Why is the allocated bandwith smaller with QAM-16 or 64 compared to e.g. QAM-4? I am not a professional so I'm looking for an intuitive explanation.
Electronic – the required bandwidth for QAM modulation
digital modulationModulation
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Alas, if highly-specialized and integrated chips don't quite meet your specs, it's impossible to upgrade just that one little section of the chip that is the weak link. Some people prefer something "I UNDERSTAND, not a sealed box of unexplained tricks made in the far east using technology I'm unlikely to be able to tweak or repair." (-- Nick Waterman, G7RZQ). You may be one of those people ;-).
As you've already pointed out, a quadrature amplitude modulation transmitter has a section that mixes 2 baseband data signals (traditionally called I and Q) into one modulated signal, and the receiver has a surprisingly similar section that decodes the modulated signal into separate I and Q data signals.
Surely there are ICs that do most or all of the job for me, but searching for "QAM encoder" doesn't give me any ICs. Perhaps there is another name for these ICs?
Historically the 2 devices in a QAM encoder or the 2 devices in a QAM decoder that "multiply 2 signals together" are each often called a "RF/IF mixer", even when it is used in a system that doesn't have a radio antenna, in order to distinguish it from "audio mixer" which acts completely differently. Analog Devices tutoral "MT-080: Mixers and Modulators" lists a bunch of specific ICs such as the AD8345 QAM encoder.
I see that at one of my favorite electronic supply websites, that QAM encoder chip is called a "quadrature modulator". I searched for "quadrature demodulator" and "quadrature modulator" at a few of my favorite electronic supply websites and found list of ICs in stock that, if I understand what you are asking, seem to meet your requirements. Analog AD8345, Linear LT5502, RFMD RF2480SR, RFMD RF2713, etc.
The chips I listed contain 2 of the "RF mixer" devices in one IC. I suppose you could use a pair of chips that each contain only a single "RF mixer" such as a pair of the popular SA612 chips (the same chip is also called the SA602, the NE602, the NE612, etc.). Or perhaps a person could build each "RF mixer" out of discrete transistors -- perhaps in the Gilbert cell configuration.
Mathematicians often call this device a "multiplier" and feed a sine wave into one and a cosine wave into the other, in order to make the math simpler.
Ham radio operators often build IQ decoders that use square waves instead of sine waves, such as the Tayloe decoder, in order to make construction easier.
While I agree that it would be nice if a QAM decoder could decode I and Q signals all the way down to DC, many systems fake it. They appear as if they can produce a constant, solid green color over the entire screen -- apparently a fixed DC level -- while internally they take the I and Q signals and immediately throw away the low-frequency components through AC-Coupled Video Signals and somehow magically restore the "right" DC level at a later stage.
DC restoration, synchronizing the receiver frequency and phase to the transmitter frequency and phase, and restoring amplitude path loss with automatic gain control, are almost always handled in a separate part of the system from the actual modulator and demodulator. As you already know, NTSC and PAL hide extra information in the horizontal retrace and vertical retrace time intervals to make the receiver's job possible/easier. I suppose you could define a special time where the transmitter sets I to +MAX and Q to 0, and other special times with the 8 other combinations of I and Q with +MAX, 0, and -MAX. Then the receiver could use the non-image information it sees at its I and Q outputs at those times to help with frequency synchronization, phase synchronization, automatic gain control, and DC restoration.
QAM really has nothing to do with PSK at all. It's just two channels of ASK that are combined together using two subcarriers that have a fixed phase shift between them.
To prevent ISI in an ASK signal, you need to be able to transmit at a minimum the main "lobe" of the power spectrum with linear phase shift. The lobe is defined by the first nulls in the PSD, which occur at ± the symbol rate/2, for a total bandwidth that's equal to the symbol rate.
So, yes, you're on the right track: The bandwidth in Hz is equal to the symbol rate, which is equal to the bitrate divided by the number of bits per symbol, or 5.
Best Answer
NxN-QAM means Quadrature Amplitude Modulation and it is a modulation scheme where the transmitted signal is the "mix" of two quadrature carriers whose amplitude is digitally modulated independently so as to give N different possible amplitude levels per each carrier. Therefore 64-QAM is 8x8-QAM, for example.
The total bandwidth of such a signal is proportional to the baud rate \$\dfrac 1 T\$ where T is the symbol time, i.e. the time needed to transmit a symbol. Note that each simbol in NxN-QAM carries \$\log_2(NxN)\$ bits of information, therefore 64-QAM carries 6 bit of information per symbol, whereas 4-QAM carries only 2 bit per symbol and 16-QAM carries 4.
If you consider a constant information transmission rate, i.e. a constant bit rate, you can see that increasing the number of bits per symbol makes the symbol time increase, hence the required bandwidth decreases.
To be more explicit, imagine you have to transmit a message with a bit rate of 64kbit/s. If you use 4-QAM you can transmit 2 bit per symbol, so you need to transmit at 32kSymbols/s (32kBaud). If you use 64-QAM you can transmit 6 bit per symbol, hence your baud rate drops to ~10.6kSymbols/s (10.6kBaud). Since we said that bandwidth is proportional to baud rate you see how the required bandwidth dropped using 64-QAM for a constant bit rate.