How can the ability of lightwaves to interfere be exploited to generate laser pulses without a chirp? Explanations of EOMs: www.rp-photoni... www.fiberoptic... Frequency chirps in current-pulsed DFB lasers: www.fiberoptic...
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Great video! Could you please do another video where you would dive deeper into the matter and explain the electro-optic modulation in more detail? For e.g. what kind of electrical signals are supposed to be modulated and how say a given input signal of v(t) would get modulated into the output intensity I(t) also considering the changes in refractive index of the material etc. It would be extremely great to see a somewhat more detailed explanation of the matter. Thanks!
Thank you, I am glad that you found it helpful! An EOM can be used for turning any time dependent voltage signal into a time dependent optical signal. The main limitation is the electrical bandwidth of the EOM, which is usually around 10-15GHz for basic models and 40-60GHz for high end ones. Miniaturized EOMs used for creating pulses for commercial fiber optical telecom can go up to around 100GHz. Around 2:30 in the video, I show that the optical power is a function of the squared cosine of the phase shift induced by the applied voltage. If one knows the crystal structure of the EOM medium (typically LiNbO3), it's possible to calculate from first principles how this phase depends on the applied voltage. Please see lecture 29 in the following playlist for details. It's a great lecture series, so if you have the time, I highly recommend watching all the videos!
An EOM works by applying an electric field to an inherently birefringent crystal (typically Lithium Niobate), which imparts a phase shift on light propagating in that arm of the device. Microscopically, the applied E-field changes the distribution of electrons in the crystal lattice and thus the refractive index. Because the crystal is inherently birefringent, the index along the x- and y-axes will respond differently to the applied field. Thus, if you launch a random polarization into the EOM, the x- and y-components of the E-field will be modulated differently. If you, for example, bias the EOM so that x-polarized light is blocked, y-polarized light may still leak through. Therefore, it is best to clean up the polarization of the light first and then use PM fiber to launch it into the EOM.
@@yourfavouriteta I have a question regarding to polarization. Is the same to talk about randomly polarized light and unpolarized light? I'm begining oh those photonics thigns and I still having some doubts about this area. Thank you for your response btw
@@davidhergueta2000 The concept of "unpolarized" light confused me for a long time as well. Whenever I am in doubt about a physics concept, I have found that it's helpful to consider an illustrative experimental setup and what can be *measured* with it. For example, imagine having a perfectly stable CW laser emitting purely x-polarized light. Let's send this light into a fiber polarization controller like the one shown at 4:05 in the video and onto an ideal polarimeter, which essentially measures the amount of *energy* that the E_x, E_y, E_d, E_a, E_r and E_l components of the electric field deposit on 6 different photodiodes within a certain sampling duration. Note that power of the x-component is proportional to |E_x|^2, so the energy deposited by the x-component within the sampling duration is proportional to the integral, int_0^T(|E_x|^2 dt). Note also that a polarimeter does not directly measure the SOP of the EM field. Rather, it measures "total energy deposited" for each component, and uses the ratios of these energies to each other to compute the SOP ***under the assumption that the fields were all constant during the sampling interval***. If you "wiggle" the paddles into a random position and then take your hands off the setup, the light entering the polarimeter will have a random state of polarization (most likely elliptical), which remains constant over time. In this case, the specific SOP is "random", but to the polarimeter, the light will appear "polarized". Now imagine rotating the paddles randomly, but very slowly. As long as the sampling rate of your polarimeter is much higher than the rate at which to modify the paddles, the device will report that the E-field changes continuously, but is otherwise "polarized" at any given instant. Now imagine rotating the paddles randomly and EXTREMELY quickly. For example, you choose a new position for them every nanosecond, while the sampling rate of the polarimeter is only 1 microsecond. In this case, |E_x|^2 in the integral above will randomly take on values between 0 and some maximum value 1.000 times within the sampling interval. The same thing is true for the other 5 energy measurements. When the polarimeter compares their ratios, it will conclude that no polarization component is "dominant" and thus that the light is "unpolarized". The crucial point is that "unpolarized" is just a label we humans slap on light if its SOP changes much more quickly than we can measure (or compared to the response time of the system we are shining it on). The actual electric field vector at any location and time always has some size and orientation. It may change very quickly over time or with position in an unpredictable way, but with sufficiently advanced equipment, you will in principle always be able to measure it. Instead of a binary "polarized"/"unpolarized" distinction, I think it's more helpful to think about the polarization having a "steadiness duration", which can be compared to the "sampling time" of your measurement system. If your sampling time is very long compared to the steadiness duration, it's a bit like trying to film a dance performance using a 1 frame/s camera with a 1s exposure time; obviously, the orientation of the dancer's arms and legs will seem blurry and incoherent. However, this impression will disappear if you use a 1.000 frames/s high-speed camera with a 1ms exposure time instead!
