Its related to my previous question Modes in waveguides I have a doubt that how are different modes actually generated? I mean in Digital and Analog communication, we studied how we modulate the signal to shift its spectrum. At that point of time (Output of modulator) its only electrical signal right (Please correct me if I'm wrong)? How do we generate EM wave corresponding to this modulated signal? How do we make sure that EM wave is belonging to that specific mode which is desired? What goes on the channel is the EM wave, right or the electrical signal?
How are modes generated in a waveguide
communicationelectromagnetismwaveguide
Related Solutions
The modes of a waveguide refer to the distribution of the electrical and magnetic fields across the cross-section of the waveguide.
The electric field going to (roughly) zero at high-conductivity surfaces (like the inner and outer conductor of a coaxial line) places boundary conditions on the differential equations describing the waveguide behavior. This leads to patterns in the way the fields distribute themselves across the waveguide area. If a certain E/M field pattern can propagate along the waveguide without changing, we call that a mode of the waveguide.
Generally there will be only one propagating mode at low frequencies, and additional modes will be able to propagate as the signal frequency increases.
The waveguide modes are important because different modes tend to propagate at different rates along the z-axis of the waveguide. Generally you want to operate a waveguide at a frequency where only a single mode is supported. At higher frequencies, a single pulse input into the guide might exit the other end highly distorted due to the different propagation velocities of the different modes.
Different mode structures can also create reflections at the connection between two transmission lines, even if both transmission lines have the same characteristic impedance.
I have also heard about single and multimode fibers in optical communication. Are they the same?
Yes, it's essentially the same. Of course, the frequencies at play are very different. And the boundary conditions are created by changes in dielectric constant of the material rather than conductive surfaces.
In a "purely digital" link where you set an output to "high" and an input the other end of a line is read as "high" then the probability error is purely to do with the SNR of the line. What is the probability that a HIGH can be interpreted as a LOW? By introducing a higher level protocol with error detection and correction you effectively negate most of the SNR errors and the question is now "What is the probability that the protocol cannot correct corrupted bits?"
So yes, the CODEC (or protocol) can be used (and is used) to negate the effects of SNR-induced signal corruption.
As for the second part...
If you assume 1 bit of information is transmitted per quantization level, and 1 bit is received per quantization level, then yes, increasing the quantization level will increase the number of bits sent at any one time. However, the SNR of the transmission medium will then have a greater effect on those now smaller quantization steps, so although you reduce the quantization noise, you now increase the SNR noise.
However, if you don't assume 1 bit per quantization level, but have multiple quantization levels per bit, then you can increase the number of quantization levels and keep the overall bitrate the same, but have more detail about each bit, so can make a better informed decision about what value that bit is.
For instance, you can think of a simple digital link with 2 states (HIGH and LOW) as a 1-bit quantized system. For simplicity we'll call it 1V for HIGH and 0V for low.
Now, you could then have it that anything received >= 0.5V is a HIGH and anything < 0.5V is a LOW. That's 1 bit quantization. 0.5V would be HIGH, but 0.499999999999V would be LOW. That's an infinitesimally small margin for noise.
However, increase the receiving quantization to 2 bits, say, would give you more detail. It would give you 4 voltage levels to consider - 0V, 0.33V, 0.66V and 1V.
You could now say that anything > 0.66V is a HIGH, and anything less than 0.33V is a LOW. You have now introduced a "noise margin". Anything that falls between those values is discarded as noise. The bitrate remains the same, but the overall SNR has fallen.
Then of course you can add a "schmitt trigger" to it (or software equivalent), whereby you toggle the value depending on a transition. When the input rises above 0.66V you see the value as HIGH, and keep it as HIGH. Only when it then drops down below 0.33V do you then switch it to LOW.
For systems where you have discrete voltage levels you could sample them at a higher resolution, and the line-induced noise would occupy the least significant bits of that sampled value. Discarding the noisy bits down to the resolution of the sent data can then reduce the noise in the system. Also taking multiple samples and averaging them, which in effect cancels the random noise out, (known as "oversampling") can reduce the noise as well.
None of those techniques affect the bitrate as such since you're not adding any extra information to the sent values.
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Best Answer
Key point: when we talk about the modes of a waveguide, we are often using that as shorthand for the propagating modes of the waveguide. However, the waveguide also has other modes, called evanescent modes. The evanescent modes don't carry energy along the waveguide, they just die out exponentially along the z-direction. Which is why those aren't normally the modes we mean when we talk about the modes of the waveguides. However, the evanescent modes are important to answering your question.
Let's say we place a feed antenna into a rectangular waveguide:
If we drive the antenna from outside, it will produce certain E and H field patterns.
We can express these fields as a superposition of the modes of the waveguide, so long as we consider both the propagating and the evanescent modes. This tells us how much of the input energy couples to the propagating modes and how much to the evanescent modes. Whatever part of the signal energy that corresponds to the propagating modes will propagate along the waveguide. The energy coupled into the evanescent mode effectively bounces back and forth. Some of it gets absorbed by conductive losses in the walls of the waveguide. But a larger part of it probably gets re-coupled to the feed antenna and appears as a reflection back up the feed line to the signal source.