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dBW in S-unit and power strenght
Hi,
A question about dBW and power at receive. Among scales used in power measurement, there is the signal strength or noise level estimation, also known as the "dB below W" (dBW or SDBW). Its relation is dBW = 10 Log P, where P is the power expressed in watt : That means that 100 W is 20 dBW into 50 ohms. But usually, in propagation program (VOACAP, etc), the power strength at receive expressed in dBW is far below such values. I read somewhere the next equivalences : - (minus) 93 ~ S9+10, -103 = S9, -127 = S5, -151 = S1. With -93 dBW for a S9+10 signal at receive, that 'd mean that the power 'd be only P(W) = 10^ (dBW/10) = 0.7 watt ? IMHO this power is much to low... What is wrong in this relations (or in my interpretation) ? Thanks in advance Thierry, ON4SKY http://www.astrosurf.com/lombry |
But usually, in propagation program (VOACAP, etc), the power strength at
receive expressed in dBW is far below such values. I read somewhere the next equivalences : - (minus) 93 ~ S9+10, -103 = S9, -127 = S5, -151 = S1. One S unit referencing 6 dB. Some receivers has meter in dB over 1 micro V. With -93 dBW for a S9+10 signal at receive, that 'd mean that the power 'd be only P(W) = 10^ (dBW/10) = 0.7 watt ? "W" mean level in Watt. IMHO this power is much to low... What is wrong in this relations (or in my interpretation) ? P log * 10 = dbW Reverse calculation: dbW / 10 10^x = P -93 / 10 10^x = 5,011e-10 = 0,0000000005011 Watt Regards, Ralf -- Vy 73 es 55 de Ralf, DL2MRB E-Mail: Board: www.hamradioboard.de |
"Ralf Ballis - DL2MRB" wrote in message ... But usually, in propagation program (VOACAP, etc), the power strength at receive expressed in dBW is far below such values. I read somewhere the next equivalences : - (minus) 93 ~ S9+10, -103 = S9, -127 = S5, -151 = S1. One S unit referencing 6 dB. Some receivers has meter in dB over 1 micro V. With -93 dBW for a S9+10 signal at receive, that 'd mean that the power 'd be only P(W) = 10^ (dBW/10) = 0.7 watt ? "W" mean level in Watt. IMHO this power is much to low... What is wrong in this relations (or in my interpretation) ? P log * 10 = dbW Does not exist ! But well dBW = 10 Log P, this is not the same excepting reading from right to left, Hi! Your explanation is not one. refer my link for a correct explanations. Anyway, how do you calculate the dBW on 20m with such a formula. Someone calculated dBW on 10m, so we must be able to do the same at 20m... But how ? Thierry Reverse calculation: dbW / 10 10^x = P -93 / 10 10^x = 5,011e-10 = 0,0000000005011 Watt Regards, Ralf -- Vy 73 es 55 de Ralf, DL2MRB E-Mail: Board: www.hamradioboard.de |
On Wed, 15 Sep 2004 12:56:16 +0200, "Thierry" to answer direct see
http://www.astrosurf.com/lombry/post.htm wrote: But usually, in propagation program (VOACAP, etc), the power strength at receive expressed in dBW is far below such values. I read somewhere the next equivalences : - (minus) 93 ~ S9+10, -103 = S9, -127 = S5, -151 = S1. With -93 dBW for a S9+10 signal at receive, that 'd mean that the power 'd be only P(W) = 10^ (dBW/10) = 0.7 watt ? IMHO this power is much to low... What is wrong in this relations (or in my interpretation) ? Hi Thierry, S-9 is the reference point for a 50µV signal into the receiver's 50 Ohm input. This is -73dBM (level below one milliwatt). Translating to levels below one Watt is performed by subtracting another 30 (the milli) dB - hence -103dBW. Power is a curious thing. Transmitter efficiency as measured by the target population that receives the signal is pitiful. If every person on the planet had a radio that picked up your 100W signal with S-9 level, you would be pumping 99.9% of your 100W into the void without anyone missing it anywhere. Efficiency 0.1% Given that you are NOT heard by everyone (they don't care to listen), then transmitter efficiency plunges at least 7 or 8 more orders of magnitude. Efficiency 0.00000001% 73's Richard Clark, KB7QHC |
"Thierry" to answer direct see http://www.astrosurf.com/lombry/post.htm
wrote: Does not exist ! But well dBW = 10 Log P, this is not the same excepting reading from right to left, Hi! OK, hi... Your explanation is not one. refer my link for a correct explanations. Taka a look at: http://hyperphysics.phy-astr.gsu.edu.../sound/db.html Anyway, how do you calculate the dBW on 20m with such a formula. Someone calculated dBW on 10m, so we must be able to do the same at 20m... This calculations aren't depend on frequency or wavelength. Regards, Ralf -- Vy 73 es 55 de Ralf, DL2MRB E-Mail: Board: www.hamradioboard.de |
Thierry wrote:
