Antenna and EM Modeling with MATLAB Antenna Toolbox. Sergey N. Makarov
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      Example 1.2

      Calculate and plot the average acquired antenna power when the generator with a periodic waveform υg(t) = υg(t + T) is characterized by the rms voltage and generator resistance given by

      (1.5c)

      Solution: We use Eq. (1.5b) and average it over period T to obtain average power

. The result has a form:

      Then, we plot the average antenna power as a function of the load resistance. The short MATLAB script given below accomplishes the task:

      Rg = 50; % Generator resistance, Ohm Vg = 9; % Generator rms voltage, V RA = [0.01*Rg:0.01*Rg:10*Rg]; % Load resistance PA = RA*Vg^2./(Rg + RA).^2; % Average antenna power plot(RA, PA); grid on; title('Average antenna power, W') xlabel('Load resistance, Ohm')

      However, it must be clearly stated that no more than 50% of the total generator power can be extracted even in this best case. This statement makes sense if we again examine the circuit in Figure 1.1 with two equal resistors. We see that the power delivered to each resistor is obviously the same. Since Rg is internal resistance, half of the total power is spent to heat up the generator.

      Note:

      The power maximum in Figure 1.2 is relatively flat over the domain Ra > Rg; however, the power drops sharply when Ra < < Rg. This last condition should be avoided if at all possible. The corresponding example is given below.

      Example 1.3

      A transmitting antenna in a radio handset features a monopole antenna. It is connected to a sine wave generator that has the same basic form as Figure 1.1 with an internal (generator) resistance of 50 Ω. The antenna has the radiation resistance of 50 Ω (which generates power loss in terms of electromagnetic radiation); its loss resistance is zero. The antenna, when properly matched to the power source, will radiate 50% of the total power. Now, a young RF engineer decides to “modify” the handset by cutting the monopole antenna and leaving only one‐third of its length, so that the antenna's radiation resistance is reduced to one‐ninth of its original value. How does this affect the radiated signal?

      Solution: For a periodic AC signal the average antenna power is given by Eq. (1.5d), that is

      (1.8)

      The ratio of the two power levels for the two antenna configurations does not depend on generator voltage, i.e.

      (1.9)

      Thus, for the shorter antenna we will only achieve about 36% of radiated power compared to the original handset. In practice, this estimate becomes even much worse due to the appearance of a very significant antenna reactance as explained in the following text.

      After the antenna has been matched to the generator, the legitimate question to ask is how to find the antenna efficiency, E. Since Ra = Rr + RL > Rr, only a part of the power delivered to the antenna is really radiated; another part is dissipated in the antenna itself and makes it a heater.

      The corresponding equivalent circuit of the antenna with losses includes two resistors in series: the radiation resistance Ra and the loss resistance RL. Since the same current i(t) flows through both resistors at any time instant, we can find the radiation efficiency or simply the efficiency of the antenna in the form of the СКАЧАТЬ