Jun 11, 2005 #1 W wcz Member level 2 Joined Jul 17, 2004 Messages 49 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 582 Given X(s) = [s+1]/[s^2 + 5s + 7] What is V(s) if a) v(t) = x(3t-4) u(3t-4) b) v(t) = d^2 x(t)/dt^2 The given answer are a) V(s) = [s+3] / [s^2 + 15s + 63] e^-4s/3 b) V(s) = [13s+28] / [s^2 + 5s + 7] I got different solutions for both. Could anyone show me the way to get the answers? Thanks.
Given X(s) = [s+1]/[s^2 + 5s + 7] What is V(s) if a) v(t) = x(3t-4) u(3t-4) b) v(t) = d^2 x(t)/dt^2 The given answer are a) V(s) = [s+3] / [s^2 + 15s + 63] e^-4s/3 b) V(s) = [13s+28] / [s^2 + 5s + 7] I got different solutions for both. Could anyone show me the way to get the answers? Thanks.
Jun 12, 2005 #2 Z zorro Advanced Member level 4 Joined Sep 6, 2001 Messages 1,130 Helped 357 Reputation 712 Reaction score 298 Trophy points 1,363 Location Argentina Activity points 8,913 Hi wcz, a) v(t) = x[3(t-4/3)]*u[3(t-4/3)], and provided that x(t)=0 for t<0, then x(t)=x(t)*u(t) and v(t) = x[3*(t-4/3)] . We have: a scaling in time by 3: Let y(t)=x(3*t) Y(s)=1/3 * X(s/3) a translation in time by 4/3: v(t) = y(t-4/3) V(s) = Y(s)*exp(-4*s/3); This is the same as the given answer. b) It should be Y(s) = s^2 * X(s), with a different result than the given answer. Regards Z
Hi wcz, a) v(t) = x[3(t-4/3)]*u[3(t-4/3)], and provided that x(t)=0 for t<0, then x(t)=x(t)*u(t) and v(t) = x[3*(t-4/3)] . We have: a scaling in time by 3: Let y(t)=x(3*t) Y(s)=1/3 * X(s/3) a translation in time by 4/3: v(t) = y(t-4/3) V(s) = Y(s)*exp(-4*s/3); This is the same as the given answer. b) It should be Y(s) = s^2 * X(s), with a different result than the given answer. Regards Z
Jun 12, 2005 #3 W wcz Member level 2 Joined Jul 17, 2004 Messages 49 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 582 Thanks Z. I'm having another problem in solving Inverse Laplace Transform. Could anyone show me the way to solve the Inverse laplace transform of the following Q, X(s) = [ s^2 - 2s + 1 ] / [ s^2 ( s^2 + 4) ]
Thanks Z. I'm having another problem in solving Inverse Laplace Transform. Could anyone show me the way to solve the Inverse laplace transform of the following Q, X(s) = [ s^2 - 2s + 1 ] / [ s^2 ( s^2 + 4) ]
Jun 13, 2005 #4 S steve10 Full Member level 3 Joined Mar 26, 2002 Messages 175 Helped 32 Reputation 64 Reaction score 0 Trophy points 1,296 Location Los Angeles (Chinese) Activity points 2,538 Factorization gives: -1/(2s)+1/(4s^2) +(s/2+3s/4)/(s^2+4) and check the formulas of laplace transformation.
Factorization gives: -1/(2s)+1/(4s^2) +(s/2+3s/4)/(s^2+4) and check the formulas of laplace transformation.
Jun 13, 2005 #5 A amraldo Advanced Member level 4 Joined Aug 29, 2004 Messages 1,183 Helped 145 Reputation 290 Reaction score 37 Trophy points 1,328 Location Egypt Activity points 5,880 steve10 said: Factorization gives: -1/(2s)+1/(4s^2) +(s/2+3s/4)/(s^2+4) and check the formulas of laplace transformation. Click to expand... & i agree with that but i think factorization will give -1/2s +1/4s^2 + (s/2 + 3/4 )/(s^2+4)
steve10 said: Factorization gives: -1/(2s)+1/(4s^2) +(s/2+3s/4)/(s^2+4) and check the formulas of laplace transformation. Click to expand... & i agree with that but i think factorization will give -1/2s +1/4s^2 + (s/2 + 3/4 )/(s^2+4)
Jun 13, 2005 #6 S steve10 Full Member level 3 Joined Mar 26, 2002 Messages 175 Helped 32 Reputation 64 Reaction score 0 Trophy points 1,296 Location Los Angeles (Chinese) Activity points 2,538 Thanks, amraldo. That was a typo.
Jun 15, 2005 #7 S suvendu Full Member level 3 Joined Oct 10, 2004 Messages 167 Helped 16 Reputation 32 Reaction score 3 Trophy points 1,298 Activity points 2,121 you can take help of ziemer and tranter book and see laplace transform chapter.