Let's say we have a radar where the transmitter and receiver are placed together, and form an angle theta with a target object moving with velocity v.
I understand that the Doppler shift in this case is given by:
where c is speed of light, and fs is source of frequency. My question is what happens when we receive a multi-path reflection instead of a direct reflection back from the target.
My thinking is that the Doppler shift will now depend on the signal from TX to the moving target, and then from moving target to the wall, and so the observed frequency will now look something like this:
Is this expression correct?
Best Answer
I don't believe that either of your two expressions for the Doppler shift expressions is correct. Let's look at the first problem - without reflecting wall - first.
At time \$t=0\$, let's assume that the distance from the radar to the target is \$d\$. Then, the roundtrip delay for the signal emitted by the radar is \$\tau(0)=2 d/c\$. Because of this delay, the phase of the reflected signal when it arrives back at the radar is \$\phi(0) = -2\pi f_s \tau(0) = -4\pi f_s/c d = -4\pi d/\lambda\$, where \$\lambda\$ is the wavelength of the signal.
At time \$t=dt\$, the target has moved by \$v \cdot dt\$. However, the distance from the radar to the target has reduced only by \$v \cos(\theta) \cdot dt\$ to \$d-v \cos(\theta) \cdot dt\$; the movement in the perpendicular direction, \$v \sin(\theta) \cdot dt\$ does not affect the distance if \$d\$ is large enough - that's the far field assumption. Proceeding as above, the phase of the incoming signal is \$\phi(dt) = -4\pi f_s/c (d-v \cos(\theta) \cdot dt)\$.
Then, the Doppler shift is \$1/(2\pi)\$ times the rate of change in the phase \$\phi\$, i.e., $$ f_D = \frac{1}{2\pi} \lim_{dt \rightarrow 0} \frac{\phi(dt) - \phi(0)}{dt} = 2 f_s \cdot v/c \cdot \cos(\theta). $$ and the observed frequency is \$f_{abs} = f_s + 2 f_s v \cos(\theta)\$.
For the case with the reflection, first transform the problem geometry as follows to simplify the analysis. Treat the reflecting wall like a mirror and move the receiver to its mirror image location, i.e., on the other side of the wall. Then, instead of looking at the reflection of the wall, let the ray that is reflected off the target pass through the wall to the receiver. This transformation preserves all distances, which we saw above are critical for the problem.
You indicated in your figure that the exiting angle of the ray at the target is \$\theta_2\$ and the incoming angle is \$\theta_1=\theta\$. Then, we can apply the same analysis as above and find that at time \$t=dt\$ the ray from transmitter to target is shortened by \$v \cos(\theta_1)\cdot dt\$ and the ray from target to (reflected) receiver is shortened by \$v \cos(\theta_2)\cdot dt\$. Thus, the Doppler shift becomes: $$ f_D = f_s \cdot v/c \cdot (\cos(\theta_1)+\cos(\theta_2)). $$