Q-Factor Question

[alex col]

Hello All!!

I am running a very simple model consisting of a transmitter, a receiver and a power combiner.

I am using the combiner to inject a signal generated from a CW and a filter at the receiver side.

My question is the following:

Regardless of the the CW signal being filtered or not, the Q-factor is the same. I would expect to have a higher Q when the signal was filtered (case 2).

Am I missing something?

Thank you!!

[alex col]

p.s apparently I am at the wrong thread – sorry cannot edit it, hopefully a mod will fix it 🙂

[Damian Marek]

Could you include a snapshot of the design layout? There must be some component selecting or filtering out the extra signal.

[alex col]

Thank you for the reply Damian!

Yes, as you can see there is a passband filter before the receiver. Changing the transmitting wavelength of the light source at the bottom left corner I run two different simulations. First having the “interfering” lightsource within the pass band (thus not filtered) and then moving the lightsource out of the pass band (thus filtering it out).

My question is why is the Q factor in both cases the same?

[alex col]

and the attached…

 

[alex col]

and the attached … 😀

[alex col]

Ok for some reason I cannot attach I just hope you nevertheless get it 😀

Cheers!

[Damian Marek]

Unfortunately I cannot see the attachment! What is the name of the receiver component you are using. Is it PIN diode or something else? Some components I believe include their own filtering in their component properties

[alex col]

Hello Marek, you can find the layout picture above.. so any ideas…?

No matter if the signal for the Spectral Light Source 2 is filtered or not (refer to the snapshots attached in a previous mail from Spectrum Analyzer 4) the Q-factor remains the same.

This is not what I would expect though…

[Damian Marek]

This issue is subtle and will be difficult to understand. You will notice that in the Optical Receiver component there is a Downsampling tab. You will also notice that in your attached pics of the spectrum the two optical signals (red curves) do not connect and are separate signals. (Two red curves, not one continuous one).

Now the problem, the software only has a finite number of sampled time values to represent the signal and therefor there is a maximum bandwidth that it can represent as one signal. This is the number of samples value in the Layout parameters. In OptiSystem, since optical channels are usually widely spaced in the frequency domain there is the ability to have multiple channels like the two red curves in your figure. In electrical signals there is no such concept and the entire signal must be represented as one channel. When the receiver component is calculating the electrical signal from two optical channels it must make a decision which frequency to center it’s available bandwidth and this is set to the maximum power by default. You can change this if you desire.

The solution of this problem is to increase the Samples per Bit in the Layout parameter so that the signal can be represented by one continuous curve or channel. I tried a Samples per Bit of 16384 and I changed the Downsampling rate to be 1*Sample rate in the receiver. The eye diagram is very noisy from thermal effects and the additional optical signal but I can see it appear in the RF Spectrum now.

[Damian Marek]

Here is snapshot

[alex col]

Thank you for the reply Damian!!

I followed your instructions so now I do get “one signal”. nevertheless I still get the same Q-factor regardless if I filter out or not the spectral light source witch is quite confusing… Any other idea?

[alex col]

Any other ideas guys?:)

When filtering out the “interfering signal” from the spectral light source, it makes sense that the Q-factor increases, right?

I wonder if this can be somehow simulated…

Thank you 😀

[alex col]

Hi there 😀 i am sorry for re opening this discussion but i was just wondering if there are any ideas… why do i see the same Q-factor for cases 1 and 2 (having an “interfering signal” & without the “interfering signal”)? Is there a way to simulate the effect and see the difference?

Thanks 😀

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