Exploring Beat Signals in Modulation: A Helping Hand for Niles

In summary, the conversation is about using a lock-in detector to detect absorption in a signal. The method involves modulating the phase of an EM-wave with a small amplitude of frequency Ω, which creates two sidebands at frequencies ω+Ω and ω-Ω with equal amplitudes but opposite phases. This results in a zero amplitude for the low frequency component cos(Ωt) when there is no absorption present. The discussion also touches on the concept of two sinusoids of different frequency being of opposite phase.
  • #1
Niles
1,866
0
Hi

On page 11 of this book, http://books.google.dk/books?id=Wv2OIC_SaDUC&printsec=frontcover&hl=da#v=onepage&q&f=false, it is stated that:

"The phase modulation has an additional advantage: The first two sidebands at frequencies ω+Ω and ω-Ω have equal amplitudes, but opposite phases. A lock-in detector tuned to the modulation frequency Ω therefore receives the superposition of the two beat signals between the carrier and the two sidebands, which cancel to zero if no absorption is present".

I have been trying to show that the signal is 0 when there is no absorption. What I have done is to plot the two functions
[tex]
\cos((\omega - \Omega)t) + \cos(\omega t) \\
-\cos((\omega + \Omega)t) + \cos(\omega t)
[/tex]
for some ω and Ω, but it is clear that these two functions when added do not cancel out. Have I misunderstood something here?

Thanks for any help in advance.Niles.
 
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  • #2
A lock-in will only detect terms oscillating at Ω so all other terms will drop out.
Start by expanding cos (ω±Ω)t
 
  • #3
Hi

Thanks for replying. OK, I get

cos (ω±Ω)t = cos(ωt)cos(Ωt) - sin(ωt)sin(±Ωt).

If I can only detect terms oscillating at Ω, then I get
[tex]
[\cos((\omega - \Omega)t)] - [\cos((\omega + \Omega)t) ]
[/tex]
which I by the above write as
[tex]
[\cos(\omega t)\cos(\Omega t)-\sin(\omega t)\sin(\Omega t)] - [\cos(\omega t)\cos(\Omega t)+\sin(\omega t)\sin(\Omega t)]
[/tex]
Shouldn't this yield 0?Niles.
 
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  • #4
If I lock-in on Ω, then I can do it with either cos(Ωt) or sin(Ωt). If I do it with the former, then I get 0 (after the low-pass filter), but if I use sin(Ωt), then it doesn't give 0. Is there something I have misunderstood here?Niles.
 
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  • #5
Try cos(ωt) . cos(ω+Ω)t
and keep the low frequency component.

And cos(ωt) . cos(ω-Ω)t
and keep the low frequency component.
 
  • #6
Hi

Thanks. I guess we are interested in the power, since that is what we measure. So I get

cos(ωt) . cos(ω+Ω)t = ½[cos(Ωt) + cos(2ω+Ω)] and
cos(ωt) . cos(ω-Ω)t = ½[cos(Ωt) + cos(2ω-Ω)].

Adding them and keeping terms at Ω gives cos(Ωt), so I get a nonzero DC-signal. Is the author wrong then?Niles.
 
  • #7
Niles said:
cos(ωt) . cos(ω+Ω)t = ½[cos(Ωt) + cos(2ω+Ω)] and
cos(ωt) . cos(ω-Ω)t = ½[cos(Ωt) + cos(2ω-Ω)].

Adding them and keeping terms at Ω gives cos(Ωt), so I get a nonzero DC-signal.
Not when you do it right. :smile:
 
  • #8
But cosine is an even function, so I thought that I could disregard the minus? But adding the signals with the minus still gives a nonzero DC-signal.Niles.
 
  • #9
EDIT
[strike]cos (-A) = -cos(A)[/strike] https://www.physicsforums.com/images/icons/icon4.gif [Broken]

Low frequency, not DC. :wink:
 
Last edited by a moderator:
  • #11
Niles said:
As far as I remember, then

cos(-A) = cos(A)
sin(-A) = -sin(A)
Oops! I was picturing 180-Ω, not -Ω! :redface:

So, I got the result I was looking for, but method was wrong. I'll take another look at it.
 
  • #12
Thanks. Can a possible explanation be that the constant DC-signal can become 0 by an offset? Then from this any change in any of the amplitudes will result in a nonzero signal.
 
  • #13
I intended looking at the technical publication you cited, but there's too much text to wade through searching for what I want. But I think you have provided the answer here: "The first two sidebands at frequencies ω+Ω and ω-Ω have equal amplitudes, but opposite phases."

This is what I've been looking for. We agree that if they are in phase the two products add, but I see in the case cited you indicate one starts off with opposite phase. I don't know how that arises, but you can investigate. So the situation comes to:

cos(ωt) . cos(ω+Ω)t = ½[cos(–Ωt) + cos(2ω+Ω)t] and
cos(ωt) .cos(ω-Ω)t = ½[cos(Ωt) + cos(2ω-Ω)t]

And the result is a low frequency component cos(Ωt) of zero amplitude, providing the sidebands are not attenuated unequally.
 
  • #14
I'm a little uneasy with the concept of two sinusoids of different frequency being of opposite phase--wouldn't they be in-phase for exactly as much time as they'd be opposite-in-phase? But the maths looks good, and that's the main thing. :tongue:
 
  • #15
NascentOxygen said:
I intended looking at the technical publication you cited, but there's too much text to wade through searching for what I want. But I think you have provided the answer here: "The first two sidebands at frequencies ω+Ω and ω-Ω have equal amplitudes, but opposite phases."

