Estimation of the power in a received radio signal

In summary, the current into the transmitter antenna is related to the electron drift velocity by the following equation: I=v_e\ A\ n_e\ e.
  • #1
jcap
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I would like to estimate the magnitude of a radio signal received from a transmitter by first principles:

Transmitter antenna length ##L=1## m

Transmitter antenna area ##A=1\hbox{ cm}^2##

Number of electrons per unit volume in antenna ##n_e=10^{28}##

Radiation resistance of antenna ##R_R=10\ \Omega##

Power radiated by transmitter ##P_R=1## kW

Frequency ##f=100## MHz

Joules law for the antenna: ##P_R=I^2 R_R##

Current into transmitter antenna ##I=\sqrt{\frac{P_R}{R_R}}=\sqrt{\frac{10^3}{10}}=10## A

Current ##I## in transmitter antenna is related to the electron drift velocity ##v_e## by the relationship ##I=v_e\ A\ n_e\ e##

$$v_e=\frac{I}{A\ n_e\ e}=\frac{10^1}{10^{-4}\ 10^{28}\ 10^{-19}}=10^{-4}\ \hbox{m/s}$$

Electron acceleration ##a_e=v_e f=10^{-4}\ 10^8=10^4\ \hbox{m/s}^2##

Total charge in transmitter antenna ##Q=L\ A\ n_e\ e = 10^0\ 10^{-4}\ 10^{28}\ 10^{-19}=10^5\ ##C

Assume receiver antenna is at a distance ##D=1## km

The strength of the vertical electric field at the receiver, ##E_{rec}##, is given by the far-field radiation formula

$$E_{rec}=\frac{Q\ a_e}{4\pi\epsilon_0\ c^2\ D}=\frac{10^5\ 10^4}{10^{-10}\ 10^{16}\ 10^3}=1\ \hbox{V/m}$$

If the receiver antenna also has length ##L=1## m and radiation resistance ##R_R=10\ \Omega## then the power it receives ##P_{rec}## is given by

$$P_{rec}=\frac{V_{rec}^2}{R_R}=\frac{(E_{rec}\ L)^2}{R_R}=\frac{1}{10}\hbox{W}$$

Have I got this approximately right?
 
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  • #2
Wow. I never came across an attempt to work out a link budget that way. Where did you get the idea to do it that way?
Just a reality check. You are are radiating 1kW over a sphere of 1km. (Omnidirectional antenna) What's the area of a 1km sphere? (4000000π m2) Then what effective receiving area would you need to intercept 10/1000 of the radiated power (if it works as simple as that)?

Are you thinking in terms of Power Transmission by Radio (à la Nicola Tesla) ?
 
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  • #3
If the antenna was omnidirectional at a distance of ##1## km the Energy flux should be ##10^3/(4\pi\times10^6)=10^{-4}## W/m##^2##. If the receiving antenna has length ##1## m and width ##1## cm I would have thought it should only intercept a power of ##1\times 10^{-2} \times 10^{-4}=10^{-6}## Watts.

I'm just trying to understand normal radio. What interested me in the calculation is the enormous charge of ##10^5## Coulombs that must oscillate in the transmitter antenna.
 
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  • #4
jcap said:
What interested me in the calculation is the enormous charge of 105 Coulombs that must oscillate in the transmitter antenna.
Can you say where you got the idea for approaching it in this way? I haven't come across your approach anywhere. It's not how the EM textbooks deal with radiation from an antenna just look at any source you can get hold of.
You are surprised by the high value of charge involved. An electron has 1.6e-19C and there are 6e23 atoms in a mole of the metal in the antenna, you can expect 1.6X6e4 = around 105C worth of valence electron in a 64g piece of Copper wire. That back-of-a-fag-packet calculation delivers your number, as it happens, almost exactly. All of those valence electrons are available to slosh around in the antenna.
 
