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Can someone answer or point to a link explaining why a nonlinear system causes harmonics.

According to Wikipedia

... non-linear systems generate harmonics, meaning that if the input of a non-linear system is a signal of a single frequency, then the output is a signal which includes a number of integer multiples of the input frequency.

but why?

The only explanation that I can come up with is the following.

Let's say we have a frequency fc going through a NLTI system with a span. The upper and lower part of the span, as well as all the other intermediate frequencies will be affected differently than the it's surrounding frequencies due to the NLTI system causing mixing between all different frequencies thus causing harmonics.

If this is true then sending a delta function through a NLTI system will appear to be a LTI system.

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  • \$\begingroup\$ "a NTLI system will appear to be a LTI system." An LTI is uniquely determined by feeding in the delta function. If you feed a delta function into an NLTI system, you will get a response but this response doesn't necessarily uniquely characterize the NLTI system. If your NLTI is strictly non-linear then the LTI system characterized by the response to the delta function will not be the original NLTI system. \$\endgroup\$ – SomeEE Mar 13 '14 at 4:33
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Can someone answer or point to a link explaining why a nonlinear system causes harmonics.

Consider a system with a quadratic non-linearity:

$$y = Ax + \epsilon x^2$$

Let the input \$x = \alpha_1 \cos(\omega_1 t) + \alpha_2 \cos(\omega_2t)\$ be the sum of two sinusoids.

The output is thus

$$y = A[\alpha_1 \cos(\omega_1 t) + \alpha_2 \cos(\omega_2t)]+ \frac{\epsilon}{2}[\alpha^2_1 + \alpha^2_1\cos(2\omega_1t) + \alpha^2_2 + \alpha^2_2\cos(2\omega_2t)+ \alpha_1 \alpha_2 \cos(\omega_1 t + \omega_2 t) + \alpha_1 \alpha_2 \cos(\omega_1 t - \omega_2 t)]$$

Note that we have both 2nd harmonic and intermodulation distortion present.

So, the answer to your question is the well known trigonometric identity:

$$\cos(a)\cos(b) = \frac{1}{2}[\cos(a+b) + \cos(a-b)]$$

When \$a = b\$ this becomes

$$\cos(a)\cos(a) = \cos^2(a) = \frac{1}{2}[\cos(2a) + 1] $$

since \$\cos(0) = 1\$

In other words, when we multiply sinusoids, we get sinusoids with sum and difference frequencies.

If this is true then sending a delta function through a NLTI system will appear to be a LTI system.

The delta "function" is a distribution so one cannot 'simply' write something like \$\delta^n(t)\$ and hope it has meaning. From the linked Wikipedia article:

Thus, nonlinear problems cannot be posed in general and thus not solved within distribution theory alone.

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The simple answer is that if the system response can be represented by a function and the function can be approximated by an expansion with terms beyond the linear that are not insignificant, then you get squares and cubes etc. (as in a Taylor series).

Any input can be represented as a Fourier sum, so this means squaring sines or cosines. A squared sin(t) has a cos(2t) term so frequency is doubled and you can filter out and amplify the first harmonic.

Sin(t) cubed has two terms. One with sin(t) and one with sin(3t). 4th power has cos(2t) and cos(4t). Higher order terms are usually negligible.

Radio people call it "mixing" and optics folks call it "up-converting", or they used to.

By the way, this is where all those blue and green LASER pointers come from. They use an efficient IR LASER diode and a non-linear crystal of KDP or similar material. The IR is "up-converted" to the visible. I have noticed some green ones that dim and quit in cold weather. I assumed it was from the lattice spacing in the crystal changing as it contracted and no longer matching the wavelength of the pumping diode. Not sure though.

Definition of linear is f(a) + f(b) = f(a+b).

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  • \$\begingroup\$ The definition of linearity is incorrect (or incomplete): a function is linear if f(p a + q b) = p f(a) + q f(b). \$\endgroup\$ – Lorenzo Donati supports Monica Jul 6 '14 at 22:43

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