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6.003: Signals and Systems Continuous-Time Systems September 20, 2011 1 Multiple Representations of Discrete-Time Systems Discrete-Time (DT) systems can be represented in diﬀerent ways to more easily address diﬀerent types of issues. Verbal descriptions: preserve the rationale. “Next year, your account will contain p times your balance from this year plus the money that you added this year.” Diﬀerence equations: mathematically compact. y[n + 1] = x[n] + py[n] Block diagrams: illustrate signal ﬂow paths. x[n] + Delay y[n] p Operator representations: analyze systems as polynomials. (1 − pR) Y = RX 2 Multiple Representations of Continuous-Time Systems Similar representations for Continuous-Time (CT) systems. Verbal descriptions: preserve the rationale. “Your account will grow in proportion to your balance plus the rate at which you deposit.” Diﬀerential equations: mathematically compact. dy(t) = x(t) + py(t) dt Block diagrams: illustrate signal ﬂow paths. Z t x(t) ( · ) dt + y(t) −∞ p Operator representations: analyze systems as polynomials. (1 − pA)Y = AX 3 Diﬀerential Equations Diﬀerential equations are mathematically precise and compact. r0 (t) h1 (t) r1 (t) We can represent the tank system with a diﬀerential equation. dr1 (t) r0 (t) − r1 (t) = dt τ You already know lots of methods to solve diﬀerential equations: • general methods (separation of variables; integrating factors) • homogeneous and particular solutions • inspection Today: new methods based on block diagrams and operators, which provide new ways to think about systems’ behaviors. 4 Block Diagrams Block diagrams illustrate signal ﬂow paths. DT: adders, scalers, and delays – represent systems described by linear diﬀerence equations with constant coeﬃcents. x[n] + Delay y[n] p CT: adders, scalers, and integrators – represent systems described by a linear diﬀerential equations with constant coeﬃcients. Z t x(t) ( · ) dt + −∞ p Delays in DT are replaced by integrators in CT. 5 y(t) Operator Representation CT Block diagrams are concisely represented with the A operator. Applying A to a CT signal generates a new signal that is equal to the integral of the ﬁrst signal at all points in time. Y = AX is equivalent to t y(t) = x(τ ) dτ −∞ for all time t. 6 Check Yourself + X A ẏ(t) = ẋ(t) + py(t) Y p X X + p A ẏ(t) = x(t) + py(t) Y + ẏ(t) = px(t) + py(t) Y p A Which block diagrams correspond to which equations? 1. 2. 3. 4. 7 1 5. none Check Yourself + X A ẏ(t) = ẋ(t) + py(t) Y p X X + p A ẏ(t) = x(t) + py(t) Y + ẏ(t) = px(t) + py(t) Y p A Which block diagrams correspond to which equations? 1. 2. 3. 4. 8 1 5. none Evaluating Operator Expressions As with R, A expressions can be manipulated as polynomials. Example: W + X A + Y A Z t w(t) = x(t) + x(τ )dτ −∞ Z t y(t) = w(t) + w(τ )dτ −∞ Z t y(t) = x(t) + Z t x(τ )dτ + −∞ Z t Z τ2 x(τ )dτ + −∞ −∞ −∞ W = (1 + A) X Y = (1 + A) W = (1 + A)(1 + A) X = (1 + 2A + A2 ) X 9 x(τ1 )dτ1 dτ2 Evaluating Operator Expressions Expressions in A can be manipulated using rules for polynomials. • Commutativity: A(1 − A)X = (1 − A)AX • Distributivity: A(1 − A)X = (A − A2 )X • Associativity: � � � � (1 − A)A (2 − A)X = (1 − A) A(2 − A) X 10 Check Yourself Determine k1 so that these systems are “equivalent.” + X + A A −0.7 X + Y −0.9 A A Y k1 k2 1. 