  • #5
jcap said:
I'm just trying to understand normal radio. What interested me in the calculation is the enormous charge of 10^5 Coulombs that must oscillate in the transmitter antenna.

Nothing unique about that number. Just imagine the huge amount of electrons oscillating back and forth in the many meters of household wiring in your average home.
 
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  • #6
Drakkith said:
Nothing unique about that number. Just imagine the huge amount of electrons oscillating back and forth in the many meters of household wiring in your average home.
It's not as if you can do much with them. The protons in their atoms keep them pretty much in place - despite the way that electricity can be described as an electron flow, very few electrons actually stray far from 'their' local metal atom. It's the electric forces that are the important thing.
The OP finds the number surprisingly large but it's like saying that, when you hear the noise of banging with a hammer at the far end of a battleship, through the steel, the thousands of tons of battleship makes it surprising. There's a sort of false dichotomy at work.
 
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  • #7
jcap said:
I would like to estimate the magnitude of a radio signal received from a transmitter by first principles:

Transmitter antenna length ##L=1## m

Transmitter antenna area ##A=1\hbox{ cm}^2##

Number of electrons per unit volume in antenna ##n_e=10^{28}##

Radiation resistance of antenna ##R_R=10\ \Omega##

Power radiated by transmitter ##P_R=1## kW

Frequency ##f=100## MHz

Joules law for the antenna: ##P_R=I^2 R_R##

Current into transmitter antenna ##I=\sqrt{\frac{P_R}{R_R}}=\sqrt{\frac{10^3}{10}}=10## A

Current ##I## in transmitter antenna is related to the electron drift velocity ##v_e## by the relationship ##I=v_e\ A\ n_e\ e##

$$v_e=\frac{I}{A\ n_e\ e}=\frac{10^1}{10^{-4}\ 10^{28}\ 10^{-19}}=10^{-4}\ \hbox{m/s}$$

Electron acceleration ##a_e=v_e f=10^{-4}\ 10^8=10^4\ \hbox{m/s}^2##

Total charge in transmitter antenna ##Q=L\ A\ n_e\ e = 10^0\ 10^{-4}\ 10^{28}\ 10^{-19}=10^5\ ##C

Assume receiver antenna is at a distance ##D=1## km

The strength of the vertical electric field at the receiver, ##E_{rec}##, is given by the far-field radiation formula

$$E_{rec}=\frac{Q\ a_e}{4\pi\epsilon_0\ c^2\ D}=\frac{10^5\ 10^4}{10^{-10}\ 10^{16}\ 10^3}=1\ \hbox{V/m}$$

If the receiver antenna also has length ##L=1## m and radiation resistance ##R_R=10\ \Omega## then the power it receives ##P_{rec}## is given by

$$P_{rec}=\frac{V_{rec}^2}{R_R}=\frac{(E_{rec}\ L)^2}{R_R}=\frac{1}{10}\hbox{W}$$

Have I got this approximately right?
I am very interested that you have tried this calculation.
As an engineer, I know that the electric field strength from a dipole radiating 1kW is 300V/m at 1km.
Your antenna has an effective length of 1m and this is about right for a dipole at 100MHz.
The Electro Motive Force at the antenna terminals is Field Strength x Length = 300 x 1 = 300mV.
The Potential Difference across a 75 Ohm load resistor is half that, so 150mV.
The power into the resistor in milliwatts is V^2/R = 1000 X 0.15^2/75 = 0.3.
You seem to have pulled 10 Ohms Rr out of th air. I can't see a calculation for that?
 
  • #8
tech99 said:
You seem to have pulled 10 Ohms Rr out of th air.
That put me right off the method. Radiation Resistance, for me, is the resistance component of the input impedance of an antenna. It's a good indication of the amount of the total Power that's being radiated but it's something (a result) that comes at the very end of the process. 10 Ohms is very arbitrary and I could have gone with 75 Ohms which is round about what you get from a (resonant) half wave dipole. As far as I can see, the argument pulls itself up by its own bootstraps, and being so much 'easier to follow' than the antenna theory that I learned way back, it made me uneasy. There is no reference to support it either, despite my request.
 
