How to use single-drive mach-zehnder modulator

[Zhang Gaolu]

Hello everyone, I want to use single drive mach-zehnder modulator. But do not know how to set parameters in the parameters set. And why there are two biased voltages as it is single-driven. in what is the meaning of conventional in Modulator type option. is the Modulation voltage the half-wave voltage? What are the two options in the operation mode? can anyone help me ?

[stack programer]

Hi, i had same problem i concluded this.you can see good info about transfer function mach-zehnder here :opticsystem_component_library.pdf
https://drive.google.com/open?id=0Bxkt-d32S34gblFOVnNUSlF4TUE

phi0 is 0 radians for a conventional modulator and π radians for
phase-shift modulator

so i test modulation type with bit sequence 01:
when bit sequence is 0 in phase shift modulation amplitude is high and when bit sequence is 0 amplitude is negative or low.

https://drive.google.com/open?id=0Bxkt-d32S34gaEJsVW8tRkxOOHc
but for conventional is vice versa phase shift

[Zhang Gaolu]

thank you so much

[Dr. Dhiman Kakati]

Hi Zhang Gaolu,
Do not get confused with the two bias voltage. Each MZM will have two bias voltages These are required for operating the MZM in Peak, Quadrature and null point of operation. Depending on the point of operation you have to set the voltages. the single drive means only one RF voltages. i.e. Among the two arms of Mach-Zehnder interferometer structure, RF voltage is applies at only one arm. Please go through the literature of MZM. I can suggest you go through the book “Broadband optical modulators” edited by Antao Chen and Edmund Murphy.

Half-wave voltage means the voltage required to make a phase shift of pi/2.

for the 3rd Question I would suggest you go through the book I mentioned above. Then only you will understand.

Regards,
Dhiman

[Dr. Dhiman Kakati]

Please Go through the attached Image. Here V1 and V2 are the bias voltages

Attachments:

1. 

   MZM-operatin-modes.png
   

[Zhang Gaolu]

thank you so much.i hinave already solve the problem.thank you again

[Author]

Posts