Matched filter approximation using ideal LPF

The true matched filter for a rectangular pulse is an integrate-and-dump filter, giving $(S/N)_{max} = 2E/N_0$. A simpler practical receiver uses an ideal LPF whose bandwidth $B$ can be chosen; the question is how close it can come.

Set-up. Pulse $p(t) = A$ for $|t| \le T/2$ (energy $E = A^2T$), white noise PSD $N_0/2$, ideal LPF of unit gain and bandwidth $B$. Pulse spectrum:

$$P(f) = AT\,\frac{\sin \pi fT}{\pi fT}$$

 |H(f)|
   1 +-----------+
     |           |
 ----+-----+-----+-----> f
    -B     0     B

Peak output signal (at the pulse centre $t=0$):

$$\begin{aligned} s_o(0) &= \int_{-B}^{B} AT\frac{\sin\pi fT}{\pi fT}\,df = \frac{2A}{\pi}\int_0^{\pi BT}\frac{\sin x}{x}\,dx \\ &= \frac{2A}{\pi}\,\text{Si}(\pi BT) \end{aligned}$$

Output noise power: $N_o = \dfrac{N_0}{2}\cdot 2B = N_0B$

Output SNR

$$\left(\frac{S}{N}\right)_{LPF} = \frac{4A^2\,\text{Si}^2(\pi BT)}{\pi^2N_0B}$$

Compare with the matched filter, $(S/N)_{MF} = 2A^2T/N_0$:

$$\rho = \frac{(S/N)_{LPF}}{(S/N)_{MF}} = \frac{2\,\text{Si}^2(\pi BT)}{\pi^2\,BT}$$

Effect of bandwidth

• $B$ too small: pulse energy is cut off, peak signal is small.
• $B$ too large: little extra signal but noise grows in proportion to $B$.
• An optimum exists in between.

Evaluating $\rho$ numerically:

Result: the best choice is $B \approx 0.685/T$ (about $0.7/T$), giving $\rho \approx 0.825$, i.e. only about 0.84 dB worse than the true matched filter. Hence an ideal LPF with bandwidth about $0.7 \times$ the bit rate is a good, simple approximation to the matched filter for rectangular pulses.