  • #9
tech99 said:
I am very interested that you have tried this calculation.
As an engineer, I know that the electric field strength from a dipole radiating 1kW is 300V/m at 1km.
Your antenna has an effective length of 1m and this is about right for a dipole at 100MHz.
The Electro Motive Force at the antenna terminals is Field Strength x Length = 300 x 1 = 300mV.
The Potential Difference across a 75 Ohm load resistor is half that, so 150mV.
The power into the resistor in milliwatts is V^2/R = 1000 X 0.15^2/75 = 0.3.
You seem to have pulled 10 Ohms Rr out of th air. I can't see a calculation for that?

jcap said:
I would like to estimate the magnitude of a radio signal received from a transmitter by first principles:

Transmitter antenna length ##L=1## m

Transmitter antenna area ##A=1\hbox{ cm}^2##

Number of electrons per unit volume in antenna ##n_e=10^{28}##

Radiation resistance of antenna ##R_R=10\ \Omega##

Power radiated by transmitter ##P_R=1## kW

Frequency ##f=100## MHz

Joules law for the antenna: ##P_R=I^2 R_R##

Current into transmitter antenna ##I=\sqrt{\frac{P_R}{R_R}}=\sqrt{\frac{10^3}{10}}=10## A

Current ##I## in transmitter antenna is related to the electron drift velocity ##v_e## by the relationship ##I=v_e\ A\ n_e\ e##

$$v_e=\frac{I}{A\ n_e\ e}=\frac{10^1}{10^{-4}\ 10^{28}\ 10^{-19}}=10^{-4}\ \hbox{m/s}$$

Electron acceleration ##a_e=v_e f=10^{-4}\ 10^8=10^4\ \hbox{m/s}^2##

Total charge in transmitter antenna ##Q=L\ A\ n_e\ e = 10^0\ 10^{-4}\ 10^{28}\ 10^{-19}=10^5\ ##C

Assume receiver antenna is at a distance ##D=1## km

The strength of the vertical electric field at the receiver, ##E_{rec}##, is given by the far-field radiation formula

$$E_{rec}=\frac{Q\ a_e}{4\pi\epsilon_0\ c^2\ D}=\frac{10^5\ 10^4}{10^{-10}\ 10^{16}\ 10^3}=1\ \hbox{V/m}$$

If the receiver antenna also has length ##L=1## m and radiation resistance ##R_R=10\ \Omega## then the power it receives ##P_{rec}## is given by

$$P_{rec}=\frac{V_{rec}^2}{R_R}=\frac{(E_{rec}\ L)^2}{R_R}=\frac{1}{10}\hbox{W}$$

Have I got this approximately right?
Actually I think it is
jcap said:
I would like to estimate the magnitude of a radio signal received from a transmitter by first principles:

Transmitter antenna length ##L=1## m

Transmitter antenna area ##A=1\hbox{ cm}^2##

Number of electrons per unit volume in antenna ##n_e=10^{28}##

Radiation resistance of antenna ##R_R=10\ \Omega##

Power radiated by transmitter ##P_R=1## kW

Frequency ##f=100## MHz

Joules law for the antenna: ##P_R=I^2 R_R##

Current into transmitter antenna ##I=\sqrt{\frac{P_R}{R_R}}=\sqrt{\frac{10^3}{10}}=10## A

Current ##I## in transmitter antenna is related to the electron drift velocity ##v_e## by the relationship ##I=v_e\ A\ n_e\ e##

$$v_e=\frac{I}{A\ n_e\ e}=\frac{10^1}{10^{-4}\ 10^{28}\ 10^{-19}}=10^{-4}\ \hbox{m/s}$$

