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Lecture 8 Transfer Functions and Convolution

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Lecture 8 Transfer Functions and Convolution S. Boyd EE102 Lecture 8 Transfer functions and convolution ² convolution & transfer functions ² properties ² examples ² interpretation of convolution ² representation of linear time-invariant systems 8{1 Convolution systems convolution system with input u (u(t) = 0, t < 0) and output y: t t y(t) = h(¿)u(t ¡ ¿) d¿ = h(t ¡ ¿)u(¿) d¿ Z0 Z0 abbreviated: y = h ¤ u in the frequency domain: Y (s) = H(s)U(s) ² H is called the transfer function (TF) of the system ² h is called the impulse response of the system PSfrag replacements block diagram notation(s): u y u y ¤ h H Transfer functions and convolution 8{2 Properties 1. convolution systems are linear: for all signals u1, u2 and all ®, ¯ 2 R, h ¤ (®u1 + ¯u2) = ®(h ¤ u1) + ¯(h ¤ u2) 2. convolution systems are causal: the output y(t) at time t depends only on past inputs u(¿), 0 · ¿ · t 3. convolution systems are time-invariant: if we shift the input signal u over T > 0, i.e., apply the input 0 t < T u(t) = ½ u(t ¡ T ) t ¸ 0 e to the system, the output is 0 t < T y(t) = ½ y(t ¡ T ) t ¸ 0 e in other words: convolution systems commute with delay Transfer functions and convolution 8{3 4. composition of convolution systems corresponds to ² multiplication of transfer functions ² convolution of impulse responses composition PSfrag replacements u A B y u BA y rami¯cations: ² can manipulate block diagrams with transfer functions as if they were simple gains ² convolution systems commute with each other Transfer functions and convolution 8{4 Example: feedback connection PSfrag replacements u G y in time domain, we have complicated integral equation t y(t) = g(t ¡ ¿)(u(¿) ¡ y(¿)) d¿ Z0 which is not easy to understand or solve . in frequency domain, we have Y = G(U ¡ Y ); solve for Y to get G(s) Y (s) = H(s)U(s); H(s) = 1 + G(s) (as if G were a simple scaling system!) Transfer functions and convolution 8{5 General examples ¯rst order LCCODE: y0 + y = u, y(0) = 0 take Laplace transform to get 1 Y (s) = U(s) s + 1 transfer function is 1=(s + 1); impulse response is e¡t t integrator: y(t) = u(¿) d¿ Z0 transfer function is 1=s; impulse response is 1 delay: with T ¸ 0, 0 t < T y(t) = ½ u(t ¡ T ) t ¸ T impulse response is ±(t ¡ T ); transfer function is e¡sT Transfer functions and convolution 8{6 Vehicle suspension system (simple model of) vehicle suspension system: y PSfrag replacements m k b u ² input u is road height (along vehicle path); output y is vehicle height ² vehicle dynamics: my00 + by0 + ky = bu0 + ku 0 assuming y(0) = 0, y (0) = 0, (and u(0¡) = 0), (ms2 + bs + k)Y = (bs + k)U bs + k TF from road height to vehicle height is H(s) = ms2 + bs + k Transfer functions and convolution 8{7 DC motor i θ PSfrag replacements 00 0 Jθ + bθ = ki (J is rotational inertia of shaft & load; b is mechanical resistance of shaft & load; k is motor constant) assuming θ(0) = θ0(0) = 0, k Js2£(s) + bs£(s) = kI(s); £(s) = I(s) Js2 + bs i.e., transfer function H from i to θ is k H(s) = Js2 + bs Transfer functions and convolution 8{8 Circuit examples consider a circuit with linear elements, zero initial conditions for inductors and capacitors, ² one independent source with value u ² y is a voltage or current somewhere in the circuit then we have Y (s) = H(s)U(s) example: RC circuit PSfrag