Pulse compression

Active positioning systems such as sonar and radar can benefit from a process called pulse compression, increasing resolution and detectability. It effectively exploits knowledge of the transmitted waveform to more robustly extract a signal of interest from noisy data.

Basic premise

The fundamental idea of pulse compression is to pass the transmitted signal across the received signal and find when they closely match. This can be done with cross correlation of the complex transmit and receive signals and is shown below.

Derivation

A common transmit signal is a pulsed linear frequency modulation (LFM), also known as a chirp signal. This involves the baseband signal beginning the pulse at a frequency of 0, and linearly increasing the frequency while maintaining a continuous wave. It looks something like………..

Applying the Hilbert transform to get into the complex domain allows this waveform to be expressed by the single phasor $s_{tx}\left(t\right)$,

Here, $k$ is the chirp rate, describing how fast the frequency increases, and $T$ is the duration of the pulse. The rectangle function is equal to unity between $t=0$ and $t=T$, but zero everywhere else.

If an echo is received after a time $d$ with amplitude $A$, the received waveform is given by the phasor $s_{rx}\left(t\right)$,

The rectangle function here is unity between $d$ and $d+T$ but zero elsewhere.

Cross correlation

To compute the cross correlation, the resulting waveform $r\left(t\right)$ is described by

Substituting the transmit and receive waveforms gives

where the conjugate operator has inverted the sign on the first exponential.

Since $t$ is seen only in the first rectangle and exponential function, varying $t$ varies the position of the first rectangle and not the second. Thus, values of $t$ which have no rectangle function overlap lead to the integrand being zero, as at least one of the rectangle functions is zero for all $\tau$.

However, values of $t$ with rectangle overlap generally exist for the scenarios we are interested in. In these cases, the rectangles overlap, with the intersection being the interval $\left( \text{max}\left(t, d\right), \text{min}\left(t,d\right)+T \right)$, with the conditions that $t+T \geq d$ and $d+T \geq t$. Since the integrand is zero outside this interval, the bounds can be shrunk from $\pm \infty$ to the interval. Furthermore, since inside the interval, both rectangle functions are equal to unity, they can be removed. Thus, we obtain

Expanding the quadratic terms in the exponentials and employing exponential laws yields

The first exponential is constant with respect to the integral variable $\tau$, leading to pretty simple integration:

Implementing the integral limits gives

Combining the exponential term pairs by summing the exponents leads to

Now taking the common factor of $\left(t-d\right)$ out of the exponents gives

Here we do something a little strange: We multiply the entire expression by $e^{ik\left(t-d\right)T} e^{-ik\left(t-d\right)T}$, which is equal to unity. Multiplying the second exponential inside the square brackets and applying exponential laws gives

Further simplifying,

Now we consider the minimum and maximum functions. If $t > d$, then we have $2\text{min}\left(t,d\right)+T-t-d = d-t+T$ and $t+d+T-2\text{max}\left(t,d\right) = d-t+T$. In contrast, if $d>t$, we have $2\text{min}\left(t,d\right)+T-t-d = t-d+T$ and $t+d+T-2\text{max}\left(t,d\right) = t-d+T$. In other words, if $t>d$, both expressions become $d-t+T$, but if $d>t$, both expressions become $t-d+T$. This can be simplified quite nicely by replacing all expressions with $T-\left|t-d\right|$. Therefore, we can replace the nasty expression in both exponentials with a much nicer one:

Notice both exponentials have the same argument, but with the second one being negative. With the division by $2i$, this is equivalent to a sine expression, and can be rewritten as

This is looking pretty nice, but we can still do better. Notice that the denominator is the same as the first part of the sine argument. We now multiply numerator and denominator by the rest of the sine argument:

We now have something of the form

and can convert the sinusoid to a sinc function, giving the final expression for the derivation,