Question

Consider a high-pass RC circuit shown below. Let the capacitor be initially uncharged, and the input v;(t) be applied at t= 0. v;(t) SR v.(t) (a) Obtain the differential equation governing the input and output voltages for t > 0. (b) Using trapezoidal rule convert the same into a difference equation. (c) Obtain the response v.(t) by solving the differential equation.

Answer 1

Date: 6 Toca witte Ort vott fig: high pass RC Circuit Given circuit as a high pass circuit a By wing. KVL an the loop. -Witt) + vcltturlt) = 0 z wilt)= Ucit) toret) = wilt)= Uct) + PR - - Also, q=ic le. Cwount flowing through capacitors & bott)= UR(t) que. Oletput voltage as equal to the voltage across the resistor R. We know that qc = Cduct) - 2) dt. using equation (1) 4 (2) Witt) = UclH + RC duch) dt & duct & vct - wilt) ao RC RC T

o The required differential equation, for the duct) + bclt) - Wit) — (3) at Rc Rc UO4) =(Ret)=ER - CRduct) dt. at > bott)= Rcdwelt) - Rc hley Leed (c) using Laplace transform for eq (3) we get SVC) - 660) = \ft(B) - GELO co Capaciters go anitially to unchanged. > 86c68) + BECA) - UAB Oc(s) RC RO & ico)=0 RC RC Vals & + RC = bi (s) I RC ve) - vi(8)/RC StyRC

By taking inverse laplace transform. vett) = 1 bittetric Page: Date: 1 Volt) = Rc duelt) at =Kcct futu) e tre - WH) a petlec] + ēHIRC dut) volt) = -2 HRC wilt) + etRc duilt) de at RE a Witt)= Acoswt » Volt) et RC Acosut + et RC d Acasut. - --ěURC LA cost + (-aſinet ).) RC boH) = -etRCLA cout + Ausimut! A=1 oo Given, R=800, Cotsuf, w = Leo RC = 800 Xootruf = 6x154 w = = 1666.67 rad/sec > Lott) --e 166667 9 1666.67 cos (1668.674) L + 1666.67 sin(16666ty

→ volt= -166667 ēlScc 67 Required olp voltage COS(1666:67 ) +sin(k6:47t)