0.7 2. 0.9 3. 1.6 4. 0.63 11 5. none of these Check Yourself Write operator expressions for each system. + X W A + A −0.7 Y −0.9 (1+0.7A)(1+0.9A)Y = A2 X (1+0.7A)W = AX W = A(X −0.7W ) → → (1+0.9A)Y = AW Y = A(W −0.9Y ) (1+1.6A+0.63A2 )Y = A2 X X + W A A Y k1 k2 W = A(X +k1 W +k2 Y ) Y = AW k1 = −1.6 → 12 Y = A2 X +k1 AY +k2 A2 Y (1−k1 A−k2 A2 )Y = A2 X Check Yourself Determine k1 so that these systems are “equivalent.” + X + A A −0.7 X + Y −0.9 A A Y k1 k2 1. 0.7 2. 0.9 3. 1.6 4. 0.63 13 5. none of these Elementary Building-Block Signals Elementary DT signal: δ[n]. δ[n] = 1, 0, if n = 0; otherwise δ[n] 1 n 0 • • • simplest non-trivial signal (only one non-zero value) shortest possible duration (most “transient”) useful for constructing more complex signals What CT signal serves the same purpose? 14 Elementary CT Building-Block Signal Consider the analogous CT signal: w(t) is non-zero only at t = 0. ⎧ ⎨0 w(t) = 1 ⎩ 0 t<0 t=0 t>0 w(t) 1 t 0 Is this a good choice as a building-block signal? Z t ( · ) dt w(t) −∞ The integral of w(t) is zero! 15 0 No Unit-Impulse Signal The unit-impulse signal acts as a pulse with unit area but zero width. p (t) 1 2 δ(t) = lim p (t) unit area →0 − p1/4 (t) p1/2 (t) t p1/8 (t) 4 2 1 1 − 2 1 2 t 1 − 4 1 4 16 t 1 1 − 8 8 t Unit-Impulse Signal The unit-impulse function is represented by an arrow with the num­ ber 1, which represents its area or “weight.” δ(t) 1 t It has two seemingly contradictory properties: • it is nonzero only at t = 0, and • its deﬁnite integral (−∞, ∞) is one ! Both of these properties follow from thinking about δ(t) as a limit: p (t) 1 2 δ(t) = lim p (t) unit area →0 17 − t Unit-Impulse and Unit-Step Signals The indeﬁnite integral of the unit-impulse is the unit-step. Z t u(t) = δ(λ) dλ = −∞ 1; 0; t≥0 otherwise u(t) 1 t Equivalently δ(t) A 18 u(t) Impulse Response of Acyclic CT System If the block diagram of a CT system has no feedback (i.e., no cycles), then the corresponding operator expression is “imperative.” + X + A A Y = (1 + A)(1 + A) X = (1 + 2A + A2 ) X If x(t) = δ(t) then y(t) = (1 + 2A + A2 ) δ(t) = δ(t) + 2u(t) + tu(t) 19 Y CT Feedback Find the impulse response of this CT system with feedback. x(t) + A y(t) p Method 1: ﬁnd diﬀerential equation and solve it. ẏ(t) = x(t) + py(t) Linear, ﬁrst-order diﬀerence equation with constant coeﬃcients. Try y(t) = Ceαt u(t). Then ẏ(t) = αCeαt u(t) + Ceαt δ(t) = αCeαt u(t) + Cδ(t). Substituting, we ﬁnd that αCeαt u(t) + Cδ(t) = δ(t) + pCeαt u(t). Therefore α = p and C = 1 → y(t) = ept u(t). 20 CT Feedback Find the impulse response of this CT system with feedback. x(t) + A y(t) p Method 2: use operators. Y = A (X + pY ) Y A = X 1 − pA Now expand in ascending series in A: Y = A(1 + pA + p2 A2 + p3 A3 + · · ·) X If x(t) = δ(t) then y(t) = A(1 + pA + p2 A2 + p3 A3 + · · ·) δ(t) 1 1 = (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) = ept u(t) . 