Calculate Coaxial Cable Loss & Input Impedance at 1 MHz

[nelectrode]

Hey Guys,

How do I work out Transmission loses and input impedance for coaxial cable?

I’ve been given this website to use http://fermi.la.asu.edu/w9cf/tran/

I do have the characteristic impedance of 75Ω and frequency and I am asked to work the attenuation in dB/100ft and “input impedance for a 150Ω load resistance”
So I am not sure How to complete the input data.
Cable type: RG59
Frequency in Mhz: 1
Cable length in feet: 100
Load resistance: 150
Load reactance: How I find that? Do I have to use Z=R+JX?

Thanks

 

[davenn]

hi there

all cable manufacturers have datasheets for their cable
this always includes a table showing
losses usually per 100ft or 100 metres at a given frequency, sometimes it may state a shorter length
10 metres or 30 ft

for standard RG59 the loss at 1MHz is going to be less than 1dB for 100 ft ( 33 metres)

Load resistance: 150

resistance or impedance ??

you can't put a load with an impedance of 150 Ohms on a 75 Ohm cable
the mis-match and resulting SWR is going to be horrible

Your load impedance must equal your cable impedance for best power transfer, else you will end up with a large reflected power component

Cheers
Dave

 

[tfr000]

you can't put a load with an impedance of 150 Ohms on a 75 Ohm cable
the mis-match and resulting SWR is going to be horrible

I think that's the idea here. The impedance changes depending how many wavelengths you go down the transmission line.
Old-school, done on a "Smith Chart", but I guess that Java app is supposed to do the same thing.

 

[Averagesupernova]

you can't put a load with an impedance of 150 Ohms on a 75 Ohm cable
the mis-match and resulting SWR is going to be horrible

Your load impedance must equal your cable impedance for best power transfer, else you will end up with a large reflected power component

Cheers
Dave

This depends on the length of the transmission line in relation to the wavelength. An insignificant portion of a wavelength in transmission line length will not cause any problem at all. While we are on the subject of transmission line length what about the matching of the transmitter to the transmission line? Suppose the transmitter output impedance is 150 ohms and the load is 150 ohms but connected with 75 ohm line? If the line length is a small portion of a wavelength then no problem. If it is a half wavelength, then the line will appear as if it is not there at all minus some loss. I've only scratched the surface of this subject.

 

[sophiecentaur]

It's not usual to find that both transmitter and antenna are well matched to the transmission line. The amplifying device is often operated just for 'best' power and the antenna would be matched to make sure most of that power goes up the spout. (In the same way that a 'good' battery has a very low output resistance but you would never dream of putting a 0.05Ohm (max power) load on it.)
Another reason for matching the antenna is that a mismatch produces standing waves which can the maximum voltage on the amplifier output. This can harm the device.

To deal with your particular problem. The operating frequency is 1MHz, so cable loss will be negligible for 100ft of feeder. If you look it up on the cable spec (here, for instance) you can see just how low it is at 1MHz!

If you use the Applet they give you, you just need to put in the load impedance (which is 150Ω), cable impedance and length to find the impedance that the transmitter will 'see'. I think it will give you all the answers you want. (It does the difficult bits for you.)

 

[davenn]

This depends on the length of the transmission line in relation to the wavelength. An insignificant portion of a wavelength in transmission line length will not cause any problem at all. While we are on the subject of transmission line length what about the matching of the transmitter to the transmission line? Suppose the transmitter output impedance is 150 ohms and the load is 150 ohms but connected with 75 ohm line? If the line length is a small portion of a wavelength then no problem. If it is a half wavelength, then the line will appear as if it is not there at all minus some loss. I've only scratched the surface of this subject.

the problem is we don't cut coax lengths to a specific portion of a wavelength of the freq used in a particular system. We cut the coax to what is needed to get from the TX connector to the antenna connector ( ignoring any cavity filters, circulators that may be also inline). This means the coax is always just some random length depending on the height of the antenna up the tower.
Because of this we match the TX (RX) output (input) to the coax characteristic impedance and then that of the coax to the antenna.. If we don't, then we have the problems that Sophie and I have already stated concerning standing waves ( SWR) and the resulting high levels of reflected power back to the transmitter.
This can cause 2 common problems