Electron acceleration ##a_e=v_e f=10^{-4}\ 10^8=10^4\ \hbox{m/s}^2##

Total charge in transmitter antenna ##Q=L\ A\ n_e\ e = 10^0\ 10^{-4}\ 10^{28}\ 10^{-19}=10^5\ ##C

Assume receiver antenna is at a distance ##D=1## km

The strength of the vertical electric field at the receiver, ##E_{rec}##, is given by the far-field radiation formula

$$E_{rec}=\frac{Q\ a_e}{4\pi\epsilon_0\ c^2\ D}=\frac{10^5\ 10^4}{10^{-10}\ 10^{16}\ 10^3}=1\ \hbox{V/m}$$

If the receiver antenna also has length ##L=1## m and radiation resistance ##R_R=10\ \Omega## then the power it receives ##P_{rec}## is given by

$$P_{rec}=\frac{V_{rec}^2}{R_R}=\frac{(E_{rec}\ L)^2}{R_R}=\frac{1}{10}\hbox{W}$$

Have I got this approximately right?
Actually I think this is very interesting and I thinkthe method is almost OK but you have assumed an arbitrary 10 Ohms Rr for both antennas.

jcap said:
I would like to estimate the magnitude of a radio signal received from a transmitter by first principles:

Transmitter antenna length ##L=1## m

Transmitter antenna area ##A=1\hbox{ cm}^2##

Number of electrons per unit volume in antenna ##n_e=10^{28}##

Radiation resistance of antenna ##R_R=10\ \Omega##

Power radiated by transmitter ##P_R=1## kW

Frequency ##f=100## MHz

Joules law for the antenna: ##P_R=I^2 R_R##

Current into transmitter antenna ##I=\sqrt{\frac{P_R}{R_R}}=\sqrt{\frac{10^3}{10}}=10## A

Current ##I## in transmitter antenna is related to the electron drift velocity ##v_e## by the relationship ##I=v_e\ A\ n_e\ e##

$$v_e=\frac{I}{A\ n_e\ e}=\frac{10^1}{10^{-4}\ 10^{28}\ 10^{-19}}=10^{-4}\ \hbox{m/s}$$

Electron acceleration ##a_e=v_e f=10^{-4}\ 10^8=10^4\ \hbox{m/s}^2##

Total charge in transmitter antenna ##Q=L\ A\ n_e\ e = 10^0\ 10^{-4}\ 10^{28}\ 10^{-19}=10^5\ ##C

Assume receiver antenna is at a distance ##D=1## km

The strength of the vertical electric field at the receiver, ##E_{rec}##, is given by the far-field radiation formula

$$E_{rec}=\frac{Q\ a_e}{4\pi\epsilon_0\ c^2\ D}=\frac{10^5\ 10^4}{10^{-10}\ 10^{16}\ 10^3}=1\ \hbox{V/m}$$

If the receiver antenna also has length ##L=1## m and radiation resistance ##R_R=10\ \Omega## then the power it receives ##P_{rec}## is given by

$$P_{rec}=\frac{V_{rec}^2}{R_R}=\frac{(E_{rec}\ L)^2}{R_R}=\frac{1}{10}\hbox{W}$$

Have I got this approximately right?
I am very interested that you have calculated this. I don't think conductor diameter has any effect because fewer electrons will have to accelerate more, so it balances out. I do notice that the received power is not quite correct because when you connect your 10 Ohm resistor to the receiving antenna the voltage falls to half, which gives quarter power. I think the main inaccuracy is taking 10 Ohms as an arbitrary value for Rr.

I believe the formal method for Rr is that, having found the field at 1km from knowledge of the current, frequency and antenna length, we could then find the power flux density at the same position, using PFD = E^2/377 (377 Ohms is the impedance of free space).
Then we can integrate this over a sphere and find the total power radiated (W).

The radiation resistance is a measure of the power radiated for a given current. Rr = W/I^2.
(I also made a slip when I said 300mV/m at 1 km for 1 kW radiated, because this is for a monopole over perfect ground, but for a dipole in free space it should be 221mV/m. An easy formula to use is E = (7 sqrt P)/d where E is the field strength in V/m, P is in watts and d is in metres).

