# What is signal modulation?

1. Oct 20, 2012

### Aziza

In our lab we are to measure speed of light by making a light signal travel through optical fiber. At one end there is the transmitter, which will output the light signal into the fiber, and it will also "modulate" the signal. According to my lab manual, "The reason that the signal is made to modulate is because that gives us a reference point for observing the time delay between signals because the delay will show up as a phase shift between the two modulations. It would be very hard to observe the time delay between two constant dc signals"

I have absolutely no clue what that entire sentence means at all.....

From what I understand of the rest of the manual, I know that this "modulated signal" will appear on an oscilloscope. Also at the other end of the fiber there will be connected a receiver, which will display the signal coming in from the fiber on the oscilloscope. So there should be two signals: a modulated signal and the actual light signal (i think..actually, does modulation somehow change the actual light signal? or does it just output its own signal?) but the sentence I quoted above speaks of two modulated signals, not one...but there shouldnt be three signals total.....I am very confused and cant find a simple explanation anywhere...

2. Oct 20, 2012

### Studiot

Why do you think your lab manual says this?

Have you ever measured the thickness of a thin oil or soap film or glass plate by interference fringes?

3. Oct 20, 2012

### Aziza

i have no clue why it says this, that is what I am trying to understand.

And ive done problems on thin films/interference but never actually did this particular experiment

4. Oct 20, 2012

### sophiecentaur

Modulation simply means varying some aspect of a continuous signal. Amplitude modulation would involve changing the brightness somehow (electronically or with a shutter etc.) perhaps just turning it into pulses.
Why do you need modulation? Well, you would need to know how long a particular portion (or pulse, perhaps) of the light beam was taking to reach your receiving equipment. You have to mark it in some way because a continuous beam of light would not indicate the time taken for the light to travel across the gap.

It looks as though the equipment you are using has a smooth (sinusoidal) modulation of the amplitude. If you look at the brightness variations with a light sensor (photo diode into an oscilloscope display, for instance) you will see a sine wave at the modulating frequency. If you have one sensor at the source and another at some fair distance away, the two sinewaves, viewed on a scope, will not coincide exactly. The time difference between the rising edges on the two displayed waveforms will correspond to the time taken for the light to travell across the gap. 1 nanosecond for every foot travelled, which is not a lot over the sort of distances in a lab, which is why you will be using a hundred metres or so of optical fibre.

5. Oct 20, 2012

### Aziza

Ohh ok i think i see...and it says that the light is modulating at 300kHz...this means that a modulating signal at 300kHz is added to my continuous light signal? This is the modulated signal?

6. Oct 20, 2012

### Aziza

Wait no it also says that "a voltage signal of the 300kHz modulation is used to modulate the light source"...???

7. Oct 21, 2012

### sophiecentaur

The 300kHz modulating signal has to be a "voltage signal" because that's how we control things these days. (electronically)
When you use the word "add" in your description of modulation, although it gets the idea across, in fact the process of modulation is really a multiplication - because the intensity of the light is proportional to the instantaneous value of the modulating signal. If you are 'into maths' you may see how the modulated light signal will have a value, at time t of
A(t) = A (1+ sin(2∏.ft))
where A is the mean amplitude and f is the frequency of the modulation. The amplitude will vary smoothly between 0 and 2A over the modulation cycle.

8. Oct 21, 2012

### Aziza

But what exactly is a 'voltage signal'? Is it just my modulating wave of frequency 300kHz? And why is the intensity of light proportional to instantaneous value of modulating signal? I thought the modulating signal is independent of my light wave? And could you please explain how you derived that formula?
So if my modulating signal is M(t)=Asin(2∏fmt) and my light signal is L(t)=Bsin(2∏flt), where I'm assuming the modulating signal's amplitude is different form the light signal amplitude (although I have no clue if this should be the case), then if I multiple the two waves I get ABsin(2∏fmt)sin(2∏flt)...so I'm guessing from here it is possible to get the amplitude but idk how..?

sorry for all these questions and thanks so much in advance!!

9. Oct 22, 2012

### sophiecentaur

It's a varying voltage. If you want to vary the amplitude (intensity) of a light beam you can 'wiggle a knob' manually, put a shutter in the way or use an electronic signal to do it with a light modulator.

Have you looked up "Amplitude Modulation" anywhere? If you look at the wavefrom resulting from your expression, you don't get normal amplitude modulation. You get what is sometimes referred to as Suppressed Carrier DSB modulation. That particular form of modulation (plot it out for yourself) involves a phase reversal of the carrier wave (the light, in this case) half way through the modulation cycle. How would you plan to achieve that to an ordinary beam of light? The reason I quoted the formula for AM is that it gives the correct output and it is easy to produce that form of AM.

I wonder just how much you have read around this subject. Wikki and all the other sites are full of information about this topic, with Maths and Pictures to back it all up. I think you need to move on from 'Question and Answer' to 'work it out for yourself' now.

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