How Many Photons Are in Each Bit of a 10 Gb/s Optical Signal?

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SUMMARY

The discussion focuses on calculating the number of photons in each bit of a 10 Gb/s optical signal transmitted at a wavelength of 1550 nm with an average optical power of 10 mW. The energy per photon is calculated as 1.28 x 10-19 Joules, while the energy per bit is determined to be 10-11 Joules. The participants clarify that the number of photons per bit can be computed using the formula: number of photons = energy per bit / energy per photon, leading to a result of 78 million photons per bit. The conversation emphasizes the importance of understanding the average power and bitrate relationship in optical communication systems.

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Homework Statement


An optical communication system operating at λ = 1550nm is transmitting pulses at 10 Gb/s. The magnitude of the optical pulses is the same. Calculate the number of photons received within each bit. Assume that the received average optical power is 10 mW.

Homework Equations

The Attempt at a Solution


I calculated the energy per photon Ep = hc/λ ---> 1.28x10-19
then i calculated the energy per bit Eb = Pτ ----> (.01)(10-9) = 10-11

In class we did a similar example where the number of photons = Eb/Ep

I do that here and get 78 million... that seems way too big to me. Can someone explain what I did wrong if its wrong.
 
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You are given average optical power ... and a bitrate.
You are asked to find the number of photons in just one bit.

Energy per photon = good start.I'd have put 1.28e-16 mJ/photon.
But it may be easier to work out the photons per milliJoule.

Now you need the amount of energy per bit.
You have 10mW = 10mJ/s in 10Gb/s pulse rate ... so how many mJ per Gb? How many mJ per bit?
 
Further to Simon's comments, it occurs to me that a bit can be a 0 or a 1. Perhaps one of these corresponds to non-transmission of any photons for 0.1 ns.
 
haruspex said:
Further to Simon's comments, it occurs to me that a bit can be a 0 or a 1. Perhaps one of these corresponds to non-transmission of any photons for 0.1 ns.
I do not think we need to get into the details of the unspecified coding scheme. We are given an "average optical power" which I would interpret as the average power when transmitting an arbitrary bit pattern.

In the real world, we might be using phase encoding, group encoding, scrambling polynomials or other techniques. The value of any particular bit will have little or no correlation with the number of photons received. [I'm not familiar with the encoding techniques for, e.g. 10 gig SONET beyond what google can provide with five minutes of surfing]
 
Simon Bridge said:
You are given average optical power ... and a bitrate.
You are asked to find the number of photons in just one bit.

Energy per photon = good start.I'd have put 1.28e-16 mJ/photon.
But it may be easier to work out the photons per milliJoule.

Now you need the amount of energy per bit.
You have 10mW = 10mJ/s in 10Gb/s pulse rate ... so how many mJ per Gb? How many mJ per bit?
Ok this is where I'm messing up. He did this in class but now that I'm lookin back at my notes I'm having trouble following it.
 
haruspex said:
Further to Simon's comments, it occurs to me that a bit can be a 0 or a 1. Perhaps one of these corresponds to non-transmission of any photons for 0.1 ns.
He mentioned this in his example. That's where the 10-9 in my energy per bit came from. But I guess I'm applying it incorrectly.
 
nmsurobert said:
He mentioned this in his example. That's where the 10-9 in my energy per bit came from. But I guess I'm applying it incorrectly.
It's 10Gbps, not 1Gbps.
 
nmsurobert said:
Ok this is where I'm messing up. He did this in class but now that I'm lookin back at my notes I'm having trouble following it.
Don't try to copy what's in your notes, just use the maths you already know.
You have 10mW = 10mJ/s in 10Gb/s pulse rate ... so how many mJ per Gb? How many mJ per bit?
If you have 10 (bagged) carrots per box and 10 bags per box, how many carrots are in each bag?
If you have 10 x per y and 10 z per y, how many x per z?
 
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