Question

Problem 5: (20 points) Given the following circuit VIN L Vout R a) Write the loop equation for the circuit, and then find the transfer function Vout(s) H(S) = Express H(s) so that the coefficient of the s in the denominator is 1. Vin(s) b) If Vin(t) is 20u(t), find out(t). c) If Vin(t) is 20u(t) find the final value using the final value theorem (show your work), d) Assume (R/L) = 10, if the input is Vin(t) = 100 sin(10t) what is the output, Vou(t)? Hint DO NOT USE LAPLACE TRANSFORMS FOR PARF D.

Answer 1

Given circuit is, con L Yout Vin R a) Circuit in s-domein, IS Vout (s) Vo (3) SL w R the circuit is I it the curent flowing through Apply loop inequation, - Vin(s) + SL. IS) + R. 10) = 0 1 (5) - Vin (s) RASL and at Node Vout(s), Vour (s) 2 I (5) R. equate 6 and

Vout (s) \(s) R R+ Vout (3) R Vin() RASL R R HS) - RASL L(+5) (RIL) H (8) ts 6.) Given Vin(t) = 20 ult) Laplace Transform, Apply Vio ) 20 S From R Vin (s) Vout (s)- RASL R 11 but 20 S RAS Yout (5) = 20(RID) (s+B) S

By applying partial fractions, Hout (S) = 20 E. st SIR Apply inverse loplace Transform, . Vour (t) = 20 [i-e ters]. 1) (0) Given Vn ()= 20 ult). It IZ final value theorem. Vout (t) = s. Your(s) to too final value, for finding Consider equation 4 S. s. Vout(s)= sto 20 (RI) S. (st 2) "S70 17. S- 20 (RIL) (5+ 20 (RIL) (0+2) 20 (RL) CRILE = 20

d.) Given R 510; Vin (t)- 100 sin(lot). LTI system As input is a asinusoidal asinusoidal input for a be a will also sinusoidal output with the output phase. in Amplitude and change 15 Vour (1)= (100xA). sin llot+o). 2 we From equation ③ know the system transfer function, is, H() (RIL) S+(RI) S 10 HIS). sto 10 H(W) = jutio in polar form, 10 M(W) = Tw' tione L-Tan (4) Therefore / A= w²110²

0 -Tant ( For the the given input, WE 10) 10 ri st 110²t10² D= -Tant (6) = -Tan (1) = -1/14 (6) -45°. Substitute I and d in equation ☺ Vat (t) = (100% ta). sin (rot -4,5 Vout (t) = 100 Sin Clot-65) va Vout (t) = 5052 sin (lot-45)
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