Signals - Integration of Heavyside Step & Dirac Delta Functions

[SpaceDomain]

Homework Statement

<br /> \int_{-\infty}^{\infty}{u(t)e^{-t}(\delta(t+1)+\delta(t-1))dt<br />

Homework Equations

<br /> \int_{-\infty}^{t}{u(t)dt = \left\{\begin{array}{cc}0,&amp;\mbox{ if }<br /> t&lt; 0\\t, &amp; \mbox{ if } t&gt;0\end{array}\right.<br /><br /> \int_{-\infty}^{\infty}{f(t)\delta(t-a)dt} = f(a)<br />

The Attempt at a Solution

<br /> \int_{-\infty}^{\infty}{u(t)e^{-t}(\delta(t+1)+\delta(t-1))dt<br />

<br /> = \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t+1)dt}<br /> + \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t-1)dt}<br />

I obviously could use the second relevant equation if the u(t) term was not in these integrals.

I am stuck. Could someone point me in the right direction?

 

[SpaceDomain]

<br /> = \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t+1)dt}<br /> + \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t-1)dt}<br />

<br /> = \int_{0}^{\infty}{e^{-t}{\delta(t+1)dt}<br /> + \int_{0}^{\infty}{e^{-t}{\delta(t-1)dt}<br />

Is that correct?

 

[vela]

Homework Statement

<br /> \int_{-\infty}^{\infty}{u(t)e^{-t}(\delta(t+1)+\delta(t-1))dt<br />

Homework Equations

<br /> \int_{-\infty}^{\infty}{u(t)dt = \left\{\begin{array}{cc}0,&amp;\mbox{ if }<br /> t&lt; 0\\t, &amp; \mbox{ if } t&gt;0\end{array}\right.<br />

The upper limit of the integral should be t, not ∞.

<br /> \int_{-\infty}^{\infty}{f(t)\delta(t-a)dt} = f(a)<br />

The Attempt at a Solution

<br /> \int_{-\infty}^{\infty}{u(t)e^{-t}(\delta(t+1)+\delta(t-1))dt<br />

<br /> = \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t+1)dt}<br /> + \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t-1)dt}<br />

I obviously could use the second relevant equation if the u(t) term was not in these integrals.

I am stuck. Could someone point me in the right direction?

Why does the presence of u(t) stop you? What's the definition of the Heaviside step function?

 

[vela]

<br /> = \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t+1)dt}<br /> + \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t-1)dt}<br />

<br /> = \int_{0}^{\infty}{e^{-t}{\delta(t+1)dt}<br /> + \int_{0}^{\infty}{e^{-t}{\delta(t-1)dt}<br />

Is that correct?

Yes.

 

[SpaceDomain]

<br /> <br /> = \int_{0}^{\infty}{e^{-t}{\delta(t+1)dt}<br /> + \int_{0}^{\infty}{e^{-t}{\delta(t-1)dt}<br /> <br />

<br /> <br /> = [e^{-t}]_{t=-1}<br /> + [e^{-t}]_{t=1}<br /> <br />
[e^{-t}]_{t=-1} = 0 because t=-1 is out of the limits of integration.
[e^{-t}]_{t=-1} <br /> + [e^{-t}]_{t=1} <br /> = \frac{1}{e}<br />


Does that look right?

 

[vela]

<br /> <br /> = \int_{0}^{\infty}{e^{-t}{\delta(t+1)dt}<br /> + \int_{0}^{\infty}{e^{-t}{\delta(t-1)dt}<br /> <br />

<br /> <br /> = [e^{-t}]_{t=-1}<br /> + [e^{-t}]_{t=1}<br /> <br />



[e^{-t}]_{t=-1} = 0 because t=-1 is out of the limits of integration.



[e^{-t}]_{t=-1} <br /> + [e^{-t}]_{t=1} <br /> = \frac{1}{e}<br />


Does that look right?

Yes and no. You figured out the answer correctly, but what you wrote isn't correct. For one thing, the first integral isn't equal to [e^{-t}]_{t=-1} since [e^{-t}]_{t=-1}=e. Second, [e^{-t}]_{t=-1} is always equal to e; it's never equal to 0. What you should have written is simply that the first integral is equal to 0 because the delta function is zero over the interval of integration.