Like you, I was amazed to discover the colossal free charge in a metal; the slightest movement creates very large effects. It seems that the remaining positive ion is very heavy so it doesn't accelerate significantly. When there is no accelerating force, the static fields of the electrons and the positive ion outside the metal must be very large, but by superposition they cancel. That is not to say they are not there, and as soon as acceleration of the electrons occurs, we see the radiated field component arising from the distortion of its static field.
 
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  • #10
tech99 said:
I don't think conductor diameter has any effect because fewer electrons will have to accelerate more, so it balances out.
Interestingly, the thickness of a dipole wire will affect the bandwidth. Fat dipoles tend to be used in wide band multi-element arrays because they match better. They don't need to be as fat as the ones below to be worth using.
1595019452765.png
Image of fat ( cage) dipoles from Wikipedia
 
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  • #11
I think there are many misconceptions with this approach. First, you need to compute or model the true radiation resistance. For that you need to define the antenna type (someone assumed it’s a dipole but you don’t say). Second, RF current flows on the skin of the wire, which you’ve completely ignored. Finally, current is highly nonuniform along the antenna length for a dipole, it is maximal at the terminals becoming zero at the ends. The Method of Moments Expands the distribution in sines and cosines. Without these, your numbers can’t be meaningful.
 
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  • #12
sophiecentaur said:
Interestingly, the thickness of a dipole wire will affect the bandwidth. Fat dipoles tend to be used in wide band multi-element arrays because they match better. They don't need to be as fat as the ones below to be worth using.
View attachment 266498Image of fat ( cage) dipoles from Wikipedia

Interesting thread. How does the thickness affect the bandwidth? After high enough frequency doesn't the signal travel mostly on the surface (skin depth)?
 
  • #13
First of all can I mention that Jcap was just doing a ball park calculation for illustration and it has been very interesting and useful.
The skin effect is small for ordinary antennas - we don't have to worry about it for engineering purposes.
The antenna also has an effective length which is less than the actual length because the current tapers away towards the ends. For a dipole the effective length is about 2/pi times the actual length, so Jcap was near the mark when he suggested 1m effective length for a dipole at 100MHz, which is 3m wavelength.
And of course there are countless different types of antenna in the World and Jcap was not trying to study them all.
Regarding conductor diameter and bandwidth, the main issue here is that the conductor is resonant, so we start to see unwanted reactance when we move away from the designed frequency.The antenna conductor can be considered as a transmission line resonator where the characteristic impedance, and hence the Q, is lowered by using fat conductors. By using low characteristic impedance the Q is lower and the bandwidth is increased. Hence the use of fat conductors.
The reason we see resonant action in an antenna is that when we accelerate electrons, not all the energy is radiated. The electron also acquires considerable kinetic energy. It is this unradiated energy which is deposited in the antenna capacitance. Half a cycle later it leaves the capaitance and moves to the antenna inductance, where it is stored in the antennas local magnetic field. It is these storage places, electric and magnetic, which represent the strong induction fields which we find close to the antenna. In a typical case, only 10% or the power is radiated each cycle, the remainder being stored.
 
  • #14
A further note about skin effect. I intended to mean that the skin resistance is small for ordinary antennas and can be ignored for ordinary engineering purposes. However it is true that the currenr flows in a very thin skin on the surface. This does not affect radiation, because the numer of free electrons involved in the radiation process does not alter the amount of radiation. Radiation depends on acceleration and charge, so if we reduce the charge, the electrons accelerate more for a given current and this maintains the status quo.
 
  • #15
The phenomenal amount of charge calculated is nearly totally balanced by the charges of the ions in the copper crystal lattice, a well-studied effect. The tiny imbalance can be calculated by boosting to the reference frame of the moving charges. The Coulomb force resulting from the charge imbalance between current and the Lorentz-contracted ions gives rise to what is called the magnetic field, and it is tiny. A correct calculation must incorporate Maxwell's equations and relativity.
 
