Dirac Delta

May 2008
138
0
Construct the convolution of g(t) and k(t)

\(\displaystyle g(t) = e^t, k(t) = \delta(t-2) \)

I first write

\(\displaystyle \int_0^t \delta(\tau-2)e^{t-\tau} d\tau \)

Can I make the same move I would make with the Heaviside step function and rewrite the integral as..

\(\displaystyle \int_2^t e^{t-\tau} d\tau \)

due to the fact that the function would be off while t < 2 ?

If not, can i use the sifting property and just evaluate \(\displaystyle e^{t-\tau}\) with t = 2, = \(\displaystyle e^{t-2} \). Or do the limits play a part here?

Would I also have to throw the heaviside in after that solution? Resulting in \(\displaystyle e^{t-2}H(t-2) \)

I know I'm pulling from properties about the Heaviside and Dirac, just a little confused.

Thanks
 
Feb 2007
666
199
USA
Sometimes these functions popping in and out can be confusing, but just remember that the equalities you remember should not change when you do these problems. The sifting property applies here as much as it would in any situation involving an integral of the product of the dirac function and an arbitrary function:

\(\displaystyle \int_0^t \delta{(\tau -2)}e^{t-\tau }~d\tau = g(t;\tau = 2) = e^{t-2}\)
 
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May 2008
138
0
Sometimes these functions popping in and out can be confusing, but just remember that the equalities you remember should not change when you do these problems. The sifting property applies here as much as it would in any situation involving an integral of the product of the dirac function and an arbitrary function:

\(\displaystyle \int_0^t \delta{(\tau -2)}e^{t-\tau }~d\tau = g(t;\tau = 2) = e^{t-2}\)
Perfect thanks for the confirmation!

As with the heaviside step function, this result would not kick in until t> 2. It that also the case here?

I ask because I am now to graph this function over 0<t<5, and am not sure if it is as simple as just graphing \(\displaystyle e^{t-2}\)
 
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Feb 2007
666
199
USA
Perfect thanks for the confirmation!

As with the heaviside step function, this result would not kick in until t> 2. It that also the case here?

I ask because I am now to graph this function over 0<t<5, and am not sure if it is as simple as just graphing \(\displaystyle e^{t-2}\)
I've edited this a few times, this is the final edit.

Be careful here, the integral is varying tau, not t. So, you're asking if the integral's lower limit can be changed to 2 instead. The answer is yes. Would it make a difference? No. The sifting property would render the integral result to be exactly the same. The Heaviside function is the same way, except the reason why you alter the limits is because you do not have to eliminate intervals on both sides of the constant, just on one. So one side of the interval is still needed to solve the problem, whereas in the Dirac Delta version, you are just shaving every value in the interval off except the constant at which the Dirac Delta has a nonzero value, so there is no need to concern yourself with what happens anywhere else but at that particular point.

The result of the integral is a function of time, so all you have to do is graph it from t = 0 to t = 5. You must remember that convolution integrals vary a dummy variable, not t itself. So the function that results from the convolution is a function of time.
 
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