# 0Hz output from a non-linear system

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#### ThermalRunaway

##### New Member
Hi everyone. Considering the addition of the following two discrete signals:

x1 = sin (2pi . 1 . nts) // 1 Hz sinusoid of peak amplitude 1
x2 = sin (2pi . 3 . nts) // 3 Hz sinusoid of peak amplitude 1

I have been investigating the theoretical output from a system that adheres to the following equation:

y = x_squared

when I apply the addition of x1+x2 to it.

I have proven that the system is non-linear because the theoretical output is a bunch of components that were not present in the original signals, as follows:

0Hz @ amplitude 1
2Hz @ amplitude 0.5
4Hz @ amplitude -1
6Hz @ amplitude -0.5

I calculated these with the help of trigonometric identities (and a text book )

I understand all of this, but... what about the 0Hz component? What does this look like in practice? Would it be seen as a D.C. level?

Brian

#### Speakerguy

##### Active Member
Yes, it would be DC.

#### RCinFLA

##### Well-Known Member
Yes, there is a D.C. component. Squaring any A.C. signal will create a D.C. component.

#### MrAl

##### Well-Known Member
Hi Thermal,

Did you check your results before making your conclusion? I dont think you did.
Here's why...

Define the following constants:
w1=2*pi*1
w2=2*pi*2
w3=2*pi*3
w4=2*pi*4
w6=2*pi*6

If we add a 1Hz sine to a 3Hz sine and square the result we get this:
ya=(sin(w1*t)+sin(w3*t))^2

and at some random check point t=0.346 we get this:
ya=(sin(w1*t)+sin(w3*t))^2=1.123667 approximately,

and if we add a 2Hz sine with amplitude 0.5 and a 4Hz sine with amplitude -1
and a 6Hz sine with amplitude -0.5 along with a dc component of 1, we get this:
yb=1+0.5*sin(w2*t)+(-1)*sin(w4*t)+(-0.5)*sin(w6*t)

and at that same random time check point t=0.346 we get this:
yb=1+0.5*sin(w2*t)+(-1)*sin(w4*t)+(-0.5)*sin(w6*t)=-0.362966 approximately.

Since ya(0.346) is not equal to yb(0.346) then we can say that the two
equations are not the same. This means you need to go over your work again
to see where it is not working.

Did you use Fourier? That will show you the components best.
I believe your conclusion is correct, but you should get the equations right
before you conclude anything.

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#### ThermalRunaway

##### New Member
Speakerguy79, RCinFLA: Thanks for your repsonse.

MrAl:

Thanks also for your reponse. I'm not sure we're on the same page here, though.

I am talking about what happens when I apply the addition of two discrete sine waves (in this case, of 1Hz and 3Hz both of amplitude 1) to a non-linear system. The output of the non-linear system is the square of the input.

The theory says that a non-linear system will result in frequency components at the output that were not present in the original applied signals (inter-modulation distortion). I wanted to find out what those components would be. To do this I made use of a trigonometric identity that says when you multiply two sine waves together (which is what you're doing when you square something - you multiply it by itself) you get the following:

sine A . sine B = (cos (A-B)/2) - (cos (A+B)) / 2

I did this calculation for the 1Hz and the 3Hz sine waves, and found that they resulted in a D.C. component and an A.C. component. The latter A.C. component was not present in the original signals, but had been created as a consequence of the non-linear act of squaring the system.

Finally I used the following algebraic equation:

(a+b)^ = a^ + 2ab + b^.

I already had answers for A^ and B^ in my previous calculations, so all I had to do to finish this off was find an answer for 2ab. I used the same trigonometric identity used previously to do this.

From all of that work I ended up with the frequency components listed in my original question. I think they are correct. I'm quite happy with the result, but I wanted clarification that the 0Hz component was infact a D.C. level.
However, I'm happy to be proven wrong if you can show me different?

Brian

Last edited:

#### MrAl

##### Well-Known Member
Hi everyone. Considering the addition of the following two discrete signals:

x1 = sin (2pi . 1 . nts) // 1 Hz sinusoid of peak amplitude 1
x2 = sin (2pi . 3 . nts) // 3 Hz sinusoid of peak amplitude 1

I have been investigating the theoretical output from a system that adheres to the following equation:

y = x_squared

when I apply the addition of x1+x2 to it.

I have proven that the system is non-linear because the theoretical output is a bunch of components that were not present in the original signals, as follows:

0Hz @ amplitude 1
2Hz @ amplitude 0.5
4Hz @ amplitude -1
6Hz @ amplitude -0.5

I calculated these with the help of trigonometric identities (and a text book )

I understand all of this, but... what about the 0Hz component? What does this look like in practice? Would it be seen as a D.C. level?

Brian

Hi again,

Ok, when you say x1+x2 that means you add the two signals.
What you seem to want to do is to multiply the two signals, which would
be equivalent to using a modulator with the two signals, which is also
sometimes called "mixing" the two signals.
Then, it appears that you want to apply that resulting signal to the
squaring circuit and then analyze the components output from that.

This would be equivalent to:
ya=(sin(w1*t)*sin(w3*t))^2

Now when you 'mix' two signals yes you get the sum and difference
frequencies 2 and 4, but when you square THAT you should also
get double frequencies 2, 4, 6 and 8, which would mean you should see
components of 2Hz, 4Hz, 6Hz, and 8Hz. If any of these are missing
from the output equation then there is still an error.
Another little hint is that the DC component is in fact present, but
it is not equal to 1.

Also, you can easily check (and you really need to do this) at least
one random point with the original equation ya above and your
resulting equation. Since you are saying that your resulting equation
is still the same (as in my previous post) you still do not have the
right answer, so if you want to prove this for sure you have to look
at your equations again. Sorry to inform you of this, but i assume
you want to know so that in the future you can get it more accurate.

BTW, your most recent post seems to be missing exponents or something as
a^ makes no sense, and i assume you meant to write a^2.

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