# Convolution of u(t) and cos(t)

1. Jul 13, 2011

### Jncik

1. The problem statement, all variables and given/known data
Hello, I'm revising this summer for signals and systems and I came across this convolution

$$cos(t)*u(t)$$

2. Relevant equations

having two signals x(t) and h(t), where x(t) is the input signal and h(t) the impulse response

the output y(t) is given by $$y(t) = x(t)*h(t) = \int_{-\infty}^{+\infty}x(\tau)h(t-\tau)d\tau$$

3. The attempt at a solution

$$cos(t)*u(t) = \int_{-\infty}^{+\infty}cos(\tau)u(t-\tau)d\tau$$ (1)

u(t-τ) is 1 for t-τ>0 => τ<t

hence
(1) is
$$\int_{-\infty}^{t}cos(\tau)d\tau = sin(\tau)|_{\tau->t, \tau->-\infty} = sin(t) - sin(-\infty) = sin(t) - undefined in [-1, 1]$$

now, I had 2 other integrals in the same exercise where the result was just sin(t), and I thought this was supposed to be sin(t) as well but I get this undefined number and I think it's wrong because I've never calculated an integral that gave me an undefined number, at least in signals and system course..

after searching in my book it had a similar integration

$$\int_{-\infty}^{+\infty}u_{0}(t)cos(t) dt$$

$$u_{0}(t)$$ is just u(t)...

now again using the same method I would have

$$\int_{-\infty}^{+\infty}u_{0}(t)cos(t) dt = \int_{0}^{+\infty}cos(t) dt = sin(t)|_{t->0,t->+\infty} = sin(0) - sin(+\infty) = -sin(+\infty)$$

which is undefined again

but he finds this result

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

I'm wondering, what am I doing wrong and why does he find such result?

2. Jul 14, 2011

### vela

Staff Emeritus
Is that exactly what's written in the book? It doesn't look correct to me.

3. Jul 14, 2011

### Jncik

yes I'm using oppenheim's book(2.20 exercise) about signals and systems and I saw this at the solution manual.. I may be looking at the wrong exercise, but Now I'm looking at it again and I don't think so...

so is my solution correct on this one?

I don't see why he uses this '0' in the notation, if I remember correct when we have a system

y(t) = x(t)

if the system is not time invariant we use this notation in order to show that the impulse response for example h2(t) comes for input δ(t-2)

and hence the convolution would be

$$\int_{-\infty}^{+\infty}x(\tau)*h_{\tau}(t) d\tau$$

now I can't link these two facts for the notation he uses for u0.. but I don't think it affects the solution

4. Jul 14, 2011

### Jaynte

It seems strange to me too. It feels like the result should be 0.

$$\int_{-\infty}^\infty u(t-\tau)*cos(\tau)d\tau=\int_{t-T}^t cos(\tau)d\tau=sin(t)-sin(t-T)=0$$
since
$$u(\tau)=0 \qquad \tau>t$$
$$u(\tau)=1 \qquad \tau\leq t$$
$$T=2\pi$$
Sinus is periodical, thats why i integrate over one period instead of infinity. The
results should be the same = 0 for all periods.

Im not sure if i'm right. Anyone?

Maybe he have used distribution theory to solve that.

5. Jul 14, 2011

### Jncik

thanks, this looks correct to me.. but I'm not 100% sure either because

according to the solution, the integration of d(t) will give us 1.....

6. Jul 14, 2011

### Jaynte

Solutions aren't always right in books :)

I'm not 100% but if you think in the discrete case:
$$\sum_{m=-\infty}^\infty u(m)*cos(n-m)=\sum_0^{\infty} cos(n-m)=0$$
This sum will be zero for all n, since the cos function is symmetric around the x-axis. Equally amount of positive cos values as negative cos values.