Hi, A question about dBW and power at receive. Among scales used in power measurement, there is the signal strength or noise level estimation, also known as the "dB below W" (dBW or SDBW). Its relation is dBW = 10 Log P, where P is the power expressed in watt : That means that 100 W is 20 dBW into 50 ohms. But usually, in propagation program (VOACAP, etc), the power strength at receive expressed in dBW is far below such values. I read somewhere the next equivalences : - (minus) 93 ~ S9+10, -103 = S9, -127 = S5, -151 = S1. With -93 dBW for a S9+10 signal at receive, that 'd mean that the power 'd be only P(W) = 10^ (dBW/10) = 0.7 watt ? IMHO this power is much to low... What is wrong in this relations (or in my interpretation) ? Your calculation of 0.7 W is wrong by many orders of magnitude. dBW = 10 Log(P) -93 = 10 Log(P) -9.3 = Log(P) Alog(-9.3) = P ALog (-9.3) = 5.012 * 10^-10 W (5.0118723362727228500155418688495e-10 by win2k calculator) So the problem is much, much worse than you thought. Try this scenario:- You are in Belgium, I'm in the west of the U.K. about 400 km away. You transmit using 100W and I get an S9+10 signal. All seems reasonable? Say this is on 80 metres so both our antennas are omnidirectional. My antenna will only receive power proportional to the solid (stereo) angle it subtends at your QTH. So the power of 5e-10 W seems about right to me. Just think how many other 80 metre dipoles would fit in around that 400 km circle. Then all the ones that would fit onto a 400km radius hemisphere. They would all get a tiny share of your 100 W!!! Receivers are very, very sensitive or radio wouldn't work at all. A signal of 1 microvolt into 50 ohms is the same as a power of 2 x 10^-14 W !!! (P = V^2/R) Hope this helps. vy 73 Andy, M1EBV |
"Thierry" to answer direct see http://www.astrosurf.com/lombry/post.htm wrote in message ... Hi, A question about dBW and power at receive. Among scales used in power measurement, there is the signal strength or noise level estimation, also known as the "dB below W" (dBW or SDBW). Its relation is dBW = 10 Log P, where P is the power expressed in watt : That means that 100 W is 20 dBW into 50 ohms. But usually, in propagation program (VOACAP, etc), the power strength at receive expressed in dBW is far below such values. I read somewhere the next equivalences : - (minus) 93 ~ S9+10, -103 = S9, -127 = S5, -151 = S1. With -93 dBW for a S9+10 signal at receive, that 'd mean that the power 'd be only P(W) = 10^ (dBW/10) = 0.7 watt ? IMHO this power is much to low... What is wrong in this relations (or in my interpretation) ? Thanks in advance Thierry, ON4SKY http://www.astrosurf.com/lombry As others pointed out, your calculation is wrong and that .7 W is way too large for received power. We receive powers in the -123 dBm (-153 dBW) range easily and that's ok However, it is not clear from your post just what you want to do/ask/learn. What is your number too low for? What are you wanting? -- Steve N, K,9;d, c. i My email has no u's. |
I 've got te answer
I wonder first if dBW depended or not on the QRG below 30 MHz. It doesn't (excepting that over 30 MHz there is less noise temperature and often transverter in front of TX showing a 20 dB gain, and this figure is thus till less). In addition dBm and co are defined by IARU recommendations where I found all equivalence dBm-S-point-dBW. Thanks for your answers but I never imagine that at receive the true power was so low. I add well a signal report and equivalent in dB or dBW but I never really make this relation. Thierry, ON4SKY "Steve Nosko" wrote in message ... "Thierry" to answer direct see http://www.astrosurf.com/lombry/post.htm wrote in message ... Hi, A question about dBW and power at receive. Among scales used in power measurement, there is the signal strength or noise level estimation, also known as the "dB below W" (dBW or SDBW). Its relation is dBW = 10 Log P, where P is the power expressed in watt : That means that 100 W is 20 dBW into 50 ohms. But usually, in propagation program (VOACAP, etc), the power strength at receive expressed in dBW is far below such values. I read somewhere the next equivalences : - (minus) 93 ~ S9+10, -103 = S9, -127 = S5, -151 = S1. With -93 dBW for a S9+10 signal at receive, that 'd mean that the power 'd be only P(W) = 10^ (dBW/10) = 0.7 watt ? IMHO this power is much to low... What is wrong in this relations (or in my interpretation) ? Thanks in advance Thierry, ON4SKY http://www.astrosurf.com/lombry As others pointed out, your calculation is wrong and that .7 W is way too large for received power. We receive powers in the -123 dBm (-153 dBW) range easily and that's ok However, it is not clear from your post just what you want to do/ask/learn. What is your number too low for? What are you wanting? -- Steve N, K,9;d, c. i My email has no u's. |
