- This topic has 4 replies, 2 voices, and was last updated 8 years, 4 months ago by .
Viewing 4 reply threads
Viewing 4 reply threads
- You must be logged in to reply to this topic.
Optiwave software can be used in different industries and applications, including Fiber Optic Communication, Sensing, Pharma/Bio, Military & Satcom, Test & Measurement, Fundamental Research, Solar Panels, Components / Devices, etc..
Download our 30-day Free Evaluations, lab assignments, and other freeware here.
Optiwave software can be used in different industries and applications, including Fiber Optic Communication, Sensing, Pharma/Bio, Military & Satcom, Test & Measurement, Fundamental Research, Solar Panels, Components / Devices, etc..
Download our 30-day Free Evaluations, lab assignments, and other freeware here.
Hello. When we transmit 2 optical side-bands having a frequency spacing of 60 GHz through a fiber, at the end how the PIN photo diode detects this signal as 60 GHz Rf signal?
Io(t)= R |Eo(t).Eo(t)*|, Where R is the reponsivity , Eo(t) is the optical field and Eo(t)* is the complex conjugate of the field.
thanks for the answer. but it can’t makes me clear . isn’t it represent the amplitude of the RF spectrum. If an optical spectrum contains two sidebands at 193.07 and 193.14 THz, then when PIN detects it Why it place the rf spectrum at 60 GHz ? this i want to know , why in 60GHz? not in 193.07 and 193.14 THz ? Thanks in advance .
this is the output .
can you please show your circuit diagram? The frequency separation between the 193.07 and 193.13 THz is 60 GHz. thats why you are getting 60 GHz RF signal.
Notifications