Question regarding decibels

[danrogers]

Just been given an assignment regarding decibels. I understand what a decibel is and how its used and I've done the rest of the assignment but im stuck on the following...

''Explain the advantage of using decibels rather than
ratios. In your discussion you should explain the term dBm.''

From some reasearch it looks like dBm is decibel per milliwat, but I'm a bit lost on the differences between using decibels instead of ratios?

Thanks kindly for any help

 

[BrownOut]

From some reasearch it looks like dBm is decibel per milliwat, but I'm a bit lost on the differences between using decibels instead of ratios?

That wouldn't make any sense. From what I can remember, dBm is the ratio of power to 1 milliwatt, expressed at a logarithm, and is dimentionless. So, for example, 100mW would be expressed as:

10log(100mW/1mW)

The advantage is large values can be examined in a "compressed" scale. This sometimes simplifies the analysis of a quantity if it's value changes as a function of some other quantity. As an example, power output of an amplifier changes over the input's signal frequency, and so plotting the output power on a log scale more clearly shows the "flatband" portion of the power, as well as the nature of the power rolloff.

 

[JimB]

''Explain the advantage of using decibels rather than ratios.
In your discussion you should explain the term dBm.''

dBm is a power measurement, the reference level is 1mW.
So a transmitter with a power of 100mW is said to have an output of +20dBm.

Why use dBs rather than ratios?
Consider a point to point radio link, just making up a few numbers:

The transmitter has an output power of 10 watts, +40dBm.
The receiver needs a minimum signal of -125dBm (about 0.2µV into 50 ohms) for an acceptable signal.
At each end of the link the antenna has a gain of +12dB, and the feeder from the antenna to the Tx/Rx has a loss of -3dB.
The radio path loss is -160dB.
Will this link work?

Calculating:
At the transmitter end, the effective radiated power is

+40 -3 +12 = +49 dbm

The signal power at the receiver input is

+49 - 160 +12 -3 = -102dBm

The receiver needs at least -125dBm, so there is a (-102 - (-125)) = 23dB "Fade Margin".

The link will work very well.

Now you try the same calculation using ratios!
(Because I dont want to even try!).

If the teacher likes this answer, you own me a bacon sandwich!

JimB

 

[dknguyen]

Writing shorter numbers is all fine and dandy but that's only a minor inconvenience. But a bigger issue of convenience is when calculating cumulative loss for things like long distance cables when you are given a Loss Per Distance (and other such scenarios like RF transmissions where you have different propogation environments).

Imagine if you were given a linear loss of 0.5 loss per km (50% per km). It's cumulative so the loss in a segment of wire is on top of the loss of the segment that came before it, etc. So after 1km the signal strength is 50% of the input, after 2km it is 25%, after 3km it is 12.5%, etc. It's an decaying exponential relationship.

If it was given in -3dB/km, you just multiply it by the number of km since the logarithm. So after 10km, you have -30dB of loss.

It seems trivial on it's own like this. But if you are given a total resultant loss and need to calculate the loss factor required to travel a given distance, or need to calculate the distance you can travel with a given loss factor it becomes a massive pain in the ass.

Take this example:

Linear:
[Total Loss] = [Loss/Distance]^[Length]
dB:
[Total Loss] = [Loss/Distance]*[Length]

Which equation would you rather solve for if you had to figure out loss/distance or length? Obviously dB is going to be way faster since it only needs a division but the linear gains will either require roots or logarithms. But now imagine if you have multiple cable segments connected to each other:

Linear:
[Total Loss] = [Loss/Distance of Cable A]^[Length of A] + [Loss/Distance of Cable B]^[Length of B]
dB:
[Total Loss] = [Loss/Distance of Cable A]*[Length of A] + [Loss/Length of Cable B]*[Length of B]

Now what if cables A and B had different loss/distance but the one with better loss/distance was more expensive? What if you had to get a certain total loss while spanning over a certain distance with minimum cost? You'd have to take a derivative of the equation, equate the derivative to zero, and then get Length of A and Length of B on opposite sides of the equal sign to get a relationship between the two. With linear gains (or losses in this case), you'd have a bunch of derivatives of exponentials, and then you'd have to isolate the exponents of multiple terms using logarithms. On the other hand, with dB, it's the simplest of derivative problems and the simplest of albebraic manipulations.

 

[danrogers]

Thank you for all the info, far clearer in my head now and logarithms do make far more sense for some of those examples!

Bacon sandwiches all round

 

[erendesh]

On top of above comments,
You can also add and subract dB values rather than multiply. E..g. in 2 stage amplifier it it has gain of 1000 and 100. Then total gain is 1000*100 but in terms of dB gain in 1000+100 ( 10log1000 = 1000 and 10log100 is 100) = 1100. So easier to manipulate and as mentioned above, compact.

Sandesh

 

[davedecibel]

That wouldn't make any sense. From what I can remember, dBm is the ratio of power to 1 milliwatt, expressed at a logarithm, and is dimentionless. So, for example, 100mW would be expressed as:

10log(100mW/1mW)

The advantage is large values can be examined in a "compressed" scale. This sometimes simplifies the analysis of a quantity if it's value changes as a function of some other quantity. As an example, power output of an amplifier changes over the input's signal frequency, and so plotting the output power on a log scale more clearly shows the "flatband" portion of the power, as well as the nature of the power rolloff.

As a general rule of thumb I'd have to agree.

Indeed Bacon sandwiches all round I say, yum