How can I calculate how much stronger a 16 dB gain is compared to 9 dB?

Question

I know that the decibel scale is logarithmic, so that a gain of 10dB would mean a 10x increase in power output. So 16dB would be approx 40x (?) and 9dB would be close to ten, perhaps 9x. So a gut feeling would say that a 16dB antenna would give me about 4-5x the power output of a 9dB antenna.

But how would I calculate this exactly?

Throwing around numbers like log 16/log 9 or log 16 - log 9 does not give me numbers anywhere near the right ballpark, it seems, so I could use a nudge in the right direction.

Answer 1

You shouldn't be taking the logarithm of the dB values. The dB values are already logarithms, which is to say they are exponents (realizing that logarithms are exponents made logarithm arithmetic in general so much easier for me to remember and understand; $\log_x(ab) = \log_x(a) + \log_x(b)$ is not just derivable from $x^{m+n} = x^m\cdot x^n$, it's exactly the same rule written in two different ways, with $a = x^m, b = x^n$).

Thus we need to raise something to the powers of $9$ and $16$, and then we can take the ratio of the two. Alternatively, once we have the correct base, we can just raise that to the power of $16-9 = 7$ to get the same answer.

We need to find that base, and to help us we have a hint: An increase of $10$ in dB value means a ten-fold increase in power. So, if we call our base $x$, then for any dB value $a$, we have $$ x^{a+10} = 10x^a\\ x^{a}\cdot x^{10} = 10x^a\\ x^{10} = 10\\ x = \sqrt[10]{10} $$ So there you have it. The ratio of power output between the two is $$ \frac{\sqrt[10]{10}^{16}}{\sqrt[10]{10}^{9}}\approx 5 $$ which is entirely within your ballpark.

Of course, actual dB values are calculated using a lot of constants here and there. I don't remember the exact formula, but I believe it could be a nice and easy exercise to show that they all cancel out in this problem and don't change the final result.

Answer 2

It is worth remembering that a decibel is a deci-bel, that is a $\frac{1}{10}$ of a Bel. Deci is a prefix alongside centi, mili, micro, etc but not used so often.

A Bel is a logarithmic measure of the ratio of two quantities. Most commonly used with power but not necessarily only for power. A ratio expressed in bels is the logarithm to base 10 of the ratio. So, a ratio of 1Bel is $10$ times bigger, 2Bel is $10^2 = 100$ times bigger, etc.

This unit is often inconveniently big and the decibel is much more popular. So, $10dB = 1B$. This is why $10dB$ represents $10 \times$ bigger. The two $10$s is sort of a coincidence and confusing because $20dB$ does not represent $20 \times$ bigger. $20dB = 2B$ so it represents $10^2 = 100$ times bigger.

It is useful to remember that a ratio of $2$ in decibels is very close to $3dB$. This is because $log_{10}(2) = .30103$. This is very useful for quick estimates. Hence $6dB$ is $4 \times$ bigger and $9dB$ is $8 \times$ bigger.