H Thread Starter hamza324 Joined Jul 10, 2011 33 Sep 28, 2011 #1 i just want to verify the answer i am getting is correct or not. -y'' + 3y' = 3x + 5x'' now h(t) = - 18 e^(-3t) - 5 using some other method i got h(t) = - 16 e^(-3t) - 3 can some bdy help me out plzz. thnx
i just want to verify the answer i am getting is correct or not. -y'' + 3y' = 3x + 5x'' now h(t) = - 18 e^(-3t) - 5 using some other method i got h(t) = - 16 e^(-3t) - 3 can some bdy help me out plzz. thnx
t_n_k Joined Mar 6, 2009 5,455 Sep 28, 2011 #2 Using Laplace \(-y^{''}+3y^'=3x+5x^{''}\) Transforming \(-s^2Y(s)+3sY(s)=3X(s)+5s^2X(s)\) \(\frac{Y(s)}{X(s)}=\frac{5s^2+3}{-s^2+3s}\) \(\frac{Y(s)}{X(s)}=\frac{-5s^2-3}{s(s-3)}\) \(\frac{Y(s)}{X(s)}=\frac{-5s}{s-3}-\frac{3}{s(s-3)}\) Inverse Transform for impulse response h(t) \(h(t)=-5(\delta(t)+3e^{3t})+(1-e^{3t})\) \(h(t)=-5\delta(t)+1-16e^{3t}\) which is an unusual result in that the function increases unbounded with time. Are you sure of the original equation? Clearly you now have three different results.
Using Laplace \(-y^{''}+3y^'=3x+5x^{''}\) Transforming \(-s^2Y(s)+3sY(s)=3X(s)+5s^2X(s)\) \(\frac{Y(s)}{X(s)}=\frac{5s^2+3}{-s^2+3s}\) \(\frac{Y(s)}{X(s)}=\frac{-5s^2-3}{s(s-3)}\) \(\frac{Y(s)}{X(s)}=\frac{-5s}{s-3}-\frac{3}{s(s-3)}\) Inverse Transform for impulse response h(t) \(h(t)=-5(\delta(t)+3e^{3t})+(1-e^{3t})\) \(h(t)=-5\delta(t)+1-16e^{3t}\) which is an unusual result in that the function increases unbounded with time. Are you sure of the original equation? Clearly you now have three different results.
H Thread Starter hamza324 Joined Jul 10, 2011 33 Sep 28, 2011 #3 Hi i got your answer..... but my professor told me slightly different procedure.... i am getting the answer above except the impulse function in the asnwer.. i.e h(t) = - 16 e^(3t) + 1 but missing that impulse thing.. i have attached the file...plz give it a look Attachments prob 2.pdf 364.1 KB Views: 9
Hi i got your answer..... but my professor told me slightly different procedure.... i am getting the answer above except the impulse function in the asnwer.. i.e h(t) = - 16 e^(3t) + 1 but missing that impulse thing.. i have attached the file...plz give it a look
t_n_k Joined Mar 6, 2009 5,455 Sep 28, 2011 #4 hamza324 said: i am getting the answer above except the impulse function in the asnwer.. i.e h(t) = - 16 e^(3t) + 1 but missing that impulse thing.. Click to expand... Yes - blink and you'll miss it. One would have to ensure that the boundary conditions are met. At t=0 your answer is h(0)=0. What will you see in practice? I suspect an impulse output y(t) at t=0 is likely, due to the presence of the second derivative of the input x(t).
hamza324 said: i am getting the answer above except the impulse function in the asnwer.. i.e h(t) = - 16 e^(3t) + 1 but missing that impulse thing.. Click to expand... Yes - blink and you'll miss it. One would have to ensure that the boundary conditions are met. At t=0 your answer is h(0)=0. What will you see in practice? I suspect an impulse output y(t) at t=0 is likely, due to the presence of the second derivative of the input x(t).
H Thread Starter hamza324 Joined Jul 10, 2011 33 Sep 28, 2011 #5 i did the changed at h(0) at t=0 but answer isn't matching.. but i understand your point so i think i would have to add this impulse at the end once i get the h(t) = - 16 e^(3t) + 1 expression.....am i right..
i did the changed at h(0) at t=0 but answer isn't matching.. but i understand your point so i think i would have to add this impulse at the end once i get the h(t) = - 16 e^(3t) + 1 expression.....am i right..
t_n_k Joined Mar 6, 2009 5,455 Sep 28, 2011 #6 By definition, the delta impulse function only exists at t=0. Elsewhere it is zero so the function reduces to h(t) = - 16 e^(3t) + 1 This is still an increasing value function with time given the positive exponential index. Last edited: Sep 28, 2011
By definition, the delta impulse function only exists at t=0. Elsewhere it is zero so the function reduces to h(t) = - 16 e^(3t) + 1 This is still an increasing value function with time given the positive exponential index.