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Phase of a down converted IF

时间:04-04 整理:3721RD 点击:
Hello All,

I had a question regarding the phase relation between the RF input and IF output of a down converting mixer. I am attempting to make a device which can measure the phase or delay of a particular DUT at X-band. I attached a simplified block diagram of the circuit below.



I am using a directional coupler to sample the forward the and reverse traveling wave. I am using the sampled forward traveling wave as a phase reference to compare and calculate the phase change seen in the sampled reverse traveling wave. For testing purposes i am using a sliding short as the DUT because I thought it would be an easy / controlled way to vary the phase ±180°. As a note, path lengths for the LO pumps are the same so the LO phase should be equal.

As i moved the sliding short I expected to see the phase of the IF test signal vary ± 180° compared to the IF reference. Due to finite directivity I also expected the amplitude of the IF singles to vary.



Above is a plot showing the maximum phase shift observed between the IF reference and test signal. For this test the RF = 8.4 GHz, LO = 8.399 GHz creating an IF = 1 MHz. Despite moving the sliding short 180° I only observed a 22 ns delay or ~8° of phase difference in the IF test signal between the two short positions. Long story short, should i actually expect to see 180 degrees of phase shift between the down converted IF test and reference signals? Not sure if its coincidence, but the measured IF delay is equivalent to a ~λ/4 delay at 8.4 GHz since my sliding short has an air dielectric.

Sorry for the super long post, hope it makes sense,
Thanks in advance,
Sami

Time delay of 22ns that you get at 1MHz with 8° phase shift, is exactly what should be using the equation:

Time delay (sec) = [Phase Shift (°)] / [360 x frequency (Hz)]

Thanks for the response,

That part I understand. I suppose my whole post can boil down to: does a 22ns delay at the 1 MHz IF correspond to a 180 phase difference at the 8.4 GHz RF since 22ns*8.4 GHz ~ 180.

How so? 1/(8.4GHz*4)=30ps, not 22ns. Quarter wavelength in air should be less than 1cm at 8.4GHz, are you sure you did your math right?

No, the phase shift at the IF and RF ports should be the same, not the time delay.


YES.. Mixing to a lower frequency allows Doppler shift over a wider range with less cycle skipping. conversely, using a 180 deg phase detector, direct conversion gives more sensitivity.

But using an IF allows easier high Q filtering for easier high SNR

Hey all Thanks for the help.

I have a follow up question. The real circuit design is a bit different than the block diagram in the original picture. There is a RF multiplixer between the directional coupler and sliding short. I was only able to see a maximum phase shift of 8 degrees with the multiplexer in place. However, when I remove the multiplexer and connect the sliding short directly to the coupler I can see the IF vary the full 180 degrees when i move the sliding short a half wavelength.



When I have the multiplexer why do the reflections caused by any discontinuities mask the phase shift created by the sliding short? The Multiplexer does present a significant insertion loss -4 dB. I would also assume if I calibrate the system using a short open load standard I will be able to account for the multiplexer causing problems?

I might think the directivity of the MUX is poor? Which waveguide model?

I am am unhappy with the built in time delay of the 8.4 GHZ down the length of the coupler. I feel that this phase shift should be compensated for at the 1 MHZ before trying out other mismatches.
Frank

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