  • #16
I believe the formula used by Jcap is the Larmor Formula for the power radiated by an accelerated charge. According to my reference book, this formula is a good approximation provided the velocity is very small compared with the speed of light. (u/c)^2 <<1.
Electromagnetic Vibrations, Waves and Radiation. George Bekefi and Alan H Barrett, MIT.
 
  • #17
Joshy said:
Interesting thread. How does the thickness affect the bandwidth? After high enough frequency doesn't the signal travel mostly on the surface (skin depth)?
For a thin wire dipole, the Impedance passes from largely negative reactance, through pure resistance to positive reactance. See fig2 in this link For a fat dipole that reactance curve has a smaller slope around the resonance frequency and that widens the bandwidth over which you have largely a resistive component.
 
  • #18
sophiecentaur said:
For a thin wire dipole, the Impedance passes from largely negative reactance, through pure resistance to positive reactance. See fig2 in this link For a fat dipole that reactance curve has a smaller slope around the resonance frequency and that widens the bandwidth over which you have largely a resistive component.
That curve looks like normal transmission line theory, but the figure doesn't show (or it's not apparent to me) how the thickness of the metal affects its impedance; I also don't understand why skin depth doesn't apply to "ordinary" antennas. Is the frequency just too low? If it requires much more detail then I'm willing to start a new thread as I don't want to hijack this one, but maybe OP might be interested since the whole idea of skin depth is that the current does not distribute equally across the cross-section... at what point would their approach break?

edit:

Found the above to be true in my textbooks although I felt the explanation fell short with "It is an important general principle that the thicker the dipole, the wider is its bandwidth." My question isn't so much about whether or not it's true, but more of how or why? Would this still apply at higher frequencies?
 
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  • #19
tech99 said:
A further note about skin effect. I intended to mean that the skin resistance is small for ordinary antennas and can be ignored for ordinary engineering purposes. However it is true that the currenr flows in a very thin skin on the surface. This does not affect radiation, because the numer of free electrons involved in the radiation process does not alter the amount of radiation. Radiation depends on acceleration and charge, so if we reduce the charge, the electrons accelerate more for a given current and this maintains the status quo.
100 MHz has significant skin effect. Also 1 meter is suboptimal length for radiation, your Rr should be closer to 200 ohms. Never heard of a transmitter radiating anything but heat. FCC rates transmitters by max power to final amplifier. So a 1 KW transmitter will generate ~500 watts RF. Because the antenna is unmatched, about 2/3 of the power is rejected (look up SWR) back to the transmitter. (Typically the Tramsmitter would fry at this point.) Recieved power at an antenna is in decibels or microvolts. Because Radio front ends are voltage amplifiers. "The Antenna Handbook" by the Amateur Radio Relay League (ham radio org.) would be a good book for you to get familiar with.
 
  • #20
Joshy said:
That curve looks like normal transmission line theory, but the figure doesn't show (or it's not apparent to me) how the thickness of the metal affects its impedance;
The effective C and L of a thick cylinder are not the same as that of a thin wire so you are dealing with a different transmission line. The R at resonance is still about the same though. But this takes us beyond the first layer of antenna theory. I would bet that the original work on antennae was very experimental and the theory 'caught up with' that later. The relationship between wire thickness and intercepted power from the wave is not obvious. An infinitely thin wire shouldn't actually intercept any power, thinking intuitively.
shjacks45 said:
100 MHz has significant skin effect.
The skin effect at just 1Mz can be significant enough for the coils in MF receivers to have been wound with Litz wire (a set of very fine, insulated wires, twisted together instead of a single conductor - a much higher surface area with Litz). But in that case, the, albeit low resistance was high enough to affect the Q factor of filters. Engineering is all about the numbers in each application.
shjacks45 said:
Because Radio front ends are voltage amplifiers.
Normal Radio Comms tends to suffer more from interference than receiver noise. In situations where noise is more relevant, front end design is more angled towards choosing the optimum input impedance of the device to maximise the power relative to the internally generated noise.
 
