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I have a seemingly simple question but I've been doing a lot of searching and can't find an answer.

What is a simple or standard way to achieve a 90 degree phase shift of an RF signal in the frequency range of ~1 MHz up to a few 10s or 100s of MHz? In the minicircuits catalog I see Hybrid couplers as low as 25 MHz so I would say 25 MHz and up is covered (though I don't know the cost) so the question can be narrowed down to how to achieve a 90 degree phase shift for a signal below 25 MHz?

In particular, in my specific application I have a signal which can be fixed in the range of 1-3 MHz that I would like to shift by 90 degrees. The 90 degree shifter doesn't need to necessarily need to work over a very large frequency range.

Ideas that I think would work but I don't know how well:

  • putting in a time delay with a length of cable. This is a bit inconvenient because it's a long length of cable (~10 m or more) and I don't have incremental control over the phase shift.
  • Put the signal through the stop band of a filter. This will give it a 90 degree phase shift but unfortunately it will also necessarily suppress the signal amplitude by a lot.. I've only thought about the case for a simple single pole filter
  • Digitizing the signal, somehow implement the phase delay digitally, then re-synthesize an analog output.

It seems like this should be something not too difficult, but in terms of commercial solutions I have only found products for higher frequency ranges. I can't tell if I can't find what I'm looking for because it's something so simple it's just not sold or if I'm searching for the wrong things or if there is some genuine difficulty with what I need. Any tips are appreciated!

I am asking for potential use in a modulation transfer spectroscopy system in which I need to generate a phase modulated (at ~3 MHz) signal at 80 MHz and then subsequently demodulate to extract the phase quadrature of a new signal generated at the modulation frequency. Both the phase modulation and phase quadrature detection could be done with the assistance of a 90 degree phase shifter and mixers.

edit: Clarification. I do NOT need a wideband phase shifter. I am curious about a general technique which could be used to phase shift a narrow band signal whose carrier may be anywhere in the range from 1-100 MHz. That is, say I have a narrowband 3 MHz signal. How can I give this a 90 degree phase shift? Say I have a narrowband 80 MHz signal. How can I give this a 90 degree phase shift? I do not need a wideband phase shifter. Apologies for the confusion. If different techniques are suitable for different frequencies within the range I have indicated then that is ok. I am at present most interested in a way to give a 90 degree phase shift to a signal which is like 3 MHz.

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  • \$\begingroup\$ How precise must the phase-shifting be? How stable? \$\endgroup\$ May 3, 2018 at 5:25
  • \$\begingroup\$ It is hard for me to quantify that. It really probably only needs to be precise to 10-15 degrees. In terms of stability it should be ok if it doesn't drift out of that range on any time scale. It would be annoying to have to tune it every day or even week for example but it could be dealt with if there was an easy knob to turn. This is not for any sort of clock application. Ideally the phase shift would be tunable. In theory I want a 90 degree phase shift but in practice other circuit elements introduce uncontrolled phase shifts which may need to be compensated. \$\endgroup\$
    – Jagerber48
    May 3, 2018 at 14:25

5 Answers 5

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If you are happy with feeding a signal into a "block" and getting two signals out that are 90 degrees apart may I recommend this: -

enter image description here

Select R = R1 = R2 = \$\sqrt{\frac{L}{C}}\$. This ensures that OUT1 and OUT2 have a phase differential of 90 degrees across all frequencies. The phase relationship between OUT1 and OUT2 is always 90 degrees but amplitudes do change with frequency as with any high-pass or low-pass fliter.

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  • \$\begingroup\$ This seems nice but I will need to check out how much the different signals might be attenuated for each path. I suppose it will depend on the loads which I am attaching ta OUT1 and OUT2. These will likely be 50 ohm loads. \$\endgroup\$
    – Jagerber48
    May 2, 2018 at 17:02
  • \$\begingroup\$ @Jagerber48 You don't have to load this, directly, in fact, it's recommended that you buffer them, or even make them active. \$\endgroup\$ May 2, 2018 at 17:05
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    \$\begingroup\$ @aconcernedcitizen please don't pre-empt this. Of course the OP can load the outputs with 50 ohm. In fact the loads can replace the resistors providing that the correct ratio of L:C is used to make \$\sqrt{L/C}\$ = 50. \$\endgroup\$
    – Andy aka
    May 2, 2018 at 17:08
  • \$\begingroup\$ Yes with subsequent buffering and/or amplification I see that this could be a general solution. \$\endgroup\$
    – Jagerber48
    May 2, 2018 at 17:18
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In my previous answer, I suggested using a PLL to lock onto a single tone signal; you commented, and that warrants a new answer, that:

My signal is not wideband. I am saying suppose I have a narrowband signal at a given frequency somewhere in the range of 1-100 MHz. How can I then give that narrow band signal a phase shift. It is ok if the phase shift device needs to be tuned or constructed a differently for different frequencies within the range.

Anything having any significant bandwidth (i.e. anything but a slowly changing tone) can't be dealt with a PLL; you could do something like using an ADC to digitize the signal and then digitally do the phase shift, but really, that might not work either, due to latency restraints: to shift a 1 MHz signal by 90° will require a latency of at least half a 1 MHz period; in that time, 50 periods of your 100 MHz signal would have passed. If you're OK with that latency, you can go digital (but that's an expensive solution).

Else, detect the center frequency (might be absolutely non-trivial), and use adjustable phase shifters (PIN diode ICs, for example), and make it so that the phase shift is approximately 90° around the center frequency.