1) as Sophie stated it can damage the transmitter output stage
in a lot of transmitter equip, protection is provided by measuring the reflected power level and it is used to lower the TX output power level till the problem is resolved

2) That reflected signal mixes in the TX output stages to produce intermodulation products that then get sent to the antenna and transmitted and can cause all sorts of hell with other comms services in the area

Matching TX, Xmission line and antenna is always the best method

Dave

 

[Baluncore]

In theory, a load mismatch is not line length dependent if the source impedance is matched to the line. That is because there will be no energy re-reflected again when returned to the source, since the source is correctly matched to the line.

Real sources such as transmitters will reflect the energy, because the output stage will be adjusted to a different impedance that reflects all the reverse energy. The transmatch will be adjusted to minimise absorption of energy by the transmitter by maximising forward power.

 

[sophiecentaur]

In theory, a load mismatch is not line length dependent if the source impedance is matched to the line. That is because there will be no energy re-reflected again when returned to the source, since the source is correctly matched to the line.

Real sources such as transmitters will reflect the energy, because the output stage will be adjusted to a different impedance that reflects all the reverse energy. The transmatch will be adjusted to minimise absorption of energy by the transmitter by maximising forward power.

I wondered what you were getting at there but I see what you mean; it's a strictly theoretical comment. A well terminated source will kill reflections but at what cost to the requirements for the transmitter design? It's hard enough to build a transmitter that will provide the right power at the right cost and efficiency, without also requiring it to look like 50Ω .

 

[sophiecentaur]

@davenn
To add to your list:
Tx line reflections will also affect the frequency response (short lines) and introduce echoes / ringing (long lines) which is particularly relevant to analogue TV and tall masts.

 

[Baluncore]

The OP was a transmission line question. The RG59 cable with Zo = 75Ω suggests a receive system with a probable maximum power of one mW. There is no need for a transmitter. Academic exercises like this are often based on theory, not practice.

It's hard enough to build a transmitter that will provide the right power at the right cost and efficiency, without also requiring it to look like 50Ω

That may be true for an inexperienced amateur, but it is not true for a competent professional.

[Generalisation] Amateurs place a tuner between the transmitter and the line, then call it an antenna tuner, but it is really a transmatch. Professionals tune the line to the antenna so as to avoid reflected energy on the line. [/Generalisation]

 

[sophiecentaur]

That may be true for an inexperienced amateur, but it is not true for a competent professional.

If a transmitter is to operate at high efficiency, it will need to have an output impedance that is not matched to the line. If it were matched, the highest efficiency (just in RF terms) would be 50%. That would hardly be a "professional" design.

 

[nelectrode]

Hey,

Thanks a lot, but as Baluncore stated this is purely a theoretical question,

so I basically have to use the Java app to work out my values.

I do know the following only:

Type: RG-59
Characteristic impedance: 75 Ω
frequency: 1Mhz

have to find:

Matched attenuation(dB/100ft): ?
Matched attenuation (dB/100m):?
input impedance for 150Ω load resistance: ?so on the app at http://fermi.la.asu.edu/w9cf/tran/

I have:

Cable type: RG-59
Frequency: 1Mhz
cable length in feet: 100
Load resistance: Do I have to put 150Ω here?
Load reactance: I thought it is going to be 75Ω but that's seems to be wrong.

Thanks

 

[sophiecentaur]

Load reactance will be zero if you're told 150 Ohms (i.e. no reactance in series)
That applet seems to do all you need.

 

[nelectrode]

That's what I thought but the values don't really match the one here:

http://www.embedded.com/print/4402915

So what is input Impedance for a 150Ω load resistance?

Because the app gives a Z and Z0?

