Consider the following geometry:

radar geometry

Where $\bar{s} = (s_x,s_y)$ is an object of interest, and $\bar{r_c}$ is the location of the radar.

Let the echo delay time of a radar pulse to the scene center be $t_c = \frac{2}{c}|\bar{r_c}|$. Likewise, the echo delay time to the object of interest is $t_s = \frac{2}{c}|\bar{r_s}|$

I would like to find the time difference quantity $(t_c - t_s)$.

An approximation that ignores wavefront curvature is as follows:

$$(t_c - t_s) = \frac{2}{c}(s_x sin(\alpha) - s_y cos(\alpha))$$

I have tried to prove the above equation but haven't been able to figure out how they come up with that result. Some help would be appreciated.

Thank you.


1 Answer 1


In this coordinate system, the radar position vector is given by $$ \mathbf{r}_c = \sin(\alpha)\hat{\mathbf{x}} - \cos(\alpha)\hat{\mathbf{y}} $$ (scaling to unit length), and the vector $\mathbf{r}_s = \mathbf{r}_c - \mathbf{s} = (\sin(\alpha) - s_x)\hat{\mathbf{x}} - (\cos(\alpha) + s_y)\hat{\mathbf{y}}$.

We are interested in the difference in the lengths of these vectors: $|\mathbf{r}_c| - |\mathbf{r}_s| = 1 - |\mathbf{r}_s|$. The time difference follows by multiplying this by $2/c$. Now, \begin{align*} |\mathbf{r}_s| &= \sqrt{(\sin(\alpha) - s_x)^2 + (\cos(\alpha) + s_y)^2} \\ &= \sqrt{(1 + |\mathbf{s}|^2) - 2(s_x\sin(\alpha) - s_y\cos(\alpha))}, \end{align*} where I've used $\sin^2(\alpha) + \cos^2(\alpha) = 1$ and $|\mathbf{s}|^2 = s^2_x + s^2_y$.

Now my assumption (which may be incorrect) is that the approximation of ignoring the wavefront curvature is the same as saying that $|\mathbf{s}| \ll |\mathbf{r}_c| = 1$. Using this and expanding the square root as a Taylor series gives $$ |\mathbf{r}_s| \approx 1 - (s_x\sin(\alpha) - s_y\cos(\alpha)), $$ and so $|\mathbf{r}_c| - |\mathbf{r}_s| \approx s_x\sin(\alpha) - s_y\cos(\alpha)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.