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harryhu Junior Member Offline Posts: 11	IIP3 calculation with unequal two-tone inputs Apr 12^th, 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.
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aaron_do

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Re: IIP3 calculation with unequal two-tone inputs
Reply #1 - Apr 12^th, 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

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harryhu

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Re: IIP3 calculation with unequal two-tone inputs
Reply #2 - Apr 13^th, 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

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aaron_do Senior Fellow Offline Posts: 1398	Re: IIP3 calculation with unequal two-tone inputs Reply #3 - Apr 13^th, 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
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philcorb

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Re: IIP3 calculation with unequal two-tone inputs
Reply #4 - Apr 13^th, 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

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harryhu

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Re: IIP3 calculation with unequal two-tone inputs
Reply #5 - Apr 14^th, 2012, 6:54am

aaron_do wrote on Apr 13^th, 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 (f_IM3), f₁ and f₂ are P_IM3, P₁ and P₂, respectively.

If we use the fundamental tone at f₁ in the calculation,
for log-magnitude response,
P₁ = x₁ + b₀
P_IM3 = x₁ + 2x₂ +b₁
We can have IIP3 = (P₁-P_IM3+2x₂)/2;

However, if we use the fundamental tone at f₂ in the calculation,
for log-magnitude response,
P₂ = x₂ + b₀�
P_IM3 = x₁ + 2x₂ +b₁
We can have IIP3� = (P₂-P_IM3+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

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harryhu

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Re: IIP3 calculation with unequal two-tone inputs
Reply #6 - Apr 14^th, 2012, 7:03am

philcorb wrote on Apr 13^th, 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.

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aaron_do

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Re: IIP3 calculation with unequal two-tone inputs
Reply #7 - Apr 14^th, 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

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philcorb

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Re: IIP3 calculation with unequal two-tone inputs
Reply #8 - Apr 16^th, 2012, 7:13am

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

philcorb wrote on Apr 13^th, 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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