lecture notes, me 340 fall 2013 21Lesson 4 – Limits, transfer functions, and impulse responsesIn the previous analysis, we considered the scaled pulsegα(t) :=u(t) − u( t − α)α(89)for α > 0. The Laplace transform of the unit step equalsL(u(t))(s)=!∞0e−stu(t) dt =!∞0e−st= −1se−st""""∞0=1s(90)and, therefore,L(gα(t))(s)=αs(91)As α → 0, the right-hand side approaches 1. On the other hand,limα→0L(gα(t))(s)=L(δ(t))(s) (92)We conclude thatL(δ(t))(s)= (93)and say that the function family in (89) converges in a weak sense toδ(t) as α → 0.Figure 25: Weak convergence of thefamily of scaled pulses to δ.More generally, suppose that gα(t) denotes a family of functionsparameterized by some parameter α such thatlimα→α0L(gα(t))(s)= (94)We say that the function family converges in a weak sense16to the16Strong convergence correspondsto pointwise convergence, i.e., for everyvalue of t.Dirac delta function δ(t) as α → α0.Example 26. Consider the family of functionsgα(t) := αe−αt(95)Since=!t0αe−(s+α)tdt = (96)converges to 1 as α → ∞, we conclude that the family in (95) converges in aweak sense to the Dirac delta function δ(t) as α → ∞ .!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 22Figure 26: Weak convergence of thefamily of exponentials to δ.For a family gα(t) of functions parameterized by α whose Laplacetransforms are given by Gα(s), the convolution theorem states thatL#!t0f (τ)gα(t − τ) dτ$(s)=L(f (t) ∗ gα(t))(s)=F(s)Gα(s) (97)where F(s) is the Laplace transforms of f (t). If the family gα(t) con-verges in a weak sense to the Dirac delta function δ(t) as α → α0,thenlimα→α0L(f (t) ∗ gα(t))(s)=== (98)i.e., that1717This requires continuity of the Laplacetransform.limα→α0(99)We express this observation by the following key identity:δ(t) ∗ = (100)Example 27. In a previous example, we found the relationshipVac= Vab+ RCdVabdt⇔dVabdt+ αVab= αVac(101)for α =1RC, between the total voltage drop Vacacross a resistor and ca-pacitor in series and the voltage drop Vabacross the resistor. The generalsolution to this equation is given byVab(t)= (102)!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 23It follows thatdVabdt(t)=−αVab(0)e−αt+ αVac(t) − α2!t0Vac(τ)e−α(t−τ)dτ= α%Vac(0) − Vab(0)&e−αt+ (103)where the second equality follows by integration by parts. By the analysisabove, it follows that as α → ∞, i.e., for RC % 1, and t > 0,dVabdt(t) → (104)i.e., that the rate of change of the voltage drop across the resistor approx-imates the rate of change of the applied voltage closely. It follows that thevoltage drop over the capacitor ..Example 28. In a previous example, we found the relationshipVac= α!t0Vbc(τ) dτ + Vbc(105)where Vac,Vbc, and α have the identical meaning as above. In this case,!t0Vbc(τ) dτ == − α!t0#!τ0Vac(t&) dt&$e−α(t−τ)dτ (106)where the second equality follows by integration by parts. Here, the integralterm equals#!τ0Vac(t&) dt&$∗ αe−αt(107)wherelimα→0L (αe−αt)(s)=limα→0= 0(108)It follows that as α → 0, i.e., for RC ' 1,!t0Vbc(τ) dτ → (109)i.e., that the integral of the voltage drop across the capacitor approximatesthe integral of the applied voltage closely. It follows that the voltage dropover the resistor ..lecture notes, me 340 fall 2013 24A system whose output (response) o(t) to an input i(t) is given bythe convolution h(t) ∗ i(t) of some function h(t) with the input, is aconvolution (system). It follows from this definition that convolu-tions are linear system whose output only depends on the past input,i.e., they are causal.Figure 27: A convolution system.Example 29. Suppose that the input i(t) to a convolution is given by afunction gα(t), such that gα(t) converges in a weak sense to the Dirac deltafunction δ(t) as α → α0. It follows thatlimα→α0o(t)= (110)i.e., that h(t) is the (unit) impulse response of this component.Example 30. Consider the outputs of the following block diagram, wherethe output of the delay line is zero for 0 ≤ t < a.Figure 28: Delay line followed by aconvolution.Figure 29: Convolution followed by adelay line.The convolution of a delayed signal is a delayed convolution of the signal.We say that convolutions are time-invariant.lecture notes, me 340 fall 2013 25The Laplace transform H(s) of the impulse response h(t) of aconvolution is the system’s transfer function.Example 31. The impulse response of an integrator with zero initial condi-tion is the unit step, since!t0i(τ) dτ =!t0i(τ)u(t − τ) dτ = (111)The transfer function is given by H(s)=L(u(t))(s)=1/s.Example 32. The impulse response of a delay line is the time-shifted Diracdelta function δ(t − a), since!t0δ(τ − a)i(t − τ)u(t − τ) dτ ='τ = t&+ adτ = dt&(=(112)The transfer function is given by H(s)=L(δ(t − a))(s)=e−αs.Example 33. The impulse response of an integrator with zero initial condi-tion and feedback is e−αt, since o(t)=i(t) ∗ e−αt.Figure 30: An integrator with feedbackis a convolution.The corresponding transfer function is given byH(s)=L%e−αt&(s)=1s + α(113)Example 34. Now consider the following block diagram.Figure 31: An integrator with feedbackand feedforward is a convolution.Let I( s) := L(i(t))(s) and O(s) := L(o(t))(s). It follows thatO(s)=βI(s)+1s(I(s) − αO(s))(114)lecture notes, me 340 fall 2013 26from which we obtainO(s)==I(s) (115)The system is thus equivalent to a convolution with transfer functionH(s)=βs + 1s + α= L%βδ(t)+&(s) (116)The impulse response is thus given by an impulse scaled by β added to anexponentially decaying signal with amplitude 1 − αβ and time constantα−1.Example 35. Now consider the following block diagram.Figure 32: A double integrator withfeedback is a convolution.Let I( s) := L(i(t))(s),X(s)=L(x(t))(s), and O(s) := L(o(t))(s). Itfollows thatO(s)=1sX(s), X(s)=1s(I(s) − αX(s) − βO(s))(117)from which we obtainO(s)== I(s) (118)The system is thus equivalent to a convolution with transfer functionH(s)=1s2+ αs + β(119)In Lesson 5, we use convolution integrals to compute the correspondingimpulse