Sine wave with variable frequency

1. Feb 15, 2010

kvtb

I'm trying to do something very simple...
I'd like to have a sine function, where the frequency is controlled by a separate frequency function

Something like this:

g(t) = sin(2*pi*f(t)*t)

Assume that
f(t) = 20*exp(-2*t)+4

I would expect a sine wave that starts at 24 Hz and then slowly slows down to 4 Hz.
But when I plot it, I get some strange results for t in [0.5, 2], see attached image. The frequency for those values is very low. How's that possible? Is it my plotting program that acts funny? How can I get the expected (described) behaviour?

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2. Feb 15, 2010

flatmaster

3. Feb 15, 2010

es1

I haven't done this myself yet but I suspect when you expand from exp to trig form you'll find you have a beat.

20*exp(-2*t) is about 4 around t=0.8.

http://en.wikipedia.org/wiki/Beat_(acoustics [Broken])

Last edited by a moderator: May 4, 2017
4. Feb 15, 2010

Redbelly98

Staff Emeritus
This is weird. Do you still get this effect if you zoom the plot to 0.5<x<2?

It looks like the software (mathcad?) is correctly interpreting the sine argument as radians, as evidenced by the 4 Hz oscillation for x>2.5.

This does not look like beating, the amplitude stays constant.

Last edited by a moderator: May 4, 2017
5. Feb 15, 2010

Redbelly98

Staff Emeritus
Update: I plotted this function in Excel, and get the same plot! I'm looking more closely at this, stay tuned.

6. Feb 15, 2010

flatmaster

Perhaps using the same x in the f(x) expression and Sin expression is confusing. I would try to explicitly plug in your exponential f(x) into one long expression as a function of x only. Don't forget brackets around your f(x) expression so it distributes correctly.

7. Feb 15, 2010

Redbelly98

Staff Emeritus
Okay, I have figured it out.

Generally, we have
y = sin[θ(t)]​
I.e., the argument of the sine can be thought of as a time-dependent phase angle.

The angular frequency is
ω = dθ/dt​
Differentiate the sine argument in the OP, and one finds
ω 2π f(t)​

8. Feb 15, 2010

es1

I am not sure I follow RedBelly98. I think you need an explanation for why the frequency is lower than 4 in the range 0.5 to 1.5. I think the only explanation is there must be positive interference between the two frequencies, which would be a beat. But I could be wrong.

I think you can expand:

Sin(2pi(20Exp(-2t)+4)t) = 2Cos(pi(20Exp(-2t)t+pi4t)Sin(pi(20Exp(-2t)t+pi4t) = 2Cos( (f1-f2)/2 t)Sin( (f1+f2)/2 t)

Then solve for f1 and f2 (assuming there is a solution which is why I could be wrong).

I suspect it's right because 20Exp(-2t) is about 4 near 1 so one would expect the beat frequency to be very low around this number which is also what we observe in the plot.

It's not in the range -2,2 like in the wikipedia article because the original function was wrapped in a sine which scaled it to the range -1,1.

Last edited: Feb 15, 2010
9. Feb 15, 2010

es1

Also, it seems the amplitude is not necessarily constant.
Try altering the function slightly,
Sin(2pi(40Exp(-2t)+4)t)
this amplitude is definitely not constant.
Darn, I might have to find some time to try to solve this one... :)

10. Feb 16, 2010

elibj123

What RedBelly was saying is that the extension from constant to varying frequency is not done by multiplication but differentiation.

Just like the differential slope of a curve A(t)*t is not A(t) but [A(t)*t]' (I wrote the function in this form to deliver the point)

The "differential frequency" of a sine wave, is the derivative of its argument rather than the coefficient of t.

Try plotting the function sin(-10*exp(-2t)+4t) and check if it gets you the desired result.

11. Feb 16, 2010

Redbelly98

Staff Emeritus
Okay. Suppose we have a spinning object, and we are given it's orientation θ as a function of time, not necessarily a linear function. How would you determine the rotation rate, ω?
You differentiate the sine argument in the OP,
2π*[20*exp(-2*t)+4]*t​
and find that the result is less than 2π*4 rad/s, or 4 Hz, over some range.

12. Feb 16, 2010

kvtb

Very interesting discussion.

When I plot sin(2*pi*(-10*exp(-2t)+4t)), I get exactly what I want/expected.

But... elibj123, how did you get to this: -10*exp(-2t)+4t ?
When I calculate [f(x)*x]' I get something else, namely:
$$-40\,x\,{e}^{-2\,x}+20\,{e}^{-2\,x}+4$$

edit: aaah, it is the antiderivative of f(x)! :-)

13. Feb 16, 2010

es1

Ok. I get the phase and phase velocity argument. That makes a lot of sense.

When I take the derivative I find that the min frequency is at t=1 and the phase velocity is ~1.3 which matches the plot very nicely.

Thanks.