7. Jul 14, 2011

### Ecthelion

Hmm... here's how I'd go about it -

$$y(t) = \int\limits_0^t h(t-\tau)q(\tau) \mathrm{d}\tau$$

where $$h(t) = 1$$ and the limits have been adjusted accordingly. Then,

$$y(t) = \int\limits_0^t cos(\tau)\mathrm{d}\tau$$ Which of course leads to the simple integration of:

$$y(t) = sin(t)$$

Which seems a bit trivial... but, I think we can use the Laplace Transform to take a look at this as well. Both the FT and LT are used when looking at differential and integral equations, which I think this qualifies as. If we consider that the convolution in the time domain is the same as multiplication in the Frequency domain, then taking use of these LT pairs:

$$u(t) = \frac{1}{s}$$ and $$cos(\alpha\,t) = \frac{s}{s^2 + \alpha^2}$$

Then simply multiplying these two:

$$Y(s) = \frac{1}{s} \frac{s}{s^2 + \alpha^2} = \frac{1}{s^2 + \alpha^2}$$

But since $$\alpha = 1$$ and the LT $$sin(t) = \frac{\alpha}{s^2 + \alpha^2}$$ we can say that:

$$y(t) = sin(t)$$

I dunno... these are my thoughts. I definitely think your book's final last two terms for the answer are incorrect though.

8. Jul 14, 2011

### Jaynte

How did you get $$u(t-\tau)=u(t)$$?

Of course Laplace is the best to use. Works every time :)

9. Jul 14, 2011

### Ecthelion

So true.... *sigh* haha

But anyways, perhaps it'd be easier to think about getting the answer if we did the shifting in the cosine term:

$$y(t) = \int\limits_0^t cos(t - \tau)\mathrm{d}\tau$$

Now using a couple trig identities....

$$y(t) = \int\limits_0^t cos(\tau)cos(t) + sin(t)sin(\tau) \mathrm{d}\tau$$

$$y(t) = [cos(t)sin(t) - sin(t)cos(t)] - [0 - sin(t)]$$

Then things start canceling out all prettily...

$$y(t) = cos(t)sin(t) - sin(t)cos(t) + sin(t)$$

Yielding, once more:

$$y(t) = sin(t)$$

Perhaps that's a little clearer? I'm not one hundred percent sure on this but the solution feels right and looks right at the moment, lol.

Last edited: Jul 14, 2011
10. Jul 14, 2011

### Jncik

thanks that looks nice but I still have one more question

since $$cos(t)*u(t) = \int_{-\infty}^{+\infty} cos(\tau - t)u(\tau) d\tau$$

and u(τ) is 0 for τ<0 shouldnt it be

$$\int_{0}^{+\infty}cos(\tau - t)d\tau = \int_{0}^{+\infty}cos(t-\tau)d\tau$$

11. Jul 14, 2011

### Ecthelion

Well if you think about convolution - what it's doing graphically - you are shifting one of the waveforms over the other. This is the tau, but the 't' is the so-called 'limit' you're integrating to when moving one of the said waveforms across the other - presumably to infinity, which of course makes this system have an unbounded response as it should be.

If this is too vague let me know.

12. Jul 14, 2011

### vela

Staff Emeritus
I dug my old linear systems analysis book out of storage to refresh my memory, and I ran across this notation as well. In this convention, u0(t) is the Dirac delta function, u-1(t) is the unit step function, etc. In this light, that integral makes sense
$$\int_{-\infty}^\infty u_0(t) \cos t \, dt = \int_{-\infty}^\infty \delta(t)\cos t\,dt = \int_{-\infty}^\infty \delta(t)\cos 0\,dt = \int_{-\infty}^\infty \delta(t)\,dt = 1$$

13. Jul 14, 2011

### vela

Staff Emeritus
Ecthelion is actually looking at the convolution of (cos t)u(t) and u(t). These one-sided functions are what you implicitly use when working with Laplace transforms. Using them avoids the difficulties you're running into, but it also means you're not solving the original problem as given.

14. Jul 14, 2011

### Ecthelion

Ahhh... I was not aware of that notation at all... definitely makes a lot more sense now, good find.

15. Jul 15, 2011

### Jaynte

What do you mean with "implicitly use"? Laplace works for all t. (both sides)

And when you use a unit step u(t) the function y(t)=cos(t)*u(t) will be one sided?

16. Jul 15, 2011

### vela

Staff Emeritus
Laplace transforms usually are defined with an integral from 0 to infinity. (See, for example, the second paragraph of http://mathworld.wolfram.com/LaplaceTransform.html.) Consequently, when you use them to solve a problem, you only find the solution for t≥0.

You could use the bilateral Laplace transformation. In that case, however, the transform
$$F(s) = \frac{s}{s^2+\omega^2}$$
for instance, corresponds to the function $f(t)=(\cos \omega t) u(t)$, which is 0 for t<0.