A linear time-invariant circuit has the transfer function H(s)= (l6)/ (s 2 +10s

Question

A linear time-invariant circuit has the transfer function
H(s)= (l6)/ (s2 +10s + 16)
This circuit has been at rest for t < 0. For t > 0, x (t)=10.
1.1 Derive Y(s) - the Laplace transform of the output.
1.2 Derive y(t)
1.3 Suppose the input is x(t) [2 exp (-2 t)] u(t)
Derive the forced component of the response y (t)
1.4 Suppose the input is x(t)= 10 cos 4t, derive the steady state
output
1.5 Identify the amplitude and phase of this output

Explanation / Answer

H(s) = 16/(s^2 + 2(s)(5) + 25 -9) = 16/((s+5)2 +(-3)2) 1.1)As we know that Y(s) = X(s).H(s) = 10/s . 16/((s+5)2 + (-3)2) =160/s((s+5)2 + (-3)2) 1.2)Now y(t) = inverse laplace (Y(s)) = inverse laplace (-120. (-3)/s((s+5)2 + (-3)2) )= -120 e-5t sin (-3t) = 120 e-5t sin3t u(t) 1.3)Now the input is 2 e-2t u(t) and we have to find the forced component of y(t) Y(s) = 2/s(s-2) .16/((s+5)2 +(-3)2) using partial fractions to solve , 2/s(s-2) .16/((s2+10s +16) = A/s + B/(s-2) + (Cs2 +Ds+E)/(s2 +10s +16) A=32/(-2)(16) =-1 B=32/(2)(40) =2/5 Applying decomposition method 32= A(s-2)(s2 +10s +16) +Bs(s2 +10s +16) +(Cs2 +Ds+E)(s2-2s) 32= (As-2A)(s2 +10s +16) +Bs3 +10Bs2 +16Bs +Cs4 -2Cs3 +Ds3 -2Ds2 +Es2 -2Es 32= As3 +10As2 +16As -2As2 -20As -32A +Bs3 +10Bs2 +16Bs +Cs4 -2Cs3 +Ds3 -2Ds2 +Es2 -2Es C=0; 32=(A+B-2C+D)s3 + (8A+10B-2D+E)s2 +(-4A+16B-2E)s -32A =>-1 +2/5 +D=0 => D=1-2/5 = 3/5 Also -8+4-6/5 +E=0 => E=4+6/5 = 26/5 so we have 2/s(s-2) .16/((s2+10s +16) = A/s + B/(s-2) + (Cs2 +Ds+E)/(s2 +10s +16) =-1/s + 2/5(s-2) + (3s/5 +26/5)/(s2 +10s +16) so inverse laplace will be -u(t) +(2/5) e-2t + 29s/5((s+5)2 +(-3)2 ) -u(t) +(2/5) e-2t +29/5 e-5t cos(3t) 1.4) steady state response=? x(t) = 10 cos(4t) steady state response = H(s).X(s) = 10 s/(s2 +16) . 16/((s+5)2 +(-3)2) = 160 s /(s2 +16)((s+10s +16) now using partial fractions to solve it we will have =160 s /(s2 +16)((s2+10s +16) = (As+B) /(s2 +16) + (Cs2 + Ds +E)/(s2+10s +16) 160s =(As+B)(s2 +10s+16) + (Cs2+Ds+E)(s2 +16) 160s = As3 + 10As2 + 16As +Bs2 +10Bs +16B +Cs3 +16Cs2 +Ds3 +10 Ds2 +16Ds +Es2 +10Es +16E 0s3 +0s2 +160s +0 =(A+C+D)s3 + (10A+B+16C+10D+E)s2 +(16A+10B+16D+10E)s + (16B+16E) A+C+D =0; B+E=0; 10A +B+16C+10D+E =0 => 10A+10D +16C =0; 16A+10B+16D+10E =160 => 16(A+D) =160 => A+D=10 => C=-10 10(A+D) +B+E=160 =>10(A+10-A) +B+E=160 =>B+E =60 solve these equation urself v) once the laplace equation is determined ,then amplitude and phase can easily be determined

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