"Richard Clark" wrote in message ... On Wed, 15 Sep 2004 12:56:16 +0200, "Thierry" to answer direct see http://www.astrosurf.com/lombry/post.htm wrote: ... Hi Thierry, S-9 is the reference point for a 50µV signal into the receiver's 50 Ohm input. This is -73dBM (level below one milliwatt). Translating to levels below one Watt is performed by subtracting another 30 (the milli) dB - hence -103dBW. Power is a curious thing. Transmitter efficiency as measured by the target population that receives the signal is pitiful. If every person on the planet had a radio that picked up your 100W signal with S-9 level, you would be pumping 99.9% of your 100W into the void without anyone missing it anywhere. Efficiency 0.1% Given that you are NOT heard by everyone (they don't care to listen), then transmitter efficiency plunges at least 7 or 8 more orders of magnitude. Efficiency 0.00000001% You really put the finger on something that I ignored during 25 years of ham radio, hi !... I know the correspondence between power expressed in dB or dBm, dBW but I never see that under that angle... Thanks Thierry, ON4SKY 73's Richard Clark, KB7QHC |
Thierry, ON4SKY wrote:
"With -93 dBW for a S9+10 signal at receive, that`d mean that the popwer`d be only P (W) = 10^ (dBW/10) = 0.7 watt? I`m familiar with received carrier power expressed in dBm. -93 dBW is a number, 30 dB less if expressed in dBm, or -123. This is a strong signal for a narrow-band receiver. Across 50 ohms, 1 watt is sq rt PR, or 7.07 volts. 93 dB is about 45,000 times, as a voltage ratio, so a received voltagw of -93 dBw is about 0.00016 volts, or 0.16 millivolts, or 160 microvolts. S9+10dB is a strong signal. So is 160 microvolts. I`ve read that each "S-unit" is 6 dB. S-1 would be 18 dB less than S9+10 db, and that`s a voltage ratio of 7.94. S-1 would be about 20 microvolts, and well above the threshold of most receivers. Best regards, Richard Harrison, KB5WZI |
At HF it's all very simple.
To summarise - The S-meter is essentially a 50-ohm power or wattmeter. It is correctly calibrated with a 50-ohm signal generator with its open-circuit volts set to 100 microvolts. The S-meter should then read S9. Therefore, at the standard S9, the input to the receiver is 50 microvolts across 50 ohms, corresponding to a power input of 50 picowatts. The standard S-unit = 6 dB. So for each change of one S-unit the input voltage halves (or doubles). At S-zero the input voltage is 50/512 = 0.1 microvolts, which is roughly equal to the the internal noise level of a good receiver with a bandwidth of a few KHz.. Theoretically this is about the noise level you should get when the antenna is disconnected. At S9 + 40dB the input voltage is 50*100 microvolts = 5 millivolts. Some S-meters may indicate as high as S9 + 60dB. The input voltage is then 50 millivolts which is about the overload point of a very good receiver. So the whole scale is calibrated logarithmically, with S9 being about half-way along it, and with 54 dB below S9 and 40 or 60dB above S9. The input voltage range is from from 0.1 microvolts to 5 or 50 millivolts. ============================== However, all meters have indicating errors. The problem arises because of the difficulties and great expense in designing and manufacturing receivers with an agc meter system which can accommodate a signal level range of 54 + 60 = 114 dB. Economics invariably rules the roost. (It helps to have very high receiver gain and attenuators at or near the receiver input.) Fortunately modern receivers all tend to have errors of the same sort and sign. So amateurs using different manufacturer's receivers will exhange very similar signal strength reports. These errors congregate at the very small signal end of the range. Meter calibration begins to go wrong at around or below S4 or S5. Meter readings of, say, S1 or S2 may actually be appropriate to power levels of S3 or S4. A meter reading of S-zero may be appropriate to a power level of S2 or S3. That is, at very small signal levels the S-meter underestimates signal power level. But it's not of great consequence. At HF, to which the foregoing applies, signals are usually below the noise and QRM levels anyway. --- Reg. |
The distortion in S-meter scale-shape at very small signal levels, below
about S5, could be corrected very easily just by changing scale markings. The meter scale would then end at about S2 instead of S-zero. And the meter would indicate correctly as a wattmeter over the whole length of its scale. But S-meter manufacturers are reluctant to do this. They consider it would spoil the linear appearance of the scale at the low end and so lose customers. How foolish they are! Keen radio amateurs can change their S-meters into accurate instruments over the whole scale just by scraping off existing scale markings and re-calibrating with a fine-nibbed pen and black indian ink. But on second thoughts, to save yourself the trouble, when using the meter it's even easier just to visualise and remember where S3 should be marked on the meter scale. --- Reg, G4FGQ |
Nice tutorial, Reg I printed it out for ref.
