Dear Zahra,

Ninad has touched the answer. The data rate depends on the double sided bandwidth and the modulation type. In order to make the answer clear assume that you have a carrier fc and you modulated this carrier by a data stream fb and the type of digital modulation is 4 QAM. Then according to the theory of the digital modulation techniques the double sided bandwidth that is the two side bands around the carrier 

will be equal to fb/2. For 16 QAM it will be fb/4 and so on. Assuming now that the resonance frequency coincides with the carrier of the input modulated wave that will be amplified by the amplifier then the bandwidth BW of the amplifier will set the maximum limit on the data rate such that BW of the amplifier= fb max/2 in case of 4-QAM. This results in the relation fbmax=2 BW,

In principle the bandwidth of the amplifier must accommodate twice the baseband bandwidth. This is the rule. The baseband bandwidth is related to the bit rate and type of modulation.

So, it is the pass band width of the amplifier that limits the bitrate of the data. The reactive elements in the circuit together with the parasympathetic capacitances that limit the BW of the amplifier.

Best wishes 

Tuning frequency is related to choke by equation fc = 1/2*pi*sqrt(L1*C) here once your highest frequency is controlled and hence data rate can be determined by sampling theorem. max data rate = f/2 and this f and fc is related to type of modulation. e.g. f(max)=fc 

Dear Zahra,

Ninad has touched the answer. The data rate depends on the double sided bandwidth and the modulation type. In order to make the answer clear assume that you have a carrier fc and you modulated this carrier by a data stream fb and the type of digital modulation is 4 QAM. Then according to the theory of the digital modulation techniques the double sided bandwidth that is the two side bands around the carrier 

will be equal to fb/2. For 16 QAM it will be fb/4 and so on. Assuming now that the resonance frequency coincides with the carrier of the input modulated wave that will be amplified by the amplifier then the bandwidth BW of the amplifier will set the maximum limit on the data rate such that BW of the amplifier= fb max/2 in case of 4-QAM. This results in the relation fbmax=2 BW,

In principle the bandwidth of the amplifier must accommodate twice the baseband bandwidth. This is the rule. The baseband bandwidth is related to the bit rate and type of modulation.

So, it is the pass band width of the amplifier that limits the bitrate of the data. The reactive elements in the circuit together with the parasympathetic capacitances that limit the BW of the amplifier.

Best wishes 

And you can continue beyond the previous answers. If you know the power output and the bandwidth of the RF power amplifier, and you also have knowledge of noise power in the RF channel you will be using, you will need the attenuation along the path, and now you can get a theoretical limit of the channel capacity (data rate) independent of modulation type.

In other words, Dr. Zekry explains how the bit rate possible from that RF power amp will depend on what modulation you use. Those computations will give you answers which provide higher and higher bit rate, based on the constellation. So for instance, all else equal, 16-QAM (4 bits per symbol) would provide twice the bit rate of 4-QAM (2 bits per symbol), and 256-QAM (8 bits per symbol) will double that 16-QAM bit rate, if the symbols are transmitted at the same rate in each case. The bandwidth determines the maximum symbol rate, and the modulation constellation will determine the bit rate which can be carried.

But in the real world, you won't be able to play this game forever. Noise gets in the way. In practice, with noise in the circuit (literally), there may be ambiguity in decoding the symbols. The finer and finer differences in phase and amplitude will become obscured by noise.

So Shannon's equation relates capacity of the channel to its bandwidth and signal to noise ratio, no matter what modulation type or error correction code you want to use. It provides the theoretical maximum capacity, within that one propagation path.

So take a look at how simply this expression will answer your question.

Channel capacity (b/s) = BW (Hz) * logbase2(1 + S/N)

where S and N are expressions of signal power and noise power, in a power ratio. Not expressed in dB, but as a simple ratio. You have the bandwidth of your class E amp. You will also have the power output of your class E amp, and the gain offered by the transmit antenna. You need an idea of the noise power in the real world RF channel you plan to use, noise in the transmitter and in the receiver, the attenuation caused by distance and obstacles in the propagation path to the receiver, and the gain of the receive antenna. With this information, you can obtain the S/N ratio at the receiver itself.

Shannon's equation allows you to trade off complexity of modulation against how elaborate your error correction code is, for instance. Shannon won't care. It just gives a theoretical limit of capacity.