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What is DP-QPSK? - FO4SALE.COM

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Fiber Optics For Sale Co.

Fiber Optics For Sale Co.

11 жыл бұрын

www.fiberoptics...
www.fiberoptics...
DP-QPSK stands for Dual Polarization Quadrature Phase Shift Keying. It is a fiber optic digital modulation technique which uses two orthogonal polarizations of a laser beam, with QPSK digital modulation on each polarization.
QPSK can transmit 2 bits of data per symbol rate, DP-QPSK doubles that capacity, so DP-QPSK can transmit 4 bits of data per symbol rate.
So in order to transmit 100 Gbps data, we just need to transmit 25 Gig to 28 Gig symbols per second, since each symbol represents 4 bits of data. In this way, the electronics just have to work in the 25 to 28 GHz range, instead of 100 GHz directly. This is within current electronics' capability and is much more cost effective.
Before we explain DP-QPSK in more detail, we have to understand BPSK and QPSK first.
BPSK stands for Binary Phase Shift Keying. BPSK changes the phase of a sin wave to represent one bit of data. So it transmits only 1 bit of data per symbol rate, either 1 or 0.
This picture shows how BPSK works. This is the digital data stream, +1 volt represents binary bit 1, and -1 volt represents binary bit 0. This is the carrier sin wave, it is a continuous sin wave with no phase jump at all.
However, when the digital data stream and the carrier sin wave is timed together, the carrier sin wave's phase is modulated based on current data.
So if the current data is 1, the sin wave has no phase shift, and the phase is 0 degree, the wave is sin(ωt).
If the data is 0, represented by a -1 volt, the sin wave's phase is shifted by 180 degree, and now it becomes --sin(ωt), since it is flipped upside down comparing to sin(ωt).
This can be concluded from the equations here. So sin(ωt) is basically a sin wave with amplitude A equals 1, and phase φ equals 0°, and --sin(ωt) is a sin wave with amplitude A equals 1, and phase φ equals 180°.
BPSK can be expressed in a constellation diagram. Here the horizontal axis is called In-Phase, and the vertical axis is called Quadrature Phase. The distance from the origin is the amplitude A, and the angle from horizontal axis is the phase. So here data 1 is represented by an amplitude of 1, and a phase of 0°. Data 0 is represented by an amplitude of 1, and a phase of 180°.
In the last slide, when we showed the BPSK constellation diagram, we talked about in-phase and quadrature phase. So what does that mean?
In-phase means no phase shift, so as shown in this graph, the sin wave starts from a initial phase of 0°, so we say the sin wave is in-phase.
Quadrature phase means a 90° phase shift. So when we move the sin wave by 90°, we get the cosine wave, and we call the cosine wave is quadrature phase.
But you can also look at it the opposite way, the cosine wave is in-phase at a initial phase of 0°, and you can get the sin wave by moving the cosine wave by 90°. The point is that sin wave and cosine wave are in quadrature state.
Why is this important? Since we will use this concept and both sin wave and cosine wave in DPSK -- Quadrature Phase Shift Keying.
Quadrature Phase Shift Keying uses the quadrature concept, since it uses both sin wave and cosine wave to represent digital data.
QPSK is basically two BPSK used in parallel, since each BPSK transmits 1 bit of data per symbol rate, so QPSK transmits 2 bits of data per symbol rate.
Let's say that you have the data 10, the serial to parallel converter splits this data into two paths, odd numbered digit, which is 1 in this case, is split into the In-phase path, and even numbered digit, which is 0 in this case, is split into the Quadrature phase path.
The local oscillator generates a cosωt wave. The cosωt wave is directly timed with the data in the In-phase path, and this produces the I signal. In this case, digital data 1 produces cosωt wave.
The cosωt wave generated by the local oscillator is phase shifted by 90°, thus produces a sinωt wave. The sinωt wave is then timed with the digital data in the Quadrature Phase path, and this produces the Q signal. In this case, digital data 0 produces --sinωt wave.
Finally, the I signal and the Q signal is linearly added together to produce the complete QPSK signal, in this case, this is a cosωt-sinωt wave.
In the In-Phase path, cosωt represents data 1, and --cosωt represents data 0.
In the Quadrature Phase path, sinωt represents data 1, and --sinωt represents data 0.

Пікірлер: 20
@MonirMorshed
@MonirMorshed 11 жыл бұрын
Dear Mr. Colin: Your explanation is so wonderful and concise that I really really appreciate your deep understanding and generosity for sharing to us. Good be on you.
@zamedo1
@zamedo1 4 жыл бұрын
Thank you very much for this - this was very well done. Excellent job.
@superlynx98
@superlynx98 4 жыл бұрын
THANK YOU! Finally an explanation that makes sense
@freud82
@freud82 10 жыл бұрын
Great video, very well explained and well structured. Thanks !
@az0304
@az0304 4 жыл бұрын
Thank you. Useful video
@newbie3566
@newbie3566 11 жыл бұрын
man you are good i like most of your videos .. very useful
@TheStarttman
@TheStarttman 11 жыл бұрын
Great Summary. Thanks!
@asifpatan
@asifpatan 7 жыл бұрын
Thank you very much!
@ashishdas8824
@ashishdas8824 8 жыл бұрын
Very Nice Video. Can You please launch a Video on Adaptive Equalization and talk about Taps, Stages and Iterations in details ? Thank You Ashish
@chilldog1234567890
@chilldog1234567890 10 жыл бұрын
5:49,that makes perfect sense.
@sangeethamalar18
@sangeethamalar18 10 жыл бұрын
is it possible that mixled line rates (10/40/100Gbps) in wdm network with advanced modulation using optisystem
@nickthanthrige2978
@nickthanthrige2978 8 жыл бұрын
Thanks man!
@GOWTHAMKRISHGowthamShanmugaraj
@GOWTHAMKRISHGowthamShanmugaraj 8 жыл бұрын
nice lecture
@RandomDavidInDaWorld
@RandomDavidInDaWorld 10 жыл бұрын
This is a great video! thanks!
@hobomnky
@hobomnky 4 жыл бұрын
soooooo helpful
@ErickMrema19
@ErickMrema19 10 жыл бұрын
Thanks , Good video ... i like it
@chilldog1234567890
@chilldog1234567890 10 жыл бұрын
I think your slides at 7:08 is wrong: -cos(t) + sin(t) = sqrt(2) * sin(t + 7/4 * pi) cos(t) - sin(t) = sqrt(2) * sin(t + 3/4 * pi) Could you double check on this? Great video nonetheless
@sunnyboy866
@sunnyboy866 5 жыл бұрын
@Aquarius: You are right. Great video nonetheless
@sharathkumar4774
@sharathkumar4774 5 жыл бұрын
But according to conrellation diagram, where x axis is cost , y axis is sint, addition of -cos and sin should result is 2 quadrant, am i wrong?
@formanite373
@formanite373 5 жыл бұрын
Salute u man
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