# How to handle delta function after finding the impulse response?

I am pretending that laplace does not exist because I am being tested on these concepts separately.

Essentially, I have solved for the step response of a first order circuit and found it to be:

$$v_{c}(t)=\left(\frac{-5}{99}e^{-5t}+\frac{106}{99}e^{\frac{-t}{20}}\right)u(t)$$

I'm only dealing with LTI systems so I know that the impulse is the derivative of the step, but I will be left with some terms attached to the delta function, and some terms attached to the unit step function.

$$h(t)= \left( \frac{-5}{99}e^{-5t}+\frac{106}{99}e^{\frac{-t}{20}} \right)\delta(t) + \left( \frac{25}{99}e^{-5t}+\frac{53}{990}e^{\frac{-t}{20}} \right)u(t)$$

If I want to use the impulse response in the convolution integral, how do I handle these delta terms to make it less... convoluted? Do they reduce to a constant?

• A property of the delta function is $$\small \int_0^\infty f(t)\delta(t)=f(0)$$ Hence the convolution integral will include the constant value $$\small \frac{-5}{99}+\frac{106}{99}=\frac{101}{99}$$
– Chu
Oct 24, 2020 at 10:40

You don't have to worry about $$\\delta(t)\$$ since the integral of it results in $$\u(t)\$$. Even integrating it alone gives $$\\int_0^x{\delta(\pm t)\text{d}t}=2u(x)-1\$$. So whatever convolutions you'll have with $$\h(t)\$$ will include the step function in the result. BTW, the derivative is with $$\-\frac{53}{990}\$$ in the 2nd term.
Since $$\\delta(t) = 0\$$ when $$\t \ne 0\$$, you should be able to replace:
$$f(t)\cdot\delta(t)$$ with $$f(0)\cdot\delta(t)$$