This is what I've been looking for. We agree that if they are in phase the two products add, but I see in the case cited you indicate one starts off with opposite phase. I don't know how that arises, but you can investigate. So the situation comes to:

cos(ωt) . cos(ω+Ω)t = ½[cos(–Ωt) + cos(2ω+Ω)t] and
cos(ωt) .cos(ω-Ω)t = ½[cos(Ωt) + cos(2ω-Ω)t]

And the result is a low frequency component cos(Ωt) of zero amplitude, providing the sidebands are not attenuated unequally.
Ah, now I see it. OK, so the author was right after all, I have to give him that.

NascentOxygen said:
I'm a little uneasy with the concept of two sinusoids of different frequency being of opposite phase--wouldn't they be in-phase for exactly as much time as they'd be opposite-in-phase? But the maths looks good, and that's the main thing. :tongue:
One can show rigorously that when modulating the phase of an EM-wave of frequency ω with a small modulation amplitude of frequency Ω, two sidebands occur out-of-phase with frequency ω+Ω and ω-Ω, respectively. It is mentioned here (in the beginning), e.g.: http://en.wikipedia.org/wiki/Pound–Drever–Hall_technique#PDH_readout_function

It most likely says so in the book too, but I haven't found it yet. I haven't read that many pages in it, because I was caught up with this. If I find it, I'll let you know on which page.

Thanks for helping out with this.

Best wishes,
Niles.
 
  • #16
Yes, I found it now, it's page 11 as you said. But I'm still trying to picture a graph of two sinewaves of different frequency being 180° out of phase. Won't they be constantly drifting through 0→360° relative phase? While they may start off in-phase, after a while they'll be precisely 180° out of phase.
 
  • #17
NascentOxygen said:
Yes, I found it now, it's page 11 as you said. But I'm still trying to picture a graph of two sinewaves of different frequency being 180° out of phase. Won't they be constantly drifting through 0→360° relative phase? While they may start off in-phase, after a while they'll be precisely 180° out of phase.

You are asking a very good question, because I thought about that very same issue. Intuitively I (still) don't really see why it cancels out, since the sidebands are not at the same frequency. This was the reason why I wanted to calculate it explicitly.

I'm logging of now, but I'll check back tomorrow.

Regards,
Niles.
 
  • #18
Certainly the sidebands themselves aren't of the same frequency, so additively they are never going to cancel. But when each is multiplied (aka mixed) with the carrier frequency, among the products of that process are two low frequency components each of frequency Ω. In this phase modulation, one of those low-frequency components is in anti-phase to the other.
 
  • #19
NascentOxygen said:
I'm a little uneasy with the concept of two sinusoids of different frequency being of opposite phase--wouldn't they be in-phase for exactly as much time as they'd be opposite-in-phase? But the maths looks good, and that's the main thing. :tongue:

It is because the sidebands are the results of a carrier being passed through a phase modulator where is is mixed with the IF, i.e. they are the products of the same process which means that they maintain a phase relationship when the signal is downmixed again in the lock-in.

I use a variation of this technique in my experiment (although not for spectroscopy), and I can assure you that the math not only looks good but also works in practice:approve:
 
  • #20
NascentOxygen said:
In this phase modulation, one of those low-frequency components is in anti-phase to the other.
I'm coming to the conclusion that there is a bit of poetic license here. When the author says that the upper sideband is out of phase with the lower sideband, I think he probably being conservative with words, and what he really means is that the modulation carried on that upper sideband, when demodulated returns a demodulated signal which is in antiphase with the demodulated signal recovered from the lower sideband.

In which case, any mathematics is moot.
K650F.png


If my hunch is right, this probably arises so often in communications engineering that it is a well-understood shorthand. Let's see if anyone can support me on this. :tongue2:
 

1. What is modulation and how does it relate to beat signals?

Modulation refers to the process of varying a signal in order to transmit information. It is commonly used in communication systems and can involve changing the amplitude, frequency, or phase of a signal. Beat signals are created when two signals with slightly different frequencies are combined, resulting in a periodic variation in amplitude. Modulation can be used to create beat signals, which can then be used to transmit information.

2. How can exploring beat signals in modulation be helpful for Niles?

Exploring beat signals in modulation can be helpful for Niles in a variety of ways. It can lead to a better understanding of communication systems and how information can be transmitted. It can also help Niles in designing and improving communication technologies, such as radios and cell phones, by optimizing the modulation techniques used. Additionally, studying beat signals can have practical applications in fields such as music and acoustics.

3. What are some real-world applications of beat signals in modulation?

Beat signals in modulation have numerous real-world applications. They are used in communication systems, such as AM and FM radio, to transmit audio signals. They are also used in radar systems to detect the presence and distance of objects. In addition, beat signals are used in music to tune instruments and create harmonies. Other applications include medical imaging, sonar, and seismology.

4. What are the mathematical principles behind beat signals in modulation?

The mathematical principles behind beat signals in modulation involve the concept of interference. When two signals with slightly different frequencies are combined, they interfere with each other, resulting in a beat signal with a frequency equal to the difference between the two original frequencies. This can be represented mathematically using trigonometric functions, such as sine and cosine, and can be explained using the principles of Fourier analysis.

5. How can understanding beat signals in modulation contribute to advancements in science and technology?

Understanding beat signals in modulation can contribute to advancements in science and technology in various ways. It can lead to the development of new and improved communication systems, such as more efficient and reliable wireless networks. It can also be applied in other fields, such as music and acoustics, to improve techniques and technologies. Additionally, studying beat signals can lead to a deeper understanding of the principles of wave behavior and interference, which have applications in many areas of science and engineering.

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