 

[sophiecentaur]

Z0 should be 75 Ohms but the loss per foot in the cable (this is a real cable they're discussing) will introduce some effective reactance, I guess. It seems to be about 2%, which is quite a lot but I think that's what it's about. RG59 isn't high quality cable. Put 75 Ohms into the 'user defined' option cable parameters and it gives you the ideal lossless cable with Z0 of 75 Ohms
The Zin is what you will see, looking into the transmitter end of the cable when you put your 150 Ohms at the other end. The line acts as a transformer to change the load impedance, as seen at the transmitter.
Applets are sooo useful - as long as yo have an idea what they're doing for you.
You guys have it so easy these days. All we had was a hand calculator or, before that, a Smith chart and china graph pencil,

 

[Baluncore]

If a transmitter is to operate at high efficiency, it will need to have an output impedance that is not matched to the line. If it were matched, the highest efficiency (just in RF terms) would be 50%. That would hardly be a "professional" design.

I disagree. Under what restrictive conditions is your statement true. Please explain.
Maximum power transfer occurs when the transmitter is matched to the line.
Are you actually suggesting matching by using a resistive network?

 

[sophiecentaur]

I disagree. Under what restrictive conditions is your statement true. Please explain.
Maximum power transfer occurs when the transmitter is matched to the line.
Are you actually suggesting matching by using a resistive network?

This can be confusing but, if your source impedance is the complex conjugate of the load impedance then you do get maximum power. Butttttt - you dissipate half of the power in the source resistance. (The melting anode or collector of your output stage) That is the 'downside' consequence of the maximum power theorem. Doing it is not a good idea when you have a 500kW transmitter. (Class C with 90% efficiency is favourite when the modulation system is suitable)

You don't need to use a resistive network for the matching (that would be plain loopy). But any form of matching for maximum power will have the above effect. It's not done in any system that I can think of except for distribution amplifiers, where the power is low and the marked gain needs to be relied on.
That goes for audio power amps too, they are usually designed as a voltage source.

 

[Baluncore]

nelectrode.
Are you neglecting velocity factor?

Attenuation is simply specified by the physical length of the cable.
Phase shift is specified by electrical length = physical length / velocity factor.

Attenuation.
Your source will provide energy that is attenuated along the cable to the load.
The mismatch will reflect energy.
The reflected energy will be attenuated on it's way back to the source.

Input Impedance.
The line length must be corrected for velocity factor, maybe about 66% for a 75 ohm foam line.
The electrical line length, (phase shift), can then be calculated from the transit time.
The reflected wave and forward wave phase combine to form the apparent Zin.

 

[sophiecentaur]

The applet works all that out for you. (Effective length etc.)
If you accept that the Z0 is the value of termination to eliminate reflections then the effect of loss in the feeder is to introduce the requirement for a bit of reactance. Mr Smith and his chart doesn't allow for that to be included easily, although I'm sure it can be done by someone really familiar with using the chart.

 

[Baluncore]

This can be confusing but, ...

Yes, I think you are confused.

The reactive component does not waste real power. It is a “reactive” recycling of circulating energy.
RF energy efficiency of close to 100% is possible with a matched and non-reactive transmission line.

Playing reflective ping-pong with energy on a lossy line can be very expensive. By avoiding reflected energy on the line, the transmitter will see a non-reactive load and so does not have to dedicate supply current to the untwisting of output stage phase.

Resistive “Pi” or “T” pads are often used when coupling broadband microwave signal modules to reduce the reactive effect of other module's mismatch, so they stabilise the module and it's specifications. I agree resistive pads should never be used to match power transmitters.

I'm glad I don't need to worry about a 500kW transmitter. I only have a total of 75 kW here and I do my best not to use it. Unfortunately, if I run the anodes too cool, the “getter” on their surface does not function so well and the cathode emission begins to fall.