2 6 21 CT Feedback We can visualize the feedback by tracing each cycle through the cyclic signal path. x(t) + y(t) A p y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t) 1 1 = (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) 2 6 y(t) 1 t 0 22 CT Feedback We can visualize the feedback by tracing each cycle through the cyclic signal path. x(t) + y(t) A p y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t) 1 1 = (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) 2 6 y(t) 1 t 0 23 CT Feedback We can visualize the feedback by tracing each cycle through the cyclic signal path. x(t) + y(t) A p y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t) 1 1 = (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) 2 6 y(t) 1 t 0 24 CT Feedback We can visualize the feedback by tracing each cycle through the cyclic signal path. x(t) + y(t) A p y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t) 1 1 = (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) 2 6 y(t) 1 t 0 25 CT Feedback We can visualize the feedback by tracing each cycle through the cyclic signal path. x(t) + y(t) A p y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t) 1 1 = (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) = ept u(t) 2 6 y(t) 1 t 0 26 CT Feedback Making p negative makes the output converge (instead of diverge). x(t) + A −p y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t) 1 1 = (1 − pt + p2 t2 − p3 t3 + · · ·) u(t) 2 6 27 y(t) CT Feedback Making p negative makes the output converge (instead of diverge). x(t) + y(t) A −p y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t) 1 1 = (1 − pt + p2 t2 − p3 t3 + · · ·) u(t) 2 6 y(t) 1 t 0 28 CT Feedback Making p negative makes the output converge (instead of diverge). x(t) + y(t) A −p y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t) 1 1 = (1 − pt + p2 t2 − p3 t3 + · · ·) u(t) 2 6 y(t) 1 t 0 29 CT Feedback Making p negative makes the output converge (instead of diverge). x(t) + y(t) A −p y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t) 1 1 = (1 − pt + p2 t2 − p3 t3 + · · ·) u(t) 2 6 y(t) 1 t 0 30 CT Feedback Making p negative makes the output converge (instead of diverge). x(t) + y(t) A −p y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t) 1 1 = (1 − pt + p2 t2 − p3 t3 + · · ·) u(t) 2 6 y(t) 1 t 0 31 CT Feedback Making p negative makes the output converge (instead of diverge). x(t) + y(t) A −p y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t) 1 1 = (1 − pt + p2 t2 − p3 t3 + · · ·) u(t) = e−pt u(t) 6 2 y(t) 1 t 0 32 Convergent and Divergent Poles The fundamental mode associated with p converges if p < 0 and diverges if p > 0. X + A Y p p<0 p>0 y(t) y(t) 1 1 0 t 0 33 t Convergent and Divergent Poles The fundamental mode associated with p converges if p < 0 and diverges if p > 0. X + A Y p Im(p) Convergent Divergent Re(p) Re(p) 34 CT Feedback In CT, each cycle adds a new integration. x(t) + y(t) A p y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t) 1 1 = (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) = ept u(t) 2 6 y(t) 1 t 0 35 DT Feedback In DT, each cycle creates another sample in the output. X + Y p0 Delay y[n] = (1 + pR + p2 R2 + p3 R3 + p4 R4 + · · ·) δ[n] = δ[n] + pδ[n − 1] + p2 δ[n − 2] + p3 δ[n − 3] + p4 δ[n − 4] + · · · y[n] n −1 0 1 2 3 4 36 Summary: CT and DT representations Many similarities and important diﬀerences. y[n] = x[n] + py[n − 1] ẏ(t) = x(t) + py(t) X + A Y X + Y p p Delay A 1 − pA 1 1 − pR e pt u(t) pn u[n] 37 Check Yourself Which functionals represent convergent systems? 