# Homework Help: Finding fundamentals of a wave from a product of two sinusoids

1. Dec 13, 2011

### elsamp123

1. The problem statement, all variables and given/known data
Find the amplitude, frequency and period of a particle whose distance from the origin is described by
x=4cos(5t)sin(8t)

2. Relevant equations

3. The attempt at a solution
Would sin x - sin y = 2 cos [(x+y)/2] sin [(x-y)/2] be a worthwhile expression to try?

2. Dec 13, 2011

### Simon Bridge

Try it and see :)

Do you understand the motion that is described by the equation?

3. Dec 13, 2011

### elsamp123

I know that it should be a standing wave kind of motion right? If so, then how would I get the x and y to form the parts to it?

4. Dec 13, 2011

### Simon Bridge

Only in the sense that the particle is not moving laterally.
What you have is a particle wobbling away and then wobbling back.
The graph looks like a sine wave drawn by someone whose hand is shaking.
Plot the function, you'll see.

You know what (x+y)/2 and (x-y)/2 are equal to - now you have simultaneous equations.

5. Dec 13, 2011

### elsamp123

Wow! I see what you're saying! so then the (x+y)/2 part would be the avg. angular frequency? I hope I'm not wasting your time with all these silly questions..

i know how in a regular standing wave, the (x+y)/2 part is the phase freq and the (x-y)/2 is the envelope freq.. is that the same here too?

6. Dec 14, 2011

### Simon Bridge

With a standing wave on a string (say) you get an overall wavelength depending on the length of the string and a frequency that the string moves up and down. This example has only one point moving up and down.

It's more like the case with beats - look again at the original expression: the amplitude of a sine wave is itself a sine wave. With beats you normally have one with a much longer period than the other. This one looks like a wiggly sine wave.

You could also think of it as a carrier wave with a signal on it.

It is really useful to be able to plot these things.

7. Dec 14, 2011

### rude man

'Beats' are the result of the linear addition of two signals of frequencies close to each other. A piano tuner listens to beats between his tuning fork and the piano key. This is a linear process (please, no quibbles about the human ear being slightly non-linear! :grumpy: There are no new frequencies generated.

This particular case is one of modulation, as you say a carrier modulated by a second signal. In fact, it's what's called 'double-sideband, suppressed-carrier' modulation since the two original frequencies both disappear. New frequencies are generated at the sums and differences of the two original frequencies.

My hint would be: sin(x)cos(y) = 1/2sin(x+y) + 1/2sin(x-y)

8. Dec 14, 2011

### Simon Bridge

Yeah - it's not beats - but "like" that.
It's mathematically the same process, a linear combination of two sine waves, just the separation of the frequencies leads to differences phenomenologically. In physics classes we like to reinforce the unity of physical law by making connections between phenomena.

You do get different frequencies with beats though - the beat frequency and the pitch frequency are both different from that of either component.

for eg - 110Hz and 104Hz produces a 107Hz tone with an amplitude varying at 6Hz.
Acoustic beats are a lot of fun.

Still - using the product-to-sum formula is quicker :)

I'm kinda interested to know how OP is expected to answer the questions in the first post though. I mean, it does not have a well-defined amplitude or frequency. I'm guessing there is context missing.

Note: by the phasor method ;) you get one phasor spinning around on the end of the other one... since they are not in lock-step.

Last edited: Dec 14, 2011
9. Dec 14, 2011

### rude man

10. Dec 14, 2011

### Simon Bridge

The attached plots are linear-superpositions of sine waves of the form:

$$\psi = \sin{(2\pi f_1)}+\sin{(2\pi f_2)}$$

... f1=110Hz and f2=104hz ... shows beats.

... f1=110Hz and f2=5hz .... shows the signal modulation.

I know about super-het receivers, I used to build them. The whole point of heterodyning is to generate an intermediate (carrier) frequency, one which is more conveniently processed.

I learned electronics first by hanging out around the back-door of a radio-engineering shop as a kid. I have a deep respect for engineers as a result so I'm wondering if perhaps we are just taking past each other by using the language differently.

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11. Dec 15, 2011

### rude man

12. Dec 15, 2011

### Simon Bridge

Of course - though I'd write[1]:

$$\sin{2\pi f_1 t} + \sin{2\pi f_2 t} = 2\cos{\left ( 2\pi \frac{f_1 - f_2}{2} t\right )}\sin{\left ( 2\pi \frac{f_1 - f_2}{2} t\right )}$$... which agrees with the plot - you'll see that at t=0 the envelope amplitude is a maximum while the overall amplitude is zero.

