Hi all,

I have a doubt related to the calculation of powers with intermodulation distorsion.
Given an example (from my professor notes) for a non linear device:

\(
x(t) = \cos(\omega t)\\
y(t) = k_1x(t) + k_2x^2(t) = k_1\cos(\omega t) + \frac{k_2}{2}(1 + \cos(2\omega t))
\)

it writes the expressions for the powers of the fundamental frequency and the second harmonics, expressed in dB are:

\(
P_0(\text{dB}) = G(\text{dB}) + P_i(\text{dBm})\\
P_2(\text{dB}) = G_2(\text{dB}) + 2 P_i(\text{dBm})\\
\)

where

\(
G(\text{dB}) = 20 \log_{10}(k_1)
G_2(\text{dB}) = 10 \log_{10} \left (\frac{k_2^2}{2} \right )
\)

I don't understand the result of \(G_2(\text{dB})\).
Shouldn't be

\(
G_2(\text{dB}) = 10 \log_{10} \left (\left (\frac{k_2}{2} \right )^2 \right )
\)

?

Could someone clarify where I am wrong ?
Thank you in advance

 

This is the way I tried to solve it:
Neglecting the DC component:

\(
y_2(t)= \frac{k_2 A^2\cos(2 \omega t)}{2}\\
P_\text{2}(\text{dBm}) = G_2(\text{dB}) + 2 P_i(\text{dBm})\\
= 20\log_{10}(k_2) + 20 \log_{10}(A^2) - 20 \log_{10}(2) + 2 P_i(\text{dBm})\\
G_2(\text{dB}) = 10\log_{10}(k_2^2) - 10 \log_{10}(2^2) \qquad A = 1\\
= 10 \log_{10}\left (\frac{k_2^2}{4} \right ) = 20 \log_{10}\left (\frac{k_2}{2} \right )
\)

But this of course leads to the wrong result, since the correct one should be

\(
G_2(\text{dB}) = 10\log_{10}\left ( \frac{k_2^2}{2}\right )
\)

So I don't understand what I am doing wrong.

@bertus , is it possible to move the thread into the homework section ?
Thank you.