# Laplace transform breaks

Discussion in 'Math' started by logicman112, Feb 21, 2011.

1. ### logicman112 Thread Starter Active Member

Dec 27, 2008
69
3
If we have a linear system with the following differential equation:
d2y/dt2+7*dy/dt+12*y=dx/dt+5*x
So H(s) = (s+5)/((s+4)*(s+3)) = -1/(s+4)+2/(s+3)--> h(t) = (-e^(-4*t)+2*e^(-3*t))*u(t)
(Please calculate the unit impulse response by Laplace transform and verify the result by yourself)
Why this answer does not satisfy the differential equation?
h(t) = (-2*e^(-4*t)+3*e^(-3*t))*u(t)

2. ### logicman112 Thread Starter Active Member

Dec 27, 2008
69
3
please look at the following equation and Laplace transform gives the correct result this time:
d2y/dt2+3*dy/dt+2*y(t) = dx/dt+5* x(t)--> H(s) = (s+5) / [(s+2)*(s+1)]-->
h(t) = ( -3*e^(-2*t)+ 4*e^(-t) ) * u(t)

3. ### Vahe Member

Mar 3, 2011
75
9
I think your results are all correct. And in the first post, the "right answer" seems to be incorrect. I cannot find anything wrong with your approach or your results.

I re-calculated the results using the method of residues and got the same results.

Cheers,
Vahe

4. ### logicman112 Thread Starter Active Member

Dec 27, 2008
69
3
The right answer satisfies the differential equation. Please give that a try. It is strange why the impulse response does not satisfy the differential equation while it is its answer by Laplace transform.

5. ### steveb Senior Member

Jul 3, 2008
2,433
469
It's not clear why you say the method gives the wrong answer because it seems to satisfy the diff equation.

Last edited: Mar 6, 2011
6. ### logicman112 Thread Starter Active Member

Dec 27, 2008
69
3
If h(t)=(-e^(-4*t)+2*e^(-3*t))*u(t) is the impulse response, as its
Laplace transform suggests, so it must satisfy
"d2y/dt2+7*dy/dt+12*y=dx/dt+5*x" so(using the chain rule and our input
is the unit impulse function, delta(t) ):

y(t) = (-e^(-4*t)+2*e^(-3*t))*u(t) and x(t) = delta(t)

dy/dt = (4*e^(-4*t)-6*e^(-3*t))*u(t)+(-e^(-4*t)+2*e^(-3*t)) * delta(t)
d2y/dt2 = (-16*e^(-4*t)+18*e^(-3*t))*u(t)+[4*e^(-4*t)-6*e^(-3*t)+4*e^(-4*t)-6*e^(-3*t)]*delta(t)+
(-e^(-4*t)+2*e^(-3*t))*d(delta(t))/dt

d2y/dt2+7*dy/dt+12*y = (-e^(-4*t)+2*e^(-3*t))*d(delta(t))/dt
+ [8*e^(-4*t)-12*e^(-3*t)-7*e^(-4*t)+14*e^(-3*t)]*delta(t)+
[-16*e^(-4*t)+18*e^(-3*t)+28*e^(-4*t)-42*e^(-3*t)-12*e^(-4*t)+24*e^(-3*t))*u(t)=
(4*e^(-4*t)-6*e^(-3*t)) d(delta(t))/dt +
[e^(-4*t)+2*e^(-3*t)]*delta(t)+0*u(t) --->

d2y/dt2+7*dy/dt+12*y = (-e^(-4*t)+2*e^(-3*t)) *d(delta(t))/dt +
[e^(-4*t)+2*e^(-3*t)]*delta(t) = d(delta(t))/dt+5*delta(t)

so the coefficient of delta(t) is 3 in the left side while it is 5 in
the other side! and it seems the equality can not be satisfied.

If (-e^(-4*t)+2*e^(-3*t))*u(t) is the impulse response why it does not
satisfy the differential equation?

7. ### steveb Senior Member

Jul 3, 2008
2,433
469
Ah, I see. You are worried about the point t=0. The equation is OK everywhere but there. I'm not a mathemetician, but I would not expect to have things well behaved at this singularity. The point t=0 becomes a nasty place for differential equations when an input impulse function (and in this case even its derivative) enter. The system impulse response helps you reconstruct the response to other less-nasty input functions, which would then have agreement at t=0.

Maybe there is a mathematican, or signals and systems expert, that can shed better light on this, but basically I wouldn't worry about singular points if you are just interested in engineering application of the theory.

From a practical point of view, the impulse response is nearly the response you would get from a very narrow input pulse with area of one. In that sense, it satisfies the system differential equation.