24-08-2012, 12:16 PM
Hi Alan,
Yes. Completely understood
Yes, I agree...
No.
You don't need me to tell you that you've made a potential divider, with R on top, and Zin below. When these two resistances are equal, Vout is half Vin. Which is 6dB down in voltage terms. Or, if you prefer, quarter-power.
Let's put some numbers in - nice simple numbers:
Assume the output from the source was 8V RMS. This is a nice number, because it gives 8 watts in an 8 ohm system...
Let's assume that you've put an 8 ohm resistor in series with the 8 ohm Zin. Now, the voltage across the input terminals of your power meter will be 4V RMS. What is 4 squared over 8? Drum-role please. 2 watts!. That's quarter-power.
I've just had a quick scan of the manual, and noted the equation at the bottom of page 10:
P = (V squared) / (4 * Re)
Substituting the values I gave above, (8 squared) divided by (4 times 8), does indeed give 2 watts. Funnily enough, that equation could be written (V squared over R) divided by 4
I think that if you'd observed the voltage as well as the power, you would have tumbled to this much sooner
Being a broadcast engineer that rarely gets involved with PA or RF, I do decibels in voltage terms. I find it much easier. Loudspeakers are rated in terms of watts, not dBW. But the end result is always the same...
In the day job, I teach fundamental audio. Which involves a lot of work with decibels, and I see folk really struggle with them. We have tutorial questions, practical exercises and exam questions that ask relatively searching questions about decibels and their application, and it's something I've had to practice a lot. On the face of it, it's simple. It's like percentages, but you just have to remember to press the Log button on your calculator; if you understand percentages, you'll be fine. But I quickly learnt that most people don't understand percentages!
Part of the problem is the fact there are two equations:
dB = 10 Log (P2 / P1)
dB = 20 Log (V2 / V1)
Of course, it doesn't take much to prove that these equations are saying the same thing:
dB = 10 Log ( ( (V2 squared) / R2) / ( (V1 squared) / R1) )
As the Rs are generally the same - e.g. 50 Ohms in the RF world, they cancel:
dB = 10 Log ( (V2 squared) / (V1 squared) )
dB = 10 Log ( (V2 / V1) squared)
dB = 20 Log (V2 / V1)
Of course, I know you'll know this, but others might be reading...
The base-10 log of 2 is very close to 0.3. The log of 1/2 is -0.3. From these, we can see that a doubling or halving of power is +3dB or -3dB. But a doubling or halving of voltage is +6dB or -6dB. To double the power, you would have to increase the voltage by 1.414 (the square-root of 2).
Back to your problem, to get a half-power reading, you need to insert a value or R that causes a voltage loss of x0.707, not x0.5. I reckon that you need something like 3.3 ohms to give half-power...
That's not quite the 5 ohms you've got, but I wonder how you were monitoring the voltage as you changed the resistance? Were you using a voltmeter or a 'scope? I wonder if the source was distorting as it supplied the load current? Just because it has a particular source impedance, it doesn't mean it can necessarily supply the necessary current to drive a matching load to full output. After all, audio amplifiers have output impedances measured in fractions of an ohm... All that said, the power levels were pretty low - so perhaps there was some noise present? Which gets me back to my original contribution to this thread: I'd want to use a 'scope to see what is going on.
The scope for inaccuracy is almost limitless, but I would be taking voltage measurements with a trusted digital meter with an RMS converter, and checking the Marconi readings against the calculated results. After all, the Marconi is a volt meter with a low input impedance!
I hope this helps to clear things up.
Mark
(24-08-2012, 10:10 AM)AlanBeckett Wrote: No Mark, I don't think so.
The object of the exercise was to check/test the calibration of the Marconi. and it's Input Impedance.
Yes. Completely understood
Quote:I set the Output of the J-2 to give 9dB on the Marconi scale, then inserted Resistance to reduce that reading to 6dB, ie half Power, while keeping the source voltage constant. That's half Power in my book, and that's what the Marconi claims to measure.
Yes, I agree...
Quote:The inserted Resistance should then equal the (Real Part) of the Marconi Input Impedance.
No.
You don't need me to tell you that you've made a potential divider, with R on top, and Zin below. When these two resistances are equal, Vout is half Vin. Which is 6dB down in voltage terms. Or, if you prefer, quarter-power.
Let's put some numbers in - nice simple numbers:
Assume the output from the source was 8V RMS. This is a nice number, because it gives 8 watts in an 8 ohm system...
Let's assume that you've put an 8 ohm resistor in series with the 8 ohm Zin. Now, the voltage across the input terminals of your power meter will be 4V RMS. What is 4 squared over 8? Drum-role please. 2 watts!. That's quarter-power.
I've just had a quick scan of the manual, and noted the equation at the bottom of page 10:
P = (V squared) / (4 * Re)
Substituting the values I gave above, (8 squared) divided by (4 times 8), does indeed give 2 watts. Funnily enough, that equation could be written (V squared over R) divided by 4
Quote:I should perhaps of measured the Input Voltage to the Marconi as a check, but I didn't. However, as I said, I was checking the Marconi, not my 'scope or AVO, which I trust.
I think that if you'd observed the voltage as well as the power, you would have tumbled to this much sooner
Quote:In what way does your methodology differ?
Being a broadcast engineer that rarely gets involved with PA or RF, I do decibels in voltage terms. I find it much easier. Loudspeakers are rated in terms of watts, not dBW. But the end result is always the same...
In the day job, I teach fundamental audio. Which involves a lot of work with decibels, and I see folk really struggle with them. We have tutorial questions, practical exercises and exam questions that ask relatively searching questions about decibels and their application, and it's something I've had to practice a lot. On the face of it, it's simple. It's like percentages, but you just have to remember to press the Log button on your calculator; if you understand percentages, you'll be fine. But I quickly learnt that most people don't understand percentages!
Part of the problem is the fact there are two equations:
dB = 10 Log (P2 / P1)
dB = 20 Log (V2 / V1)
Of course, it doesn't take much to prove that these equations are saying the same thing:
dB = 10 Log ( ( (V2 squared) / R2) / ( (V1 squared) / R1) )
As the Rs are generally the same - e.g. 50 Ohms in the RF world, they cancel:
dB = 10 Log ( (V2 squared) / (V1 squared) )
dB = 10 Log ( (V2 / V1) squared)
dB = 20 Log (V2 / V1)
Of course, I know you'll know this, but others might be reading...
The base-10 log of 2 is very close to 0.3. The log of 1/2 is -0.3. From these, we can see that a doubling or halving of power is +3dB or -3dB. But a doubling or halving of voltage is +6dB or -6dB. To double the power, you would have to increase the voltage by 1.414 (the square-root of 2).
Back to your problem, to get a half-power reading, you need to insert a value or R that causes a voltage loss of x0.707, not x0.5. I reckon that you need something like 3.3 ohms to give half-power...
That's not quite the 5 ohms you've got, but I wonder how you were monitoring the voltage as you changed the resistance? Were you using a voltmeter or a 'scope? I wonder if the source was distorting as it supplied the load current? Just because it has a particular source impedance, it doesn't mean it can necessarily supply the necessary current to drive a matching load to full output. After all, audio amplifiers have output impedances measured in fractions of an ohm... All that said, the power levels were pretty low - so perhaps there was some noise present? Which gets me back to my original contribution to this thread: I'd want to use a 'scope to see what is going on.
The scope for inaccuracy is almost limitless, but I would be taking voltage measurements with a trusted digital meter with an RMS converter, and checking the Marconi readings against the calculated results. After all, the Marconi is a volt meter with a low input impedance!
I hope this helps to clear things up.
Mark
