Impulse Response using Dirac(t)

Consider the following code segment used to generate the step response

   syms y t 
   y=dsolve('D2y+5*Dy+6*y=heaviside(t)','y(0)=0','Dy(0)=0','t');

Now, the impulse response can be determined (and plotted) using

   ir = diff(y);
   subplot(211)
   ezplot(ir,[0,5]);
   title('Impulse response from Step reponse');

However, when I try to find (and plot) the impulse response using the following code, it gives the same shape as before, but with (almost) half the magnitude.

   x=dsolve('D2y+5*Dy+6*y=dirac(t)','y(0)=0','Dy(0)=0','t');
   subplot(212)
   ezplot(x,[0,5]);
   title('Impulse response From dirac')

Although I personally think that the answer must be the same, can anyone explain why it isn't the case??

Answer 1

[Teja Muppirala]

I think this is a very good question. Here is my explanation, but maybe someone can give a better one.

In this equation:

dsolve('D2y+5*Dy+6*y=dirac(t)','y(0)=0','Dy(0)=0','t');

Dy is actually discontinuous at the origin. The way the initial conditions are interpreted is that Dy(0) means the average of Dy(0+) and Dy(0-).

Here (0+) means the limit from the positive side, (0-) means the limit from the negative side. If you look at x around the origin, you will indeed see that the derivative is actually discontinuous at the origin. At (0-) it is negative, and at (0+) it is positive with the same magnitude so that the AVERAGE value is zero, It's a little different than how it is defined for Laplace transforms, where initial conditions are taken at (0-).

Try this for example:

dsolve('Dy = dirac(t)', 'y(0) = 0')

And you will see that it is not exactly the step function, but rather a step function that is shifted down by one half to make the average value at y(0) = 0.

ans = heaviside(t) - 1/2

In fact these two evaluate to be the same:

dsolve('D2y+5*Dy+6*y=dirac(t)','y(0)=0','Dy(0)=0','t');
diff(dsolve('D2y+5*Dy+6*y=heaviside(t)-1/2','y(0)=0','Dy(0)=0','t'));

Note, also, that the following two expressions where I simply added a delay will also give you the same answers, because in this case there is no discontinuity at the t = 0 and the initial conditions are in agreement regardless of which definition you use:

dsolve('D2y+5*Dy+6*y=dirac(t-1)','y(0)=0','Dy(0)=0','t');
diff(dsolve('D2y+5*Dy+6*y=heaviside(t-1)','y(0)=0','Dy(0)=0','t');

[Ultrazord]

Nice one.

So, I'm inferring that there is no workaround for the differential equation with the delta function, given that the delay is 0.

Is it so???

[Teja Muppirala]

There is a very easy workaround. Use limits to enforce your boundary conditions:

x = dsolve('D2y+5*Dy+6*y=dirac(t)','y(e)=0','Dy(e)=0','t');

x = limit(x,'e',0,'left')

ezplot(x,[-1 5])