University of Illinois Fall 2022ECE210 / ECE211 - Homework 5SolutionProblems:1. Sign acknowledging you will abide by this course’s and the University’s Academic Integritypolicies or face sanctions for not doing so. These policies include, among others, plagiarism:representing the words or ideas of others as your own; cheating: using unauthorized materialsand/or information. The possible sanctions include, among others, point reduction, reducedletter grades and an F in the course. If your solution upload does not include your signature,your homework will NOT be graded, resulting in a zero.sign:2. Consider the following circuit with R = 4√3 Ω, f(t) = 2 cos(ωt) V and v (0−) = 2 V.It is known that for t > 0, v(t) = Ae−4t√3+12cos (4t) +√32sin (4t) V.(a) Write the ODE that governs this system for t > 0 in terms of C, v(t), and ω.(b) Determine the value of C.(c) What is the value of ω?(d) If v (t) is the zero-state response, what is the value of A?(e) Identify vtr(t), the transient component of v(t).(f) Identify vss(t), the steady-state component of v(t).(g) What is steady-state phasor V ?Solution :(a) Set KCL at the loop and combine with the current-voltage relation of the capacitor,f(t) =iCR + v(t)iC= Cdv(t)dt⇒ 2 cos(ωt) = 4√3Cdv(t)dt+ v(t)(b) −1RCt = −4√3t and C =116(c) ω = 4(d) If v(t) is zero-state, then v(0) = 0 =⇒ A = −12Page 1 of 4(e) To find transient components, we need first to solve the full solutionv(0) = Ae−0√3+12cos (0) +√32sin (0)2 = A +12A =32=⇒ v(t) =32e−4t√3+12cos (4t) +√32sin (4t)The transient components should vanish as t → ∞, vtr(t) =32e−4t√3(f) The steady components remain as t → ∞, vss(t) =12cos (4t) +√32sin (4t)(g) vss(t) =12cos (4t)+√32sin (4t) =12cos (4t)+√32cos (4t − π/2) and therefore V =12+√32e−jπ/2=12−j√323. Consider the following circuit with R = 4 Ω, L =√3 H, i (0−) = 1 A and let f(t) = 2e−t/√3A for t > 0.For t > 0, obtain:(a) the zero-state current through the inductor, iZS(t),(b) the zero-input current through the inductor, iZI(t),(c) the transient current through the inductor, itr(t),(d) the steady state current through the inductor, iss(t), and(e) the total current through the inductor, i(t).Solution(a) Set KCL at the top node and combine with inductor current-voltage relation,f(t) =vL(t)R+ iL(t)vL(t) = LdiL(t)dt⇒diL(t)dt+4√3iL(t) =4√32e−t/√3Let particular solution ip(t) = K e−t/√3, and plug in the differential equation−1√3K +4√3K =4√32K =83Let the homogeneous solution ih(t) = A e−4√3tand the full solution itotal(t) = ih(t) + ip(t) = A e−4√3t+83e−t/√3For zero-state solution, we have itotal(0) = 0 and solve for A we have A = −83andiZS(t) = −83e−4√3t+83e−t/√3Page 2 of 4(b) For zero-input solution, we have f(t) = 0 and thus ip(t) = 0. Solve for A we haveitotal(0) = A = i(0−) = 1A = 1thereforeiZI(t) = 1 e−4√3t(c) For transient solution or steady state solution, we have to look at the full solution that isitotal(t) = iZI(t) + iZS(t)= −53e−4√3t+83e−t/√3and the transient solution isitr(t) = −53e−4√3t+83e−t/√3(d) There is no steady state solutioniss(t) = 0(e) Full solutionitotal(t) = −53e−4√3t+83e−t/√34. Let x(t) =Re{3ejπ3e−j3t}. Express x(t) in terms of a cosine function.Solutionx(t) = Re{3ejπ3e−j3t} = Re{3ej(−3t+π3)}=3Re{cos(−3t +π3) + j sin(−3t +π3)}= 3 cos(−3t +π3)= 3 cos(3t −π3)5. Determine the phasor F of the following co-sinusoidal functions f (t):(a) f(t) = −√3 cos(3t −π3),(b) f(t) = 4 sin(−4t −π2).Solution(a) F = −√3 e−jπ/3=√3 ej2π/3(b) f(t) = −4 sin(4t + π/2) = −4 cos(4t) ⇒ F = −46. Determine the cosine function f(t) with the frequency ω = 3radscorresponding to the following phasors:(a) F = 2e−jπ3,(b) F = −√3 − j√3,(c) F = −2 − j√3 + 3ejπ3.Solution(a) f(t) = 2 cos(3t −π3)(b) F = −√3 − j√3 =√6 e−j3π4⇒ f(t) =√6 cos(3t −3π4)Page 3 of 4(c) F = −2 − j√3 +32+ j3√32= 1 ej2π3⇒ f(t) = cos(3t +2π3)7. Use the phasor method to express the following signals f (t) as single cosines:(a) f(t) = 3 cos(4t) − 4 sin(4t).(b) f(t) = 3 cos(3t) + 3 cos(3t − π/2).Solution(a) F = 3 + 4j = 5 ej arctan(4/3)⇒ f(t) = 5 cos(4t + arctan(4/3))(b) F = 3 + 3 e−jπ/2= 3√2 e−jπ4⇒ f(t) = 3√2 cos(3t − π/4)8. Use the phasor method to obtain the steady state solution, vss(t), to the following differential equation:cos(t/4) = v(t) +164√3dv(t)dt.SolutionThe differential equation can be rewritten in phasor form as1 = v +164√3j14vv =√32e−j arctan(1/√3)Therefore the steady state solution is vss(t) =√32cos(t/4 − arctan(1√3))Page 4 of