4.1 Sonar receiver model

Active sonar receiver model

Consider the sonar receiver processing chain shown below:

The sonar array input is heterodyned, low-pass filtered, sampled and scaled to generate a discrete time set of samples of the noise plus signal.

The input noise $n(t)$ is a real-valued, wide sense stationary random process with power spectral density $P_{\text{nn}}(f)$ . Because $n(t)$ is wide sense stationary, the power spectral density and the autocorrelation function are related by

$R_{\text{nn}}(\tau )=\underset{-\infty }{\overset{\infty }{\int }}{P}_{\text{nn}}(f)e^{-\mathrm{j2\pi f\tau }}\text{df}$

We assume that $n(t)$ is zero mean. A complex carrier $e^{\mathrm{j2\pi }\text{ft}}$ is applied to the random process $n(t)$ resulting in a complex random process $z(t)\text{.}$ The mean value of $z(t)$ is given by:

$E\{z(t)\}=E\{n(t)e^{\mathrm{j2\pi }\text{ft}}\}=E\{n(t)\}{e}^{\mathrm{j2\pi }\text{ft}}=0$

The covariance of $z(t)$ is given by

$E\left\{z(t)z^{}(s)\right\}=e^{\mathrm{j2\pi f}(t-s)}E\left\{n(t)n(s)\right\}=e^{\mathrm{j2\pi f}(t-s)}{R}_{\text{nn}}(t-s)$ ,

which shows that $z(t)$ is wide sense stationary as well.

$z(t)$ is passed through a band-pass filter to produce $x(t)$ . The frequency response of the band-pass filter is assumed to be low-pass with a bandwidth of $\text{B/2}$ Hertz. That is we will assume that the filter transfer function $H_{\text{BPF}}(f)$ is given by:

The resulting $x(t)$ is a wide sense stationary random process with zero-mean and power spectral density:

, (1)

Where we have assumed that the bandwidth of the receiver is small relative to the center frequency of the signal we are trying to detect, . The power spectral density of $x(t)$ can then be approximated by the power spectral density of the noise near $f_0$ :

If $x(t)$ has a power spectral density given by Eq-1, then the autocorrelation function of $x(t)$ becomes:

$R_{\text{xx}}(\tau )=E\left\{x(t)x^{}(t+\tau )\right\}=\underset{-B/2}{\overset{B/2}{\int }}\frac{N_0}{2}{e}^{\mathrm{j2\pi f\tau }}\text{df}=\frac{N_0}{2}\frac{\text{sin}\mathrm{\pi B\tau }}{\text{πτ}}$

Note that $R_{\text{xx}}(0)=\frac{N_{0}B}{2}$ .

Now if we choose a sampling interval $\mathrm{\Delta t}=1/B$ ; then the samples at $\mathrm{k\Delta t}$ have an autocorrelation given by

$E\{x(\mathrm{k\Delta t})x^{}(\mathrm{l\Delta t})\}=\frac{N_0}{2}\frac{\text{sin}(\mathrm{\pi B}(k-l)\mathrm{\Delta t})}{\pi (k-l)\mathrm{\Delta t}}=\frac{{\text{BN}}_0}{2}{\delta }_{\text{kl}}$ ,

Hence $x(\mathrm{k\Delta t}),k=\mathrm{0,1,}\text{.}\text{.}\text{.}$ is a discrete time, wide sense stationary, white noise with intensity $\frac{{\text{BN}}_0}{2}$ .

For matched filtering applications, we scale the output of the Analog to Digital conversion process by $\mathrm{\Delta t}=1/B$ to conserve the signal energy over a time interval $T\text{.}$ This creates the discrete time process $y_k=x(\mathrm{k\Delta t})\sqrt{\mathrm{\Delta t}}$ .

To see this, consider that

$E\left\{\underset{0}{\overset{T}{\int }}{\mid x(t)\mid }^2\text{dt}\right\}=\underset{0}{\overset{T}{\int }}E\{{\mid x(t)\mid }^2\}\text{dt}=\underset{0}{\overset{T}{\int }}{R}_{\text{xx}}(0)\text{dt}=\frac{{\text{BTN}}_0}{2}$

And

$E\left\{\sum _{k=1}^{k=\frac{T}{\mathrm{\Delta t}}}{\mid y(k)\mid }^2\right\}=\sum _{k=1}^{k=\frac{T}{\mathrm{\Delta t}}}E\left\{{\mid y(k)\mid }^2\right\}=\sum _{k=1}^{k=\frac{T}{\mathrm{\Delta t}}}\frac{{\text{BN}}_0}{2}\mathrm{\Delta t}=\frac{{\text{BTN}}_0}{2}$ .