I'm coming from a software defined radio background, so: Use a mixer to mix your band-limited signal down to a fixed frequency (an intermediate frequency (IF), this is a superhet, then, or to complex baseband, that's a direct receiver), and then mix it back up with a single tone and a 90° shifted version of that tone. (That, by the way, is effectively a quadrature mixer, and you'd be building a signal that's equivalent to a baseband signal with \$\Re(s)=\Im(s)\$. Whatever that is good for.)

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If you need a phase shift across a band of frequencies, then you need a Hilbert transformer. If you say, in your third point, that you are willing to do so, then I'm afraid that's your only way. Any passive, or active filter will only add a 90 deg phase shift at a certain frequency. Unless you have the option to tune that frequency, a Hilbert transformer is what you're looking for.

Here's an example of what three frequencies will look like:

ht

In your case, you'll need some DSP or some sort. Again, if you don't have the possibility to tune the frequency in an analog way.

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  • \$\begingroup\$ Any passive, or active filter will only add a 90 deg phase shift at a certain frequency. that is a very feeble thing to imply. My answer's circuit will produce two outputs that are 90 degrees apart across the full spectrum. \$\endgroup\$
    – Andy aka
    May 2, 2018 at 16:58
  • \$\begingroup\$ @Andyaka Yes, but at what amplitude? Otherwise a simple differentiator will do the same thing. A Hilbert transformer solves that. At any rate, it's one variant of an answer. \$\endgroup\$ May 2, 2018 at 17:00
  • \$\begingroup\$ See my comment and edit for clarification. I am not looking for a wideband phase shifter. Only a way to phase shift a narrowband signal which may fall in the indicated range. It is ok if I need a different tuning or different device for each frequency. \$\endgroup\$
    – Jagerber48
    May 2, 2018 at 17:00
  • \$\begingroup\$ @Jagerber48 Then a HT will be too expensive schematic-wise. It would also induce a necessary delay (see the picture), which may get in the way. In this case, Andy's solution, seems simple enough to be able to tune it to a desired frequency by using a twin potentiometer and properly chosen L and C. \$\endgroup\$ May 2, 2018 at 17:04
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    \$\begingroup\$ @aconcernedcitizen there you go again - it doesn't need to be specifically tuned to a frequency. Once set (and loaded correctly) it will produce 90 degrees from DC to infinity. \$\endgroup\$
    – Andy aka
    May 2, 2018 at 17:11
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putting in a time delay with a length of cable. This is a bit inconvenient because it's a long length of cable (~10 m or more) and I don't have incremental control over the phase shift.

You don't get a constant phase shift that way at all – the phase shift at \$f_0\$ will be \$\frac1{10}\$ of that at \$10f_0\$ and

What is a simple or standard way to achieve a 90 degree phase shift of an RF signal in the frequency range of ~1 MHz up to a few 10s or 100s of MHz?

Uhhhhh, that's more than ultra-wideband, by common definitions, because the bandwidth that you want to cover is much higher than the center frequency (of ca 50 MHz).

So, this is actually hard to do in the analog domain, and there's no real "simple" general way. However, Andy showed up the appropriate way to go:

By combining two filters with "complementary" bode phase plots, you can get two signals that are 90° shifted over a large range of frequencies.

Things get a lot easier if your input isn't actually wideband, but a single tone at any single point in time: Then, you could just as well use a PLL to produce a 90° shifted version of that, but that doesn't work if your input actually is more than a clean tone.

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  • \$\begingroup\$ Sorry, I was unclear in my question. My signal is not wideband. I am saying suppose I have a narrowband signal at a given frequency somewhere in the range of 1-100 MHz. How can I then give that narrow band signal a phase shift. It is ok if the phase shift device needs to be tuned or constructed a differently for different frequencies within the range. \$\endgroup\$
    – Jagerber48
    May 2, 2018 at 16:55
  • \$\begingroup\$ Narrowband will really not be enough, without loss of generality: a PLL can only lock onto a tone. \$\endgroup\$ May 2, 2018 at 16:59
  • \$\begingroup\$ See upcoming answer from me. \$\endgroup\$ May 2, 2018 at 17:03
  • \$\begingroup\$ single tone is fine. I await upcoming answer! \$\endgroup\$
    – Jagerber48
    May 2, 2018 at 17:04
  • \$\begingroup\$ single tone doesn't need a new answer: simply use a PLL with a 90° shift. It's what is used in any quadrature mixer to convert the LO to the LO+90° shifted version and thus is the very heart of modern digital comms. \$\endgroup\$ May 2, 2018 at 17:08
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The existing method of producing narrow band quadrature RF signals in Instrument Design over a wide rage spanning many decades is performed at a higher frequency and then mixed down. This so that the quadrature Local Oscillator (LO) only has to be generated over an octave with traditional RF methods above twice the maximum frequency rather than span many decades.

As it happens... I just found such a VNA instrument that fits your range.

This article describes a low-cost vector network analyzer that operates from 200 kHz to 100 MHz, and connects to a personal computer using a USB 1.1 interface.

The Analog Devices AD9854 DDS produces the quadrature outputs.

It is clocked by a 24 MHz sine wave; it then internally multiplies it up to 288 MHz with an on-chip PLL. This 288 MHz internal signal clocks the DDS frequency-generation circuits and the digital-to-analog (DAC) circuits on the chip. Significant aliasing of the output signal is observed on an oscillo scope even at an output frequency of less than 100 MHz. The two low-pass filters, one on I and one on Q, remove most of these aliasing and stair-stepping artifacts and produce clean sinewaves in phase quadrature.

But I suggest you shop around for a used commercial VNA for calibrated results that meet your spectroscopy requirements.

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