Signal to Noise Ratio ? - Page 2 - UK Vintage Radio Repair and Restoration Discussion Forum

ppppenguin · 7th Jul 2006, 7:50 am

The original use of dB was (I think) in telephone communications in circuits with matched impedance. When impedances are matched the power gain and voltage gain are the same number of dB. As Sam correctly said, power gain is proportional to square of voltage gain but this only makes sense when the impedance is the same at input and output.

If you have an amp with high Z input and low Z output and it feeds a high Z load the voltage gain may be considerable but the power gain irrelevant and probably negligible.

GMB's amp with low voltage gain and high power gain clearly has a lot of current gain. A typical audio power amp would be rather like this, delivering a lot of current into a low impedance load. The voltage gain is 2 (or 6dB). The power gain is lots.

Dickie · 7th Jul 2006, 8:01 am

A voltage gain of 2 and a power gain of 1,000,000 implies a gain of 60dB.

It alos implies a substantial change in impedance from input to output (I can't be bothered to work out the maths) which is where the confusion arises. To describe a voltage ratio in terms of dBs the impedances must be same.

A dB is a ratio of one power to another. It is simply convenient to measure power by using a voltmeter, and assume that the impedance doesn't change.

ppppenguin · 7th Jul 2006, 8:21 am

Quote:

Originally Posted by Dickie

A dB is a ratio of one power to another. It is simply convenient to measure power by using a voltmeter, and assume that the impedance doesn't change.

This is strictly correct. For better or worse we now often use dB simply to indicate a voltage ratio regardless of impedance. This can cause confusion but it's far too late to enforce the original meaning.

GMB · 7th Jul 2006, 9:43 am

Yes, I think the only valid answers are 60 or "undefined" and the latter might be more appropriate. The RF engineers have an easy time because of everything often being 50 ohm.

I vomit over the 20log(V/V) usage because it contains an assumption and is expressing the answer in power ratio units for something that was never measured as actual power and that is the problem - you don't always know when it has been abused like that. For an s/n you can see that in a real situation a db(V/V) and true dB could diverge considerably.

I find it necessary to ask people what they mean by dB because it is so often either abused or just miscalculated. (When dBs are applied to things that don't have a simple connection with power it gets even worse!)

Skywave · 7th Jul 2006, 11:21 pm

Hi to you all!

I've had a few urgent matters to attend to, which I why my "replying contribution" to this saga is late!

In the meantime, I see that GMB's Q. on "amplifier gain" has been answered by others, and, in effect, have given my reply! Thanks to you all for saving me the trouble!

However, to get back on Thread - since we are now well OT - all this "confusion over dBs" started with the following quote (measurement of S/N):

"The normal method is to feed in a modulated signal (usually 1KHz 30% AM), and measure the output level, then switch off the modulation and measure the output level. The two values give the signal to noise, usually expressed in dB." (Obviously, the author means the log RATIO of the two values . . . But let's not split hairs, gentlemen!)

Which, in turn, brought the response from GMB about "voltage dBs or power dBs".

etc . . . . . . .

My understanding is:

With reference to the original quote (as above), the load impedance stays constant for the two measurement conditions. So, if the two levels are measured in voltage, then the "dBs" will be calculated using the "20 log" rule. If, however, the two levels are measured in power, then the "10 log" rule applies.
You get the same "no. of dBs" in either case.
Hence my comment "dBs are dBs" - which I stated with respect to the original quote.

However, when "dBs" are used for referring to gain / loss - such as in amplifiers - then things get a little muddy - as other contributors have commented. Unless the source and load impedances are the same, quoting "voltage dBs" is a meaningless phrase.

For myself, I have had my original Q. (which opened this Thread) satisfactorily answered.

Therefore, I believe that this Thread has served its purpose and should now be closed.

My sincere thanks go to all those who made helpful & constructive contributions.

