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My background is in radar where it's quite common to use chirped excitation signals (i.e. pulse compression).

I have noticed that this method seems to not be used in medical ultrasound. Why not?

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  • \$\begingroup\$ Signal degradation through flesh not air? \$\endgroup\$
    – Solar Mike
    Sep 20, 2020 at 20:49
  • \$\begingroup\$ Who's to say medical ultrasound doesn't use chirped excitation - \$\endgroup\$
    – D Duck
    Sep 20, 2020 at 21:07
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    \$\begingroup\$ I mean, I have yet to find any ultrasound ICs that use chirps, I couldn't find any chirped commercial systems, and there seems to be relatively little literature on the topic. I'm not saying it doesn't exist at all, but it seems rare and I was hoping for some insight as to why. \$\endgroup\$ Sep 20, 2020 at 21:18
  • \$\begingroup\$ Distance to be measured and wavelength of sound vs ditto for radar. \$\endgroup\$
    – Andy aka
    Sep 20, 2020 at 21:18
  • \$\begingroup\$ @ColinMarcus By "pulses", do you mean a simple pulse at the carrier with no modulation on it? \$\endgroup\$
    – Envidia
    Sep 21, 2020 at 2:58

1 Answer 1

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There seems to be a fairly large number of papers on the subject of pulse compression in medical ultrasound according to Google.

The main reason to use pulse compression (ie using chirps) is to increase the average transmitted power to increase SNR but it does come with its own set of limitations, such as increasing the minimum range response and ambiguities in the presence of doppler.

It is used with radar because the available amplifiers that can provide high-quality output are limited in power (especially with semiconductor PA) but even TWTs can't provide the peak power that magnetrons do. Magnetrons however can't provide the signal quality needed for sophisticated beam-forming and don't integrate well with modern electronics.

If the transducers can provide adequate SNR without using compression, there is not much reason to use it.

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  • \$\begingroup\$ Chirps, or any other modulation on a waveform, is primarily used to achieve a desired range resolution for a given pulse width...SNR considerations usually come second. In pulsed-Doppler, this is done with a matched filter, where either a simple pulse or a chirp will give you the same SNR. Regarding Doppler, chirps are great because they are Doppler-tolerant. \$\endgroup\$
    – Envidia
    Sep 21, 2020 at 3:03
  • \$\begingroup\$ @Envidia It’s actually to do with the bandwidth. For a given maximum transmitting power, a chirp will provide a higher bandwidth. In order to distinguish two close scatters one needs to be able to separate two signals that are close together in the time domain. Thus shorter signals are preferable. Hence one needs a high bandwidth signal by the uncertainty principle, i.e. the narrower the signal is in the time domain the broader it is in the frequency domain. The higher bandwidth also allows for easier decoding. Remember that Shannon’s channel capacity is proportional to the bandwidth. \$\endgroup\$
    – user110971
    Sep 21, 2020 at 4:19
  • \$\begingroup\$ @user110971 I know it's about the bandwidth. Short simple pulses are preferable for better range resolution if you're willing to live with less average power on target. Pulse compression decouples these two requirements by introducing modulation so you can have good range resolution and still keep the longer pulse. \$\endgroup\$
    – Envidia
    Sep 21, 2020 at 4:50
  • \$\begingroup\$ @Envidia - the longer pulse required when using pulse compression restricts the minimum range of operation. That may be significant for medical use where the probe is in contact with tissue. \$\endgroup\$ Sep 21, 2020 at 14:10
  • \$\begingroup\$ @KevinWhite "Long" is relative depending on the application. I use "long" here to mean a pulse width that would be out of reach for a desired range resolution (and blind-range as you've said). You also aren't "required" to have a long pulse width in pulse compression, even though that's what you're shooting for in the first place. I tried not to make my statement application specific. \$\endgroup\$
    – Envidia
    Sep 21, 2020 at 16:35

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