Question

lebi, 03134\. The Assume that a circuit has the impulse response of h(t)= { 10, elsewhere input x(t) to the circuit is given by x(t)=6u(t), where u(t) is the unit step function. Assume that the circuit output is given by y(t). a) (20 pts) Use the convolution integral to derive the expressions for the output y(t) for several different intervals of t. (from 0 to + ). b) (5 pts) In summary, write the expression for y(t) with respect different intervals of t. Plot the curve for the output y(t). c) (13 pts) How you can obtain the output y(t) using the S-domain method and by the Laplace Transform? For the given x(t) and h(t), obtain the expressions for the X(s) and H(s). Using the inverse Laplace Transform obtain y(t). (Hint: You can re-write the above h(t) as follows: h(t)=e[u(t)- uſt - 4)], then obtain its Laplace Transform, H(s)).

Please help to answer this question and show the work clearly because I want to learn. I will give thumbs up for sure, thank you very much!!!

Answer 1

h(t)=( est 70 ostsy Elsewhere r(t)= Guct) THET) . x(t-1) for to de y(t) = (x(+_^)( (7) dr fetto: common area is zero yet=0 fortso & tau OLELY yit) = See other ylt) = (1-est) fit >4 4 . 6 .&ct=j bet der = CET)* = [-33] yet) al fart >4

octy Y(t) = fart> 4 HES) = Ice hH)= ē uc) - e tuct-4) het) = estuct - Eclt-tilt-u) with = -24 -6lt-u) bluct) - e e ult-u) -24-45 St6 (5+6) xicts= 6ucts XES) = YO3) = xes) HES) .-24.45 yes a ese to be S(5+6) ple)- ststo) = te A = Pes) xs/s=o= ut B = pcs) $+6)/s=-6 =

eus tes) = oeste) 65 Best 6) = be reus t 66+6) -6(t-u) Apply yet) = inverse laplace transform. -6t gult) - e ult) - euct-u) + e ult=4)ě tu+ estuctum 9#) = ult) - estuch y(t) = uct) obtect) + ēstuct-u) YES = uct) si-e6t] + estuct-u) the response yt). 19 yetyr -->t response is same as past @ Just we twritten in another form both are Eveivalent, at t=4 it reaches to 90+) = 1, for t>4 approssimately YCH) = u(t) = 1. so both are same