University of Illinois Spring 2016ECE210 / ECE211 - Homework 05Solution1. Consider the following circuit with v (0)=1V . For t>0, obtain:cos(4t)V4 3 Ω1 Fv16(a) the zero-state voltage across the capacitor’s terminals, vZS(t),(b) the zero-input voltage across the capac it or ’ s terminal s, vZI(t),(c) the transient voltage across the c apac it or ’ s termi nal s, vtr(t),(d) the steady state voltage across t h e capacitor’ s termi n als , vss(t), and(e) the total voltage across the capaci t or ’ s terminal s, v(t).Solution:First we are going to find the ODE that rules the circuit. Applying KVL givescos (4t)=v + vR= v + R ⇥ Cdvdtwhich yields to t h e following ODE:cos (4t)=v +4p316⇥dvdt,.To solve this ODE we start from t he parti cu l ar solutionvp(t)=A cos (4t)+B sin (4t) .Taking the derivative yieldsdvp(t)dt= 4A sin (4t)+4B cos (4t) .Substituting v(t) anddv( t)dtinto the ODE, we obtaincos (4t)=A cos (4t)+B sin (4t)+p34⇥ [4A sin (4t)+4B cos (4t)]This i d entity imposes th e two constraintsA + Bp3 = 1 and B Ap3=0,yieldingA =14and B =p34.Therefore, the particular solution isvp(t)=14cos (4t)+p34sin (4t) V.The homogeneous solution for the RC circuit has the form vh(t)=K1et/⌧,with⌧ = RC =p34s. Addingthe h omogen eou s and the particular solutions yields to the general solution:v(t)=vh(t)+vp(t)=K1et/⌧+14cos (4t)+p34sin (4t)V.Page 1 of 3(a) From th e general soluti on we can calculate the zero-state response, i.e. the solution t h at satisfies v(0)=0:0=K1+14) K1= 14.Therefore the zero-state solution isvZS(t)=14e4tp3+14cos (4t)+p34sin (4t)V.(b) The zero-input response can be obtained applying the initial condition to the homogeneous solution, i.e.vh0=1V=K1e0) K1=1V.Therefore the zero-input response isvZI(t)=e4tp3V.(c) From the total response we notice that the transient response isvtr(t)=34e4tp3V,(d) since that is the component of v(t) that vanishes as t !1. The remainder i s the steady state response,i.e.vss(t)=14cos (4t)+p34sin (4t)V.(e) The total voltage across the capacitor is the sum of the zero-input and the zero-state responses . Also, itis th e sum of transient response and steady state response:v(t)=vZI(t)+vZS(t)=vtr(t)+vss(t)=34e4tp3+14cos (4t)+p34sin (4t)V.2. Consider the following problems:(a) Show thatej3t+ej3t4=12cos(3t).(b) Expressej3tej3t8jin te r ms of a cosine function.(c) Given that 4j⇣ejt3 ejt3⌘= A cos(!t + ✓), find A>0, !>0 and ✓ 2 (⇡,⇡].(d) Express Re{2ej⇡6ej5t} in terms of a cosine function.Solution:(a)ej3t+ ej3t4=14(cos (3t)+j sin (3t) + cos (3t)+j sin (3t))=14(cos (3t)  j sin (3t) + cos (3t)+j sin (3t)) =14(2 cos (3t)) =12cos (3t)(b)ej3t ej3t8j=18j(cos (3t)+j sin (3t)  cos (3t)  j sin (3t))=18j(cos (3t)  j sin (3t)  cos (3t)  j sin (3t)) =18j(2j sin (3t)) = 14sin (3t) ,but we know that the sin(x) function leads the cos(x) function by ⇡/2 rad., i.e. sin(x)=cos(x  ⇡/2),for all real x. Therefore14sin(3t)=14cos(3t  ⇡/2).Page 2 of 3(c) 4j⇣ejt3 ejt3⌘= 4j(cos(t3)+jsin(t3)  cos(t3)  jsin(t3)) = 4j(cos(t3)+jsin(t3)  cos(t3)+jsin(t3)) = 4j(2jsin(t3)) = 8sin(t3)=8cos(t3 ⇡/2),so A =8,! =13,✓ = ⇡2(d) Re{2ej⇡6ej5t} =Re{2ej(5t+⇡6)} =Re{2cos5t +⇡6+ j sin5t +⇡6} = 2 cos5t +⇡63. Determine the phasor F of the following co-sinusoidal functions f (t):(a) f(t)=p3 cos(3t ⇡6),(b) f(t)=4sin(4t ⇡2).Solution:(a) f(t)=p3 cos(3t ⇡6)=p3 cos3t + ⇡ ⇡6=p3 cos3t +5⇡6. The magnitude of the cos in e signalis |F | =p3. The phase shift is \F =5⇡6rad. Therefore th e phasor is: F =p3ej5⇡6=p3\150o.(b) f(t)=4sin(4t ⇡2) = 4 cosp3 t ⇡2⇡2+2⇡. In this case |F | = 4, \F = ⇡ rad. Therefore thecorresponding phasor is: F =4ej⇡=4\180o.4. Find the c osi ne functi on f (t)withthefrequency! =5radscorresponding to the fol l owing phasors:(a) F =2ej⇡3,(b) F = p3  jp3,(c) F = 2  jp3+3ej⇡3.Solution:(a) |F | = 2, \F = ⇡3rad. Therefore, f (t) = 2 cos5t ⇡3.(b) F = p3  jp3=qp32+p32ej(⇡+arctan⇣p3p3⌘=p6ej3⇡4. Therefore f (t)=p6 cos5t 3⇡4(c) F = 2  jp3+3ej⇡3= 2  jp3 + 3 cos(⇡3)+j3sin(⇡3)=2+32+ j⇣3p32p3⌘= 12+ jp32.Thereforef(t)=r14+34cos 5t + ⇡  arctan p31!!= cos✓5t +2⇡3◆.5. Use the phasor method to express the following signals f(t) as si n gle cosines:(a) f(t) = 3 cos(4t)  4sin(4t).(b) f(t) = 3 cos(3t) + 3 cos(3t  ⇡/2).Solution:(a) f(t)=4sin(4t) + 3 cos(4t) = 4 cos4t +⇡2+ 3 cos (4t). In phasor form F =4e+j⇡2+3=j4+3=5ej(arctan(43)), whi ch yields the time-domain form:f(t) = 5 cos✓4t + arctan✓43◆◆⇡ 5 cos (4t +0.9273) ,with amplitude 5 and phase shift arctan43⇡ 0.9273 rad = 53.1301o.(b) g(t) = 3 cos(3t) + 3 cos(3t  ⇡/2). In phasor form F =3ej⇡2+3=j3+3=3p2ej⇡4,whichyieldsf(t)=3p2 cos⇣3t ⇡4⌘,with amplitude 3p2 and phase shift ⇡4rad.Page 3 of