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Autocorrelation for power signals is defined by $$R_x(\tau)=\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^Tx(t)x^*(t-\tau)dt\tag{1}$$ Is it true that for periodic signals $(1)$ can be computed by $$R_x(\tau)=\frac{1}{T_0}\int_{T_0}x(t)x^*(t-\tau)dt\tag{2}$$What are the other ways for computing $R_x(\tau)$? In this question, it's showed that $$|X(f)|^2 =\sum_{n=-\infty}^\infty |c_n|^2 \delta (f-nf_0)\tag{3}$$ But I don't think that's a valid derivation since, as it's pointed by Matt L, it involves product of $\delta(f)$'s which is meaningless. So is it possible to prove $(3)$ by other means or it's a false statement? And, in the end, can we conclude that for a periodic power signal $$(\mathcal{F}R_x)(f) = S_x(f) = |X(f)|^2 = \sum_{n=-\infty}^\infty |c_n|^2 \delta (f-nf_0)$$ where $S_x(f)$ is PSD, holds?

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  • $\begingroup$ Do you think the derivation is wrong simply because of the use of deltas? The derivation is correct...and deltas are absolutely useful as a measure. A "measure" here being how applying a delta to a function and integrating (or summing) yields useful results. $\endgroup$ – Envidia Oct 15 '20 at 21:26
  • $\begingroup$ @Envidia I said that in a mathematical point of view. As far as I know, in the distribution theory $\delta^2(x)$ is undefined. $\endgroup$ – S.H.W Oct 15 '20 at 21:42
  • $\begingroup$ So in (3), where is the product between deltas occurring? None of these deltas are overlapping in the summation. The only thing the deltas are being multiplied with are their respective $c_n$ term. $\endgroup$ – Envidia Oct 15 '20 at 21:47
  • $\begingroup$ @Envidia You are right. I was just looking for a more rigorous proof. Intuitively it makes sense as you said. $\endgroup$ – S.H.W Oct 15 '20 at 21:54
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The OP states that for a (deterministic) power signal $x(t)$, the autocorrelation function is defined as $$R_x(\tau) = \lim_{T\to \infty}\frac{1}{2T}\int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt\tag{1}$$ and then wonders whether in the case when $x(t)$ happens to be a periodic signal with period $T_0$, then it is true that $$R_x(\tau) = \frac{1}{T_0}\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt ??\tag{2}$$ The answer is Yes, for the reasons given below.

Let us consider the integral $\int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt$ in $(1)$ when $x(t)$ is periodic with period $T_0$ and $T = nT_0+\alpha$ where $0\leq \alpha < T_0$. Note that the integrand also has period $T_0$.

  • When $\alpha=0$, the range of integration is over $2n$ periods of the integrand and so \begin{align} \int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt &= \int_{-nT_0}^{nT_0} x(t)x^*(t+\tau) \,\mathrm dt\\ &= 2n\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt. \end{align} It follows that $$\frac{1}{2nT_0}\int_{-nT_0}^{nT_0} x(t)x^*(t+\tau) \,\mathrm dt = \frac{1}{T_0}\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt.$$
  • The astute reader will have seen that if $\alpha=\frac{T_0}{2}$, the range of integration is over an interval of length $(2n+1)T_0$ (that is, over $2n+1$ periods of the integrand) and so we have $$\frac{1}{2(n+1)T_0}\int_{-(n+\frac 12)T_0}^{(n+\frac 12)T_0} x(t)x^*(t+\tau) \,\mathrm dt = \frac{1}{T_0}\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt$$ also.
  • For $\alpha \in \left(0,\frac 12\right)$ (or $\alpha \in \left(\frac 12,1\right)$, the integral is over an integer number of periods (say $N$) of the integrand plus an interval $\beta T_0$ of length less than $\frac 12T_0$ and so \begin{align} \int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt &= N\int_0^{T_0} x(t)x^*(t+\tau)\,\mathrm dt + \Delta \end{align} where $\Delta$ denotes whatever the contribution is from the integration of $x(t)x^*(t+\tau)$ over that little interval of length $\beta T_0 < \frac 12 T_0$. But then we have that \begin{align} \frac{1}{2T}\int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt &= \frac{1}{(N+\beta)T_0}\left[N\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt + \Delta \right]\\ &= \frac{N}{N+\beta}\cdot \left[\frac{1}{T_0}\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt\right] + \frac{\Delta}{(N+\beta)T_0}. \end{align} But $(1)$ says that we need to take the limit of the above as $T \to \infty$ which we can do by letting $N\to\infty$ and so we see that in all cases,

If $x(t)$ is a deterministic periodic power signal, then its autocorrelation function $$R_x(\tau) = \lim_{T\to \infty}\frac{1}{2T}\int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt\tag{1}$$ can be computed via $$R_x(\tau) = \frac{1}{T_0}\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt, \tag{2}$$ which is what the OP asks for.

