Signals - Integration of Heavyside Step & Dirac Delta Functions

[SpaceDomain]

Homework Statement

$$\int_{-\infty}^{\infty}{u(t)e^{-t}(\delta(t+1)+\delta(t-1))dt$$

Homework Equations

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

The Attempt at a Solution

$$\int_{-\infty}^{\infty}{u(t)e^{-t}(\delta(t+1)+\delta(t-1))dt$$

$$= \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t+1)dt} + \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t-1)dt}$$

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]

$$= \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t+1)dt} + \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t-1)dt}$$

$$= \int_{0}^{\infty}{e^{-t}{\delta(t+1)dt} + \int_{0}^{\infty}{e^{-t}{\delta(t-1)dt}$$

Is that correct?

 

[vela]

Homework Statement

$$\int_{-\infty}^{\infty}{u(t)e^{-t}(\delta(t+1)+\delta(t-1))dt$$

Homework Equations

$$\int_{-\infty}^{\infty}{u(t)dt = \left\{\begin{array}{cc}0,&\mbox{ if } t< 0\\t, & \mbox{ if } t>0\end{array}\right.$$

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

$$\int_{-\infty}^{\infty}{f(t)\delta(t-a)dt} = f(a)$$

The Attempt at a Solution

$$\int_{-\infty}^{\infty}{u(t)e^{-t}(\delta(t+1)+\delta(t-1))dt$$

$$= \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t+1)dt} + \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t-1)dt}$$

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]

$$= \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t+1)dt} + \int_{-\infty}^{\infty}{u(t)e^{-t}{\delta(t-1)dt}$$

$$= \int_{0}^{\infty}{e^{-t}{\delta(t+1)dt} + \int_{0}^{\infty}{e^{-t}{\delta(t-1)dt}$$

Is that correct?

Yes.

 

[SpaceDomain]

$$= \int_{0}^{\infty}{e^{-t}{\delta(t+1)dt} + \int_{0}^{\infty}{e^{-t}{\delta(t-1)dt}$$

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


Does that look right?

 

[vela]

$$= \int_{0}^{\infty}{e^{-t}{\delta(t+1)dt} + \int_{0}^{\infty}{e^{-t}{\delta(t-1)dt}$$

$$= [e^{-t}]_{t=-1} + [e^{-t}]_{t=1}$$



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



$$[e^{-t}]_{t=-1} + [e^{-t}]_{t=1} = \frac{1}{e}$$


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.