Handwrite and step by step solutions Consider the following ODEs modeling causal LTI systems at initial rest, withinput x(t) and output y(t).System 1:y(t) + 4ªy(t) + 8y(t) =x(t) + 3x(t).d²System 2: y(t) + 4⁄y(t) + 8y(t) =x(t) — 3x(t).dt(a) Is System 1 stable? Is it invertible with a causal stable inverse?(b) Is System 2 stable? Is it invertible with a causal stable inverse?(c) Find an explicit time domain expression for the response of System 1 to the input x(t)5 sin(2t - ).(d) Repeat (c) for System 2. How do the two responses differ?=

According to the question, Analysis of the system-1: The differential equation of the system is given as, d2dt2yt + 4ddtyt + 8yt=ddtxt + 3xt ;(1) Since output yt is depend on the present value of the input. Hence, the system is causal.(2) Let x(t)=1, Then the expression for the output is,d2dt2yt + 4ddtyt + 8yt= 3, ..............iLet x(t)=-1, Then the expression for the output is,d2dt2yt + 4ddtyt + 8yt=- 3, ..............iiSince different input leads different output. Hence, the system is Invertible.(3) By taking the Laplace transform of the given differential equation,d2dt2yt + 4ddtyt + 8yt=ddtxt + 3xt;We get,s2Ys+4sYs+8Ys=sXs+3Xs;The transfer function of the system is calculated as,Hs=YsXs=s+3s2+4s+8;Hs=YsXs=s+3s+22+22;By taking the inverse Laplace transform, we getht=L-1ss+22+22+3s+22+22;h(t)=e-2t[cos (2t) + 32sin(2t)]Since the impulse response of the system is bounded, and it it always provide bounded output in the response of the bounded input.Hence, the system is stable. (a) The given system is stable system. It also causal and invertible system.Analysis of the system-2: The differential equation of the system is given as, d2dt2yt + 4ddtyt + 8yt=ddtxt - 3xt ;(1) Since output yt is depend on the present value of the input. Hence, the system is causal.(2) Let x(t)=1, Then the expression for the output is,d2dt2yt + 4ddtyt + 8yt=- 3, ..............iLet x(t)=-1, Then the expression for the output is,d2dt2yt + 4ddtyt + 8yt=3, ..............iiSince different input leads different output. Hence, the system is Invertible.(3) By taking the Laplace transform of the given differential equation,d2dt2yt + 4ddtyt + 8yt=ddtxt - 3xt;We get,s2Ys+4sYs+8Ys=sXs-3Xs;The transfer function of the system is calculated as,Hs=YsXs=s-3s2+4s+8;Hs=YsXs=s-3s+22+22;By taking the inverse Laplace transform, we getht=L-1ss+22+22-3s+22+22;h(t)=e-2t[cos (2t) - 32sin(2t)]Since the impulse response of the system is bounded, and it it always provide bounded output in the response of the bounded input.Hence, the system is stable. (b) The given system is stable system. It also causal, and invertible system.(c) Calculation for the output of the system-1: xt=5 sin2t-π6;After expanding the above expression, we getxt=5sin2t.cosπ6-cos2t.sinπ6;xt=532sin2t-52cos2t;After taking the Laplace transform, we getXs=532.2s2+4 - 52.ss2+4;Xs=-5s + 1032s2+4The output of the system-1 is written as,Ys=Xs.Hs;Ys=-5s + 1032s2+4×s+3s2+4s+8;After taking the inverse Laplace, we get the output yt as,(c) Calculation for the output of the system-2: xt=5 sin2t-π6;After expanding the above expression, we getxt=5sin2t.cosπ6-cos2t.sinπ6;xt=532sin2t-52cos2t;After taking the Laplace transform, we getXs=532.2s2+4 - 52.ss2+4;Xs=-5s + 1032s2+4The output of the system-2 is written as,Ys=Xs.Hs;Ys=-5s + 1032s2+4×s-3s2+4s+8;After taking the inverse Laplace, we get the output yt as,