Why do I get only odd harmonics appearing? This not only happens at saturation of the transfer curve, but also in the linear portion of the curve.

I make a input signal that has too high power that it will saturate the transfer curve. The transfer curve below is for Amplitude modulation, I also have a transfer curve of the output phase change for each input signal level.

This is basically how I do it

My input signal, x(t), is the sum of two sine waves

For each point on the transfer curve above I apply Amplitude and Phase modulation by using this array

modulating_phasor[] = (power_out / power_in) * exp(i*phase_deviation)

Then I look at my input signal and lookup where its value matches the input power of the transfer curve and use that location in the modulating_phasor array

output_signal[] = x[t] * modulating_phasor[location]

This makes the output signal complex array so I take the absolute value and for parts that should be negative, I multiply that location on by -1 by finding the correct location on input signal.

I cant upload my code... but you get the idea of what I am doing right?

Now this is my output and no matter what my input level on my signal is i.e. whether I am in the non-linear part or linear part, I always get odd harmonics. I've also tried a different transfer curve. Only get odd again.

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  • $\begingroup$ You say that "the input is the sum of two sine waves"; but the "time domain input signal" plot looks like AM, and the "time domain input signal spectrum" looks like a single sine wave. Can you clarify? $\endgroup$
    – MBaz
    May 23, 2019 at 14:59
  • $\begingroup$ This code is this f = (200:1:201); for c=1:length(f) sinusoids(c,:) = cos(2*pi*f(c).*n) ; end signal=sum(sinusoids); $\endgroup$ May 23, 2019 at 15:10

3 Answers 3


An ideal amplifier would have a transfer characteristic of $f(x)=Ax$: the input signal comes out amplified and otherwise undistorted. A real amplifier will deviate from this and go into saturation. We could model it by a polynomial $f(x) = \sum_{n} a_n x^n$. Now, what we would still expect is that the amplifier treats positive and negative values the same way, i.e., it is symmetric in the sense that $f(-x) = -f(x)$. This leads to the requirement $a_n = 0$ for all even $n$.

Now, one can show that raising a sinusoidal with frequency $f_0$ to the $n$-th power generates harmonics up to order $n$, however, preserving their oddity:

  • $n = 2$ generates frequencies 0 and $2f_0$
  • $n = 3$ generates frequencies $f_0$ and $3f_0$
  • $n = 4$ generates frequencies 0, $2f_0$ and $4f_0$
  • $n = 5$ generates frequencies $f_0$, $3f_0$ and $5f_0$

and so on. It's a consequence of the binomial theorem and quite easy to show (see, e.g., here).

In short: A transfer function that is odd symmetric will generate only odd harmonics. This is why you are seeing only those.

  • $\begingroup$ Shouldn’t it be $f(-x)=-f(x)$ to get odd symmetry and to put even $n$ coefficients zero in your answer? $\endgroup$
    – user135626
    Feb 14 at 0:19
  • $\begingroup$ Yes, of course! Thanks, good catch! I edited my reply. $\endgroup$
    – Florian
    Feb 14 at 10:55

Your question is a bit confusing (see my comment), but I'll mention here some general properties of non-linear systems that might be relevant.

The simplest way to characterize the response of a memoryless non-linear system is with a power series. If the input is $x(t)$, then the output is $$ y(t) = a_0 + a_1 x(t) + a_2 x^2(t) + a_3 x^3(t) + \ldots $$ where the coefficients $a_i$ depend on each particular system.

Systems such as amplifiers that have a linear region around 0, but then saturate, typically have small or zero $a_0, a_2, a_4,\ldots$ and relatively large $a_1, a_3, a_5,\ldots$. Then, if the input is $\cos(2\pi f_0 t)$, the output has terms of frequency $f_0$, $3f_0$, $5f_0$, etc. This may explain what you are seeing.

If the input is the sum of two sinusoids of frequencies $f_0$ and $f_1$, then you'll see terms with the sum and difference of the frequency, some of first order: $f_0+f_1$, $f_0-f_1$, and some of third order: $2f_0 \pm f_1$, $2f_1 \pm f_0$, etc. These are called "intermodulation products".

If the input is an AM signal, then you get even more complicated interactions. However, you can derive them all from the power-series model above, with a bit of algebra and trigonometry.

  • $\begingroup$ Using this method doesn't allow applying non linear phase modulation. Basically I needed to create a modulating AM and PM phasor that amplifies each point of the input signal. The phasor is built from the amplifier transfer curves. $\endgroup$ May 23, 2019 at 15:14
  • 1
    $\begingroup$ What I see in your plots is an AM input and a clipped AM output. The power-series model explains why you see odd harmonics in that case. I don't understand the non-linear phase part of your question; I'd suggest writing the equations or inlining your code to make yourself clearer. $\endgroup$
    – MBaz
    May 23, 2019 at 15:29

I don't completely understand what you're doing with the lookup part, but if you're multiplying this transfer function in the time domain, then I think it makes sense that you're adding harmonics. If you look at the frequency response of a triangular wave (https://en.wikipedia.org/wiki/Triangle_wave), it is made up of decaying odd harmonics, so even if x(t) were constant, I'd expect you'd still see the harmonics. If you set x(t) to a constant, you could see what the frequency response of the transfer function itself is. I believe applying a triangular window function (https://en.wikipedia.org/wiki/Window_function#Triangular_window) would have a similar effect.


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