EE 350 Problem Set 5 Cover Sheet Fall 2014Last Name (Print):First Name (Print):ID number (Last 4 digits):Sectio n:Submission deadl ines:• Turn in the written solutions by 4:00 pm on Tuesday October 14 in the homework slot outside 121 EE East.Problem Weight Score21 1522 1523 1524 1825 1926 18Total 100The solution submitted for grading represents my own analysis of the problem, and not that of another student.Signature:Neatly print the name(s) of the students you colla bo rated with on this assignment.Reading assignment:• Lathi Chapter 3, Sections 3.1 through 2.3• Priemer Chapter 5, Section 12.3.5Exam II is scheduled for Thursday, October 16 from 8:15 pm to 1 0:15 pm in 108 Forum (all sections). The secondexam covers material from Problem Sets 4 and 5, and Recitations 5 through 8. The date and location of a reviewsession will be announced on the EE 350 web page.Problem 21: (15 points)Using the relationshipf(t) ∗ δ(t −T ) = f(t − T ) (1)derived in l ecture and in Problem 17, part 1 , evaluation of the convolution integraly(t) = f(t) ∗ h(t) =Z∞−∞f(τ)h(t − τ )dτis simple when either f(t) or h(t) is a sum of weighted impulses. This problem extends this result to the case whereeither the derivative o f either f(t) or h(t) y ields a sum of weighted impulses.1. (5 points) As as an example of the utility of equation (1), suppose that y(t) = f(t) ∗ h(t) wheref(t) = 2δ(t − 3)h(t) = e−|t|.Determine y(t), and sketch f(t), h(t), and y(t) on a single plot.2. (10 points) Now suppose we apply the inputf(t) = u(t) − u(t − 1)to a LTI system that has the i mpulse response functionh(t) = u(t −2).Neither f(t) or h(t) is expressed directly as a sum of weighted impulses.• (2 points) Find a n expression fo r dh/dt in terms of an impulse and sketch dh/dt.• (2 points) Let g(t) denote the response of a system with impulse response dh/dt to the input f, that isg(t) = f(t) ∗dhdt.Calculate and sketch g(t).• (2 points) Using the result from Problem 17 part 4, show the zero-state response of the system withimpulse response h(t) to the input f(t) can be expressed asy(t) =Zt−∞g(τ)dτ.• (4 points) Using the last two results, calculate and sketch y(t).Problem 22: (15 points)In lecture it was shown that when the im pulse response f unction h (t) is a causal signal, then the system is causal.Conversely, if the system impulse response is noncausal then the system is noncausal. To illustrate this importantconcept, consider two LTI systems that are represented by the impulse response functionsSystem 1 : h1(t) = e−tu(t)System 2 : h2(t) = etu(−t).1. (4 points) Sketch h1(t) and h2(t), and specify whether or not each impulse response function is a causal ornoncausal signa l.2. (2 points) Specify whether or not each system is casual.3. (9 points) Using convolution, find and sketch the zero-state response of each system to the inputf(t) = e−tu(t)By comparing a sketch of each zero-state response to a sketch of the input f(t), explain why which system, ifany, is noncausal.Problem 23: (15 points)1. (10 points) Consider two linear time-invariant systems whose impulse responses h(t) are specified as• h1(t) = e−|t|• h2(t) = u(t)Classify each system, corresponding to the impulse functions considered above, as either BIBO stable or notBIBO stable. In order to receive credit, justify your answer.2. (5 points) The system shown in Figure 1 is composed of four LTI systems whose i mpulse response functionsare h1(t), h2(t), h3(t), and h4(t). The input to the overall system i s f(t), and the output is y(t). Using thedistributive, commutative, and associative properties of convolution, represent the composite system by a singleblock with an impulse response function h(t) so thaty(t) = f(t) ∗ h(t),and express h(t) in terms o f h1(t), h2(t), h3(t), and h4(t).Figure 1: System comprised of several subsystems specified by their impulse response function.Problem 24: (18 points)1. (6 points) Figure 2 shows an RL circuit with input f(t) and output y(t).Figure 2: RL circuit.(a) (1 point) Sketch the circuit in the phasor domain by replacing the inductor with its impedance represen-tation.(b) (3 points) Using circuit analysis techniques, show that the frequency response function isH(ω) =˜Y (ω)˜F (ω)=Kωτ + 1.Specify the DC gai n, K, and time constant, τ , in terms of the parameters RL, L, and R.(c) (2 points) Determine the magnitude, |H(ω)| and phase angle,6H(ω) of the frequency response f unctionin part (b) in terms of the parameters RL, L, and R.2. (5 p oints) In part 1 you obtained the frequency response function by applying circuit analysis techniquesto the phasor domain representation of the circuit In a pplications other tha n electronic circuits, for examplecontrol systems engi neering, it is necessary to obtain the frequency response function directly from the ordinarydifferential equation that describes the input-output behavior of the system. Consider a system with inputf(t), output y(t), and the ODE representationQ(D)y(t) = P (D)f(t)dnydtn+ an−1dn−1ydtn−1+ ···+ a1dydt+ aoy(t) = bmdmfdtm+ bm−1dm−1fdtm−1+ ···+ b1dfdt+ bof(t).Suppose that the sinusoidal inputf(t) = Fmcos(ωt + θf) = Ren˜F eωtogenerates the sinusoidal steady-state responsey(t) = Ymcos(ωt + θy) = Ren˜Y eωto,where the phasor representation of f(t) and y(t) are˜F = Fmeθf˜Y = Ymeθy,respectively.(a) (3 points) Substitutingf(t) = Re{˜F eωt}y(t) = Re{˜Y eωt}into the ODE representation, show that˜F and˜Y satisfy(ω)n˜Y + an−1(ω)n−1˜Y + ··· + a1ω˜Y + ao˜Y = bm(ω)m˜F + bm−1(ω)m−1˜F + ···+ b1ω˜F + bo˜F .(b) (2 points) Using the la st result, show that the frequency response function can be written asH(ω) =˜Y˜F=P (ω)Q(ω),and find an expression for H(ω) in terms of the parameters an−1, an−2, . . . , a1, a0and bm, bm−1, . . . , b1, bo.3. (7 points) Figure 3 shows an RLC circuit with input f(t) and output y(t).Figure 3: RLC circuit.(a) (2 points) Using basic circuit analysis l aws such as KVL and KCL, show that the ODE representation ofthe RL C circuit is¨y +RL+ RL˙y +1LCy =1LCf.(b) (2 points) Using the result from part 2, find the frequency response function H(ω) of the RLC circuit interms of the parameters RL, L, R, and C, and express your answer in the standard formH(ω) =bm(ω)m+ bm−1(ω)m−1+ ···+ b1(ω) + b0(ω)n+ an−1(ω)n−1+ ···+ a1(ω) +
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