# What happens when you induce two currents of differing wavelength on the same line?

If I induce two currents of differing wavelengths on a single conductive body, say an iron pipe for example, how can I calculate what the resulting signals wavelength will be?

Is it as simple as taking the root mean square of the frequency of the signals? Any ideas for calculating the voltage of the final signal?
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Example:
Signal 1: 60hz at 100v
Signal 2: 75hz at 100v

RMS frequency sqrt(((60^2)+(75^2))/2) =  67.9153
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Thanks  a bunch!

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6 years ago
Alright, so you've got two sinusoidal voltage sources, and you've go them wired in series so that the voltage across both is the sum of their individual voltages.

Just to get this party started,  I am going to assume that:

y1(t) = A1*cos(2*pi*f1*t) = A1*cos(2*pi*t/T1)
y2(t) = A2*cos(2*pi*f2*t) = A1*cos(2*pi*t/T2)

where T1=1/f1 and T2=1/f2

The signal you're interested in is the sum: y(t)=y1(t)+y2(t)

There's no phase difference, but if you want you can go back over this later and put one in, if you want to make the math harder.  I'm guessing an arbitrary phase differce between y1(t) and y2(t) won't affect the results.

Your first question is about the frequency of y(t), or equivalently it's period. I am going to assume I can find some integer number m of periods of the first signal, equal to some other integer number n of periods of the second signal. This way there will be some "big period" T12 that applies to y(t).  If I wait wait T12 amount of time, then y(t) will back where it started, i.e:

m*T1 = n*T2 = T12   and y(t) = y(t + T12)

For the frequecies f1=60Hz and f2=75Hz, I think the smallest integers I can pick are m=4 and n=5, and the smallest period is T12=(1/15) s. f12=15 Hz

4*T1 = 4*(1/60) = (1/15) = 5*(1/75) = 5*T2

Anyway that's the answer to your first question. Your second question was about the RMS amplitude of y(t).  To figure that out you have to know how to calculate RMS amplitude, and the way you do that is:

(1) Find the square of the signal.
(2) Find the time-averaged value of the squared signal over one period, i.e. integrate the sqaured signal over one period and divide by T.
(3) Take the square root.

I'm not actually going to do the math for you, but I'll sumarize some important results. I'll use the <> angle bracket notation to represent the operation of "time-averaged over one period".  <>^(1/2) is the same as RMS().

For a signal that is just one sinusoid, the RMS amplitude is (1/2)^(1/2) =0.70711 times the peak amplitude.

<(y1(t))^2> = (1/2) * A1^2

RMS(y1(t)) = (1/2)^(1/2)*A1

For a signal that is the sum of two sinusoids, the amplitudes of the individual sinusoids sum in a Pythagorean way.

<(y1+y2)^2> = (1/2)*A1^2 + (1/2)*A2^2

RMS(y1+y2) = (1/2)^(1/2)*(A1^2+A2^2)^(1/2)
RMS(y1+y2) = (RMS(y1) + RMS(y2) )^(1/2)

For example if y1(t) has an RMS amplidue of 100 V, and y2(t) also has an RMS amplitude of 100 V, then the RMS amplitude of y(t)=y1(t)+y2(t) will be 141 V. That is:
(100^2 + 100^2)^(1/2) = 141.42...

If you want to do the math yourself, the "double-angle" and "product-to-sum" trigonometric identities are helpful for reducing those cosine-squareds into something easier to integrate.  Also you will find that a whole bunch of these integrals actually reduce to zero when integrating over one, or an integer number of periods. That helps too.

I didn't quite believe that result about RMS amplitudes adding in a Pythagorean  way either, but I checked it with Octave, using some crude numerical integration, and it seems to agree.  Also I made some nice graphs for you. Print em' out and put 'em on the fridge or whatever.
;-)

octave.exe:105> sumof2cosines
meany1 = -6.0822e-015
meany2 = -8.3939e-015
meany = -1.6674e-014
meany1squared = 10000.0
meany2squared = 10000.0
meanysquared = 2.0000e+004
RMSy1 = 100.000
RMSy2 = 100.000
RMSy =  141.42

Full res graphs:
https://www.instructables.com/file/F0EZYPEGJ2834YG/
https://www.instructables.com/file/FR0I51MGJ27ZGVO/
6 years ago
How do you know it's series? I asked "series or parallel" and got no answer.
6 years ago
Honestly I'm just guessing at what the questioner is describing. I'm guessing that the underlying operation is addition, and one way to add  two signals together is by stacking two ideal AC voltage sources on top of each other.

I admit this might not be the best way to implement adding two signals in the real physical world, like with real signal generators with a common physical ground.

By the way, putting two, ideal, on-paper, voltage sources in parallel leads to a contradiction, so to do the parallel connection I had to add some resistors. The answer is pretty much the same. I'm basically adding two signals together. The underlying operation is addition.

