Question

6. Given the D.E: y = 9y' + 20y = r(t) y(0) = 10 y'(O) = 2 that describes a circuit with input r(t). To find the impulse response of the system, h(t) you would: Y(S) i. Find = H(s) including the initial conditions, then find h(t) by taking the R(S) inverse Laplace transform. ii. Find Y(s) = H(s) with the initial conditions set to zero , then find h(t) by taking R(S) the inverse Laplace transform. iii. Give up since there is not enough information to find h(t).

Answer 1

We can not find the impulse response by considering the initial condition...Because we are getting the equation which is not easier to solve..

So Without considering the Initial conditions we can get the impulse response...

Considering the Initial Conditions...

solo- Griven differential eqreation y = .94'+ 20y = rit) yo) = 10, yo =2 Initial Conditions Initial conditions are given so we need to Consider then and Solve for bch. Basics: - LT Yet ky YES) y'et) kysyls) -710). y (t) Ly s'ycs) - 5 410) – y'10) Using these Com be written as y = ayl+20y =rt) y – qy'-2oy ret) Laplace transplorn on both Applying sides (s'yes) - sy.co) – y'o)) – g(syls) – ylö) - 2049 -R(S) -(0)

get impulse response. 08 - 2+90 -105 sels). Applying initial Conditions Ś'y(s) – 5 (10) - 2) +9 (sy(s) – 10) – 20403) = RIS) $ '40s) - - 954(s)— 204(S) -2 +90 POS - RIS). 465) ($95-20) YCS) ģ95-20) -88-los RCS) RIS 88 tios Cs) ROS) 5-95-20 Res) } kind of We comint Solve this we can't egreations

Without Considering the Initial Conditions

S42 -9465)—20408) = R(S). we need to make initial condition Hence zeso then Basics- yit E YCS) y't) LY SYL) y"(t) 2544). y " qy' +20y rity rcti y" _ay' -2oy Applying Leplace transforms Ycs) C$$_95-20). = R(S) - YCS) Sas-20 Rcs) HCS) YCS2 Ris) 9 His s'-95-20

H(S) Šas-20 HIS) (S-10-844) (st1844) Asing partial fractions A B + S-10.844) (5+1.844) s-losun St1.844. AS +A(1.844) +B5-B(10844) " Coefficients ATB (a) 1.844.A - 70.84413 -1 -(b) Solving (m) Eco 'A=0.0786 B=-0.0788.

Then 0.0788 0.0788 H(S) - (5-10-844) (st1844) Applying Biverse deplace HS) L bit) at LT s-a -a ld I LY sta they 0.0788 O 0788 HOS) St1.844 S-10.844 I.LT F 18444 10:8446 bits = 0.0788 e 0.018se
So without Considering initial conditions finding h(t) by taking Inverse Laplace we get the Impulse response...

so option (ii) is my choice...

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