Hi, I am trying to take the inverse laplace transform from (Signal and System
with Matlab Application by Steven T. Karris) I(s) = (s^2+2*s-1)/(2*s^3+9*s^2+6*s+3) I use the residue command >> n = [1 2 -1]; >> d = [2 9 6 3]; >> [r,p,k] = residue(n,d) r = 0.24020 + 0.00000i 0.12990 + 0.23250i 0.12990 - 0.23250i p = -3.81700 + 0.00000i -0.34150 + 0.52570i -0.34150 - 0.52570i k = [](0x0) from this I got the inverse laplace transform of *0.24*exp(-3.8*t)+exp(-0.34*t)(0.26*cos(0.53*t)-0.46*sin(0.53*t)* In the book, he got answer of *0.48*exp(-3.8*t)+exp(-0.34*t)(0.52*cos(0.53*t)-0.92*sin(0.53*t) * which is just a factor of 2 from my solution. In the book he did not use the residue command. He did it partial fraction expansion: (s^2+2*s-1)/(s+3.817)*(s^2+0.683*s+0.393) = r1/(s+3.817)+(r2*s+r3)/(s^2+0.683*s+0.393) found r1 = 0.48, r2 = 0.52, r3 = -0.31, thus 0.48/(s+3.817)+(0.52*s-0.31)/(s^2+0.683*s+0.393) He got the answer as (a scale factor of 2 from my solution). 0.48*exp(-3.82*t)-0.93*exp(-0.34*t)*sin(0.53*t)+0.53*exp(-0.34*t)*cos(0.53*t) But which one is correct? I checked both ways of doing it and they seem to be correct. What am I doing wrong here? Thanks for the help in advance. Anthony -- Sent from: http://octave.1599824.n4.nabble.com/Octave-General-f1599825.html |
On 2018-07-15 11:03 AM, aheak123 wrote:
> Hi, I am trying to take the inverse laplace transform from (Signal and System > with Matlab Application by Steven T. Karris) > > I(s) = (s^2+2*s-1)/(2*s^3+9*s^2+6*s+3) > > I use the residue command >>> n = [1 2 -1]; >>> d = [2 9 6 3]; >>> [r,p,k] = residue(n,d) > r = > > 0.24020 + 0.00000i > 0.12990 + 0.23250i > 0.12990 - 0.23250i > > p = > > -3.81700 + 0.00000i > -0.34150 + 0.52570i > -0.34150 - 0.52570i > > k = [](0x0) > > from this I got the inverse laplace transform of > > *0.24*exp(-3.8*t)+exp(-0.34*t)(0.26*cos(0.53*t)-0.46*sin(0.53*t)* > > In the book, he got answer of > > *0.48*exp(-3.8*t)+exp(-0.34*t)(0.52*cos(0.53*t)-0.92*sin(0.53*t) > * > > which is just a factor of 2 from my solution. In the book he did not use > the residue command. > > He did it partial fraction expansion: > > (s^2+2*s-1)/(s+3.817)*(s^2+0.683*s+0.393) = > r1/(s+3.817)+(r2*s+r3)/(s^2+0.683*s+0.393) > > found r1 = 0.48, r2 = 0.52, r3 = -0.31, thus > > 0.48/(s+3.817)+(0.52*s-0.31)/(s^2+0.683*s+0.393) > > He got the answer as (a scale factor of 2 from my solution). > 0.48*exp(-3.82*t)-0.93*exp(-0.34*t)*sin(0.53*t)+0.53*exp(-0.34*t)*cos(0.53*t) > > But which one is correct? I checked both ways of doing it and they seem to > be correct. What am I doing wrong here? I'm not sure. I think you're asking a math question rather than an Octave question. One way to check the two solutions would be to compute the (forward) Laplace transform and check which one gives I(s). >> pkg load symbolic >> t >> B = 0.48*exp(-3.8*t) + exp(-0.34*t)*(0.52*cos(0.53*t)-0.92*sin(0.53*t)) (Note this will give lots of warnings because of those decimal places, but should be close enough for a rough test). >> laplace(B) ans = (sym) ⎛ 2 ⎞ 4⋅⎝12500⋅s + 24895⋅s - 12384⎠ ───────────────────────────────────────── ⎛ 3 2 ⎞ 5⋅⎝10000⋅s + 44800⋅s + 29805⋅s + 15067⎠ Note that is very close to I. But if you do this with your 0.24*exp()..., the result is also off by a factor of two. best, Colin |
Hi Colin, thanks for the reply. I am using Octave, but basically asking a
math question. I did a forward laplace transform on my time domain solution and it came pretty close to the original I(s). The answer from the book is off by a factor of 2. Regards, aheak123 -- Sent from: http://octave.1599824.n4.nabble.com/Octave-General-f1599825.html |
On 07/17/18 21:17, aheak123 wrote:
> Hi Colin, thanks for the reply. I am using Octave, but basically asking a > math question. I did a forward laplace transform on my time domain solution > and it came pretty close to the original I(s). The answer from the book is > off by a factor of 2. > I think I sent a reply to Colin, not the list. Maple 2018 agrees with p and r. > lprint(evalf(soln)); .2401970034*exp(-3.816996842*t)+(.1299014983-.2324973824*I)*exp((-.3415015788-.5256955164*I)*t)+(.1299014983+.2324973824*I)*exp((-.3415015788+.5256955164*I)*t) |
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