University of Illinois Spring 2016ECE210 - Homework 13Solutions1. Consider the following zero-state input-output relations for a variety of systems. In each case, determine whetherthe system is zero-state linear, time invariant, and causal.(a) y(t) = f(t − 1) + f(t + 1).(b) y(t) = 5f(t) ∗ u(t).(c) y(t) = δ(t − 4) ∗ f(t) −Rt−2−∞f2(τ)dτ.(d) y(t) =Rt+2−∞f(τ)dτ.(e) y(t) =Rt−2−∞f(τ2)dτ.Solution:(a) Proving linearity: Let the input be f(t) = af1(t) + bf2(t). Then, following the input-output relation given,the output isy(t) = f(t − 1) + f(t + 1)= af1(t − 1) + bf2(t − 1) + af1(t + 1) + bf2(t + 1)= a (f1(t − 1) + f1(t + 1)) + b (f2(t − 1) + f2(t + 1))= ay1(t) + by2(t),where y1(t) and y2(t) are the outputs of inputs f1(t) and f2(t) respectively. Consequently, the system islinear.Proving if time-invariant: Let the input be f1(t) = f(t − t0). Then the output isy1(t) = f1(t − 1) + f1(t + 1)= f((t − 1) − t0) + f((t + 1) − t0)= y(t − t0).Therefore, the system is time-invariant.An alternative way to analyze this LTI system is recognizing that we can express the output as a convolutionbetween the input and a system impulse response asy(t) = f(t) ∗ (δ(t − 1) + δ(t + 1)) .Hence, the system is LTI (linear-time-invariant)We recognize that the output y(t) depends on future values of the input f(t + 1). Hence, the system isnoncausal.(b) Since the output is a convolution between the input and a system impulse response, the system is linear-time-invariant (LTI).Since the output y(t) does not depend on future values of the input f (t), the system is causal. (The systemimpulse response h(t) = 5u(t) is right-sided.)(c) Proving linearity: Let the input be af1(t) + bf2(t). Then the output isy12(t) = δ(t − 4) ∗ (af1(t) + bf2(t)) −Zt−2−∞(af1(t) + bf2(t))2dτ= aδ(t − 4) ∗ f1(t) + bδ(t − 4) ∗ f2(t) − a2Zt−2−∞f21(t)dτ−2abZt−2−∞f1(t)f2(t)dτ − b2Zt−2−∞f22(t)dτ6= ay1(t) + by2(t).Page 1 of 5Clearly the system is nonlinear.Proving if time-invariant: Let the input be f1(t) = f(t − t0). Then the output isy1(t) = δ(t − 4) ∗ f1(t) −Zt−2−∞f21(τ)dτδ(t − 4) ∗ f(t − t0) −Zt−2−∞f2(τ − t0)dτ= [δ(t − 4) ∗ f(t)]|t=t−t0−Zt−t0−2−∞f2(τ)dτ= y(t − t0)Therefore , the system is time-invariant.The output y(t) does not depend on future values of the input f(t), since both δ(t−4)∗f (t) andRt−2−∞f2(τ)dτdepends on past value of f (t) only .Thus the system is causal.(d) Since we can express the output as a convolutiony(t) =Zt+2−∞f(τ)dτ = f(t) ∗ u(t + 2),the system is LTI.The output y(t) depends on future values of the input f (t), sinceRt+2−∞f(τ)dτ integrates f(t) until t + 2.So the system is noncausal.