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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.

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1 Answer 1

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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 $|\mathbf{r}_c| - |\mathbf{r}_s| \approx s_x\sin(\alpha) - s_y\cos(\alpha)$.

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