lecture notes, me 340 fall 2013 27Lesson 5 – Second-order response, inverse Laplace TransformsThe transfer function for a double integrator with (negative) feedbackis given byH(s)=1s2+ αs + β=1s − γ11s − γ2(120)whereγ1,2:= −α2±!(121)and α, β "= 0. We say that the system has poles18at s = γ1and18These poles live in the complex plane.They are interesting objects of studyin complex analysis, e.g., in residuecalculus.s = γ2, respectively.Recall thatL"e−at#(s)=1s + a(122)It follows thatH(s)=L"eγ1t#(s) ·L"eγ2t#(s) (123)Suppose that γ1"= γ2. By the convolution theorem, theis then given byh(t)=eγ1t∗ eγ2t=$t0dτ=$t0eγ2t+(γ1−γ2)τdτ==eγ1t− eγ2tγ1− γ2(124)Substitution then yieldsh(t)=e−α2t%α24− β(125)We distinguish between two cases corresponding to different signs ofthe expression under the radical.Example 36. A double integrator with feedback is overdamped if α2> 4β.As seen in Fig. 33, the unit impulse response shows an initial increase fromzero and a subsequent decrease back to zero without any overshoot.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 28Figure 33: The impulse response of anoverdamped double integrator withfeedback.Example 37. A double integrator with feedback is underdamped if α2<4β. In this case19,19Remember the relationship sinh(jθ)=j sin θ.sinh%α24− βt%α24− β=%β −α24t%β −α24(126)Figure 34: The impulse response of anunderdamped double integrator withfeedback.The unit impulse response is an exponentially decaying harmonic withamplitude1%β −α24(127)time constant 2/α, angular frequency%β −α24, and no phase shift.When γ1= γ2= γ,eγ1t∗ eγ2t= teγt(128)Example 38. A double integrator with feedback is critically damped ifα2= 4β. In this case, the impulse response is given byh(t)=te−α2t(129)The response of a critically damped double integrator with feedback to a!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 29Figure 35: The impulse response of acritically damped double integratorwith feedback.unit step input is then given byh(t) ∗ u(t)=$t0τe−α2τu(t − τ) dτ=== (130)whose Laplace transform equals H(s)/s.Suppose that h(t) is the impulse response of a convolution corre-sponding to the transfer function H(s), such that h(t)=0 for t < 0.The distributional derivative20of h(t) is then the impulse response20For piecewise-continuous functions,the distributional derivative equalsthe regular derivative at points ofcontinuity plus Dirac delta functionsshifted to points of discontinuity andscaled by the size of the discontinuities.of a convolution with transfer function sH(s).Example 39. The unit step h(t)=u(t) is the impulse response of aconvolution corresponding to the transfer functionH(s)=L(u(t))(s)= (131)For all t, the distributional derivative of the unit step equals zero plusthe Dirac delta function δ(t), since the discontinuity in u(t) is of size1 and occurs at t = 0. The corresponding transfer function is given byL(δ(t))(s)=1, which equals s/s = sH(s).Example 40. The transfer function H(s)=β/(s + α) corresponds to theimpulse response h(t)=βe−αtfor t > 0, whose distributional derivativeequals −αβe−αt+ βδ(t), since the discontinuity in h(t) is of size β andoccurs at t = 0. The corresponding transfer function equalsβ −αβs + α= (132)i.e., sH(s).!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 30Example 41. Consider the block diagram in Fig. 36. Let I(s) := L(i(t))(s) ,X(s)=L(x(t))(s), and O(s) := L(o(t))(s). It follows thatX(s)=1sO(s), O(s)=1s(I(s) − αO(s) − βX(s))(133)from which we obtainO(s)== I(s) (134)Figure 36: A convolution with anintegrator in the feedback loop.The system is thus equivalent to a convolution with transfer functionH(s)=ss2+ αs + β(135)We obtain the corresponding impulse response by computing the distri-butional derivative of the impulse response corresponding to the transferfunction"s2+ αs + β#−1.Example 42. Consider the transfer functionH(s)=s2+ s − 2(s + 3)(s + 1)2(136)We obtain the corresponding Inverse Laplace Transform by several suc-cessive convolution integrals, as followsL−1&1(s + 1)2'(t)= =te−t.(137)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 31L−1&1(s + 3)(s + 1)2'(t)=e−3t∗=$t0τe−τe−3(t−τ)dτ= (138)whose first and second distributional derivatives equal14(−3 e−3t+ e−t(3 − 2t))(139)and14()(140)The corresponding impulse response then equals14(9e−3t+ e−t(2t − 5))+14(−3 e−3t+ e−t(3 − 2t))− 214(e−3t+ e−t(2t − 1))= e−3t− te−t(141)as is also easily verified from the definition of the Laplace transform.For a second-order, scalar, linear, homogeneous initial-value prob-lem with time-independent coefficients,y$$+ αy$+ βy = 0, y(0)=y0, y$(0)=y$0(142)application of the Laplace transformation to the differential equationyieldss2Y(s) − sy0− y$0+ αsY(s) − αy0+ βY(s)=0(143)where Y(s)=L(y(t))(s). It follows thatY(s)= (144)The impulse response is thus identical to the free response withy0= , y$0= (145)Example 43. The free response of a critically damped double integratorwith feedback is obtained from a linear combination of the impulse responsete−αt/2and its distributional derivative:(y$0+ αy0)te−α2t+ y0()=12e−α2t()(146)which reduces to the impulse response when y$0= 0 and y0=