I am studying matched filters and the sense of signal to noise ratio. In a question I was given, stated that:

Show that a matched filter of a signal $s(t)$ maximizes the SNR of the output on a receiver of AWGN with power spectral density $$S_n(f) = \frac{N_0}{2}.$$

After they show that that happens when we sample at $t=T$.

I was given a step of this which stated that the power of the noise is

\begin{align} P_n &= \mathbb{E} \left[ \int n(t_1) h(T-t_1) dt_1 \int n(t_2) h(T-t_2)dt_2\right]\\ &= \int \int \frac{N_0}{2}\delta(t_1-t_2) h(T - t_1) h(T-t_2) dt_1 dt_2. \end{align}

Can anyone explain why that is? Which is the starting thought, or formula that yields to different times $t_1$ and $t_2$?


1 Answer 1


Can anyone explain why that is? Which is the starting thought, or formula that yields to different times $t_1$ and $t_2$?

The noise power $P_N(t)$ at any given time $t$ at the output of any filter (whether matched or not) is, by definition, $\mathbb E[(N(t))^2]$ where $N(t)$ is the noise in the output at time $t$. If the input noise is $n(t)$ and the filter is a bounded-input-bounded-output (BIBO) filter (as all matched filters are) with impulse response $h(t)$, then $N(t)$ is given as the familiar convolution integral $$N(t) = \int_{-\infty}^\infty n(\tau)h(t-\tau) \,\mathrm d\tau.$$ If $\{n(t)\}$ is a zero-mean process (as white Gaussian noise is for most readers though there is a minority who feels otherwise), then so is $\{N(t)\}$ a zero-mean process. Thus, $P_N(t) = \mathbb E[(N(t))^2]$ is given by
$$P_N(t) = \mathbb{E} \left[ \int n(\tau_1) h(t-\tau_1) d\tau_1 \int n(t_2) h(t-\tau_2)dt\tau_2\right].$$ We DSPers would be perfectly happy to write just $\tau$ instead of $\tau_1$ and $\tau_2$ but those nitpickers over on math.SE and stats.SE have severe objections so let's keep them happy, OK? So, combine the product of the two integrals into a double integral and take the expectation operator inside the double integral (as is allowed for BIBO filters) to get \begin{align} P_N(t) &= \mathbb{E} \left[ \int n(\tau_1) h(t-\tau_1) \, d\tau_1 \int n(\tau_2) h(t-\tau_2) \,d\tau_2\right]\\ &= \mathbb{E} \left[ \int \int n(\tau_1)n(t_2)h(t-\tau_1) h(t-\tau_2) \,d\tau_1 \, d\tau_2\right]\\ &= \int \int \mathbb E[n(\tau_1)n(\tau_2)]h(t-\tau_1) h(t-\tau_2) \,d\tau_1 \, d\tau_2\\ &= \int \int \frac{N_0}{2}\delta(\tau_1-\tau_2) h(t-\tau_1) h(t-\tau_2) \,d\tau_1 \, d\tau_2 \end{align} which sure looks like the integral that is puzzling the OP. $\tau_1$ and $\tau_2$ are variables of integration that disappear when the integrals are evaluated, e.g. by computing the antiderivative and plugging in the limits. Let's go ahead and do the complete calculation. We write the double integral as a nested integral and go from there.
\begin{align} P_N(t) &= \int \int \frac{N_0}{2}\delta(\tau_1-\tau_2) h(t-\tau_1) h(t-\tau_2) \,d\tau_1 \, d\tau_2\\ &= \frac{N_0}{2} \int \left[\int \delta(\tau_1-\tau_2) h(t-\tau_1) \,d\tau_1 \right] \,h(t-\tau_2) \, d\tau_2\\ &= \frac{N_0}{2} \int h(t-\tau_2)\cdot h(t-\tau_2) \, d\tau_2\\ &= \frac{N_0}{2} \int |h(\tau_2)|^2 \, d\tau_2. \end{align} The result also shows that the noise power at the output of the filter is constant, and so the exact sampling instant $T$ is a mere distraction.


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