EE 102 spring 2001-2002 Handout #15Lecture 8Transfer functions and convolution• convolution & transfer functions• properties• examples• interpretation of convolution• representation of linear time-invariant systems8–1Convolution systemsconvolution system with input u (u(t)=0, t<0)andoutput y:y(t)=t0h(τ)u(t − τ) dτ =t0h(t − τ)u(τ) dτabbreviated: y = h ∗ uin the frequency domain: Y (s)=H(s)U(s)• H is called the transfer function (TF) of the system• h is called the impulse response of the systemblock diagram notation(s):uyuy∗ hHTransfer functions and convolution 8–2Properties1. convolution systems are linear:forallsignals u1, u2and all α, β ∈ R,h ∗ (αu1+ βu2)=α(h ∗ u1)+β(h ∗ u2)2. convolution systems are causal:theoutput y(t) at time t depe nds onlyon past inputs u(τ), 0 ≤ τ ≤ t3. convolution systems are time-invariant:ifweshift the input signal uover T>0, i.e.,apply the inputu(t)= 0 t<Tu(t − T ) t ≥ 0to the system, the output isy(t)= 0 t<Ty(t − T ) t ≥ 0in other words: convolution systems commute with delayTransfer functions and convolution 8–34. composition of convolution systems corresponds to• multiplication of transfer functions• convolution of impulse responsesuucompositionyyABBAramifications:• can manipulate block diagrams with transfer functions as if they weresimple gains• convolution systems commute with each otherTransfer functions and convolution 8–4Example: feedback connectionuGyin time domain,wehavecomplicatedintegral equationy(t)=t0g(t − τ)(u(τ) − y(τ)) dτwhich is not easy to understand or solve . . .in frequency domain,wehaveY = G(U − Y );solveforY to getY (s)=H(s)U(s),H(s)=G(s)1+G(s)(as if G were a simple scaling system!)Transfer functions and convolution 8–5General examplesfirst order LCCODE: y+ y = u, y(0) = 0take Laplace transform to getY (s)=1s +1U(s)transfer function is 1/(s +1);impulse response is e−tintegrator: y(t)=t0u(τ) dτtransfer function is 1/s;impulse response is 1delay: with T ≥ 0,y(t)= 0 t<Tu(t − T ) t ≥ Timpulse response is δ(t − T );transferfunction is e−sTTransfer functions and convolution 8–6Vehicle suspension system(simple model of) vehicle suspension system:uymkb• input u is road height (along vehicle path); output y is vehicle height• vehicle dynamics: my+ by+ ky = bu+ kuassuming y(0) = 0, y(0) = 0,(andu(0−)=0),(ms2+ bs + k)Y =(bs + k)UTF from road height to vehicle height is H(s)=bs + kms2+ bs + kTransfer functions and convolution 8–7DC motoriθJθ+ bθ= ki(J is rotational inertia of shaft & load; b is mechanical resistance of shaft&load;k is motor constant)assuming θ(0) = θ(0) = 0,Js2Θ(s)+bsΘ(s)=kI(s), Θ(s)=kJs2+ bsI(s)i.e.,transferfunction H from i to θ isH(s)=kJs2+ bsTransfer functions and convolution 8–8Circuit examplesconsider a circuit with linear elements, zero initial conditions for inductorsand capacitors,• one independent source with value u• y is a voltage or current somewhere in the circuitthen we have Y (s)=H(s)U(s)example:RCcircuituyRCRCy(t)+y(t)=u(t),Y(s)=11+sRCU(s)impulse response is L−111+sRC=1RCe−t/RCTransfer functions and convolution 8–9to find H:writecircuit e quations in frequency domain:• resistor: v(t)=Ri(t) becomes V (s)=RI(s)• capacitor: i(t)=Cv(t) becomes I(s)=sCV (s)• inductor: v(t)=Li(t) becomes V (s)=sLI(s)in frequency domain, circuit equations become algebraic equationsTransfer functions and convolution 8–10example:vinvout1Ω1Ω1Ω1F1Fv−v+let’s find TF from vinto vout(assuming zero initial voltages for capacitors)• by voltagedivider rule, V+= Vin11+1/s= Vinss +1• current in lefthand resistor is (using V−= V+):I =Vin− V−1Ω=1 −ss +1Vin=1s +1VinTransfer functions and convolution 8–11• I flows through 1F1Ω,yielding voltageVin1s +1(1)(1/s)1+1/s= Vin1(s +1)2• finally we have Vout= V−− Vin1(s +1)2= Vins2+ s − 1(s +1)2so transfer function isH(s)=s2+ s − 1(s +1)2=1−1s +1−1(s +1)2impulse response ish(t)=L−1(H)=δ(t) − e−t− te−twe havevout(t)=vin(t) −t0(1 + τ)e−τvin(t − τ) dτTransfer functions and convolution 8–12Interpretation of convolutiony(t)=t0h(τ)u(t − τ) dτ• y(t) is current output; u(t − τ) is what the input was τ seconds ago• h(τ) shows how much current output depe nds on what input was τseconds agofor example,• h(21) big means current output depe nds quite a bit on what input was,21sec ago• if h(τ) is small for τ>3,theny(t) depends mostly on what the inputhas been over the last 3 seconds• h(τ) → 0 as τ →∞means y(t) depe nds less and less on remote pastinputTransfer functions and convolution 8–13Graphical interpretationy(t)=t0h(t − τ)u(τ) dτto find y(t):• flip impulse response h(τ) backwards in time (yields h(−τ))• drag to the right over t (yields h(t − τ))• multiply pointwise by u (yields u(τ)h(t − τ))• integrate over τ to get y(t)Transfer functions and convolution 8–14h(τ)u(τ)u(τ)h(−τ)h(t1− τ)h(t2− τ)h(t3− τ)ττττττy = u ∗ ht1t2t3Transfer functions and convolution 8–15Examplecommunication channel, e.g.,twistedpair cableuy∗himpulse response:0 2 4 6 8 1000.511.5thadelay≈ 1,plussmoothingTransfer functions and convolution 8–16simple signalling at 0.5 bit/sec; Boolean signal 0, 1, 0, 1, 1,...0 2 4 6 8 1000.510 2 4 6 8 1000.51ttuyoutput is delayed, smoothed version of input1’s & 0’s easily distinguished in yTransfer functions and convolution 8–17simple signalling at 4 bit/sec; same Boolean signal0 2 4 6 8 1000.510 2 4 6 8 1000.51ttuysmoothing makes 1’s & 0’s very hard to distinguish in yTransfer functions and convolution 8–18Linear time-invariant systemsconsider a system A which is• linear• time-invariant (commutes with delays)• causal (y(t) depends only on u(τ) for 0 ≤ τ ≤ t)called a linear time-invariant (LTI) causal systemwe have seen that any convolution system is LTI and causal; the converseis also true: any LTI causal system can be represented by a convolutionsystemconvolution/transfer function representation gives universal description forLTI causal systems(precise statement & proof is not simple . . . )Transfer functions and convolution