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Consider white gaussian noise $\omega(t)$ that has passed through and ideal lowpass filter with bandwidth $B$. This filter has the impulse response

$$h(t) = 2B\text{sinc}(2Bt)$$

Sampling the filter output at time instants $t_1$ and $t_2$ yields two Gaussian random variables, $w_1$ and $w_2$ (with zero mean). I don't understand the following derivation of the variances $E[w_1^2], E[w_2^2]$ and covariance $E[w_1w_2]$, and would like some help with that:

We have that for any $t_1$ and $t_2$, $$E[w_1w_2] = \sigma^2 \int_{-\infty}^\infty h(\tau - t_1)h(\tau - t_2) d\tau = \\ 2B\sigma^2 \int_{-\infty}^\infty \sqrt{2B}\text{sinc}(2B(\tau-t_1))\sqrt{2B}\text{sinc}(2B(t_2-\tau)) d\tau = \\ 2B\sigma^2 \int_{-B}^B e^{-j2\pi f(t_1-t_2)} df =\\ 2B\sigma^2\text{sinc}(2B(t_1-t_2)). \quad (1.)$$ The third step of $(1.)$ follows from Parseval's theorem together with the facts that $$\mathcal{F}\{\sqrt{2B}\text{sinc}(2B(\tau-t_1)) \} = \begin{cases} \sqrt{\dfrac{1}{2B}}e^{-j2\pi f t_1}, & |f| \leq B \\ 0, & \text{otherwise.} \end{cases} \quad (2.)$$ $$\mathcal{F}\{\sqrt{2B}\text{sinc}(2B(t_2 - \tau)) \} = \begin{cases} \sqrt{\dfrac{1}{2B}}e^{-j2\pi f t_2}, & |f| \leq B \\ 0, & \text{otherwise.} \end{cases} \quad (3.)$$ $$\mathcal{F}\{\sqrt{2B}\text{sinc}(2B(t_1 - t_2 - \tau)) \} = \begin{cases} \sqrt{\dfrac{1}{2B}}e^{-j2\pi f (t_1-t_2)}, & |f| \leq B \\ 0, & \text{otherwise.} \end{cases} \quad (4.)$$ Consequently,
$E[w_1^2] = E[w_2^2] = 2B \sigma^2$...

I don't understand the third step. The version of Parseval's theorem that has been used in the book is the one that states that the energy in the time and frequency domain is the same
$$\int_{-\infty}^\infty |x(t)|^2 = \int_{-\infty}^\infty |X(f)|^2$$.

I don't see how this comes into play here since there is no function squared in $(1.)$? I wonder if there might be some other version of Parseval's theorem that is referred to, like the one at wikipedia, which I'm not familiar with.

Secondly, I have a harder time to see why equation $(4.)$ might be relevant, as compared to equations $(2.)$ and $(3.)$ which are at least the fourier transforms of expressions that exist in $(1.)$. If anything I think it would be relevant in the fourth, final step in $(1.)$: $$ 2B\sigma^2 \int_{-B}^B e^{-j2\pi f(t_1-t_2)} df =\\ 2B\sigma^2\text{sinc}(2B(t_1-t_2)) $$ but there's no $\tau$ involved here so it confuses me still.

Any help with understanding the derivation cited is appreciated, thanks for reading.

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The inverse Fourier transform of $H(f) = \operatorname{rect}\left(\frac{f}{2B}\right)$, the transfer function of the ideal lowpass filter of bandwidth $B$ Hz, is, as the OP says, $h(t) = 2B\operatorname{sinc}(2Bt)$. Now, the power transfer function of this ideal lowpass filter (ILPF) of bandwidth $B$ Hz is $|H(f)|^2$ to which we apply low-level math such as $0\times 0 = 0$ and $1\times 1 = 1$ to conclude that the power transfer function of the ILPF of bandwidth $B$ is also $\operatorname{rect}\left(\frac{f}{2B}\right)$. Thus, the power spectral density of the noise process at the output of the ILPF is $\sigma^2|H(f)|^2 = \sigma^2\operatorname{rect}\left(\frac{f}{2B}\right)$ and since high-power math (the Wiener-Khinchine theorem) says that the inverse Fourier transform of the power spectral density is the autocorrelation function, we get that the noise process at the output of the ILPF of bandwidth $B$ Hz has autocorrelation function $R(t) = \sigma^2\cdot(2B)\cdot\operatorname{sinc}(2Bt)$ which tells us that the process has mean $0$ (because $\lim_{t\to\infty}R(t) = 0$) and that the variance is $R(0) = 2B\sigma^2$ as the OP's text claims it is.


