2
$\begingroup$

I am trying to find the correlation between the signals $u(t)$ and $\sin(t)[u(t)-u(t-2)]$

The correlation function $C(t) = \int^{\infty}_{-\infty} u(\tau+t)\sin(t)(u(t)-u(t-2))d\tau$

This is my progress so far:

$u(t+\tau) = 1\rightarrow \tau>-t , 0\rightarrow \text{elsewhere}$

So for $t\ge 0$ the graph looks like this:

But I dont get the correct result , I get something which over time can go above 1.What's wrong with my analysis?

$\endgroup$
2
  • 1
    $\begingroup$ Hi, welcome to DSP.SE! Are you sure about the limits of your integral? If so, why do you think the output cannot go above $1$? Also what do you think should be the range of your output? $\endgroup$ Sep 15, 2023 at 21:02
  • $\begingroup$ Your integrand is also wrong. That should be $\sin(\tau) ...$ not $\sin(t) ...$ $\endgroup$
    – Hilmar
    Sep 16, 2023 at 11:12

1 Answer 1

3
$\begingroup$

OP has the correct idea, but have overlooked some details. Note that both functions have limited support. Then, the correlation is \begin{align} C(t)&=\int_{-\infty}^{\infty} u(\tau+t)\sin(\tau)\big[u(\tau)-u(\tau-2)\big]\,d\tau\\ &=\int_{-\infty}^{\infty} \sin(\tau)\underbrace{u(\tau+t)\big[u(\tau)-u(\tau-2)\big]}_{v(\tau)}\,d\tau\,. \end{align}

This means that the integral limits depend on $v(\tau)$ and not only on $x(\tau)$ as OP did. We have that $$v(\tau)=u(\tau+t)\big[u(\tau)-u(\tau-2)\big]=\begin{cases}1&\tau\in\big[0,2\big)\text{ and } \tau\geq-t\\0&\text{otherwise}\end{cases}\,\,. $$ In other words, if $-t\geq2$, $v(\tau)=0\,\forall\tau$. Equivalently, $$v(\tau)=u(\tau+t)\big[u(\tau)-u(\tau-2)\big]=\begin{cases}1&\tau\in[\max\{0,-t\},\max\{2,-t\}\big)\\0&\text{otherwise}\end{cases}\,\,. $$

Now we can substitute the integration limits in $C(t)$ and obtain $$\begin{align} C(t)&=\int_{\max\{0,-t\}}^{\max\{2,-t\}} \sin(\tau)\,d\tau=-\cos(\tau)\bigg|_{\tau=\max\{0,-t\}}^{\tau=\max\{2,-t\}}\\ &=\cos\big(\max\{-t,0\}\big)-\cos\big(\max\{-t,2\}\big)\\ &=\cos\big(\min\{t,0\}\big)-\cos\big(\min\{t,-2\}\big)\,, \end{align}$$ where I use the fact that $\max\{-x,-y\}=-\min\{x,y\}$ and that cosine is an even function.

This result can also be rewritten in terms of the step function if required, noting that $$C(t)=\begin{cases}0&t<-2\\\cos(t)-\cos(2)&t\in[-2,0)\\1-\cos(2)&t\geq0\end{cases}$$ Thus, one possible expression (as there are multiple) is $$C(t)=u(t+2)\big[\cos(t)-\cos(2)\big]-u(t)\big[\cos(t)-1\big]\,.$$

$\endgroup$
1
  • $\begingroup$ You read my mind. $\endgroup$
    – Peter K.
    Feb 20 at 1:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.