**1 Answer**

written 19 months ago by |

**Solution:**

The matched filter and the correlation function. The output of the matched filter is not a replica of the input signal.

However, from the point of view of detecting signals in noise, preserving the shape of the signal is of no importance.

If it is necessary to preserve the shape of the input pulse rather than maximize the output signal-to-noise ratio, some other criterion must be employed.

The output of the matched filter may be shown to be proportional to the input signal cross-correlated with a replica of the transmitted signal, except for the time delay t1.

The cross-correlation function R (t) of two signals y(λ) and s(λ), each of finite duration, is defined as The output yo(t) of a filter with impulse response h(t) when the input is yin (t) = s(t) + n(t) is If the filter is a matched filter, then h(λ) = s(t1 - λ) and Eq. above becomes ,

$ R(t)=\int_{-\infty}^{\infty} y(\lambda) s(\lambda-t) d \lambda $

$ y_0(t)=\int_{-\infty}^{\infty} y_{\mathrm{in}}(\lambda) h(t-\lambda) d \lambda $

$ y_0(t)=\int_{-\infty}^{\infty} y_{1 n}(\lambda) s\left(t_1-t+\lambda\right) d \lambda=R\left(t-t_1\right) $

Thus the matched filter forms the cross correlation between the received signal corrupted by noise and a replica of the transmitted signal.

The replica of the transmitted signal is "built in" to the matched filter via the frequency-response function. If the input signal yin (t) were the same as the signal s(t) for which the matched filter was designed (that is, the noise is assumed negligible), the output would be the autocorrelation function.

The autocorrelation function of a rectangular pulse of width τ is a triangle whose base is of width 2τ.