The second plot is more like what OP has.

$$2\sin(x)\cos(y) = \sin(x+y) + \sin(x-y)$$... I can show you that this is the same as for the beat equation, put:

u=x+y
v=x-y
x=(u+v)/2
y=(u-v)/2

substitute back:
$$\sin(u)+\sin(v)=2\cos\big ( (u-v)/2 \big ) \sin\big ( (u+v)/2 \big )$$... that's what I mean by "mathematically the same thing".

The resulting phenomina are very different but the same general process (linear superposition of two waves) can give rise to each. The formula OP asked about would give the same decomposition.

This should not be controversial - all periodic functions can be obtained by linear superposition of sine waves.

But it looks like we are having a difference of nomenclature. For eg. "beat frequency" is jargon: it's defined as "twice the envelope frequency" - because that is the rate, in acoustics, that the volume of the tone increases and decreases. You can look it up - Halliday and Resnick (your edition) point this out, I think, within a page of where you cited. (Or do I have this back-to-front? <checks> nope. " the beat frequency equals the difference between the frequencies of the two tuning forks"[2]) But don't just go by them - it's in common use as in here, here and here.

We say "envelope frequency" because the term "carrier wave" is not really a good description of what is happening (also from H&R - same chapter). You are not getting any disagreement with me here. Thing is, the math does not care what names we give the forms or how we experience them.

Perhaps an illustration:
If you set up signal sources to feed into an oscilloscope, you can change continuously from beats to carrier forms just by changing the frequency of one source. This is a standard experiment for electrical engineers at the University of Auckland, I've personally demonstrated this thousands of times and seen it done by hundreds of students a year. Have you ever tried this? (H&R do a nice demo with tuning forks - varying the frequency with bits of plasticine, but you get to see the continuous change better with electronic forms.)

------------------------
[1] I left out the stupid t in last post - nice of you not to quibble.
[2] Halliday, Resnick and Walker Fundamentals of Physics 7th ed ch17

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13. Dec 15, 2011

### Simon Bridge

I suspect this is what you are talking about:

The example: $x=4\cos(5t)\sin(8t)$, if I took the "like beats" description literally I's think of the cosine factor as the "envelope" (beat frequency 10rad/s) and the sine factor as the "tone". But if I plot them I get:

... where the blue is the waveform, the green is the cosine term and the red is half the sine term.
Clearly the green line cannot be seen as an "envelope" of anything - it's nonsence. Yet the math is consistent.

If I separate the waveform into components, the situation is clearer.

... here the blue is still the waveform, the green line is the slow term and the red one is the fast term.
Notice the amplitudes are (sort of) half the waveform's?
I artificially halved the amplitude of the red wave in the first plot to make it clearer - but this time I left everything.

We'd tend to think of the green one as the carrier wave and the red one as the signal - though there is no special reason it cannot be the other way around. In fact the situation in the problem is ambiguous: we are not told this is for radio or an electric circuit of any kind. It could be a cork bobbing on water - in which case the roles of carrier and signal make no sense. It's just sometimes useful to have something to visualize.

Heterodyne mixing is usually done non-linearly in a radio circuit (I used to use MOS-FETs - valves and transistors are also non-linear) - but it can also be done linearly. It's just these components are so handy. The mixing can still be described by linear superposition - it's just not so simple as above. We usually think of the oscillator signal as heterodyning the carrier wave - though, physically, it modifies both waves at the same time (algebraically the order of the sum/product doesn't matter). It's just that we recover the unmodified signal without fuss so it's a handy mental picture. So this is where you may get the impression there is no frequency modification (of the signal) by the on-board oscillator.

Note: it seems it's possible to heterodyne light - that must be fun: quantum heterodyny! Technically this should be possible for anything in the situation where it displays wave-like properties.

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14. Dec 15, 2011

### rude man

15. Dec 15, 2011

### Simon Bridge

I'll concede that you can start with a mixed signal and decompose it, and you can start with two signals and mix them. These are different physical processes. But they are described by the same mathematical equations... that's all I'm saying.

The fact I can change one into the other by a linear process shows that mixing can be described by a linear superposition. The mixing and adding equations are the same kind of mathematical object.

I have produced several experimental situations as well as the math to demonstrate that these are equivalent.