  • #21
Joshy said:
That curve looks like normal transmission line theory, but the figure doesn't show (or it's not apparent to me) how the thickness of the metal affects its impedance; I also don't understand why skin depth doesn't apply to "ordinary" antennas. Is the frequency just too low?

The skin effect occurs over distances which in metals is much smaller than the free space wavelength of the radiation. As far as the boundary value problem is concerned the skin depth is normally negligible. Where it does come into play is with ohmic losses.
 
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  • #22
sophiecentaur said:
Found the above to be true in my textbooks although I felt the explanation fell short with "It is an important general principle that the thicker the dipole, the wider is its bandwidth." My question isn't so much about whether or not it's true, but more of how or why? Would this still apply at higher frequencies?
Why is bandwidth greater with a thicker conductor? If you consider the antenna as a transmission line, a thicker conductor gives a lower characteristic impedance (Zo). This is because the capacitance, which shunts the line, is greater for the thicker conductor. When the transmission line resonator moves away from resonance, the reactance it presents is less with the lower Zo. It is similar to an LC circuit with high C.
This lower reactance is in series with the radiation resistance of the antenna, so that the feed point impedance is less disturbed.
For example, a wire which is an 0.2 of a wavelength long (72 degrees) has a radiation resistance of, say, 20 Ohms. If it is thin and has a Zo of 1000 Ohms then it will present a capacitive reactance of Zo/tan 72 = 333 Ohms. So the feedpoint impedance is 20 - j333. On the other hand a large tube having a Zo of 100 Ohms wil present a capacitive reactance of Zo/tan 72 = 33 Ohms, so the feed point impedance is now 20 - j33. If we have a 20 Ohm resistive source then the latter case is a far better match.
Regarding skin resistance, it is important for inductors in high frequency tuned circuits because it decides the Q of the circuit. Skin resistance is increased because the wire is subjected to the magnetic field from adjacent turns of wire in addition to itself. But for an ordinary antenna, the radiation resistance is tens of Ohms and skin resistance might be in the order of 1 Ohm, so we do not need to make allowance for it.
 
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  • #23
tech99 said:
But for an ordinary antenna, the radiation resistance is tens of Ohms and skin resistance might be in the order of 1 Ohm, so we do not need to make allowance for it.
This last part is what I was looking for and makes sense to me. Thank you very much.

Admittedly kind of felt like a lot of the general responses although I'm not offended by it were trying to shove RF down my throat seemingly as if I was unfamiliar with it. What kind of electronics engineer without RF knowledge would ask about skin effect??
 
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  • #24
I'm surprised that no one has compared the OP result to that from the Friis transmission formula. I'll do that here, assuming that the antennas are center-fed lossless dipoles.

First, the radiation resistance of a 1 m dipole at 100 MHz is about 22 ohms. Making that adjustment in the OP increases the received power to approximately 0.5 W. We compare that to the Friis power.

For simplicity, assume that the antennas are 1.5 m long instead of 1 m, because relevant values for half-wavelength dipoles are tabulated everywhere. (The difference in directivity is only about 10% so we will be very close). Match the 1 kW transmitter perfectly to the transmitting antenna without loss, assume lossless antennas for convenience and align the wires along z with 1 km separation along x so the polarizations match. The receiving antenna is similarly matched perfectly into a resistive load.

The antenna directivity is 1.62 and this is also the gain since we are assuming no losses. The Friis transmission formula then gives the received power as 1.50e-4 W or 150 uW. That the OP is off by over 3 orders of magnitude arises from its naive treatment of current and current distribution in the antenna.
 