(e) Proving linearity: Let the input be af1(t) + bf2(t). Then the output isy12(t) =Zt−2−∞af1(τ2) + bf2(τ2)dτ= aZt−2−∞f1(τ2)dτ + bZt−2−∞f2(τ2)dτ= ay1(t) + by2(t).Clearly the system is linear.Proving if time-invariant: Let the input be f1(t) = f(t − t0), then the output isy1(t) =Zt−2−∞f1(τ2)dτ =Zt−2−∞f(τ2− t0)dτ.Clearly this is different thany(t − t0) =Zt−t0−2−∞f(τ2)dτ 6= y1(t).Therefore, the system is time-varying.Analyzing causality, we notice that y(t = 5), depends on future values of f(t), such as f(5 − 2)2= f (t =9). Since the output y(t) depends on future values of the input f (t), the system is noncausal.2. Determine the zero-state response y(t) = h(t)∗f (t) of the following LTI systems to the input f (t) = u(t)−u(t−2).(a) h(t) = u(t).(b) h(t) = e−2tu(t).(c) h(t) = e2tu(t).Solution:(a) y(t) = h(t) ∗ f(t) = u(t) ∗ [u(t) − u(t − 2)] = u(t) ∗ u(t) − u(t) ∗ u(t − 2) = tu(t) − [(t − 2)u(t − 2)].Where u(t) ∗ u(t) = tu(t) is in table 9.2 in the textbook, and by the time-shift property of convolutionu(t) ∗ u(t − 2) = [u(s) ∗ u(s)]|s=t−2.Page 2 of 5(b) y(t) = h(t) ∗ f(t) = e−2tu(t) ∗ [u(t) − u(t − 2)] = e−2tu(t) ∗ u(t) − e−2tu(t) ∗ u(t − 2).e−2tu(t) ∗ u(t) =(0, t < 0Rt0e−2τdτ =1−e−2t20 ≤ t=1−e−2t2u(t).Therefore, y(t) = e−2tu(t) ∗ u(t) − e−2tu(t) ∗ u(t − 2) =1−e−2t2u(t) −1−e−2(t−2)2u(t − 2), where we usedthe time-shift property of convolution again.(c) Similarly to part (b), y(t) = h(t) ∗ f(t) = e2tu(t) ∗ [u(t) − u(t − 2)] = e2tu(t) ∗ u(t) − e2tu(t) ∗ u(t − 2).e2tu(t) ∗ u(t) =(0, t < 0Rt0e2τdτ =e2t−120 ≤ t=e2t−12u(t).Therefore, y(t) = e2tu(t) ∗ u(t) − e2tu(t) ∗ u(t − 2) =e2t−12u(t) −e2(t−2)−12u(t − 2), where we used thetime-shift property of convolution again.3. Find the impulse responses h(t) of the LTI systems having the following unit-step responses.(a) g(t) = 5u(t − 5).(b) g(t) = t2u(t).(c) g(t) = (2 − e−t)u(t − 5).Solution:(a) If we know the unit-step response g(t), then h(t) =ddtg(t). Thereforeh(t) =ddtg(t) =ddt(5u(t − 5)) = 5 δ(t − 5).(b) h(t) =ddtg(t) =ddtt2u(t)= t2δ(t) + u(t)2t = 2tu(t), because t2δ(t) = 02δ(t) = 0.(c) h(t) =ddtg(t) =ddt((2 − e−t)u(t − 5)) = ((2 − e−t)) δ(t−5)+u(t−5)e−t=(2 − e−5)δ(t−5)+e−tu(t−5).4. If the unit-step response of an LTI system is g(t) = 6rectt−63, find the system zero-state responses to thefollowing inputs.(a) f(t) = rect(t).(b) f(t) = e−4tu(t).(c) f(t) = 2δ(t).Solution:(a) First, we need to find the impulse response:h(t) =ddtg(t) =ddt6ut −92− 6ut −152= 6δt −92− 6δt −152.Then yZS(t) = f(t) ∗ h(t) = rect(t) ∗6δt −92− 6δt −152= 6rectt −92− 6rectt −152.(b) The impulse response is the same as in part (a), h(t) = 6δt −92− 6δt −152.Therefore,yZS(t) = f(t) ∗ h(t) = e−4tu(t) ∗6δt −92− 6δt −152= 6e−4(t−9/2)ut −92− 6e−4(t−15/2)ut −152.(c) The impulse response is the same as in part (a), h(t) = 6δt −92− 6δt −152.Therefore yZS(t) = f(t) ∗ h(t) = 2δ(t) ∗6δt −92− 6δt −152= 12δt −92− 12δt −152.5. For each one of the 5 signals f(t) in parts (a), (b), (c), (d), and (e), do the followingi. Obtain its Laplace transformˆF (s).ii. Indicate the poles and zeros ofˆF (s).iii. Indicate the ROC ofˆF (s).(a) f(t) = u(t) − u(t − 8)(b) f(t) = u(t) + u(t − 8)(c) f(t) =