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I'm reading a book about Ham radio questions and it has this line:

A two-time increase in power results in a change of 3dB higher

And is followed by the these examples:

1- The power output from a transmitter increases from 1 watt to 2 watts. This is a dB increase of 3.3

2- The power output from a transmitter increases form 5 watts to 50 watts by a linear amplifier. The power gain would be 10 dB.

I don't get how these are calculated? for (1) if the power goes from 1 to 2, then there is a two-time increase in power? which means the dB increase should be 3 dB? where does the extra .3 come from?

for (2) 5 to 50 means a 10 time increase? 10/2 = 5 (number of 3 db increases), 5 * 3 = 15 dB, how is it 10?

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where does the extra .3 come from?

From a typo, probably. To be more exact, a ratio of 2:1 is very close to 3.0103 dB, but 3 is close enough for almost all practical purposes. 3.3 is incorrect.

10/2 = 5 (number of 3 db increases),

Nope. 5 3dB increases would mean 5 doublings, which is $2^5$, or $2 \times 2 \times 2 \times 2 \times 2$, or 32, not 2*5. In fact, 15dB corresponds to a ratio of about 31.62:1, which is reasonably close to 32.

A 10:1 ratio is 10dB, exactly, by definition. Everything else follows from that. You can always use the formula $d = 10 \log_{10} R$ to find out the number of dB corresponding to a ratio, or $R = 10^{\frac{d}{10}}$ to find the ratio corresponding to a number of dB.

The fact of 2:1 being very close to 3dB comes from the fact that $2^{10} = 1024$ is very close to $10^3 = 1000$. If $2^{10} \approx 10^3$, then $2 \approx 10^{\frac{3}{10}}$.

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  • $\begingroup$ Thanks for sharing the formula. $\endgroup$
    – Dan
    Commented Jul 2, 2023 at 4:04

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