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  • #25
marcusl said:
I'm surprised that no one has compared the OP result to that from the Friis transmission formula. I'll do that here, assuming that the antennas are center-fed lossless dipoles.

First, the radiation resistance of a 1 m dipole at 100 MHz is about 22 ohms. Making that adjustment in the OP increases the received power to approximately 0.5 W. We compare that to the Friis power.

For simplicity, assume that the antennas are 1.5 m long instead of 1 m, because relevant values for half-wavelength dipoles are tabulated everywhere. (The difference in directivity is only about 10% so we will be very close). Match the 1 kW transmitter perfectly to the transmitting antenna without loss, assume lossless antennas for convenience and align the wires along z with 1 km separation along x so the polarizations match. The receiving antenna is similarly matched perfectly into a resistive load.

The antenna directivity is 1.62 and this is also the gain since we are assuming no losses. The Friis transmission formula then gives the received power as 1.50e-4 W or 150 uW. That the OP is off by over 3 orders of magnitude arises from its naive treatment of current and current distribution in the antenna.
I agree with your Friis calculation assuming 1.5m dipoles. There are two sources of error in the jcap calculation as far as I can see but I still can't get it exact. I have a predicted received field strength of 338mV/m and it should be 221mV/m. (As I have mentioned, it was only intended as a ball park illustration I believe).
1) The Rr for both antennas was assumed to be 10 Ohms instead of about 70 Ohms.
2) The received power calculation used the Electro Motive Force at the receive antenna terminals instead of the Potential Difference.
Here is a corrected calculation as good as I can get it:-
W=I^2*R, so for 1kW, transmit antenna current = sqrt W/R = sqrt 1000/70 = 3.8 A
ve = 3.8/(10^-4*10^28^10^-10) = 3.8*10^-5
Electron acceleration = ve*f = 3.8*10^-5*10^8 = 3.8*10^3 m/s^2
Total charge in antenna = 10^5 C
Electric field strength at 1km,
E = 10^-5*3.8*10^3 / 10^-10*10^16*10^3
E= 0.38V/m (this is about double the correct figure)
Receiver calculation:-
Voltage across 70 Ohm resistor
Effective Length * Field strength/2
= 1 * 0.38/2
= 0.19 mV
Power in load = V^2/R
Wr = 0.19^2 / 70
Wr = 515 uW
The Friis figure is 150 uW, so it is an error of 5dB, somewhere in the assumptions about the transmitting antenna I suspect. Maybe peak or average acceleration, or something of that sort..
 

Related to Estimation of the power in a received radio signal

1. What is the purpose of estimating the power in a received radio signal?

The purpose of estimating the power in a received radio signal is to determine the strength or intensity of the signal. This information is important for various reasons, such as determining the quality of the received signal, evaluating the performance of a communication system, and determining the distance between the transmitter and receiver.

2. How is the power in a received radio signal measured?

The power in a received radio signal is typically measured in decibels (dB). This unit of measurement is logarithmic and represents the ratio of the received signal power to a reference power level. It is a convenient way to express the power of a signal, as it allows for easy comparison and calculation.

3. What factors can affect the power of a received radio signal?

The power of a received radio signal can be affected by various factors, including the distance between the transmitter and receiver, obstructions in the signal path, interference from other signals, and the quality and strength of the transmitter and receiver equipment. Weather conditions and atmospheric disturbances can also impact the power of a received signal.

4. How can the power in a received radio signal be estimated?

The power in a received radio signal can be estimated using various methods, such as using a power meter or spectrum analyzer, performing mathematical calculations based on the signal strength and distance, or using specialized software. The method chosen will depend on the specific application and the available equipment.

5. Why is it important to accurately estimate the power in a received radio signal?

Accurately estimating the power in a received radio signal is important for ensuring reliable communication and maintaining the overall performance of a communication system. It can also help in troubleshooting and identifying any issues with the signal. Additionally, accurate power estimation can aid in complying with regulatory requirements and optimizing the use of radio frequency spectrum.

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