For the important special case when $x(t)$ is a real-valued function, Eq. $(2)$ simplifies to $$R_x(\tau) = \frac{1}{T_0}\int_0^{T_0} x(t)x(t+\tau) \,\mathrm dt, \tag{3}$$ and this is the only case we consider below.


Note that so far, there has been no whiff of Fourier transforms or Fourier series or Dirac deltas or anything to do with the frequency domain, and so let us bring that stuff in. Note that a real-valued deterministic periodic power signal $x(t)$ can be represented by a Fourier series

$$x(t) = \sum_{n=-\infty}^\infty c_n\exp(j2\pi nf_0t)\tag{4}$$ where $f_0 = T_0^{-1}$ and $c_{-n}=c_n^*$, and does not enjoy a Fourier transform in the classical sense (where Dirac deltas a.k.a. impulses are not allowed). However, $x(t)$ does have a Fourier transform $X(f)$ in the generalized sense where $$X(f) = \sum_{n=-\infty}^\infty c_n\delta(f-nf_0).\tag{5}$$ In the classical theory, if $X(f)$ is the Fourier transform of $x(t)$, then $|X(f)|^2$ is the Fourier transform of its autocorrelation function $R_x(\tau)$. Does this result hold in the generalized sense too? That is, for a periodic signal $x(t)$ with Fourier series as in $(3)$ and periodic autocorrelation function $R_x(\tau)$ as in $(2)$, what is the Fourier series for $R_x(\tau)$ and the corresponding generalized Fourier transform of $R_x(\tau)$???

We have that $R_x(\tau)$ has Fourier series $\displaystyle \sum_{n=-\infty}^\infty a_n\exp(j2\pi nf_0\tau)$ where \begin{align}a_n &= \frac{1}{T_0}\int_{0}^{T_0} R_x(\tau)\exp(-j2\pi nf_0\tau)\,\mathrm d\tau\\ &= \frac{1}{T_0}\int_{0}^{T_0} \left[\frac{1}{T_0}\int_{0}^{T_0} x(t)x(t+\tau)\,\mathrm dt\right] \exp(-j2\pi nf_0\tau)\,\mathrm d\tau\\ &= \frac{1}{T_0}\int_{0}^{T_0} x(t)\left[\frac{1}{T_0}\int_{0}^{T_0} x(t+\tau)\exp(-j2\pi nf_0\tau)\,\mathrm d\tau\right] \,\mathrm dt\\ &=\frac{1}{T_0}\int_{0}^{T_0} x(t)\left[\frac{1}{T_0}\int_{0}^{T_0} x(t+\tau)\exp(-j2\pi nf_0(t+\tau))\,\mathrm d\tau\right] \exp(j2\pi nf_0t)\,\mathrm dt\\ &=c_{n}\cdot \frac{1}{T_0}\int_{0}^{T_0} x(t) \exp(j2\pi nf_0t)\,\mathrm dt\\ &= c_n\cdot c_{-n}\\ &= |c_n|^2. \end{align} Thus, for a real-valued deterministic periodic power signal $x(t)$,

$$\text{If } x(t) = \sum_{n=-\infty}^\infty c_n\exp(j2\pi nf_0t), ~~\text{then } R_x(\tau) = \sum_{n=-\infty}^\infty |c_n|^2\exp(j2\pi nf_0\tau). \tag{6}$$

