Question

1. A Consider the following nonhomogeneous differential equation: j(t) + (a - b)y(t) - aby(t) = x(t). Assume a and b are both strictly positive. The answers to nearly all of the questions below will be in terms of a and b. (a) (5 points) Is this system internally stable or unstable? Why? (b) (10 points) For arbitrary inital conditions yo and yo, write the zero-input response (ZIR) for t > 0. (c) (10 points) Derive this system's impulse response in the time domain. Are there any additional conditions we can place on a or b to ensure this system is externally (BIBO) stable? (d) (5 points) Identify one set of non-zero initial conditions that produce a bounded ZIR. Explain to the reader why the values you chose will work. (e) (5 points) Identify the set of all initial conditions that produce an unbounded ZIR. Explain to the reader why the relationship you identified produces unbounded responses.

Answer 1

Since Let's you only wont e) part take Lapuce transform . - را ر = ( واطع اعلاه وو - 870) - SY(0) -Y²60 = 0 Since xitle - ta-b) sYG) -a -b) 7/6) - abyco. 7(s) (2 + 6-b/s-ub) = Y(0) (Stab) + y (0) /6)=Y6) (sta-b) tylo] 12 Kath)s - ab 2

DATE Y() = (s-b) tu yol tycol -(S-b(sta) (S-besta XL = Y/6/5=b) taylo) tyllo) -b (sta) S-bista) (S-besta Now stb term canceld out Here we need to keep 7(0) = 0 so that the s-h term remains which leads to unboundeed response VG)= y'rol 710)=k (S-bestan) when kis any zational no ench - 0 -

y lo ek A TB Sta s-b AS- Ab ABst Bu = A = - B - Car - Abt By = IC B (at) - K B = TC. A = - K atba th Y6 = - K tic S ath) (sta) cath) co-bo y ct) ==Keat (6) t K et ult ath unbounded