Al / G8DLH

7th Jul 2006, 7:50 am	#21
ppppenguin Retired Dormant Member Join Date: Dec 2003 Location: North London, UK. Posts: 6,168	Re: Signal to Noise Ratio ? The original use of dB was (I think) in telephone communications in circuits with matched impedance. When impedances are matched the power gain and voltage gain are the same number of dB. As Sam correctly said, power gain is proportional to square of voltage gain but this only makes sense when the impedance is the same at input and output. If you have an amp with high Z input and low Z output and it feeds a high Z load the voltage gain may be considerable but the power gain irrelevant and probably negligible. GMB's amp with low voltage gain and high power gain clearly has a lot of current gain. A typical audio power amp would be rather like this, delivering a lot of current into a low impedance load. The voltage gain is 2 (or 6dB). The power gain is lots.

7th Jul 2006, 8:01 am	#22
Dickie Octode Join Date: Sep 2004 Location: St. Albans, Hertfordshire, UK. Posts: 1,478	Re: Signal to Noise Ratio ? A voltage gain of 2 and a power gain of 1,000,000 implies a gain of 60dB. It alos implies a substantial change in impedance from input to output (I can't be bothered to work out the maths) which is where the confusion arises. To describe a voltage ratio in terms of dBs the impedances must be same. A dB is a ratio of one power to another. It is simply convenient to measure power by using a voltmeter, and assume that the impedance doesn't change. __________________ Regards, Richard, BVWS member

7th Jul 2006, 9:43 am	#24
GMB Dekatron Join Date: Aug 2003 Location: near Reading (and sometimes Torquay) Posts: 3,099	Re: Signal to Noise Ratio ? Yes, I think the only valid answers are 60 or "undefined" and the latter might be more appropriate. The RF engineers have an easy time because of everything often being 50 ohm. I vomit over the 20log(V/V) usage because it contains an assumption and is expressing the answer in power ratio units for something that was never measured as actual power and that is the problem - you don't always know when it has been abused like that. For an s/n you can see that in a real situation a db(V/V) and true dB could diverge considerably. I find it necessary to ask people what they mean by dB because it is so often either abused or just miscalculated. (When dBs are applied to things that don't have a simple connection with power it gets even worse!)

7th Jul 2006, 11:21 pm	#25
Skywave Rest in Peace Join Date: Jun 2006 Location: Chard, South Somerset, UK. Posts: 7,457	Re: Signal to Noise Ratio ? Hi to you all! I've had a few urgent matters to attend to, which I why my "replying contribution" to this saga is late! In the meantime, I see that GMB's Q. on "amplifier gain" has been answered by others, and, in effect, have given my reply! Thanks to you all for saving me the trouble! However, to get back on Thread - since we are now well OT - all this "confusion over dBs" started with the following quote (measurement of S/N): "The normal method is to feed in a modulated signal (usually 1KHz 30% AM), and measure the output level, then switch off the modulation and measure the output level. The two values give the signal to noise, usually expressed in dB." (Obviously, the author means the log RATIO of the two values . . . But let's not split hairs, gentlemen!) Which, in turn, brought the response from GMB about "voltage dBs or power dBs". etc . . . . . . . My understanding is: With reference to the original quote (as above), the load impedance stays constant for the two measurement conditions. So, if the two levels are measured in voltage, then the "dBs" will be calculated using the "20 log" rule. If, however, the two levels are measured in power, then the "10 log" rule applies. You get the same "no. of dBs" in either case. Hence my comment "dBs are dBs" - which I stated with respect to the original quote. However, when "dBs" are used for referring to gain / loss - such as in amplifiers - then things get a little muddy - as other contributors have commented. Unless the source and load impedances are the same, quoting "voltage dBs" is a meaningless phrase. For myself, I have had my original Q. (which opened this Thread) satisfactorily answered. Therefore, I believe that this Thread has served its purpose and should now be closed. My sincere thanks go to all those who made helpful & constructive contributions. Al / G8DLH