Finally, what about Fourier transforms in the generalized sense? Well, we have that \begin{align} \mathscr F[x(t)] &= \sum_{n=-\infty}^\infty c_n\delta(f-nf_0)\\ \mathscr F[R_x(\tau)] &= \sum_{n=-\infty}^\infty |c_n|^2\delta(f-nf_0) \end{align} and so it is not the case that the formula $$\mathscr F[x(t)] = X(f) \implies \mathscr F[R_x(\tau)] = |X(f)|^2 \tag{7}$$ of classical Fourier analysis holds unless we are willing to assume that when we multiply out $\left|\sum_{n=-\infty}^\infty c_n\delta(f-nf_0)\right|^2$ into a doubly infinite sum over $m$ and $n$, $$c_n \delta(f- nf_0)c_m^* \delta^*(f- mf_0) = \begin{cases} |c_n|^2 \delta(f- nf_0), & \text{if}~m=n,\\ 0, &\text{if}~m\neq n, \end{cases} \tag{8}$$ holds. In particular, we must treat $|c_n|^2\delta(f-nf_0)\delta^*(f-nf_0)$ as being equal to $|c_n|^2\delta(f-nf_0)$ as stated in my previous answer that the OP has denigrated as invalid and asserted that the claim that, in this context, the square of a Dirac delta should be treated as a Dirac delta, is meaningless because $\delta^2(x)$ is meaningless. The alternative is to insist that $(7)$ holds in classical Fourier analysis only and cannot be applied in the generalized Fourier analysis that allows for Dirac deltas. You pays your money and you takes your choice.....

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  • $\begingroup$ @robertbristow-johnson Thank you for reading my answer carefully and for the edit. $\endgroup$ – Dilip Sarwate Oct 19 '20 at 3:03
  • $\begingroup$ Thanks a lot for your valuable answer. $\endgroup$ – S.H.W Oct 20 '20 at 15:57
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Assuming that the signal $x(t)$ is periodic then it can be described by its complex Fourier series as

$$x(t) = \sum_{n=-\infty}^{\infty }c_ne^{j2{\pi}nf_0t}$$

Where $c_n$ are the complex Fourier coefficients.

There is a Fourier transform pair that states

$$\mathcal{F}(e^{j2{\pi}f_0t}) = \delta(f - f_0)$$

Applying this to the Fourier series of $x(t)$ element by element you eventually get the Fourier transform $X(f)$

$$X(f) = \mathcal{F}\big(\sum_{n=-\infty}^{\infty}c_ne^{j2{\pi}nf_0t}\big) = \sum_{n=-\infty}^{\infty }c_n\delta(f-nf_0)$$

This is also a common Fourier transform pair.

If we assume that $x(t)$ is also wide-sense stationary then the Wiener-Khinchin Theorem states that the power spectral density $S_x(f)$ is given by the Fourier transform of the autocorrelation $R_{x}(\tau)$.

The power spectral density is then given by

$$S_x(f) = |X(f)|^2 = \sum_{n=-\infty}^{\infty }|c_n|^2\delta(f-nf_0)$$

Take the inverse Fourier transform using the first transform pair above and you get the autocorrelation function $R_x(\tau)$

$$R_x(\tau) = \mathcal{F^{-1}(S_x(f))} = \sum_{n=-\infty}^{\infty }|c_n|^2e^{j2{\pi}nf_0t}$$

Which is the result from the post you linked. You can take this the other way and calculate the autocorrelation first and then take the Fourier transform to yield the power spectral density.

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  • $\begingroup$ How can we calculate the autocorrelation at first? $\endgroup$ – S.H.W Oct 15 '20 at 22:53
  • $\begingroup$ @S.H.W Use the definition that you provided. Because we're dealing with complex exponentials, there's significant simplification that occurs with the integrand. Try it with a single complex exponential $e^{j2{\pi}f_0t}$ first and then try to form the total sum. Integration is linear, so the operation can be applied to each element of the summation individually which eventually leads to the final sum expression. $\endgroup$ – Envidia Oct 15 '20 at 23:11
  • $\begingroup$ Using $(2)$ I can do that but I don't know how to do that with $(1)$. Can we derive $(2)$ from $(1)$? $\endgroup$ – S.H.W Oct 15 '20 at 23:35
  • $\begingroup$ @S.H.W (1) is the more general expression, where as (2) is a simplified version when explicitly exploiting the periodic nature of a signal. In this case either one should give you the same answer. Also keep in mind that there are different convetions with (1): If the limits are taken to be from -T/2 to T/2, then you remove the factor of 2. The expression you have is over a period of 2T, hence the factor of two in the denominator. $\endgroup$ – Envidia Oct 15 '20 at 23:39

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