Band pass filter

[Dr. Dhiman Kakati]

While designing receiver a Band Pass Filter (BPF) is often used at the input end of the receiver, Which formula or mathematical expression is there for calculating the bandwidth of the filter. what are the drawback and advantages of increasing bandwidth.

Regards

[alistu]

Hi Dhiman,

The bandwidth of the filter is related to the bandwidth of the received signal that contains information, and the signal bandwidth is highly dependent on your system. In a general sense, use of the filter causes the noise in other frequencies to be filtered and therefore, improves system OSNR, at the price of the filter itself being added to the components used.

Regards

[Hamada Masri]

Hi Dhiman,
choosing bandwidth of BPF relays on the signal you need to receive it in the receiver section.

Regards,

[Dr. Dhiman Kakati]

Dear Hamada, Will you please elaborate modifying the above design files. Thanks

Regards

[Dr. Dhiman Kakati]

Oh Thanks. Suppoese I am attaching here a Design file that is based on Dual polarization Differential phase shift keying i want to know the bandwidth that will give the best result. this a part of the main system so unconnected part of other components you may ignore. Here direct detection technique is used. I need to improve the output constellation also. Thanks

Regards

[Dr. Dhiman Kakati]

That one is without Band pass filter and this one with Band pass filter. Please compare the performance. I need suggestion in this.
Constellation is better but Q-factor decreasing. Thanks.

Regards

[alistu]

i can’t see how the second constellation is better than the first one. I increased the bandwidth to 100GHz and the second implementation got better results than the first one in terms of Q-factor. So there seems to be a compromise between Q-factor and bandwidth for your system (If Q=40 is acceptable for your purpose, then you might use only 10GHz of bandwidth).

[Dr. Dhiman Kakati]

Dear Alistu,
Yes I can use the parameters that you have mentioned. But the question is without using the band pass filter the quality factor is far better than using the same. Is that mean we need to modify other parameter in the design, moreover the the is used for better performance.

I need to mention that we can go down to Q-factor of 6 in electrical domain (that) i found in one of the video lectures, in this case also we are using the electrical constellation visualizer so can we go up to that point i.e. minimum Q-factor of 6.

With regards

[alistu]

Imagine you want to use your scheme in every channel of a WDM system. Without using the filter, you have to choose a very large value as your channel spacing (more than 120GHz) or otherwise the signals from different channels overlap. But if you use a 10GHz filter, you can even have a channel spacing of 40GHz or less at the cost of getting a smaller Q-factor. So like I said, sometimes it is a compromise (you can’t have both the quality and bandwidth together).

[Dr. Dhiman Kakati]

I understand the use of Band Pass Filter in case of wavelength division Multiplexing we can extract the required frequency by using the Band pass filter having some suitable bandwidth. BPF also used for blocking unwanted frequency i.e. noise, here in the above design the filter is used for blocking the unwanted frequency component, but it is not doing so I think because there must be some specific value which we have not set.

Regards

[alistu]

It can still make the system performance better in your case I suppose. As I said, you can set the filter bandwidth value as 100GHz and the performance of the system gets slightly better in terms of the Q-factor. If you set the bandwidth to a higher value, you might even get better results.

[Dr. Dhiman Kakati]

Yes Alistu I have tried with 100 GHz and get a better result got the solution. the BPF is centered at the frequency of the CW laser but 100 GHz is a large bandwidth that is allowing more sideband to enter into the receiver. I am not getting the reason/logic behind this. it seem after transmission the signal bandwidth is increasing. Am I correct..?

Regards

[alistu]

The reason for this large bandwidth (although large is a relative word) being needed is the large bandwidth of your received signal. You can see it by an optical spectrum analyzer. The sideband entering the filter pass band contains parts of both signal and noise. So it can occur that by leaving it out of the filter pass band, the signal degrades as a result of losing some part of the signal.

[Dr. Dhiman Kakati]

Now I am Sure about. Thank you so much.

Regards

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