replacements R u C y 0 1 RCy (t) + y(t) = u(t); Y (s) = U(s) 1 + sRC ¡ 1 1 ¡ impulse response is L 1 = e t=RC µ1 + sRC ¶ RC Transfer functions and convolution 8{9 to ¯nd H: write circuit equations in frequency domain: ² resistor: v(t) = Ri(t) becomes V (s) = RI(s) ² capacitor: i(t) = Cv0(t) becomes I(s) = sCV (s) ² inductor: v(t) = Li0(t) becomes V (s) = sLI(s) in frequency domain, circuit equations become algebraic equations Transfer functions and convolution 8{10 1F example: 1­ 1­ PSfrag replacements v¡ 1F vout v+ vin 1­ let's ¯nd TF from vin to vout (assuming zero initial voltages for capacitors) 1 s ² by voltage divider rule, V = V = V + in1 + 1=s ins + 1 ² current in lefthand resistor is (using V¡ = V+): V ¡ V¡ s 1 I = in = 1 ¡ V = V 1­ µ s + 1¶ in s + 1 in Transfer functions and convolution 8{11 ² I flows through 1Fk1­, yielding voltage 1 (1)(1=s) 1 V = V ins + 1 1 + 1=s in(s + 1)2 1 s2 + s ¡ 1 ² ¯nally we have V = V¡ ¡ V = V out in(s + 1)2 in (s + 1)2 so transfer function is s2 + s ¡ 1 1 1 H(s) = = 1 ¡ ¡ (s + 1)2 s + 1 (s + 1)2 impulse response is ¡ ¡ ¡ h(t) = L 1(H) = ±(t) ¡ e t ¡ te t we have t ¡¿ vout(t) = vin(t) ¡ (1 + ¿)e vin(t ¡ ¿) d¿ Z0 Transfer functions and convolution 8{12 Interpretation of convolution t y(t) = h(¿)u(t ¡ ¿) d¿ Z0 ² y(t) is current output; u(t ¡ ¿) is what the input was ¿ seconds ago ² h(¿) shows how much current output depends on what input was ¿ seconds ago for example, ² h(21) big means current output depends quite a bit on what input was, 21sec ago ² if h(¿) is small for ¿ > 3, then y(t) depends mostly on what the input has been over the last 3 seconds ² h(¿) ! 0 as ¿ ! 1 means y(t) depends less and less on remote past input Transfer functions and convolution 8{13 Graphical interpretation t y(t) = h(t ¡ ¿)u(¿) d¿ Z0 to ¯nd y(t): ² flip impulse response h(¿) backwards in time (yields h(¡¿)) ² drag to the right over t (yields h(t ¡ ¿)) ² multiply pointwise by u (yields u(¿)h(t ¡ ¿)) ² integrate over ¿ to get y(t) Transfer functions and convolution 8{14 h(¿) u(¿) h(¡¿) u(¿) ¿ ¿ h(t1 ¡ ¿) h(t2 ¡ ¿) PSfrag replacements ¿ ¿ t1 t2 h(t3 ¡ ¿) y = u ¤ h ¿ ¿ t3 Transfer functions and convolution 8{15 Example communication channel, e.g., twisted pair cable PSfrag replacements u y ¤h impulse response: 1.5 1 h 0.5 PSfrag replacements 0 0 2 4 6 8 10 t a delay ¼ 1, plus smoothing Transfer functions and convolution 8{16 simple signalling at 0:5 bit/sec; Boolean signal 0; 1; 0; 1; 1; : : : 1 u 0.5 0 0 2 4 6 8 10 t 1 PSfrag replacements y 0.5 0 0 2 4 6 8 10 t output is delayed, smoothed version of input 1's & 0's easily distinguished in y Transfer functions and convolution 8{17 simple signalling at 4 bit/sec; same Boolean signal 1 u 0.5 0 0 2 4 6 8 10 t 1 PSfrag replacements y 0.5 0 0 2 4 6 8 10 t smoothing makes 1's & 0's very hard to distinguish in y Transfer functions and convolution 8{18 Linear time-invariant systems consider a system A which is ² linear ² time-invariant (commutes with delays) ² causal (y(t) depends only on u(¿) for 0 · ¿ · t) called a linear time-invariant (LTI) causal system we have seen that any convolution system is LTI and causal; the converse is also true: any LTI causal system can be represented by a convolution system convolution/transfer function representation gives universal description for LTI causal systems (precise statement & proof is not simple . ) Transfer functions and convolution 8{19.
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