1 1− 1 1 R2 4 1 − 14 A2 1 1 + 2R + 43 R2 √ x √ 1. x √ √ 2. x x 1 1 + 2A + 43 A2 √ √ 3. √ √ √ x 4. x √ 38 5. none of these Check Yourself 1 1− 1 R2 4 1 = 1 (1 − 1 R)(1 + 1 R) 2 2 (1 − 1 A)(1 + 1 A) 2 2 1 √ both inside unit circle left & right half-planes X 1 1 = 3 1 2 (1 + 2 R)(1 + 32 R) 1 + 2R + 4 R inside & outside unit circle X 1 1 = 3 1 2 (1 + 2 A)(1 + 32 A) 1 + 2A + 4 A both left half plane 1− 1 A2 4 = √ 39 Check Yourself Which functionals represent convergent systems? 1 1− 1 1 R2 4 1 − 14 A2 1 1 + 2R + 43 R2 √ x √ 1. x √ √ 2. x x 4 1 1 + 2A + 43 A2 √ √ 3. √ √ √ x 4. x √ 40 5. none of these Mass and Spring System Use the A operator to solve the mass and spring system. x(t) F = K x(t) − y(t) = M ÿ(t) y(t) x(t) + K M ÿ(t) A −1 K A2 Y M = K A2 X 1 + M 41 ẏ(t) A y(t) Mass and Spring System Factor system functional to ﬁnd the poles. K A2 K 2 Y MA = M K = (1 − p0 A)(1 − p1 A) X 1 + M A2 1+ K 2 A = 1 − (p0 + p1 )A + p0 p1 A2 M The sum of the poles must be zero. The product of the poles must be K/M . p0 = j K M p1 = −j K M 42 Mass and Spring System Alternatively, ﬁnd the poles by substituting A → 1 s . The poles are then the roots of the denominator. K A2 Y = MK X 1 + M A2 Substitute A → 1s : K Y M = K X s2 + M r K s = ±j M 43 Mass and Spring System The poles are complex conjugates. Im s s-plane q K M ≡ ω0 Re s q K ≡ −ω − M 0 The corresponding fundamental modes have complex values. fundamental mode 1: ejω0 t = cos ω0 t + j sin ω0 t fundamental mode 2: e−jω0 t = cos ω0 t − j sin ω0 t 44 Mass and Spring System Real-valued inputs always excite combinations of these modes so that the imaginary parts cancel. Example: ﬁnd the impulse response. K A2 K Y A A M M = = − X 1 + K A2 p 0 − p1 1 − p 0 A 1 − p 1 A M ω02 A A − = 2jω0 1 − jω0 A 1 + jω0 A A ω0 A ω0 − = 2j 1 + jω0 A 2j 1 − jω0 A ' v " ' v " makes mode 1 makes mode 2 The modes themselves are complex conjugates, and their coeﬃcients are also complex conjugates. So the sum is a sum of something and its complex conjugate, which is real. 45 Mass and Spring System The impulse response is therefore real. Y ω0 A ω0 A = − X 2j 1 − jω0 A 2j 1 + jω0 A The impulse response is ω0 jω0 t ω0 −jω0 t e = ω0 sin ω0 t ; h(t) = e − 2j 2j t>0 y(t) t 0 46 Mass and Spring System Alternatively, ﬁnd impulse response by expanding system functional. x(t) + ω02 ÿ(t) A ẏ(t) −1 ω02 A2 Y = = ω02 A2 − ω04 A4 + ω06 A6 − + · · · X 1 + ω02 A2 If x(t) = δ(t) then 3 5 2 4t 6t − + · · · , t ≥ 0 y(t) = ω 0 t − ω 0 + ω0 5! 3! 47 A y(t) Mass and Spring System Look at successive approximations to this inﬁnite series. ∞ � �l 0 ω02 A2 Y 2 2 2 2 = = ω A −ω A 0 0 X 1 + ω02 A2 l=0 If x(t) = δ(t) then ∞ � �l 0 y(t) = ω02 −ω02 A2l+2 δ(t) l=0 = ω02 t − ω04 t5 t7 t9 t3 + ω06 − ω08 + ω010 − + · · · = ω0 sin ω0 t 3! 5! 7! 9! y(t) t 0 48 Summary: CT and DT representations Many similarities and important diﬀerences. y[n] = x[n] + py[n − 1] ẏ(t) = x(t) + py(t) X + A Y X + Y p p Delay A 1 − pA 1 1 − pR e pt u(t) pn u[n] 49 MIT OpenCourseWare http://ocw.mit.edu 6.003 Signals and Systems Fall 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.