Also, you can verify that "unpolarized" is just a result of a low sampling rate by building the setup described above with the polarimeter set to a short integration time. If you use a programmable polarization controller (check my video on GPIB) you can make it change the SOP gradually at some rate. Then, by simply increasing the integrating time of your polarimeter, you will see that it eventually begins to complain about the light being unpolarized. If light can become "unpolarized" by you altering the properties of the measurement system, it cannot be a property of the light itself!
Thank you so much; Please, I have very interesting question because I'm very confused in what's the difference between digital and analog fiber transmission (after data modulation) for example by using Electro-Optical Modulator?
Hello there! From what I understand, you are asking what the difference is between using an EOM for analog signal transmission versus digital signal transmission. Consider a continuously varying electrical signal, for example one coming from an old fashioned record player that represents sound. Instead of sending it to a loudspeaker, where the changing voltage will move a membrane to produce music, we can hook the signal up directly to the terminals of an EOM. If a CW laser is sent into the EOM and the bias voltage is adjusted to the 50% transmission point, the incident analog signal will continuously change the transmitted optical power over time. A photodiode in the other end can detect the varying optical power, convert it back into an analog voltage signal and feed it to a loudspeaker to produce sound. To send the same sound signal digitally, we can use an analog to digital converter (ADC) to convert the electrical signal into a series of discrete levels. Typically, an ADC will have a number of levels corresponding to some power of 2. For example, a "3-bit ADC" will have 2^3=8 levels, meaning it discretizes the electrical signal into levels ranging from 0 to 7. In binary, this would be from 000 to 111. To transmit the digitized signal, we must select a "modulation scheme". In practice, a combination of EOM's and phase modulators is used, but let's for simplicity consider a modulation scheme with 2^6=64 power levels and no changes to the optical phase. In such a scheme, detecting, for example, 1mW of power in an incoming pulse would be interpreted as the bit string 000001, while 4mW would be 000100. Let us imagine that we want to transmit *two* samples of the electrical signal corresponding to the bit strings 001 and 010. By sending *one* optical pulse with a power of 001010=10mW, detecting it at the destination and "splitting" it back up into 001 and 010, we have effectively sent along two data points simultaneously! This is the main advantage of digital communication over analog. We discretize the signal into bitstrings, merge bitstrings into larger "symbols" and use modulation schemes altering amplitude and phase of optical pulses to send multiple bitstrings simultaneously. My most recent video shows a simple approach to doing 4QAM, where 00, 01, 10 and 11 can be sent using only changes in optical phase rather than amplitude.
@@yourfavouriteta Thank you so much, great explanation :) Please, if you could upload video for more informations about the difference between digital and analog fiber transmission! I'll so thankful to you. All the best to you;
You're welcome! Unfortunately, I no longer have lab access, but I do use an AOM in my video on measuring laser linewidth. For a theoretical explanation of AOMs and many other topics, I can recommend the RU-vid playlist "Modern Optics" by Prof. Partha Roy of IIT Kharagpur.