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Message started by harryhu on Apr 12th, 2012, 4:58am

Title: IIP3 calculation with unequal two-tone inputs
Post by harryhu on Apr 12th, 2012, 4:58am

How to calculate IIP3 in two-tone tests with unequal input amplitude? For example, in the WCDMA out-of-band test case, the two tones at 190MHz and 95MHz offset are -30dBm and -45dBm, respectively.

Title: Re: IIP3 calculation with unequal two-tone inputs
Post by aaron_do on Apr 12th, 2012, 5:28pm

Hi,


IIP3 itself is defined for equal tones. You can find IM3 for unequal tones and then use those numbers to calculate the IIP3. Anyway, it is easy to work out the numbers graphically. For instance,

for magnitude response,

fundamental = b0x1
IM3 = b1x1x22

for log-magnitude response,

fundamental = X1 + B0
IM3 = X1 + 2X2 +B1

So you can work out B0 and B1. From there, IIP3 is the value of X (X1=X2 in this case) for which fundamental=IM3. Also note that you must chose X1 and X2 carefully since the upper and lower sidebands will be different.


Aaron

Title: Re: IIP3 calculation with unequal two-tone inputs
Post by harryhu on Apr 13th, 2012, 12:34am

Hi, Aaron. Thank you for the reply. It is useful to get back to the mathematical basics.

But I have a further question here. I am wondering if the method you gave may lead to different results if we use different fundamental tones to do the calculation (fundamental = b0x1). This maybe a problem when these two fundamental tones are out-of-band and experience different gains.

Song

Title: Re: IIP3 calculation with unequal two-tone inputs
Post by aaron_do on Apr 13th, 2012, 1:54am

Hi Song,


I don't really understand your question. If you change the frequency, the results can definitely be different, especially if the gain is different at the new frequency.


regards,
Aaron

Title: Re: IIP3 calculation with unequal two-tone inputs
Post by philcorb on Apr 13th, 2012, 2:58am

Hi,

It is possible to calculate IP3 from an sstac analysis.  In which case one of the two input tone is considered small signal.

Supposing two tones are applied to a system, with Amplitudes A1 and A2 and frequencies f1 and f2 respectively.

The amplitude of the tone at 2f1-f2 (A_IM2) is determined by the amplitude A2
The amplitude of the tone at 2f2-f1 (A_IM1) is determined by the amplitude A1

OIP3 = A1 x sqrt(A2/A_IM2), in V
OIP3 = A2 x srqt(A1/A_IM1), in V

These equations can be referred to the input for IIP3

I hope this helps




Title: Re: IIP3 calculation with unequal two-tone inputs
Post by harryhu on Apr 14th, 2012, 6:54am


aaron_do wrote on Apr 13th, 2012, 1:54am:
Hi Song,


I don't really understand your question. If you change the frequency, the results can definitely be different, especially if the gain is different at the new frequency.


regards,
Aaron


Hi Aaron,

I am sorry I did not clearly express my thinking.

Supposing the amplitude of the input two tones at f1 and f2 are x1 and x2, respectively. And the output tones at 2f2-f1 (fIM3), f1 and f2 are PIM3, P1 and P2, respectively.

If we use the fundamental tone at f1 in the calculation,
for log-magnitude response,
P1 = x1 + b0
PIM3 = x1 + 2x2 +b1
We can have IIP3 = (P1-PIM3+2x2)/2;

However, if we use the fundamental tone at f2 in the calculation,
for log-magnitude response,
P2 = x2 + b0
PIM3 = x1 + 2x2 +b1
We can have IIP3’ = (P2-PIM3+x1+x2)/2;

IIP3 and IIP3’ are equal only if the gain at f1 and f2 are equal, which may not be true for all systems.

I hope the aforementioned explanation can make my question clearer. Thank you.

Best Regards,
Song

Title: Re: IIP3 calculation with unequal two-tone inputs
Post by harryhu on Apr 14th, 2012, 7:03am


philcorb wrote on Apr 13th, 2012, 2:58am:
Hi,

It is possible to calculate IP3 from an sstac analysis.  In which case one of the two input tone is considered small signal.

Supposing two tones are applied to a system, with Amplitudes A1 and A2 and frequencies f1 and f2 respectively.

The amplitude of the tone at 2f1-f2 (A_IM2) is determined by the amplitude A2
The amplitude of the tone at 2f2-f1 (A_IM1) is determined by the amplitude A1

OIP3 = A1 x sqrt(A2/A_IM2), in V
OIP3 = A2 x srqt(A1/A_IM1), in V

These equations can be referred to the input for IIP3

I hope this helps


Hi,

Although I usually use analysis like PSS, QPSS in simulations, it is interesting to apply this idea in this case. Would you kindly please provide some references for detailed reading? Thank you.

Title: Re: IIP3 calculation with unequal two-tone inputs
Post by aaron_do on Apr 14th, 2012, 6:57pm

Hi,


if the two tones are close enough, then the IIP3 for the lower and upper sideband will be the same. However, if they are not then they can be different as you pointed out. If you care about both upper and lower, then I would think you should either specify the two IIP3's separately, or only specify whichever is worse. Sometimes datasheets specify the worst case IM3 in a specific band, and sometimes they specify IIP3.


regards,
Aaron

Title: Re: IIP3 calculation with unequal two-tone inputs
Post by philcorb on Apr 16th, 2012, 7:13am


harryhu wrote on Apr 14th, 2012, 7:03am:

philcorb wrote on Apr 13th, 2012, 2:58am:
Hi,

It is possible to calculate IP3 from an sstac analysis.  In which case one of the two input tone is considered small signal.

Supposing two tones are applied to a system, with Amplitudes A1 and A2 and frequencies f1 and f2 respectively.

The amplitude of the tone at 2f1-f2 (A_IM2) is determined by the amplitude A2
The amplitude of the tone at 2f2-f1 (A_IM1) is determined by the amplitude A1

OIP3 = A1 x sqrt(A2/A_IM2), in V
OIP3 = A2 x srqt(A1/A_IM1), in V

These equations can be referred to the input for IIP3

I hope this helps


Hi,

Although I usually use analysis like PSS, QPSS in simulations, it is interesting to apply this idea in this case. Would you kindly please provide some references for detailed reading? Thank you.


I think I got this from one of Ken's papers

http://www.designers-guide.org/Analysis/intercept-point.pdf

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