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I'm currently taking a course about antennas, based on the book "Antenna Theory: Analysis and Design" by Constantine A. Balanis.

The Thevenin equivalent circuit for a transmitting antenna is presented. And it is said there that maximal power will reach the antenna when it is matched to the source impedance, and that in this case (assuming a lossless antenna), half of the power generated by the source will reach the antenna, while the other half will dissipate in the source itself.

As part of my home work I had to solve a question from the book. I got the correct answers (according to the solution manual and my teacher), but I can't understand one thing. Clearly, there is no impedance matching in the problem. So I would expect to see that less than half of the power supplied by the source will reach the antenna. However, a quick calculation shows that it is exactly the other way around! that is, more than half of the generated power is transmitted to the source. $$P_{\text{source}} = \Re[\frac{1}{2}VI^*] \approx 36\text{W}$$ $$P_{\text{rad}} = \frac{1}{2}|I_{\text{ant}}|^2R_r=21.36\text{W} > 18\text{W} = \frac{1}{2}36\text{W}$$

How can it be? Have I misunderstood something?

The question

The answer

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2 Answers 2

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Assuming that the \$100\space\text{V}\$ is the effective voltage of the source.

Well, notice that the amplitude of the input current is given by:

\begin{equation} \begin{split} \hat{\text{I}}_\text{i}&=\left|\frac{\text{V}_\text{i}}{\text{Z}_\text{i}}\right|\\ \\ &=\frac{\left|\text{V}_\text{i}\right|}{\left|\text{Z}_\text{i}\right|}\\ \\ &=\frac{\left|100\exp\left(\varphi_\text{i}\text{j}\right)\right|}{\left|50+0.625+73+42.5\text{j}\right|}\\ \\ &=\frac{100}{\sqrt{\left(50+0.625+73\right)^2+42.5^2}}\\ \\ &=\frac{800}{\sqrt{1093721}}\\ \\ &\approx0.764956\space\text{A} \end{split}\tag1 \end{equation}

Now, the power losses are given by:

$$\text{P}_{\text{R}_\text{Loss}}=\text{I}^2\text{R}_\text{Loss}=\left(\frac{800}{\sqrt{1093721}}\right)^2\cdot0.625=\frac{400000}{1093721}\approx0.365724\space\text{W}\tag2$$

The power radiated by the antenna is given by:

$$\text{P}_{\text{R}_\text{r}}=\text{I}^2\text{R}_\text{r}=\left(\frac{800}{\sqrt{1093721}}\right)^2\cdot73=\frac{46720000}{1093721}\approx42.7166\space\text{W}\tag3$$

And the total power delivered by the source is given by:

\begin{equation} \begin{split} \text{P}_\text{total}&=\left|\text{V}_\text{i}\right|\cdot\hat{\text{I}}_\text{i}\cdot\cos\left(\arg\left(\text{Z}_\text{i}\right)\right)\\ \\ &=100\cdot\frac{800}{\sqrt{1093721}}\cdot\cos\left(\arctan\left(\frac{42.5}{50+0.625+73}\right)\right)\\ \\ &=\frac{80000}{\sqrt{1093721}}\cdot\cos\left(\arctan\left(\frac{340}{989}\right)\right)\\ \\ &=\frac{80000}{\sqrt{1093721}}\cdot\frac{989}{\sqrt{1093721}}\\ \\ &=\frac{79120000}{1093721}\\ \\ &\approx72.3402\space\text{W} \end{split}\tag4 \end{equation}

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  • \$\begingroup\$ Thank you for the reply. However, I don't think your calculations are correct. Power is current times voltage, while in your calculation you say the total power is voltage times resistance. \$\endgroup\$ May 13, 2023 at 18:19
  • \$\begingroup\$ @MichaelVaispaper My mistake, I edited my answer. Now it is correct. \$\endgroup\$ May 13, 2023 at 21:26
  • \$\begingroup\$ It is, but my question remains - the power reaching the antenna is more than half the power generated by the source. \$\endgroup\$ May 14, 2023 at 0:11
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This is not any power transmission maximized case. The radiation resistance is higher than the internal resistance of the source and there's no matching network items included. The same current causes less dissipation in the source resistance. Thus more than a half of the power extracted from Vs is used in Rr. What's lost in the 0.625 ohm Rloss is so little that it doesn't change the imbalance .