 

[Averagesupernova]

I think Baluncore and sophie are barking up different trees.
-
Sophie says that an output stage with a very low Zout relative to the load will dissipate little power in said output stage compared to the load. I would say this is correct. Notice that sophie did NOT say that a reactive component was involved. Sophie also said that this is most efficient. This is where it gets questionable. Maximum power transfer does not imply most efficient. It implies that given an output that has a Zout of R, then you can load said output with a load of R and no matter how you adjust the load you will not get more power to dissipate in the load. Raising the load resistance will cause less power to be dissipated in the load but will NOT cause more power to be dissipated in the source. Lowering the load resistance will cause less power to be dissipated in the load but will cause MORE power to be dissipated in the source.
-
As I have stated in a different thread, introducing a reactive component in the load is hard on the output stage since when the most voltage is dropped across the output transistor (zero crossing) there is more current passing through the device compared to if there were a resistive load. Voltage and current are not in phase, so this naturally occurs.

 

[sophiecentaur]

If you had a 50 Ohm load connected directly to a transmitter with a 50 Ohm internal resistance, you would agree that was a perfect match (I hope). It is possible to match any load to any source impedance (at least in principle), so you could achieve the equivalent of the above - if you wanted to. But, under those conditions, you would have equal power dissipated in the transmitter as the power dissipated in the load. That is 50% efficiency and not very kind to the output device.
If you have any doubt about the Maximum Power theorem then you can look it up. I don't need to find it for you.
Very few power distribution systems use a matched arrangement (name one, in fact). Car batteries would all melt, given a 0,05 Ohm load and all the Generating Company's equipment would fly off its beds with such a high current demand (the equivalent of several times the present National Load). I could ask you what the source resistance of your mains supply is? Quick calculation: The volts probably dip by, perhaps 5V when you take 50A - so that means 0.1 Ohms. Would you connect a 0.1 Ohm load and expect the equipment to survive (without those fuses in the way?)
Whatever it is that you know about the practicalities of transmitters, you need to realize that, when you 'tune for max power', the max power you are tuning for is set by the practical limits of the amplifying device you are using and not how well it's matched.
You have stated that you 'back off' your transmitter to stop it knackering itself. That is the case for all transmitter designs. I suggest that you don't in fact, have any way of measuring the output impedance of your transmitter so you aren't in a position to assert whether or not you are providing it with a conjugate match.

There's nothing special in the principles involved, whether it's a battery, generator of transmitter.

 

[sophiecentaur]

I think Baluncore and sophie are barking up different trees.
-
Sophie says that an output stage with a very low Zout relative to the load will dissipate little power in said output stage compared to the load. I would say this is correct. Notice that sophie did NOT say that a reactive component was involved. Sophie also said that this is most efficient. This is where it gets questionable. Maximum power transfer does not imply most efficient. It implies that given an output that has a Zout of R, then you can load said output with a load of R and no matter how you adjust the load you will not get more power to dissipate in the load. Raising the load resistance will cause less power to be dissipated in the load but will NOT cause more power to be dissipated in the source. Lowering the load resistance will cause less power to be dissipated in the load but will cause MORE power to be dissipated in the source.
-
As I have stated in a different thread, introducing a reactive component in the load is hard on the output stage since when the most voltage is dropped across the output transistor (zero crossing) there is more current passing through the device compared to if there were a resistive load. Voltage and current are not in phase, so this naturally occurs.

I agree (and have stated) that a perfect match is NOT efficient. It involves losing half the power from your source.
I took the case of a Class C transmitter, which can be 90% efficient and which cannot be matched - for the above reason. The actual efficiency of output stages is more to do with the physics of the devices themselves and the need to 'control' the current flowing through them whilst avoiding too much voltage drop. But that's another issue.
I agree that reactive components are rather a red herring.
My post ,above, states the case about not matching source to load in power systems.

 

[Averagesupernova]

Yeah sophie I was thinking about car batteries, 1000 watt subwoofer amplifiers, the source impedance of a typical electrical service, etc. also. None of them worry about loading to satisfy maximum power transfer. Nor should they. In car audio, preoutputs have a Zout of a couple hundred ohms maybe up to 1000 I believe but the Zin at the amplifier is typically 10K. Interesting point there also, they don't worry about transmission line impedance either. My comment in a previous post about insignificant portions of a wavelength would apply here.

 

[Baluncore]

nelectrode, first let me apologise for the sound of battle in the background.

