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I'm new to radar. please excuse my question if it doesn't make sense to you.

In a pulsed radar signal, why is the required sampling frequency "2 times the inverse of pulse width" instead of "2 times the carrier center frequency"?

Consider a 100 ms pulse width and a 1000 Hz carrier frequency, i.e. there are 100 cycles of sinusoidal waves per pulse width.

If I use a sampling frequency of 2/(0.1 s), what would I get in the spectrum plot?

Also, is this sampling frequency (2*1/pulseWidth) the same as the fast time sampling frequency?

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    \$\begingroup\$ You don't sample the carrier, rather the envelope but for better than binary quality > 2x is required \$\endgroup\$
    – Tony Stewart EE75
    Oct 11, 2019 at 15:31
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    \$\begingroup\$ In radar, the term 2 / pw is the required receiver bandwidth (as I learned many years ago). \$\endgroup\$
    – Peter Smith
    Oct 11, 2019 at 15:58
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    \$\begingroup\$ The carrier of microwaves is needed to propagate RF reflections but measured by the envelope after signal processing. \$\endgroup\$
    – Tony Stewart EE75
    Oct 11, 2019 at 19:24
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    \$\begingroup\$ usualy the radar Transmitter Antenna is highly focused, to concentrate the power in a narrow beam, hoping to overcome the 1/Range^4 energy attenuation of the return pulse. Thus 500MHz and 5,000MHz and 50,000MHz are common frequency ranges for radars, because the dish antennas are of tolerable size. \$\endgroup\$
    – analogsystemsrf
    Oct 12, 2019 at 3:57
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    \$\begingroup\$ @Jung_Zheng The choice of carrier frequency is very important as it affects many things. One practical example is the size of components, such as an antenna. An antenna of the same gain at two largely separated frequencies will have a large impact on their size difference. An antenna with a gain of 24 dBi at 100 MHz is much larger than the same again antenna operating at 30 GHz. \$\endgroup\$
    – Envidia
    Nov 15, 2019 at 17:08

2 Answers 2

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In a pulsed radar signal, why is the required sampling frequency "2 times the inverse of pulse width" instead of "2 times the carrier center frequency"?

In a "pure" digital system, you do need to sample at a higher rate, although twice the carrier frequency won't work. You could get unlucky and sample on the zero value of the carrier, which occurs twice per cycle, and miss the return entirely.

However, radars normally work with an analog front end which produces the envelope of the return. Assuming a "perfect" envelope detector, which produces a rectangular return when it detects a pulse, if you sample at 1 times the pulse width, you could sample at the edges of the pulse and miss it completely. Instead, a good rule of thumb is to sample twice per pulse width, which is a frequency twice the inverse of the pulse width.

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  • \$\begingroup\$ None of the radars I'm familiar with use an analog front end to produce an envelope of the return. They use an analog front end to do signal conditioning, maybe some gain control, and (most importantly) mixing or down converting to create an intermediate frequency, usually I&Q in the hundreds of MHz range. This IF is then digitized and the rest of the processing and information extraction takes place in the digital domain. \$\endgroup\$
    – SteveSh
    Jan 10, 2022 at 19:57
  • \$\begingroup\$ @SteveSh - Downconverting the return produces something very similar. It should be pretty clear that in order to reliably detect the return, the converted return must provide a full period within the pulse width. However, when sampling a downconverted return you actually do need at least two samples during the pulse width. That's the minimum required to reconstruct the pulse keeping the Nyquist limit in mind. \$\endgroup\$
    – WhatRoughBeast
    Jan 20, 2022 at 2:47
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In a pulsed radar signal, why is the required sampling frequency "2 times the inverse of pulse width" instead of "2 times the carrier center frequency"?

This is the case when we are only sampling the pulse envelope. When this is the case the signal bandwidth \$B\$ is: \$B = \frac{1}{\tau}\$ where \$\tau\$ is the pulse width. By the Nyquist–Shannon sampling theorem we must sample at a rate twice the signal bandwidth. In this case this corresponds to \$2B\$ or samples spaced half a pulse width apart.

This does not necessarily apply when using a system with modulated waveforms and performing pulse compression. In this case the waveform bandwidth can be greater than the inverse of the pulse width and the sample rate must be correspondingly higher. This is no longer an envelope sampling system however.

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