3
\$\begingroup\$

Given the following square wave signal g(t) : enter image description here

I'm trying to find the \$R(\tau)\$ of this signal, but I'm confused about how to solve the integral. In the signal above, the red square wave is the shifted signal \$g(t-\tau)\$. I understand the autocorrelation function will be periodic because g(t) is periodic, and will be a triangular waveform.

$$ R(\tau) = \int_{-T/2}^{T/2}g(t)g(t-\tau)dt $$

Trying to find the first integral for \$\tau < T/2\$ (first picture) I have: $$ R_1(\tau) = \int_{-T/4}^{-T/4+\tau} -dt + \int_{-T/4+\tau}^{T/4} dt + \int_{T/4}^{T/4+\tau} -dt + \int_{T/4+\tau}^{3T/4}dt = (-\tau) + (T/2 - \tau) + (-\tau) + (T/2 - \tau) = T - 4\tau $$

For \$T/2 < \tau < T\$ I would have the inverse I suppose since it must be symetric.

I'm confused on the first integral. I'm not exactly sure I'm correct. I think I must calculate both integrals (in the picture) on one period i.e. so that the combined integral interval for \$R_1(\tau)\$ is on one period. Or is it on a half period? In which case \$R_1(\tau)\$ would be \$ R_1(\tau) = T/2 - 2\tau\$. The fact the signal is periodic is what is confusing me. I want ultimately to get one period of \$R(\tau)\$. So my question simply is: is the expression I got for \$R_{1}(\tau)\$ correct or not?

Any help will be greatly appreciated.

\$\endgroup\$
1
\$\begingroup\$

It should not be difficult to verify that \$R(\alpha) \$ is periodic with period \$T\$. Here, the value of \$ R_1(\alpha) = T- 4\alpha \$ looks correct, for \$ R_2(\alpha) \$ I calculate it to be \$4\alpha - 3T\$. Both \$R_1,\ R_2\$ define \$R(\alpha)\$ over one period, as \$R_1\$ is defined for \$0<\alpha<T/2\$ whereas \$R_2\$ is for \$T/2<\alpha<T\$.

EDIT: Here, let \$T_p\$ is the period of the your signal \$g(t)\$. Using Wikipedia, we get $$R(\alpha)=\lim _{T\to \infty}T^{-1} \int_0^{T} g(t)g(t-\alpha)dt.$$ Now this can simplified by the periodicity of the your signal \$g(t)\$ to give $$R(\alpha) =\lim _{T\to \infty}T^{-1}\sum_{k=0}^{k=T/T_p-1} \int_{k T_p}^{(k+1)T_p} g(t)g(t-\alpha)dt. $$ Now since the integral is periodic over a period \$T_p\$ we get $$R(\alpha) = T_p^{-1} \int_0^{T_p}g(t)g(t-\alpha)dt.$$

Upto a constant factor this matches with the definition you used.

\$\endgroup\$
  • \$\begingroup\$ But, more importantly, I wonder if it's even the correct way to calculate the autocorrelation of a periodic function. I'm not even sure. I wonder if it's not more appropriate to use fourier series. After all, autocorrelation by convolution I think applies to energy signals, I'm not sure it works for power signals. This is mostly the clarification I seek. \$\endgroup\$ – Yannick Oct 1 '18 at 0:52
  • \$\begingroup\$ I believe yes. Read here, en.wikipedia.org/wiki/Autocorrelation#Signal_processing . NB: The \$T\$ in your question is a time period and not the \$T\$ used in Wikipedia. \$\endgroup\$ – MUB Oct 1 '18 at 3:07
  • \$\begingroup\$ That's interesting because in their derivation for a periodic signal, they take the limit as T -> infinity, i.e. \$ \lim_{T\to \infty} \frac{1}{T} \int_{0}^{T} f(t)f(t\pm \tau) dt\$ But you said something here that raises another question. Do you mean T is not considered as a period in this equation? How should I interpret it? Sorry for asking so much... just wish to understand. \$\endgroup\$ – Yannick Oct 2 '18 at 4:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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