# How to calculate Doppler Shift for multipaths from moving objects?

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?

• I disagree. This is classic radar systems engineering problem. Most the radar systems engineers in my organization have EE backgrounds & training, not physics. Commented May 20, 2020 at 23:39
• I'm only a beginner when it comes to radar systems, however I suspect it comes down to the phase relationship of the signal, e.g. how the human ear is able to pick out what direction a signal comes from by how the human ear "colours" the phase in different ways for different frequencies. Commented May 23, 2020 at 10:53
• excuse me, but why you assume that the incident wave from the transmitter propagates via the shortest possible path to the target and only the echo returns otherwise.? A stealth aircraft can theoretically cause this.
– user136077
Commented May 23, 2020 at 11:12
• I am not saying that these are the only two reflections that I will get back. Instead, I am picking two of the many possible reflections, and want to know what their Doppler shift should be if they follow a certain geometry. Commented May 23, 2020 at 17:11
• You can work it out but it is somewhat involved. More information is required: can the object be approximated to an isotopic scatter; is the wall far enough for the far field approximation to be valid (planar waves); is it parallel to the velocity of the object; is it large enough such that diffraction effects become negligible. If the answer to the above questions is yes, then it won’t affect the Doppler shift. Commented May 25, 2020 at 10:05

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)).$$

• Thanks! I agree with your analysis, but if you take my expressions for observed frequency and calculate Doppler shift, they approximate to your expressions. For example, $f_D = f_{obs} - f_s = f_s \frac{2vcos(\theta)}{c-vcos(\theta)} \approx f_s \frac{2vcos(\theta)}{c}$ Commented May 27, 2020 at 22:17
• Similarly, for the second expression: $f_D = f_{obs} - f_s = f_s \frac{c+vcos(\theta_1)}{c-vcos(\theta_2)} - f_s \approx f_s \frac{vcos(\theta_1) + vcos(\theta_2)}{c}$ Commented May 27, 2020 at 22:20
• @Learner, that doesn't look right; for the first case $f_D = f_s \frac{2v \cos(\theta)}{c} \neq f_s \frac{2v \cos(\theta)}{c-v\cos(\theta)}$!? More generally, your $f_s \frac{c + v\cos(\theta)}{c-v \cos(\theta)}$ can't be right because $c$ is (nearly always) so much larger than\ $v$that the Doppler shift is barely noticeable. Commented May 27, 2020 at 22:49 • I'm qualifying my own comment to show that @Learner's original expression is not very far from the expression that was derived above. This relies on$\frac{1}{1-\delta} \approx 1+\delta$for small$\delta$. Then, f_s \frac{c+v\cos(\theta)}{c-v\cos(\theta)} \approx f_s \frac{c+v\cos(\theta)}{c}(1+v/c \cos(\theta)) = f_s(1+v/c \cos(\theta))^2 \approx f_s(1+2v/c \cos(\theta)). That implies a Doppler shift approximately equal to$2f_s v/c \cos(\theta)$. Commented May 28, 2020 at 0:07 • Right. It follows from the general formula$f=\frac{c \pm v_r}{c \pm v_s}f_s\$, as explained here: en.wikipedia.org/wiki/Doppler_effect. It's great that you reached the same result via another reasoning. Thanks. Commented May 28, 2020 at 3:50

I don't know the answer, but maybe I can give some thinking contribution:

It is clear to me that the frequency of the signal will not be changed by the reflection on the wall, because the wall is not moving relative to the radar. So the wall has no doppler effect contribution, i.e. it will not change the frequency.

The frequency change caused by "bouncing off" the moving target must be heavily influenced by the reflection angle of the target. This reflection angle is heavily dependent by the form of the target.

Lets consider the case of a simple target is moving towards us (theta=0).

If the target is completely not facing towards us (i.e. a thin sheet of paper faced 90° from us), then the radar signal will be reflected on the back wall and thus will not be changed at all when received: $$fobs = f_s$$

If the object is completely facing towards us, we can use the above formula a we will see the doppler shift: $$fobs = f_s*\frac{c+v}{c-v}$$

In between there must be some simple gemoetric transition rule. E.g. if the object (sheet of paper) is facing $$\alpha=45°$$ to us while approaching, our signal will bounce off the object and thus will be changed by some amount between the results from above.

It think the formula could be $$fobs = f_s*\frac{c+v*cos(\theta)}{c-v*cos(\theta)}*cos(\alpha)$$ where alpha is the angle of the target surface towards the radar.