A picture-diagram of all this is attached.
https://www.instructables.com/file/F4DGYA4GJ2837J8/
6 years ago
I just wondered if I was missing it in the opening question.
6 years ago
Jack, you really should have posted this at the top level! Once it's an embedded reply, it can't be flagged as Best Answer.
6 years ago
Thanks for the compliments, K, and also the assurance that RMS amplitudes of different frequency components in a signal really do add that way.  I think the only reason I wrote such a longish write-up is because I really did not remember how it was that RMS amplitudes like that were supposed to work.

I remember back when I was little,  I mean younger than I am today, and they introduced us to Fourier transforms (I think it was Fourier transforms) there was this notion of an "orthogonal" set of functions.  I think "orthogonal" was the word they used, and the functions themselves were just sinusoids of different frequencies.  I think the word "orthogonal" was used because the sinusoids with different frequency were, in some sense, all at right angles to one another.  That is to say there was some operation like a dot product, and whenever you'd use it on two functions of different frequency you'd get zero, and the only time this funny-product was non-zero was when both sinusoids had the same frequency.

Anyway it makes sense that things that things that are all at right angles to each other should add in a Pythagorean way, the same way orthogonal vectors like x,y, and z do.

BTW, I think I did post that answer at top level, or at least that's what it looks like from my browser.
6 years ago
(I got fooled by looking at the nesting while I was composing my reply. You did post at top level).

Orthogonal functions are a very general concept, not just for Fourier transforms. In quantum mechanics, the wavefunctions which correspond to different "states" are generally orthogonal to one another. Just as you recall, you can define the equivalent of a "dot product" (more generally called a scalar product) for functions.

Usually it's just the product of the two functions integrated over their whole domain (e.g., for Fourier components, int_-infty^+infy sin(nx)sin(mx)dx). Functions are defined as orthogonal if their scalar product vanishes. For sinusoids, you should be able to see from the definition that the integral vanishes for any n != m, but not for n==m (where it's just int sin^2(x)).

Your intuition is correct. It's exactly because the sinusoids are orthogonal that the RMS's sum in quadrature. Recall that RMS^2 = when = 0. So the RMS of f1+f2 would be sqrt(<(f1+f2)^2>); and (f1+f2) = f1^2 + 2*f1.f2 + f2^2. With orthogonal functions, f1.f2 = 0, and the expression reduces to <(f1+f2)^2> = + (since averaging is linear, = + always).
6 years ago
Hey, uh I just noticed a typo in what I wrote above, an A1 that should have been an A2. Also I found the same thing in the Octave/MATLAB script I wrote, a A1 that should be A2. It's on line 17 of "sumof2cosines.m". Anyway, if you fix that then you can get that script to work for values of A1, A2 not equal to each other. For example, A1=30 and A2=40:

octave.exe:107> sumof2cosines
meany1 = -1.7764e-015
meany2 = -3.8417e-015
meany = -4.3485e-015
meany1squared = 900.00
meany2squared = 1600.0
meanysquared = 2500
RMSy1 = 30.000
RMSy2 = 40.000
RMSy = 50
6 years ago
Hi, Jack. Great writeup! The bit about RMS's summing is quadrature is definitely true (it's fundamental to Gaussian error propagation), and fairly easy to prove from the definition (the constant terms pull out of all the integrals and cancel).
Re-design6 years ago
Are you signals AC or DC?

Are they series or parallel?

By "wavelength" you mean "RMS voltage" I assume.  OR are you actually asking what is the "frequency" of a 60 hz signal mixed with a 75 hz signal.
paulcauchon (author)  Re-design6 years ago
I suppose I'm asking if one was to take a voltage reading downstream from both sources attached to the pipe, would the voltage peak out at 200v?

And would it do so at the crest of the beat's frequency, which I'm assuming should be 15hz?

So if you attached an oscilloscope to the pipe, would it show distinct signals, one for the 60hz signal RMS 100v, one for the 75hz signal RMS 100v, and one for the beat from the interference at 15hz RMS 200v?
6 years ago

Here's another nice site.  Look at the last "movie".
6 years ago

Go here and look at question #2.  It shows some patterns on the o-scope of mixing different frequencies.

The voltage peak is different if you connect them in series or parallel.
6 years ago
depends if its series or parallel -- if you had 2 amplifiers in series with 0db gain, then the signals would add, and the amplitudes would be sometimes 2x 100v, sometimes 0, sometimes halfway.

The interference pattern in this case would be goofy looking.
orksecurity6 years ago
You don't get a single frequency out. Both frequencies will be present, superimposed on each other, and you'll hear both as well as their beat frequencies (from constructive and destructive interference).
paulcauchon (author)  orksecurity6 years ago
Would the voltage then vary as the beats fall in and out of phase but remain RMS 100v or would it be additive and jump to 200v?
6 years ago