# How can I describe a sum of sinusoids by an equivalent lowpass signal?

I consider a sinusoidal signal with fixed amplitude $A_{\rm in}$ and frequency $f_c$.

This signal is sent into a nonlinear amplifier, whose AM/AM and AM/PM characteristics are known. Using these characteristics for the given amplitude, I can determine the output amplitude $A_{\rm out}$ and the output phase $\phi_{\rm out}$.

Now however, I consider a sum of two sinusoids with different amplitudes $A_{\rm in_1}$ and $A_{\rm in_2}$ and different frequencies $f_{\rm c_1}$ and $f_{\rm c_2}$. First, I want to describe this input signal by its equivalent lowpass signal. I want to do this in MATLAB. I know the hilbert() command gives the analytic signal. So I suppose when my first sinusoid is called $x_1(t)$, the equivalent lowpass signal can be described by $\mathcal H(x_1(t)) \cdot \exp(j2\pi f_{c_1} t)$. The same is valid for the second sinusoid. But I don't know how to represent the sum by an equivalent lowpass signal. In a second step, I want to apply the AM/AM and AM/PM characteristics to the signal. But now, I have two different amplitudes.

I found this: Consider a multiple carrier input: $x(t) = \sum(A_i(t)*\cos[(\omega_c + \omega_i)* t + \theta_i (t)],i,N)$

This can be written as $x(t) = A(t)*\cos[\omega_c *t + \theta(t)]$

where $A(t) = [x_I^2(t) + x_q^2(t)]^{1/2}$

$\theta(t) = tan^{-1} [x_I(t) / x_q(t)]$
I really don't understand this equation given for the resulting phase. Why is it valid? I mean X_I and X_Q are not just real and imaginary parts of x(t), are they? I know by Euler's formula, that the cos corresponds to the real part of the exponential and the sin corresponds to the imaginary part. But (see below) X_I and X_q are both sums of cos or sin waves of different amplitudes and different phases. How can I just add them?

$X_I(t) = \sum(A_i(t) * \cos[\omega_i * t + \theta_i (t)],i,N)$

$X_q(t) = \sum(A_i(t) * \sin[\omega_i * t + \theta_i (t)],i,N)$

• What does AM/AM and AM/PM mean? – Arnfinn Dec 10 '16 at 22:44
• i think, in order to be clear about what "AM/AM" and "AM/PM" mean, you should define what the "nonlinear amplifier" does mathematically. – robert bristow-johnson Dec 10 '16 at 23:21
• Since the amp is nonlinear it's AM AM and AM PM conversions operate on the whole input. I e the equivalent amplitude and phase of the sun of two sinusoids. Not on two amplitudes and two phases. – Nir Regev Dec 11 '16 at 7:54
• AM stands for amplitude modulation, PM stands for phase modulation. The AM/AM characteristics shows how the amplitude is changed by the amplifier, depending on the input amplitude. The AM/PM characteristic shows how the phase is changed, depending as well on the input amplitude. How can I determine the equivalent amplitude and phase of the sum of two sinusoids? – user25356 Dec 11 '16 at 12:14

The sum of two sinosoids with different frequencies $f_{c1}, f_{c2}$ does not add up to a single sinosoid. Only, if both sinosoids have the same frequency, you can combine them into a single sinosoid using the equation you state at the end of your post. See also for example this post.