Sound power: What is 100mW amplified by 2dB?

[dbag123]

Homework Statement

Signal rated at 100mW is amplified by 2dB. How much is the power out of the amplifier

Relevant Equations

i=10lg(I/I0)db

100mW = 20 dBm, that is then amplified by 2dB

20dBm/2dB = 10mW so power out of the amplifier would be 100mW+10mW =110mW?

 

[jbriggs444]

Homework Statement: Signal rated at 100mW is amplified by 2dB. How much is the power out of the amplifier
Homework Equations: i=10lg(I/I0)db

100mW = 20 dBm, that is then amplified by 2dB

20dBm/2dB = 10mW so power out of the amplifier would be 100mW+10mW =110mW?

dB in general and dBm in particular are logarithmic scales. Dividing dBm by dB does not make the kind of sense you expect.

20dBm/2dB gives a result of "10". But not 10 mW. The "10" that it yields is the number of times you would need to apply a 2dB amplification to amplify a 0 dBm (1 mW) signal to 20dBm (100 mW).

Each such amplification step would multiply power by the same factor as every other step.

 

[dbag123]

allright. how do i get this problem started?

 

[jbriggs444]

allright. how do i get this problem started?

Do you know what it means to amplify something by 2dB?

Or, simpler -- do you know what it means to amplify something by 1B?

 

[dbag123]

increase the intensity levels of sound

 

[jbriggs444]

increase the intensity levels of sound

Not really. If you are learning from a textbook, you should take time to go back and look at the definition. Or use Google.

The unit is not specific to sound. It can be used for anything.

1 Bell is an amplification by a factor of 10.
2 Bells is an amplification by a factor of 100.
3 Bells is an amplification by a factor of 1000.

In general, if 0 B is the base level on [whatever quantity you are measuring], x B is $10^x$ times that much.

One dB is one tenth of a B. If 0 dB is the base level on [whatever quantity you are measuring], x dB is $10^{x/10}$ times that much.

 

[dbag123]

so a 2 dB amplification would multiply the power by 1,585

 

[jbriggs444]

so a 2 dB amplification would increase the power by 1,585

Bingo.

 

[jbriggs444]

interesting. i am not learning from a book, just some course notes, the term Bell is not mentioned anywhere in these notes.

The story I learned from my father many years ago and which I may have embellished a bit myself [so take it with a grain of salt] is that Alexander Graham Bell was studying sound and came up with a scale for intensity.

0 for inaudible silence.
1 for the softest barely audible sound
2 for something detectably louder than that
3 for something detectably louder than that
and so on.

He discovered that human hearing was approximately logarithmic. By the time you had 10 detectable increments in loudness, you'd increased signal power by a factor of ten. So each detectable increment in loudness was approximately a factor of $^{10}\sqrt{10}$ increase in power.

The name taken for the factor of 10 was Bell and the name for the smallest audible increment in loudness then becomes one tenth of a Bell -- the decibel.

By happy coincidence, 3 dB is very nearly a factor of 2. Handy engineering approximation.

 

[dbag123]

thank you very much. any chance i could borrow you for another problem? it has to do with sound dampening through an object. the question is does that object dampen the sound by 1 or by 2 due to the mediums it has to pass

 

[berkeman]

thank you very much. any chance i could borrow you for another problem? it has to do with sound dampening through an object. the question is does that object dampen the sound by 1 or by 2 due to the mediums it has to pass

Go ahead and start a new thread with your new question. Be sure to make a complete, clear problem statement (including a diagram if you have one), and show your attempt at the solution. That's the best way to get help on schoolwork questions here.