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I've consulted many books about radar working principle.
The well known formula of the resolution and alias free range extent (unambiguous range) appear referred to FMCW radar,ie

$$ \delta R = c/(2B) $$

$$ W_{r} = c/(2 \Delta f) $$ ,where $ \Delta f $ is the frequency step. The thing is that they appear 'magically', without any derivation.
I've tried through some physical relationship to give them a different meaning, but the result was inconclusive.
Could someone provide a mathematical proof/derivation of the above formula?

Thanks in advance

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The range in FMCW radars is found by generating a beat frequency after homodyning with the received signal. That beat frequency can be mapped to the target's range based on the parameters of the transmitted pulse and the resulting time delay of the target return.

Given a linear frequency-modulated (LFM) pulse with pulse length $\tau$ and swept-bandwidth $\beta$, the range mapping $R$ is given by

$$R = \frac{f_bc\tau}{2\beta}$$

Where $f_b$ is the beat frequency generated after mixing. The derivation can be found here.

In order to analyze the beat frequencies, we have to sample and perform a DFT on the homodyned signal. Let's assume that the signal is composed $N$ samples. At some sampling frequency $f_s$ and a DFT size of $N$, the frequency bin size $\Delta f$ is given by

$$\Delta f = \frac{f_s}{N}$$

Using the form of the first equation, we can take differentials and see that the range resolution $\Delta R$ is given by

$$\Delta R = \frac{\Delta fc\tau}{2\beta}$$

Using the second equation and the fact that $\tau=\frac{N}{f_s}$, then we get

$$\Delta R = \frac{f_s}{N}\frac{N}{f_s} \frac{c}{2\beta} = \frac{c}{2\beta}$$

And we get the ubiquitous equation for the range resolution of a waveform.

With these conditions, the sampling and DFT parameters make it so that the resulting frequency bins accomodate the waveform's range resolution capability exactly. In practical systems, the frequency bin sizes are commonly finer since padding is used, but this does not increase the raw capability of the waveform. In other words, it is common that the range-bin size of the DFT is finer than the range resolution that the waveform can achieve, which is fine! It's the other way around that needs to be avoided.

As for your expression for $W_r$, please clarify what $\Delta f$ is or where you sourced it. It does not look correct.

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  • $\begingroup$ $ \Delta f $ is the frequency step in the bandwidth. For example, if we transmit from 1 GHz till 5 GHz, 5 pulses equally spaced, then $ \Delta f $ is 1 GHz. Please refer to the paper 'SAR image formation toolbox for MATLAB' as a source. And thanks for the great answer ! $\endgroup$
    – teuto42
    Sep 11, 2021 at 7:59
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    $\begingroup$ Ok, that's a different type of waveform than a simple LFM pulse. What you're now referring to is stepped-FM waveforms, which need special consideration. $\endgroup$
    – Envidia
    Sep 11, 2021 at 9:02

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