....however. The 2M rigs I've measured (3) are WAY off. The +20 +40 and +6o are usually also around 6dB plus or minus 3 or 4 dB. I measured my TS-830s and it was not very close, and my 706, but don't recall the results and haven't done the TS 2000 yet. 73, Steve K9DCI "Reg Edwards" wrote in message ... At HF it's all very simple. To summarise - The S-meter is essentially a 50-ohm power or wattmeter. It is correctly calibrated with a 50-ohm signal generator with its open-circuit volts set to 100 microvolts. The S-meter should then read S9. Therefore, at the standard S9, the input to the receiver is 50 microvolts across 50 ohms, corresponding to a power input of 50 picowatts. The standard S-unit = 6 dB. So for each change of one S-unit the input voltage halves (or doubles). At S-zero the input voltage is 50/512 = 0.1 microvolts, which is roughly equal to the the internal noise level of a good receiver with a bandwidth of a few KHz.. Theoretically this is about the noise level you should get when the antenna is disconnected. At S9 + 40dB the input voltage is 50*100 microvolts = 5 millivolts. Some S-meters may indicate as high as S9 + 60dB. The input voltage is then 50 millivolts which is about the overload point of a very good receiver. So the whole scale is calibrated logarithmically, with S9 being about half-way along it, and with 54 dB below S9 and 40 or 60dB above S9. The input voltage range is from from 0.1 microvolts to 5 or 50 millivolts. ============================== However, all meters have indicating errors. The problem arises because of the difficulties and great expense in designing and manufacturing receivers with an agc meter system which can accommodate a signal level range of 54 + 60 = 114 dB. Economics invariably rules the roost. (It helps to have very high receiver gain and attenuators at or near the receiver input.) Fortunately modern receivers all tend to have errors of the same sort and sign. So amateurs using different manufacturer's receivers will exhange very similar signal strength reports. These errors congregate at the very small signal end of the range. Meter calibration begins to go wrong at around or below S4 or S5. Meter readings of, say, S1 or S2 may actually be appropriate to power levels of S3 or S4. A meter reading of S-zero may be appropriate to a power level of S2 or S3. That is, at very small signal levels the S-meter underestimates signal power level. But it's not of great consequence. At HF, to which the foregoing applies, signals are usually below the noise and QRM levels anyway. --- Reg. |
"Steve Nosko" wrote in message ... Nice tutorial, Reg I printed it out for ref. ...however. The 2M rigs I've measured (3) are WAY off. The +20 +40 and +6o are usually also around 6dB plus or minus 3 or 4 dB. I measured my TS-830s and it was not very close, and my 706, but don't recall the results and haven't done the TS 2000 yet. 73, Steve K9DCI "Reg Edwards" wrote in message ... At HF it's all very simple. To summarise - The S-meter is essentially a 50-ohm power or wattmeter. It is correctly calibrated with a 50-ohm signal generator with its open-circuit volts set to 100 microvolts. The S-meter should then read S9. While I will not debate or dispute the origional S-meter meanings, I would have to say there is almost no rig that follows that calibration. I have measured a few low band rigs and several vhf/uhf rigs. The low band units seem to be somewhat more 'calibrated' than the VHF and above rigs but most are usually way off. From my observations and the write ups in QST the S-meter and the 50 uV and all the 6 db per S-unit should not even be spoken of . We need to kill the idea of the s-unit being 6 db or anything at all. They are just numbers on an uncalibrated meter on most receivers. You might just as well mark the meter from 0 to 100 and go by whatever number the meter is showing. The meters are not linear, log, or any combination that can be depended on from one brand to another brand. Even changing bands on the same receiver will usually change the 'calibration' of the s-meter. |
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