You have provided a number of examples that have only supported my case.
You have provided an authoritative reference in a popular physics text whose authors agree with me.

So it remains with you to demonstrate your assertions.

That the mixing case is different mathematically from the adding case.
That there is no change in frequencies involved in one but there is in the other.

Of course we can certainly leave it there :)
(I still think we are in agreement - but are using the words slightly differently giving the appearance of a disagreement. In which case, we can agree to agree xD)

Well it's been exercise. At least we have demonstrated how to have a gentlemanly scientific debate without acrimony.

Since you have expressed a desire to let matter lie, I have answered your challenge as an afterward:

Afterward
A challenge!
In the spirit of illustrating how I am using the language above:

$f_{LO}=f_{RF}\pm f_{IF}$

Well that was easy.

But you want the messy details I'm guessing ... right. It's not good enough to just do this, I have to demonstrate that this comes from a linear superposition of the waveforms?
<rolls up sleeves>

In general the incoming AM signal will be something like:$$s_{rf}(t)=A[1+g(t)]\cos(2\pi f_{rf})$$ ... so g(t) is the data-stream we want to recover.
Using low-side injection - after mixing we get:
$$s_{if}=s_{rf} s_{lo} = s_{rf}\cos{(2\pi f_{lo})}$$... this is the multiplicative "mixing". So lets explicitly multiply it out.
$$s_{if}=A[1+g(t)]\cos(2\pi f_{rf})\cos{(2\pi f_{lo})}$$... I can describe this equation as the linear superposition of two waves:
$$s_{if}=A[1+g(t)]\big ( a_1 s_1(t)+a_2 s_2(t) \big ): \qquad a_1 = a_2 = 1/2; \; \; s_1 = \cos{2\pi (f_{rf}+f_{lo})}; \; \; s_2 = \cos{2\pi (f_{rf}-f_{lo})}$$

i.e. $s_{if}=\frac{1}{2}A[1+g(t)]\big ( \cos(2\pi f_0 t)+\cos(2\pi f_{if} t) \big )$

This is the IF output expression explained in terms of a linear superposition of two signals.

At this stage no electronics has been used to physically separate these waves - this is entirely a mathematical description of the actual multiplied waveform as a linear superposition. I can use this to predict that I need to design a band-pass filter to exclude f0 from the IF signal if I want a decent IF to send to demodulation.

iirc: the strategy is to tune flo so fif falls in the region passed by the BPF. The BPF has to be wide enough to pass the desired sideband of the IF signal but narrow enough to exclude the undesired sideband for the entire operating range of the reciever.[2]

This has worked on every circuit I have ever built.
Of course, things are never this ideal - real-life carrier waves are seldom a single frequency for eg. This would be an ideal mixer operating on an ideal signal.

The process by which the IF signal is arrived at in the mixer is non-linear - but the before and after picture is still described by the equations shown. You can see this in the mixer's signal response for eg. It's a bit like how conservation of momentum ignores the fine details of what happens during a collision - concentrating on the before and after pictures.

I have noticed that the mixing process is often referred in engineering books as "beating" the RF and LO signals. Accessible example - in the context of RADAR..

-----------------------------------
[1] http://www.usna.edu/EE/ee354/Handouts/Superhet_Handout.pdf
[2] this is usually considered college level electrical engineering or electro-physics - the cliff-notes version for the non-college student: A classical radio receiver tunes the entire circuit so it's resonant frequency is close to the desired incoming signal: the radio station frequency. The super-het receiver, instead, tunes the incoming signal to something close to the receiver's resonant frequency ... sort-of. The process has application in anything to do with radio - RADAR for eg and radio astronomy. Almost all modern radio communication uses this method, and it is not restricted to AM (that's just the easy case).

16. Dec 16, 2011

### rude man

1. Superposition process (linear):

sin(2πf1t) + sin(2πf2t)

Original frequencies f1 and f2 retained
No new frequencies generated
(so not a good way to generate an IF)

2. (One form of ) mixing (non-linear process, and the OP’s problem as stated):

sin(2πf1t)*sin(2πf2t) = (1/2)cos2π(f1-f2)t – (1/2)cos2π(f1+f2)t]

Original frequencies disappear
New frequencies |f1-f2| and (f1+f2) generated
Good way to generate an IF: f1 = RF, f2 = LO, |f1-f2| = IF

Q.E.D.
I don’t need any references. This is high school trig.

Cheers!