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Parsevals theorem

itos

Hi,

I want to verify parsevals theorem in Cadence:

2.387E-3        stddev(v("/gnd" ?result "pss_td"))
2.387E-3        sqrt(integ(mag(v("/gnd" ?result "pss_td"))**2)/200e-9)
3.760E-3        sqrt(integ(mag(dft(v("/gnd" ?result "pss_td") 0 200n 2048 "Rectangular" 1 "default" 1.0 )/2048)**2))

The factor 3.76/2.387 is roughly pi/2 in which case I would accept it as some weird normalization. But not exactly. According to https://en.wikipedia.org/wiki/Parseval%27s_theorem my definition is valid.

What am I missing?

Thank you!

Parents
  • itos

    I try it again with this simple testbench:

    and running pss with 5MHz fundamental (containing 21 cycles), conservative, 51 harmonics.

    stddev(v("/out" ?result "pss_td"))

    is 707.1m, as expected.

    sqrt((integ((mag(v("/out" ?result "pss_td"))**2)) / 2e-07))

    is 707.1m, as expected.

    sqrt((integ((mag(dft(v("/out" ?result "pss_td") 0 2e-07 4096 "Rectangular" 1 "default" 1.0))**2)) / 5000000.0 / 2))

    is 706.1m - 0.15% off!

    WHY?

    (I divide the result by 10M because this is fs)

Reply
  • itos

    I try it again with this simple testbench:

    and running pss with 5MHz fundamental (containing 21 cycles), conservative, 51 harmonics.

    stddev(v("/out" ?result "pss_td"))

    is 707.1m, as expected.

    sqrt((integ((mag(v("/out" ?result "pss_td"))**2)) / 2e-07))

    is 707.1m, as expected.

    sqrt((integ((mag(dft(v("/out" ?result "pss_td") 0 2e-07 4096 "Rectangular" 1 "default" 1.0))**2)) / 5000000.0 / 2))

    is 706.1m - 0.15% off!

    WHY?

    (I divide the result by 10M because this is fs)

Children
  • Andrew Beckett
    in reply to itos

    It's because you don't have enough points in the shooting interval to do the DFT accurately - so you're suffering from interpolation error. If you plot the DFT you've done, you'll see the fundamental is a bit low - and this is the root cause.

    If instead you integrated the pss_fd results, you'd get the right answer because the Fourier integral used is more accurate. Alternatively, if you set (say) maxacfreq=16G on the pss analysis and then do the DFT, it would be fine.

    Or simpler still, do a transient and set strobeperiod so that the strobe points match the sample points of the DFT - that way you eliminate the interpolation error:

    tran tran stop=2e-7 strobeperiod=2e-7/2048

    Then if I do:

    printf("Tran power=%g\n" sqrt(integ(v("out" ?result "tran")**2)/2e-7))
    DFT=mag(dft(v("out" ?result "tran") 0 2e-7 2048 "Rectangular" 1 "default" 1.0 ))**2
    printf("Tran DFT integ=%g\n" sqrt(integ(DFT)/5M/2))

    I get:

    Tran power=0.707107
    Tran DFT integ=0.707107

    Regards,

    Andrew.

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