The zero state response does not satisfy the diff equation. Why?

[logicman112]

If we have a linear system(it could be a RLC circuit) 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(which is Zero-State response) does not satisfy the differential equation?
The right answer is:
h(t) = (-2*e^(-4*t)+3*e^(-3*t))*u(t)

 

[logicman112]

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?

 

[uart]

Good question logicman112, and it's answer goes right to the heart of why we must be very careful when using the impulse function, \delta(t).

First off you need to know that the unit impulse function does not exist, it's a useful as a "limiting case" abstraction but it does not actually exist in nature. The only place the impulse function actually makes sense is inside an integral, use it any other way and you'd better be careful.

Now as bad as the unit impulse is it's derivative is infinitely worse. Imagine for example (by way of analogy) you had both sides of an algebraic equation divided by zero. It would be hard to say anything conclusive about the terms in that equation, right? Same thing with your equation that has derivatives of delta functions on both sides, it means you have to re-consider what you're doing and how you're doing it.

I suggest you look at the following integral equation and confirm that your proposed impulse response does indeed work correctly.

\frac{dy}{dt} + 7 y + 12 \int_{0}^{t} y(\lambda)\,d\lambda \,=\, x + 5 \int_{0}^{t} x(\lambda)\,d\lambda

It's easy to show that the LHS and RHS of the above are equal for x(t) = \delta(t) and y(t) =2 e^{-3t} - e^{-4t} \,\, : \, x>0, \, \left\{ y(t)=0 \,\,:\, x<0 \right\}.

 

[logicman112]

Thanks uart. As you wrote, it seems that the chain rule of derivatives becomes false for f(t)*delta(t)