Third order Intercept Point (IP3)

Thrid order intercept point(IP3) is a used to quantify weak non-linearity in an amplifier or RF system.
It is an interpolated point where the fundamental component power curve meets the third order intermod(IMD3) component power curve.

Consider a nonlinear amplifier whose input-output characteristic is represented by a 3rd order polynomial given by

(1)   \begin{equation*} v_o(t) = \alpha_o + \alpha_{\tiny 1} v_{i}(t) + \alpha_{\tiny 2} v_{i}^{\tiny 2}(t) + \alpha_3 v_{i}^3(t) \end{equation*}

where, v_o(t) is the output voltage of the amplifier
and v_i(t) is the input voltage applied to the amplifier

Generally a two tone test is used to compute IP3 for an amplifier.
The two tone input to amplifier is given by

(2)   \begin{equation*} v_{i} = A_{i,1} \cos(\varphi_{1}) + A_{i,2} \cos(\varphi_{2}) \end{equation*}

\varphi_{\tiny 1}=\omega_{\tiny 1}t+\phi_{\tiny 1} and
\varphi_{\tiny 2}=\omega_{\tiny 2}t+\phi_{\tiny 2}.

For two tone input, the output of the amplifier is given by substituting Eq.(2) in Eq.(1)

(3)   \begin{eqnarray*} v_o &=& \underbrace{\left(\alpha_{ 1}+ \alpha_{ 3}\left(\frac{3}{4}A_{i,1}^{ 2} + \frac{3}{2}A_{i,2}^{2} \right) \right)A_{i,1}\cos{\varphi_{1}} + \left(\alpha_{1}+ \alpha_{3}\left(\frac{3}{4}A_{i,2}^{2} + \frac{3}{2}A_{i,1}^{2}\right) \right) A_{i,2} \cos{\varphi_{2}}}_{fundamental} \\ & &+\underbrace{\frac{1}{2}\alpha_{2} \left(A_{i,1}^{2}\cos{2\varphi_{1}}+ A_{i,2}^{2} \cos{2\varphi_{2}} \right)}_{2^{nd}~~harmonic }\\ & &+\underbrace{\frac{1}{4}\alpha_{3} \left(A_{i,1}^{3} \cos(3\varphi_{1}) + A_{i,2}^{3} \cos(3\varphi_{2})\right)}_{3^{rd}harmonic}\\ & &+\underbrace{\frac{1}{2}\alpha_{2}A_{i,1}A_{i,2}\left(\cos(\varphi_{1}-\varphi_{2}) + \cos(\varphi_{1}+\varphi_{2}) \right)}_{2^{nd}~~order~~intermods}\\ & &+\underbrace{\frac{3}{4}\alpha_{3}\left(A_{i,1}A_{i,2}^{2}\cos(\varphi_{1}-2\varphi_{2}) + A_{i,1}^{2}A_{i,2}\cos(2\varphi_{1}-\varphi_{2})+ A_{i,1}A_{i,2}^{2}\cos(\varphi_{1}+2\varphi_{2}) + A_{i,1}^{2}A_{i,2}\cos(2\varphi_{1}+\varphi_{2})\right)}_{3^{rd}~~order~~intermods} \end{eqnarray*}

From Eq.(3), the output contain fundamental, harmonics, second order intermod and third order intermod components.
The third order intermod(IMD3) components are generated at frequencies (2\omega_1-\omega_2), (\omega_1-2\omega_2), (2\omega_1+\omega_2) and (\omega_1+2\omega_2). The first two components generally fall in the signal band.

Let us compute the IP3 using fundamental component(\omega_1) and IMD component(\omega_1 - 2\omega_2).
From Eq.(3), at IP3(where fundamental and intermod component are equal),

(4)   \begin{equation*} \left[\alpha_1 + \alpha_3 \left(\frac{3}{4}A_{i,1}^{2} + \frac{3}{2}A_{i,2}^{2} \right) \right]A_{i,1}} =\frac{3}{4}\alpha_3 A_{i,1}A_{i,2}^{2} \rightarrow \alpha_1 \approx {3 \over 4}\alpha_3 A_{i,2}^{2} \end{equation*}

The approximation (neglecting higher order terms in fundamental component) is reasonable at low power levels. At this power level, the amplitude of A_{i,2} corresponds to 3rd order intercept point A_{IP3}.

(5)   \begin{equation*} \boxed{A_{IP3} = \sqrt{{4\over 3} {\alpha_1 \over \alpha_3}} } \end{equation*}

The point is referred to input power. Hence it is called input third order intercept point or IIP3. If the point is referred to output it is called output third order intercept point or OIP3

If IIP3 or OIP3 is known then we can compute the intermod component power at any other power level.
Let’s derive this
From Eq.(3), the ratio of fundamental component to intermod component in output voltage is

(6)   \begin{equation*} { v_{o,1} \over v_{o,IMD3} } = {{\left[\alpha_1 + \alpha_3 \left(\frac{3}{4}A_{i,1}^{2} + \frac{3}{2}A_{i,2}^{2} \right) \right]A_{i,1}} \over \frac{3}{4}\alpha_3 A_{i,1}A_{i,2}^{2} } \approx { \alpha_1 \over \frac{3}{4}\alpha_3 A_{i,2}^{2} } \end{equation*}

Taking log on both sides, and rearranging

(7)   \begin{eqnarray*} 20 \log(v_{o,1}) - 20 \log(v_{o,IMD3}) &=& 20 \log\left( {A_{IP3} \over A_{i,2}} \right)^2  \\ 20 \log( {A_{IP3}) &=& 20 \log(A_{i,2}) + {1\over 2} \left[ 20 \log(v_{o,1}) - 20 \log(v_{o,IMD3}) \right] \end{eqnarray*}

Writing the equation in terms of power,

(8)   \begin{eqnarray*} IIP3 &=& P_{i,2} + {1 \over 2} \left[ P_{o,1} - P_{o,IMD3} \right] \\ &=& P_{i,2} + {1 \over 2} \Delta P_{IMD3} \\ \end{eqnarray*}

where \Delta P_{IMD3} is the IMD3 power relative to fundamental component power


An amplifier has a IIP3 of -10dBm. Compute the IMD3 power relative to fundamental at an input signal power level of -20dBm.

At input power of -10dBm(same as IIP3 point), \Delta P_{IMD3}=0.
When input power back-off by 10dB from IIP3, fundamental output power reduces by 10dB and IMD3 component power reduces by 30dB. Therefore \Delta P_{IMD3}=P_{o,1} - P_{o,IMD3}=20dB

Alternatively from Eq.(8), \Delta P_{IMD3}= 20dBc

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