Turning to the OP's specific question, I have no idea what $x(n)$ or those commas mean in $(2)$, $(3)$, and $(4)$ (they have now been edited out), or where Parseval's theorem is being applied, but in $(1)$, the fact that the sinc is an even function and a straightforward replacement of $\tau$ by $\hat{\tau}+t_1$ gives a convolution \begin{align} \int_{-\infty}^\infty h(\tau - t_1)h(\tau - t_2) \,\mathrm d\tau &= \int_{-\infty}^\infty 2B\operatorname{sinc}\big(2B(\tau - t_1)\big)\cdot2B\operatorname{sinc}\big(2B(\tau - t_2)\big) \,\mathrm d\tau\\ &= \int_{-\infty}^\infty 2B\operatorname{sinc}\big(2B(\tau - t_1)\big)\cdot2B\operatorname{sinc}\big(2B(t_2 -\tau)\big) \,\mathrm d\tau\\ &= \int_{-\infty}^\infty 2B\operatorname{sinc}\big(2B\hat{\tau}\big)\cdot 2B\operatorname{sinc}\big(2B((t_2-t_1)-\hat\tau)\big) \,\mathrm d\hat\tau\\ &= 2B\operatorname{sinc}(2B\tau)\star 2B\operatorname{sinc}(2B\tau)\bigg\vert_{\tau = t_2-t_1} \end{align} which the OP's text insists on writing as $$2B\int_{-\infty}^\infty \sqrt{2B}\text{sinc}(2B(\tau-t_1))\sqrt{2B}\text{sinc}(2B(t_2-\tau)) d\tau$$ (why a $2B$ is pulled out, I don't know). Be that as it may, the notion that "convolution in the time domain corresponds to multiplication in the frequency domain" allows us to multiply $\operatorname{rect}\left(\frac{f}{2B}\right)$, the Fourier transform of of $2B\operatorname{sinc}(2B\tau)$, and $\operatorname{rect}\left(\frac{f}{2B}\right)$, the Fourier transform of of $2B\operatorname{sinc}(2B\tau)$, to get $\operatorname{rect}\left(\frac{f}{2B}\right)$, as explained earlier. Then, \begin{align} 2B\operatorname{sinc}(2B\tau)\star 2B\operatorname{sinc}(2B\tau)\bigg\vert_{\tau = t_2-t_1} &= \mathcal F^{-1}\left(\operatorname{rect}\left(\frac{f}{2B}\right)\right)\bigg\vert_{\tau = t_2-t_1}\\ &= \int_{-\infty}^\infty \operatorname{rect}\left(\frac{f}{2B}\right) e^{j2\pi (t_2-t_1)f} \,\mathrm df\\ &= \int_{-B}^B 1\cdot e^{j2\pi (t_2-t_1)f} \,\mathrm df. \end{align} That's where the $\displaystyle \int_{-B}^B$ comes from; the limits on the inverse Fourier transform integral can be changed from $-\infty$ and $\infty$ to $-B$ and $B$ because the integrand is zero outside these limits. Since the rect has value $1$ for $f \in [-B,B]$, we recognize the OP's text's $$ \int_{-B}^B e^{-j2\pi f(t_1-t_2)} df = \int_{-B}^B 1 \cdot e^{j2\pi f(t_2-t_1)} df$$ as the inverse Fourier transform of the rect evaluated at $t_2-t_1$, the point at which we are evaluating the convolution.

Sheeeesh! The OP's text (assuming that he copied it correctly) makes a straightforward calculation as complicated and messy as possible; a maximum of obfuscation coupled with a minimum of enlightenment.

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  • $\begingroup$ Thanks, that makes sense and is a shorter line of reasoning. I had made a slight error when copying the part with the different facts (2.)-(4.), there was not supposed to be any $x(n)$ in those, that was a left over from me copying the latex. Sorry if it caused confusion. $\endgroup$ Jun 1 at 21:32

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