If a transmitter is to operate at high efficiency, it will need to have an output impedance that is not matched to the line. If it were matched, the highest efficiency (just in RF terms) would be 50%.

If you had a 50 Ohm load connected directly to a transmitter with a 50 Ohm internal resistance...

Resistance and output impedance share the same dimensions, but they are certainly not the same thing.

The difference in the barking up trees appears to come from the fact that a 50 ohm internal resistance in a low frequency power supply, is quite different to the output impedance of an amplifier driving a 50 ohm transmission line. Zout is actually the slope of the RF amplifier's load line, the ratio of the voltage to current in the output stage.

A power distribution grid needs a very low resistance so it does not vary in voltage with variable load conditions. If you could match your electric light to the grid, you would get most of the power available at the time. That would not be good.

 

[Averagesupernova]

Resistance and output impedance share the same dimensions, but they are certainly not the same thing.

The difference in the barking up trees appears to come from the fact that a 50 ohm internal resistance in a low frequency power supply, is quite different to the output impedance of an amplifier driving a 50 ohm transmission line. Zout is actually the slope of the RF amplifier's load line, the ratio of the voltage to current in the output stage.

I don't think there is a difference at all. Source impedance is source impedance. And that is the Zout of the previous stage. You claim that Zout is simply the ratio of voltage to current in the output stage. That ratio depends on how the output is loaded and the load does not define the Zout.
-
Edit: Naturally a DC power supply isn't concerned with a reactive component in it's Zout. But for the sake of argument here, reactive components are being left out of the discussion.

 

[Baluncore]

I don't think there is a difference at all.

I beg to differ.

A 1 kW RF transmitter driving a matched line does not generate >= 1 kW internally.
Class C and D amplifiers can operate at efficiencies better than 80% into matched lines.

 

[Averagesupernova]

I beg to differ.

A 1 kW RF transmitter driving a matched line does not generate >= 1 kW internally.
Class C and D amplifiers can operate at efficiencies better than 80% into matched lines.

The reason it does not generate >= 1kW internall is because it has a Zout below 50 ohms which is exactly what sophie has been saying. The load at the end of the line may be 50 ohms and the line may be 50 ohms and we can call that matched but that does not mean the Zout of the transmitter is 50 ohms. It may well be spec'd to drive a 50 ohm load, but that is irrelevant.

 

[davenn]

The reason it does not generate >= 1kW internall is because it has a Zout below 50 ohms which is exactly what sophie has been saying. The load at the end of the line may be 50 ohms and the line may be 50 ohms and we can call that matched but that does not mean the Zout of the transmitter is 50 ohms. It may well be spec'd to drive a 50 ohm load, but that is irrelevant.

and right there is the same subject that others were arguing about on another place on the net several years ago.
I vaguely remember it coming down to that the Zout of the transmitter can actually be quite a bit lower a way down ~ 20 Ohms. I have to admit that I didn't fully understand the debate way back then and even again with the same debate over the last ~ dozen posts.
The 50 Ohms label on the back of the TX is only stating the transmission line impedance required
NOT the TX output impedance, as you stated in your last sentence, Averagesupernova.

I will see if I can dig up some of the emails that flew between various participants

cheers
Dave

 

[the_emi_guy]

Let me provide a quantitative example of what Baluncore is saying:

Let's say my RF source has an impedance of 10 ohms (10 + 0j) and I am running at 100MHz.

I have these choices:

1 - Go ahead and drive the 50 ohm transmission line directly and accept that anything that reflects back from the antenna will reflect again at source.

2. Add (40 + j0) ohms in series with RF source. Source is now 50 ohms so energy reflected from the antenna will not be reflected again at source. But I have introduced significant loss in the process.

3 - Install a network at output of RF source consisting of 32nH in series and 64pf in parallel. Assuming these reactive components are lossless, we have not impacted the efficiency or created any new losses. Yet energy reflected from the antenna returning to the RF source will see this as exactly (50 + j0) ohms and there will be no reflection at 100MHz.

Option 3 has built up source impedance to 50 ohm at 100MHz without loss of power, this is a narrowband match.

