- This topic has 20 replies, 2 voices, and was last updated 9 years ago by alistu.
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September 29, 2015 at 9:15 pm #25488maha slitiParticipant
Hello,
Please have you an idea about how to filter optical signals based on their power, by using SOA for example? -
September 30, 2015 at 3:56 am #25499alistuParticipant
Hi Maha,
I personally don’t know of any component who would do this, but you might be able to make one using a fork, an optical hard limiter and an optical multiplier. Signal is connected to the fork first. The upper branch of the fork enters a hard limiter with the threshold you wish to define for your filter, upper power level of 1 W and lower power level of a very small number. Then the output signal enters the multiplier, together with the main signal from the lower branch of the fork. And the output is the filtered signal based on the power. If there was no component to do this, you could give this a try.
Regards
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September 30, 2015 at 12:24 pm #25521maha slitiParticipant
Thank you very match, I will try this solution.
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October 9, 2015 at 8:35 pm #26118maha slitiParticipant
Hi alistu,
Can I have your email please? I will send you a simulation to check it with me if it is possible. -
October 12, 2015 at 5:20 am #26245maha slitiParticipant
Hi alistu,
I have attached the simulation here, I send you a message to explain the problem that I have found.
Thank you very match for your help.-
October 12, 2015 at 9:07 am #26258alistuParticipant
Hi Maha,
The upper hard limiter and the lower hard limiter do not have the same setting. Their threshold values are not the same. So the output of the two branches is not the same and naturally, a zero signal is not yielded at the output of the substractor. Please set the same values for them and you’ll have zero at the output.
Regards
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October 12, 2015 at 9:41 am #26261maha slitiParticipant
Hi alistu,
Exactly, I need that the upper hard limiter and the lower hard limiter verify that an input signal power is in the interval [p1,p2], so the threshold of the first hard limiter is p1, and the threshold of the second is p2.
The objective is to have an output power only when the input signal is located in the concerned power interval. and, I want to obtain 0w and -100dbm in the other intervals.-
October 12, 2015 at 11:28 am #26263alistuParticipant
I did not exactly get one thing. I know you want parts of the the signal with less power than p1 to be omitted (or set to zero). What about the signal parts with optical power above the p2 threshold? Do you want to omit them as well or do you want to set them to the p2 threshold you have defined?
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October 12, 2015 at 1:12 pm #26272maha slitiParticipant
The signal parts with optical power above the p2 threshold, I want to omit them as well. So, If the signal have not a power in the interval of [p1,p2], I want to omit it.
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October 12, 2015 at 2:33 pm #26278alistuParticipant
Thank you for your answer. I think you can implement it using two blocks in series (and not in parallel like the one formerly attached). The first block simply omits the signal parts with power level below p1, whereas the second block omits the parts above p2. The second block consists of two parallel parts, with the upper part being merely the input signal to the second block and the lower part omitting the parts below p2. At the output of the second block, the two signals from the upper and lower parts are subtracted and the intended signal would be obtained.
Regards
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October 13, 2015 at 5:57 am #26324maha slitiParticipant
Hi, Thank you for your help.
I have considered the design that you have described. But, I did not obtain the objective to omit the signal if its power is not in the interval of [p1,p2]. I have attached my simulation file.-
October 13, 2015 at 6:39 am #26329alistuParticipant
The parts from the second block are not parallel. I have attached the modified file, but unfortunately it doesn’t work. The problem is that the signal magnitude gets multiplied many times instead of being subtracted. And the signal magnitude should not get increased by subtraction unless one of the signals have negative values, while no signal has negative values here.
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October 13, 2015 at 1:18 pm #26425maha slitiParticipant
Hi, I cannot open the file, can you send me a print screen?
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October 14, 2015 at 4:20 am #26484alistuParticipant
Sure! Please see the attached screenshot. As you can see, the first and the second block I was discussing in my former replies are in series, which means one is completely after the other (with only one common connection point). Nevertheless, due to the aforementioned problem I could not make it work.
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October 14, 2015 at 5:25 pm #26537maha slitiParticipant
You think that there is solution to this problem?
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October 14, 2015 at 7:28 pm #26539maha slitiParticipant
Hi,
Can you please help me to implement a matlab code that achieves the role of an optical substractor
(output = InputSignal1 power – InputSignal2 power )? I can use it instead of the optisystem optical substractor?-
October 15, 2015 at 5:57 am #26572alistuParticipant
Hi Maha,
Have you studied the OptiSystem tutorials regarding how to cosimulate OptiSystem with Matlab? If you learn about the format of the OptiSystem signals, then it will be quite easy to implement the filter itself and not the subtractor. For the main program, you just need to use two “if” commands in Matlab.
Regards
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October 15, 2015 at 7:12 am #26580alistuParticipant
Hi Maha,
I have attached the optical filter I have created using Matlab for this purpose. I suggest that you carefully and thoroughly analyze this very simple example to learn how to co-simulate Matlab with optiSystem. Also, pay attention to the fact that values chosen as upper and lower threshold correspond to the amplitude, while Optical time domain visualizer shows the power of the signal, so you have to use square root of the power levels in the Matlab component parameter settings.
Regards
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October 22, 2015 at 4:08 am #26954maha slitiParticipant
Hi AListu,
Thank you very match for your help. Excuse me for the delay 🙂
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