Convolution Problem: Solve 2e^{-t}u(t) \ast e^{t}u(-t)

[jegues]

Homework Statement

Convolve,

2e^{-t}u(t) \ast e^{t}u(-t) \quad \text{Where u(t) is the step function}

Homework Equations

The Attempt at a Solution

2e^{t}\int_{?}^{?}e^{-\tau}u(\tau) e^{-\tau}u(-t+\tau)d \tau

I'm not sure what the limits on my integral should be, the u(-t) is confusing me. Shouldn't the second function only be nonzero from -infinity to 0 and the other from to +infinity?

Thanks again!

 

[LCKurtz]

Homework Statement

Convolve,

2e^{-t}u(t) \ast e^{t}u(-t) \quad \text{Where u(t) is the step function}

Homework Equations

The Attempt at a Solution

2e^{t}\int_{?}^{?}e^{-\tau}u(\tau) e^{-\tau}u(-t+\tau)d \tau

I'm not sure what the limits on my integral should be, the u(-t) is confusing me. Shouldn't the second function only be nonzero from -infinity to 0 and the other from to +infinity?

Thanks again!

The limits are from $-\infty$ to $\infty$, but of course the step functions can cut the interval down by being zero in places. But perhaps an easier approach would be to remember that the Fourier transform $\mathcal F$ satisfies$$
\mathcal F(f * g) =\hat f (\omega)\cdot \hat g(\omega)$$So transform the functions, multiply the transforms, and invert the result. That would give $f*g$ and you can likely just use transform tables and be done with it.

 

[jegues]

The limits are from $-\infty$ to $\infty$, but of course the step functions can cut the interval down by being zero in places. But perhaps an easier approach would be to remember that the Fourier transform $\mathcal F$ satisfies$$
\mathcal F(f * g) =\hat f (\omega)\cdot \hat g(\omega)$$So transform the functions, multiply the transforms, and invert the result. That would give $f*g$ and you can likely just use transform tables and be done with it.

We haven't learned that technique yet nor do we have access to tables so is there a way we can mechanically do the integration?

How would it be done?

 

[jbunniii]

For what values of \tau is u(\tau) nonzero?

For what values of \tau is u(-t + \tau) nonzero?

 

[LCKurtz]

Homework Statement

Convolve,

2e^{-t}u(t) \ast e^{t}u(-t) \quad \text{Where u(t) is the step function}

Homework Equations

The Attempt at a Solution

2e^{t}\int_{?}^{?}e^{-\tau}u(\tau) e^{-\tau}u(-t+\tau)d \tau

I'm not sure what the limits on my integral should be, the u(-t) is confusing me. Shouldn't the second function only be nonzero from -infinity to 0 and the other from to +infinity?

Thanks again!

The limits are from $-\infty$ to $\infty$, but of course the step functions can cut the interval down by being zero in places. But perhaps an easier approach would be to remember that the Fourier transform $\mathcal F$ satisfies$$
\mathcal F(f * g) =\hat f (\omega)\cdot \hat g(\omega)$$So transform the functions, multiply the transforms, and invert the result. That would give $f*g$ and you can likely just use transform tables and be done with it.

We haven't learned that technique yet nor do we have access to tables so is there a way we can mechanically do the integration?

How would it be done?

Well, start with this:$$
2e^{t}\int_{-\infty}^{\infty}e^{-\tau}u(\tau) e^{-\tau}u(-t+\tau)d \tau$$Then, since $u(\tau)$ is zero for $\tau <0$ and 1 for $\tau > 0$ you can write it as$$
2e^{t}\int_{0}^{\infty}e^{-\tau}\cdot 1 \cdot e^{-\tau}u(-t+\tau)d \tau$$Next you know $u(\tau - t)$ is only going to be 1 if $\tau > t$, so that needs to be taken into account in the limits. When you work on it you will see it matters whether $t > 0$ or not. See if you can take it from there.