So I have given up using QPSS and QPAC to find mixer IIP3 due to convergence problem. So the alternative is to use transient analysis and input 2 sine waves and take the DFT at the output.

My method is to use a port and give two sine waves @ 2.45G and 2.451G. I run transient with maximum step of 1p. For DFT calculation I choose the range 3n to 3n+(2/2.45G).  But I am having an issue to see its spectrum displayed. I can't see two tones displayed only one. Is there anything wrong with my setup?

Thanks

I think the reason I don't see the tones displayed @ 2.45G, 2.451G and 2.449G has to do with the number of samples and the time for transient analysis.  Anyone out there can offer me some tips on how I should run transient analysis to obtain IIP3?

Thanks

Uncle Ezra,

  The fundamental tones in the simulation are 2.45G and
2.451G. The third-order intermodulational products are
(2*2.45G - 2.51G)=2.49G and (2.51G-2.45G)=2.51G. So
the FFT needs to have a fundamental frequency of 1MHz,
or the FFT needs a transient simulation that runs from
x ns to x + 1000ns.  Since the FFT displays the frequencies
from 0 to fs/2, you will need to have at least 8192 points
in the FFT, fmax=4.096G.

   The IIP3 is the third order intercept point of the
intermodulation product, 2*f1-f2 and 2*f2-f1. QPSS
analysis was developed to deal with the issue of
simulating circuits with closely spaced tones efficiently.
because this is a difficult simulation to perform with
transient analsysis.


                                                     Best Regards,

                                                        Sheldon

Uncle Ezra.

  You need to be a little careful, the point was that the
you need to display all the frequencies without aliasing
them. To display 2451MHz without aliasing means that
2451MHz > 2048 * 1MHz or 4096 points are required.
Also, the FFT displays the frequencies from 0 to fs/2 which
means that fs/2=4096, or the FFT needs 8192 points.  
To improve the accuracy of the FFT, you might want to
try using the zvcvs element to force the correct
timepoint.

  One other option to consider, depending on the version
of SpectreRF you have access to is using, Rapid IP3
analysis. It is intended for the problem of Mixer IP3
calculation.          

                                                       Best Regards,

                                                           Sheldon                                    

Sheldon explained this, but maybe this is a little clearer ...

There are two fundamental issues which impact your problem
1. You have two high frequency signals which dictate the minimum sampling rate
2. You want to resolve two closely spaced signals (in frequency) which dictates the required frequency resolution (Hz/bin)

The sampling rate (i.e. time step) dictates the highest frequency without aliasing; although, many dft functions use interpolation to permit extension of spectrum (but this is a different issue).

The FFT/DFT frequency resolution is dictated by the duration of the simulation. So if you want to resolve two-tones spaced 1 MHz then the DFT resolution has to be at least this or less. I always try to have at least 1 DFT bin between tones just to be sure the DFT is completely correct (no unexpected energy between the tones).

*** You will never be able to resolve tones that are spaced closer together than 1/tstop (where tstop represents the time duration of the simulation).

So the total number of time points (fft points) required are

N = trange/tstep = freqRange/freqResolution

where

freqResolution = 1MHz (or maybe 500kHz) = 1/trange, trange=1usec simulation stop time (plus whatever is needed for the transient to settle)

freqRange = 2*2451MHz = 1/tstep

N = 2*2451MHz/1MHz = 4902 points minimum plus however many points needed for the transient to settle.

So, let's say you have done all this correctly and simulated long enough for the initial transient to settle and captured enough time beyond that to resolve the tones. Now how do you set the time range for the DFT function?

stop time = tstop (last time point)
start time = tstop - (1/freqResolution)

of course you may need to offset this by maybe one point to get it right on the DFT grid.

Many DFT functions (like in the calculator) offer waveform interpolation when you specify the number of points to use. This is helpful when there are higher harmonics present in the time domain output since these will alias down and cause some confusion when you look at the spectrum. The interpolation allows you to extend the frequency range of the solution without simulating at even finer time steps to capture the harmonics.

As sheldon suggested, just let the simulator set tstep as it usually does to satisfy the convergence criterion. This is usually adequate for an accurate simulation solution for the fundamentals and harmonics even if tstep is not fine enough to prevent aliasing in the subsequent FFT/DFT analysis. Interpolation on the accurate solution is effective for providing you with a wider output spectrum without distorting the answer (assuming the interpolation does not produces distortion).

Finally, if there is some transient in the signal or the transient is too slow for you to wait for it to completely settle out then you can use DFT windowing functions to help suppress the DFT spectral leakage that results from taking the DFT of a signal with an aperiodic transient signal. But the cost is that you loose some frequency resolution whenever you use a window function (usually a minimum of +/- 1 bin around any frequency of interest). This means the simulation stop time would have to double, so this is only helpful if there is a long time constant in the circuit that you can't mitigate some other way.

Perhaps this was more information that you needed, but hopefully it (along with sheldon's solutions) helped answer your question.

Also, you might want to search the forum or post a separate question regarding your qpss convergence issues. Perhaps someone has some useful suggestions.