With proper matching one could extract more from Vs and then a half of it would be radiated and the rest would be dissipated in the source resistance and the loss resistance. The proper matching network makes the antenna look a 50.625 ohm resistor assuming the transmission line to the antenna is short, only about a couple of percents of the wavelength or shorter. Then Vs=100V delivers about 98.8 watts and a half of it is radiated. That's 49.4 watts - substantially more than what's radiated now.

I guess the basic reason for the confusion is some ham radio talk that you have heard with half an ear. The transmitter may well push to the antenna transmission line its full output power. Load mismatch cause a part of it to reflect back to the transmitter. The actual radiated power can then well be less than a half of what's extracted from Vs. But your case is totally different than the case there's a transmission line. Resistor Rs=50 ohm does not at all present a 50 ohm transmission line, not even as a coarsest possible approximation. In practical transmitters which feed antennas Rs is tried to be made as small as possible and there's something, maybe nonlinear or wave travelling direction dependent, which prevents the reflected wave to cause severe overvoltage or overheating in the transmitter.

Inserted after a bunch of comments: This homework problem is a pure AC circuit analysis exercise except the names of some parts are taken from antenna texts. It contains no practical transmitter. The model of practical transmitter is substantially more complex than a voltage source. Resistor Rs does not make it better. For ex. a practical transmitter contains an output filter which itself is designed for certain, typically resistive load. Then there's non-linear parts such as transistors. Their non-linearity forces us to insert the output filter to keep the signal clean enough.

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  • \$\begingroup\$ Ok, I think I'm starting to see it, Correct me if I'm wrong. There is no point talking about "impedance matching" unless a transmission line is involved, so when analyzing the equivalent Thevanin circuit there is no matching needed. However, When a transmission line is involved (as in most cases) one should consider the impedance matching. And this is why in the book they presented the circuit analysis results under the impedance matching conditions. \$\endgroup\$ May 15, 2023 at 1:03
  • \$\begingroup\$ This is generally alright, but the last paragraph is misleading. While real transmitters might not have resistors in the output path (or as much as possible), this does not give them a low output impedance. We must at this point separate what we mean by this effective circuit: do we mean the resistor to be real and physical (representing efficiency, perhaps)? Or do we want to describe the output impedance of the system (which might not have any relation to power dissipated inside the transmitter)? \$\endgroup\$ May 15, 2023 at 5:31
  • \$\begingroup\$ One could also suppose that the impedance in the problem happens to be that at the end of a transmission line and antenna system, which just happens to be somewhat inductive at whatever dimensions and frequency the system has. Whether or not a transmission line is involved, doesn't really matter at a single frequency. Cheers! \$\endgroup\$ May 15, 2023 at 5:35
  • \$\begingroup\$ @TimWilliams 1) Practical transmitters have output filters which assume certain loads, single voltage source is quite a poor model. inserting Rs does not make the model better. Without those filters any resistive load would be ok if the resistance is high enough. 2) Even at single frequency transmission line is a reactive part which fortunately causes nothing if it has matched load. But a mismatched transmission line can draw a substantial current. Think for ex the extreme case "no antenna at all" . An open circuit with no transmission line draws nothing from a voltage source. \$\endgroup\$ May 15, 2023 at 7:10
  • \$\begingroup\$ @MichaelVaispaper the comment above is also for you. Filters can cause in practical transmitters severe overvoltage or overcurrent and poor harmonic suppression problems if they have wrong load. This is also a matching problem, but not just the same as maximizing power transmission from a voltage source which has resistive series resistance. Your current homework contains no practical transmitter, it's a pure basic AC circuit theory exercise. \$\endgroup\$ May 15, 2023 at 7:24

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