What Baluncore is saying is that Option 3 is generally chosen in favor of better VSWR.

Think of it this way, we can move any point on the Smith chart to the center using a couple of reactive components at a given frequency.

On the other hand, if we need the source impedance to be 50 ohms across a wide frequency range, such as within a spectrum analyzer, then you use resistors and accept the losses (broadband match).

 

[Averagesupernova]

the_emi_guy, you are causing the output of the transmitter to see a 10 ohm load correct? Looking into the LC network? If this is your aim, here is what happens which gets up back to what sophie and I are saying is something to avoid:
-
Take a transmitter with Zout of 10 ohms as you described. Feed it into a 50 ohm load. We will say there is a voltage across the load of 10 VAC. Thevenize and you will find the thevenin voltage is 12 VAC. The load is dissipating 2 watts, the Zout of the transmitter dissipates .4 watts. Use an impedance matching network and now the Zout of the transmitter dissipates 3.6 watts. From the thevenin voltage you find this to be the case, and assuming a lossless matching network the power dissipated in the load will also be 3.6 watts. This kind of power dissipated in the output stage is considerably higher than the .4 watts when driving the load 'mismatched' which is exactly what sophie was saying is to be avoided in the first place.

 

[the_emi_guy]

Sorry, my input was not well thought out.

The output stage of an RF transmitter is significantly more complex than what I proposed. It includes a tank circuit with a carefully selected Q (too low and we get distortion from class B or C amplifiers since it provides the freewheeling when active devices are off), too high and peak currents are excessive.

Then there is the matching network. This is required since the RF amp itself wants to see a specific load, usually lower than 50 ohms for transistor amplifiers (often higher than 50 ohms for tube amps).

I *believe* that this same matching network (and it is more than two components) is designed to make the RF source look like 50 ohms for a reverse flowing signal.

I will have to try to find (or work out) a better example...

 

[Baluncore]

The transmitter output stage has an impedance transformer called a tank circuit. To cancel the inductive component of the wound transformer it has a parallel capacitor, adjustable for different frequencies.

The load line or V:I ratio of the amplifier's active element is transformed to the impedance of the transmission line by the tank.

A reflected signal returning from the output t'line is reverse transformed by the tank from a V:I ratio of 50 ohm in the t'line to the V:I ratio of the amplifier's active element, and so is not reflected.

 

[sophiecentaur]

A few basic points would not go amiss here.
Reactive components do not dissipate any power. Their effect may be to change the maximum volts of current for a given power, which may, by implication, be problematic for the equipment and cause further internal dissipation (hence the Power Factor correction which is done in AC power systems). This is not relevant to the basic issue. Specific details of tank circuits and the like, just cloud things.

Anyone who cares to look at this link can see what my point has been. It's only one of many sources which back me up in this. A transmitter (any electrical / mechanical power source) that is designed with efficiency in mind will not be matched to its load. Yes, you will get maximum power into your load for a given emf but that will demand twice as much input power (DC power supply, in the end) and it will require you to dissipate half of this power in the amplifying device.

The confusion that other contributors are showing is that they are making assumption about what their practical results are telling them. What actual evidence do they have that the output impedance of their transmitter is '50 - or whatever'? How have they measured it? Efficient transmitters tend to be non-linear so it is hard to determine internal resistance. You can't just measure Open Circuit and then put 50Ω across the output to see how may volts are lost - as you can with the mains or a car battery. I guess you could see the differential effect of 50Ω and 51Ω loads on the output volts. Have any of you done this - or equivalent?

The idea of a conjugate match is not of much significance because the feeder is assumed to be providing the transmitter with a 'good enough '50Ω' at its output.

 

[sophiecentaur]

Here are a few links on the subject of matching. Read what they are actually saying about the significance of transmitter / load matching and efficiency. They all agree with what I have been saying.
http://www.jaycar.co.uk/images_uploaded/impmatch.pdf
http://users.tpg.com.au/users/ldbutler/OutputLoadZ.htm
http://urgentcomm.com/test-amp-measurement-mag/maximum-power-transfer