3
$\begingroup$

I would like to create a pulsed RF signal that has a carrier frequency of 2 GHz by IQ up-conversion. The pulse should be 2-4 ns long. The carrier phase has to jump by 90° in the centre of the pulse. The problem is that my IQ modulator has only 200 MHz bandwidth. I work at an IF of 100 MHz, the maximum IF sampling rate is 1 GHz. Therefore, I see quite some distortions. Are there some tricks one can play to circumvent distortions e.g. by amplitude shaping or smoothening the phase change or is the bandwidth of the IQ modulator a hard boundary condition for such short pulses?

$\endgroup$
1
  • 2
    $\begingroup$ You're talking about a super broadband pulse compared to your carrier. A 2ns pulse, even without a 90$^\circ$ phase shift in the middle, has a bandwidth of at least 500MHz, and realistically probably 1GHz. At those sorts of bandwidth ratios, you probably want to be restating your question by showing us a picture of the pulse itself that you want to generate. That makes this question much more appropriate for electronics.stackexchange.com, because your biggest challenge at the moment is just generating the pulse -- and that's an electronics question, not signal processing per se. $\endgroup$
    – TimWescott
    Commented Dec 30, 2022 at 17:28

2 Answers 2

4
$\begingroup$

The choice of an IF frequency of 100 MHz limits the achievable single-sided bandwidth for a real signal centered on it to be 100 MHz, so any pulse out of the modulator would be limited in rise/fall time as given by that bandwidth, without introducing other distortion effects.

Given the 1 GHz sampling rate, and ideal IF frequency if we must have a real IF that would maximize the achievable rise / fall time would be $f_s/4$ or 250 MHz. Or even better would be a direct RF approach with two DACs and an analog modulator that can support an RF output at the desired 2 GHz directly with carrier nulling techniques used (as commonly done) to null the inevitable carrier feedthrough with such an approach. With that a maximum possible single-sided pulse bandwidth at baseband would be 500 MHz (plus margin for filtering).

This then leaves the modulator bandwidth as the limit to the achievable rise/fall time where there are actual possible tricks we can play. If the power and dynamic range allow, and if the driving signal has the excess bandwidth, you can briefly overshoot your maximum value to achieve the faster rise time desired.

A good approximation of the achievable rise/fall time is

$$t_r = \frac{0.35}{BW}$$

Where $t_r$ is the 10%/90% rise or fall time, and BW is the single sided bandwidth in Hz of the equivalent baseband signal. This is exact for a first order (single pole) system but works as a good approximation for any system with a dominant single pole and gives us a rough idea for higher order systems as well as to what the settling time will be.

Note that with a simple example of a first order system (or any system with a dominant pole) the time step response is given as:

$$y(t) = A(1 - e^{t/\tau})$$

Where $A$ is the intended final value at pulse peak and $\tau$ is the time constant of the system with a BW in Hz given as $BW = 1/(2\pi \tau$). If we solve this for 10% and 90% of $A$ we would get the relationship given above for bandwidth and rise/fall time. I show how this relationship holds up for higher order systems at this post.

From this we see with the OP's use of a 200 MHz Modulator (as well as the IF as noted) will limit the maximum single-sided bandwidth (or equivalent baseband bandwidth). If we assume this is the RF bandwidth, then the real passband signal is limited to be 100 MHz. However it is possible that this is the baseband modulation bandwidth (not clear from the post). Without doing any "tricks", the 10% - 90% rise/fall time for a 100 MHz BW is 3.5 ns, or 1.75 ns for 200 MHz BW (in which case if a 1.75 ns rise/fall time was acceptable, raising the IF to 250 MHz would be the way to go). The phase transition would also follow a similar settling time.

However here comes the trick: if we have the power and dynamic range and ability to sink and source with that full power needed, we can instead overshoot the intended final value! Note that the rise time is independent of our final value $A$. So if we were able to (if our power and dynamic range allows) instead create a pulse that were to rise to a much higher value than $A$ on its own, we will reach $A$ in a much shorter time. When we do reach $A$ in that case, we pull the system low (within the same bandwidth constraint) flattening the top of the pulse. The signal at the input to the modulator would appear as a sharply rising peak with much higher bandwidth, which assuming is still in the linear operation of the modulator would be filtered by the modulator to be a faster rising pulse at the output ultimately limited by the 1 GHz sampling spectrum to not have aliasing artifacts.

This is exactly what occurs in a PID loop as I explain further at this post, where we are able to achieve rise/fall times beyond what the physical plant would allow based on its own time constant (and nicely so since it is under loop control, so the overshoot generated in the control signal is exactly what is needed to achieve the desired output signal and transition times.) A PID rise/fall time is limited by the second lowest pole in the plant rather than it's lowest pole.

As a simplified demonstration of this, consider the baseband equivalent signal as represented below showing a 5 ns duration pulse that changes phase by 180 degrees (we could easily extend this to the I and Q pulses associated with a 90 degree rotation) in the middle of the pulse which is passed through a 100 MHz low pass filter. This predicts the envelope including the phase transition at the output of a modulator that has similar bandwidth.

filtered pulse

I created a pre-emphasis compensator resulting in approximately 12 dB of high frequency gain, which modifies the driving pulse as follows (where the red circles show the actual samples of a 1 GSps sampled signal):

pre-emphasis

Then using the pre-emphasized drive signal through the same filter results in an improved time response:

improved time response

The comparative result for this particular pre-emphasis is shown below. This assumes that we have considered the linearity and noise figure trades with operating the output of the modulator 12 dB below it's maximum linear operating range.

comparative result

With this I wanted to demonstrate the possible "trick" that can be done, and its limitations. On a practical level I would first seek to find a faster modulator solution, or do direct modulation at the 2 GHz carrier, or raise the IF bandwidth and potentially the sampling rate to achieve the target rise/fall times and phase transition time most simply.

$\endgroup$
4
$\begingroup$

Your mixer's bandwidth is a physical property of both the nonlinear part that implements the mixing as well as the strictly necessary filters to eliminate the unwanted intermodulation products.

You can't get 1 GHz through a component with 200 MHz bandwidth, no matter what tricks you play.

As a matter of fact, not even your sampling rate is high enough: for something to turn on, change phase completely after 1ns, and then turn off after another nanosecond, you will need 1 GHz bandwidth, and your IF sampling is real, thus offerring you at most 500 MHz bandwidth (at 1 GHz sampling rate).

So, you can't build the thing you want with your DAC, either.

$\endgroup$
1
  • 1
    $\begingroup$ Not to nit-pick but to note since this is interesting (and as Tim Wescott cleared up for me on an earlier post) that the mixing with an ideal mixer is not due to non-linearity but due to a linear time-variant operation. Although we can mix with non-linearities as we may be forced to do with single diodes at mmW frequencies, the common double-balanced mixer indeed uses the time varying property (switch the RF 0/180 degrees basically) to do the mixing. It's the unavoidable non-linearity in the non-ideal mixers that creates intermod products, but not the primary freq x-lation $\endgroup$ Commented Dec 30, 2022 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.