] = a1 T [x1(n)] + a2 T [x2(n)] for any arbitrary input sequences x1(n) and x2(n), and any arbitrary constants a1 and a2. Homogeneity: Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 7 / 61 Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 8 / 61 Discrete-Time LTI Systems Discrete-time Systems Discrete-Time LTI Systems The Convolution Sum Common Properties The Convolution Sum I Causal system: output of system at any time n depends only on present and past inputs I a system is causal iff Recall: y(n) = F [x(n); x(n − 1); x(n − 2);:::] 1 X for all n. x(n) = x(k)δ(n − k) k=−∞ I Bounded Input-Bounded output (BIBO) Stable: every bounded input produces a bounded output I a system is BIBO stable iff jx(n)j ≤ Mx < 1 =) jy(n)j ≤ My < 1 for all n and for all possible bounded inputs. Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 9 / 61 Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 10 / 61 Discrete-Time LTI Systems The Convolution Sum Discrete-Time LTI Systems The Convolution Sum The Convolution Sum The Convolution Sum Let the response of a linear time-invariant (LTI) system denoted T to the unit sample input δ(n) be h(n). T Therefore, δ(n) −! h(n) T 1 δ(n − k) −! h(n − k) X y(n) = x(k)h(n − k) = x(n) ∗ h(n) T αδ (n − k) −! α h(n − k) k=−∞ x(k) δ(n − k) −!T x(k) h(n − k) 1 1 for any LTI system. X T X x(k)δ(n − k) −! x(k)h(n − k) k=−∞ k=−∞ x(n) −!T y(n) Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 11 / 61 Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 12 / 61 Discrete-Time LTI Systems The Convolution Sum Discrete-Time LTI Systems The Convolution Sum Causality and Convolution Stability and Convolution For a causal system, y(n) only depends on present and past inputs values. Therefore, for a causal system, we have: It can also be shown that 1 X 1 y(n) = h(k)x(n − k) X jh(n)j < 1 () LTI system is BIBO stable k=−∞ n=−∞ −1 1 X X = h(k)x(n − k)+ h(k)x(n − k) Note: k=−∞ k=0 1 I () means that the two statements are equivalent X = h(k)x(n − k) I BIBO = bounded-input bounded-output k=0 where h(n) = 0 for n < 0 to ensure causality. Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 13 / 61 Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 14 / 61 Discrete-Time LTI Systems The Convolution Sum Discrete-Time LTI Systems The Convolution Sum PROOF PROOF For a stable system, y(n) is bounded if x(n) is bounded. What are the implications on h(n)? P1 We have: To prove the reverse implication (i.e., necessity), assuming n=−∞ jh(n)j = 1 we must find a 1 bounded input x(n) that will always result in an unbounded y(n). Recall, X jy(n)j = j h(k)x(n − k)j 1 k=−∞ X 1 1 y(n) = h(k)x(n − k) X X ≤ jh(k)x(n − k)j = jh(k)j·j x(n − k)j k=−∞ 1 1 k=−∞ k=−∞ | {z } X X jx(n)|≤Mx <1 y(0) = h(k)x(0 − k) = h(k)x(−k) 1 1 X X k=−∞ k=−∞ ≤ jh(k)jMx = Mx jh(k)j k=−∞ k=−∞ Consider x(n) = sgn(h(−n)); note: jx(n)j ≤ 1. 1 Therefore, P1 jh(k)j < 1 is a sufficient condition to guarantee: X k=−∞ y(0) = h(k)x(−k) k=−∞ 1 X 1 1 y(n) ≤ Mx jh(k)j < 1 X X = h(k)sgn(h(−(−k)))= h(k)sgn(h(k)) k=−∞ k=−∞ k=−∞ 1 and we can write: X 1 = jh(n)j = 1 X jh(n)j < 1 =) LTI system is stable n=−∞ n=−∞ Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 15 / 61 Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 16 / 61 Discrete-Time LTI Systems The Convolution Sum Discrete-Time LTI Systems The z-Transform and System Function PROOF The Direct z-Transform Therefore, 1 X jh(n)j = 1 I Direct z-Transform: n=−∞ 1 guarantees that there exists a bounded input that will result in an unbounded output, so it is X −n also a necessary condition and we can write: X (z) = x(n)z n=−∞ 1 X jh(n)j < 1( = LTI system is stable n=−∞ I Notation: Putting sufficiency and necessity together we obtain: X (z) ≡ Zfx(n)g 1 X jh(n)j < 1 () LTI system is stable Z n=−∞ x(n) ! X (z) Note: () means that the two statements are equivalent. Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 17 / 61 Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 18 / 61 Discrete-Time LTI Systems The z-Transform and System Function Discrete-Time LTI Systems The z-Transform and System Function Region of Convergence z-Transform Properties Property Time Domain z-Domain ROC Notation: x(n) X (z) ROC: r2 < jzj < r1 x1(n) X1(z) ROC1 I the region of convergence (ROC) of X (z) is the set of all values x2(n) X1(z) ROC2 Linearity: a1x1(n) + a2x2(n) a1X1(z) + a2X2(z) At least ROC1\ ROC2 of z for which X (z) attains a finite value Time shifting: x(n − k) z−k X (z) ROC, except z = 0 (if k > 0) and z = 1 (if k < 0) n −1 I The z-Transform is, therefore, uniquely characterized by: z-Scaling: a x(n) X (a z) jajr2 < jzj < jajr1 Time reversal x(−n) X (z−1) 1 < jzj < 1 1. expression for X (z) r1 r2 Conjugation: x∗(n) X ∗(z∗) ROC 2. ROC of X (z) dX (z) z-Differentiation: n x(n) −z dz r2 < jzj < r1 Convolution: x1(n) ∗ x2(n) X1(z)X2(z) At least ROC1\ ROC2 among others . Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 19 / 61 Dr. Deepa Kundur (University of Toronto) Discrete-Time LTI Systems and Analysis 20 / 61 Discrete-Time LTI Systems The z-Transform and System Function Discrete-Time LTI Systems The z-Transform and System Function Common Transform Pairs The System Function Signal, x(n) z-Transform, X (z) ROC 1 δ(n) 1 All z h(n) Z! H(z) 1 2 u(n) 1−z−1 jzj > 1 Z n 1 time-domain ! z-domain 3 a u(n) −1 jzj > jaj 1−az Z n az−1 impulse response ! system function 4 na u(n) (1−az−1)2 jzj > jaj n 1 5 −a u(−n − 1) 1−az−1 jzj < jaj n az−1 Z 6 −na u(−n − 1) (1−az−1)2 jzj < jaj y(n) = x(n) ∗ h(n) ! Y (z) = X (z) · H(z) −1 1−z cos !0 7 (cos(!0n))u(n) −1 −2 jzj > 1 1−2z cos !0+z −1 z sin !0 8 (sin(!0n))u(n) −1 −2 jzj > 1 1−2z cos !0+z −1 n 1−az cos !0 Therefore, 9 (a cos(!0n)u(n) 2 −2 jzj > jaj 1−2az−1 cos !0+a z −1 Y (z) n 1−az sin !0 H(z) = 10 (a sin(!0n)u(n) 2 −2 jzj > jaj 1−2az−1 cos !0+a z X (z) Dr.
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