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I have a signal $X(t)=\sum_{n=-\infty}^{\infty} Z_n \delta(t-n\tau)$, $Z_n$ is a random variable with equal possibility of $\pm 1$ and I know the power spectrum of this signal is $\frac {1}{\tau}$ from the solution.

I have try to calculate the power spectrum by the auto-correlation and then calulated by its Fourier transform. As following

\begin{align} R_X (t+\tau, t) &=E\left[X(t+\tau) X(t)\right]\\ &=E\left[\sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty} Z_n Z_m \delta(t+\tau-n \tau)\delta(t-m\tau)\right]\\ &=\sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty} \delta(t+\tau-n \tau)\delta(t-m\tau)\\ \end{align}

Then, I have no idea how to calulate its Fourier transfrom to get the solution as simple as $\frac {1}{\tau}$. I don't know which part is gone wrong, appreciate any help.

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  • $\begingroup$ Why is $E[Z_n Z_m] = 1$ ? I don't see how that can be if $Z_k$ is either $+1$ or $-1$ with equal probability. $\endgroup$ – Peter K. Mar 22 '13 at 1:03
  • $\begingroup$ @PeterK. $E[Z_nZ_m]$ is 1 for $n=m$, 0 otherwise. As such, the last line above is incorrect (though for more reasons than just that). I'm working on it atm. $\endgroup$ – lxop Mar 22 '13 at 1:09
  • $\begingroup$ @lxop : D'oh! Thanks! I had my stoopid had on last night. $\endgroup$ – Peter K. Mar 22 '13 at 11:47
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You have to simplify your expression in order to obtain a Dirac Comb (a.k.a. impulse train).

$$R_X(t+\tau,t)=\sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty} \delta(t+\tau-n \tau)\delta(t-m\tau) = \sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty} \delta(t-(n+1)\tau)\delta(t-m\tau)$$

Both $\sum_{m=-\infty}^{\infty} \delta(t-m\tau)$ and $\sum_{n=-\infty}^{\infty} \delta(t-(n+1)\tau)$ are actually the same signal: a Dirac Comb with $\tau$ period. The difference would be the that the ''n'' signal would be time shifted by n+1, while the ''m'' signal is shifted by m. As they are defined in $(-\infty,\infty)$, they are defined in exactly the same instants.

Applying delta's well known property: $$\delta(t-t_0)x(t)=x(t_0)$$

we can obtain that, as we intuitively know because both signals are the same, $$R_X(t+\tau,t)=\sum_{m=-\infty}^{\infty} \delta(t-m\tau)$$

From here, we just have to obtain the Fourier Transform of the Dirac Comb, which can be easily found, e.g. in Wikipedia, obtaining that all the coefficients are $\frac{1}{\tau}$.

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I'm not quite finished, so this answer is highly likely to change, but I'll show what I've got for now:

For a start, you have used the same variable $\tau$ for both autocorrelation lag and as a parameter of the original function - can't do that.

$$ R_X(\theta) = E\left[\sum_n Z_n\delta (t - n\tau + \theta) \sum_m Z_m\delta (t - m\tau)\right]\\ = E\left[\sum_n \sum_m Z_n Z_m \delta (t - n\tau + \theta) \delta (t - m\tau)\right] \\ = \sum_n \sum_m E\left[Z_n Z_m \delta (t - n\tau + \theta) \delta (t - m\tau)\right] \\ = \sum_n \sum_m E\left[Z_n Z_m\right] E\left[\delta (t - n\tau + \theta) \delta (t - m\tau)\right] $$ I can split the expectation like that because the two terms are independent - the delta terms aren't affected by what value is taken by $Z$. Now note that the expected value of $Z_nZ_m$ is 1 if $n=m$, and 0 otherwise, (also known as a Kronecker delta), so $$ R_X(\theta) = \sum_n E\left[\delta (t - n\tau + \theta)\delta(t-n\tau)\right] \\ = \begin{cases} \sum_n \int_{-\infty}^{\infty} \delta (t - n\tau) \delta (t - n\tau) dt, &\theta = 0 \\ 0, &\rm\ otherwise \end{cases} $$ which is looking like a delta too, which is good. I say good because the answer you are shooting for is $S_X(f) = \frac{1}{\tau}$, which is constant with frequency. A constant in the frequency domain is a delta in the time domain.

The problem I've struck now is that I'm not sure what to do with two deltas under an integral. I can treat it like a normal sifting situation, to get $$ R_X(0) = \sum_n \delta (n\tau - n\tau) \\ = \sum_n \delta(0) $$ but I don't know if that's correct, nor do I know how to get the $\tau$ term involved to get to the final answer... any suggestions?

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