Evaluating integrals with delta function

[nhoratiu]

Hi there!
I have a problem with one of the questions given to us in the signals and systems course. If anyone could help me I would greatly appreciate it!

Homework Statement

integral(from -infinity to +infinity) of u₀(t) * cos(t) dt

u(t) is a step function.

Homework Equations

Why is this equal to 1 and not to sin(t), for t>0 ?

The Attempt at a Solution

I thought this is equal to sin(t) but the answer manual says it's equal to 1 because the whole thing transforms into a delta function. How does it do that?

 

[arkajad]

You need to define $$u_0$$.

 

[nhoratiu]

u0(t) is a step function. its "0" from negative infinity to t=0 and "1" from t=0 to positive infinity.

 

[arkajad]

So, I guess u0 is the same as u. And, I also guess, "*" stands for the convolution?

 

[nhoratiu]

hey, the * is for multiplication. Overall we have:

integral from negative infinity to positive infinity of ( u(t) times cos(t) dt)

the solution manual says it's equal to 1. I thought it was equal to sin(t).

 

[arkajad]

I would, formally (which is, strictly speaking, illegal), do it like that:

$$\int_{-\infty}^{\infty}u(x)\cos(x)dx=\int_{-\infty}^{\infty}u(x)\sin'(x)dx=-\int_{-\infty}^{\infty}u'(x)\sin(x)dx=-\int_{-\infty}^{\infty}\delta(x)\sin(x)dx=-\sin(0)=0$$

But I do not know what your course is based on, therefore I can't help more. And probably my result is not helping you anyway.

 

[vela]

When you integrate cos x, you get sin x, but the problem is evaluating the result at the upper limit since sin x doesn't converge in the limit as x→∞. To get around this problem, you can use the convergence factor e^-λx and take the limit as λ→0⁺:

$$\int_{-\infty}^\infty u(x)\cos(x)\,dx = \lim_{\lambda\to 0^+}\int_{-\infty}^\infty u(x)\cos(x)e^{-\lambda x}\,dx$$