13
\$\begingroup\$

We just started a communications course in college and we came across SN ratio. The following is an ambiguity I am facing which my professor is unable to resolve:

Signal to Noise Ratio is the ratio of signal power to the noise power. It is often expressed in decibels. But it is a ratio of two similar quantities, so it must not have a unit right? Why then do we use decibel?

If anybody could answer this question or provide links to resources which solve it, I would be very grateful.

PS: I tried Google and Wikipedia but I could not find anything specifically related to this.

\$\endgroup\$
  • 11
    \$\begingroup\$ all decibels are ratios. all ratios are decibels. they are two ways of expressing the same thing. \$\endgroup\$ – markrages Sep 19 '13 at 14:53
  • 22
    \$\begingroup\$ If your communications professor does not understand decibels, you need to switch classes (or switch schools). I mean this seriously. \$\endgroup\$ – markrages Sep 19 '13 at 14:54
  • 2
    \$\begingroup\$ Logs are convenient to work with. Multiplications and divisions become additions and subtractions. Also, I wonder if it has to do w/ the days of yore, when calculations were performed on slide rules \$\endgroup\$ – Scott Seidman Sep 19 '13 at 15:34
  • 2
    \$\begingroup\$ @ScottSeidman the concept was probably easier for students already familiar with slide rules to grasp; but I think numbers like 10, 30, 50, 90 being easier to work with than 10, 1000, 100000, 1000000000 probably had more to do with why log scales were adopted where they were. \$\endgroup\$ – Dan is Fiddling by Firelight Sep 19 '13 at 17:06
  • 10
    \$\begingroup\$ One of the reasons that dB intuitively feels like a unit is because the most common usage of dB outside of engineering is to describe volume of sound. When we say that a sound is 30 dB we mean that the sound pressure is the unitless 10^(30/10) multiplied by a standard unit-ful quantity, namely 20 micropascals. If you don't know that then it is very easy to confuse "a 30 dB noise" that you hear with "a 30 dB signal-to-noise ratio" and think they have something to do with each other. They don't. \$\endgroup\$ – Eric Lippert Sep 19 '13 at 22:50
21
\$\begingroup\$

To express a ratio in dB, the ratio must be unit-less, since the logarithm of the ratio must be taken, so I'm not sure I understand why you're puzzled that we use dB.

dB is often used to express unit-less ratios precisely because of the properties of logarithm.

For example, multiplication becomes addition, division becomes subtraction.

Also, since the the signal my be many orders of magnitude greater than the noise, it is more convenient to express the SNR as, say, 50dB rather than 100,000.

I am puzzled because as you said SNR is a unit-less ratio, but at the same time we express it in dB... If the ratio and its logarithm both do not have a unit, what then is the dB? ".

The phrase "the SNR is 50dB" is equivalent to "10 times the log of the ratio of the signal power to noise power equals 50."

The dB is not a dimensionful unit like a unit of length or of time, it is a dimensionless unit.

The number x is a pure number just as the number \$y = 10 \log(x) \$ is though we might say that "y is just x expressed in dB".

\$\endgroup\$
  • \$\begingroup\$ I am puzzled because as you said SNR is a unit-less ratio, but at the same time we express it in dB. Don't these statements contradict each other. My point is why does this contradiction arise? Is it because of some special property of dB? Hope you get what I am saying. \$\endgroup\$ – iluvthee07 Sep 19 '13 at 14:10
  • 1
    \$\begingroup\$ @iluvthee07, no the statements don't contradict so I suspect that your understanding of dB is inchoate. The number x does not have a unit and the number 10log(x) does not have a unit. \$\endgroup\$ – Alfred Centauri Sep 19 '13 at 14:13
  • \$\begingroup\$ As I said before we are just learning this in college. I still don't get it. If the ratio and its logarithm both do not have a unit, what then is the dB? \$\endgroup\$ – iluvthee07 Sep 19 '13 at 14:18
  • 1
    \$\begingroup\$ Good answer +1, and I'd give you another for inchoate if I could. \$\endgroup\$ – placeholder Sep 19 '13 at 14:59
  • 6
    \$\begingroup\$ You can't have a dB of water, a dB long piece of string, a cubic dB of stuff... but you can have a 3dB pay-rise :D \$\endgroup\$ – John U Sep 19 '13 at 15:20
10
\$\begingroup\$

Decibels isn't a "unit" in the sense of meter, Netwons, seconds, etc. It is like percent, dozen, parts per million, and the like. Those are all ways of expressing dimensionless numbers. Decibels happens to be a way to express values on a logarithmic scale, but that doesn't change the fact that there is nothing wrong with having various "units" for dimensionless quantities.

\$\endgroup\$
6
\$\begingroup\$

Similarly, radians should not have a unit, but are still expressed as rad for clarity.

More specifically, SNR is measured in dB, because dB are convenient for the situation. dBs are convinient for the situation, as the differences of signal and noise can have a large dynamic range, that is, to be small or very large.

So the SNR of 100000V signal with 1V noise is 100000. We take the logarithm of that number and arrive at 10*log(100000) = 50dB. A much nicer number.

Or some such.


Summarizing the discussion in the comments, quantities can be

  • unitless
  • have units, that have physical significance (e.g. meters)
  • or represent units, the do not represent the physical nature of the phenomena, but describe the way we measure it mathematically (e.g. radians, logarithms etc).

The claim has been made that adding quantities, expressed in different units is always meaningless. This is the same as what I have been thought, but might me a simplification for the young learners, just entering the field. IMHO supercat or kriss should ask this topic as a separate (excellent!) question.

\$\endgroup\$
  • 1
    \$\begingroup\$ I think you should use log not ln to calculate decibels \$\endgroup\$ – Andy aka Sep 19 '13 at 14:04
  • 1
    \$\begingroup\$ Natural logarithm is actually used for unit called neper (Np). \$\endgroup\$ – AndrejaKo Sep 19 '13 at 16:11
  • 1
    \$\begingroup\$ @kriss, yea, but by definition, the angle of radians is the ratio of the arc, divided by the radius. I am getting confused now! \$\endgroup\$ – Vorac Sep 20 '13 at 9:15
  • 1
    \$\begingroup\$ You can take the sine of one radian. That in itself is proof of the fact that they are unit-less. To convince yourself, look at the Taylor expansion of sin(x). If x has a unit, you're calculating x-(x^3/6). \$\endgroup\$ – MSalters Sep 20 '13 at 10:09
  • 1
    \$\begingroup\$ @Vorac: as far as I understand radians are dimensionless, but not unitless. Both are not exactly synonyms. Radian are always a mesure of an angle, which is physically defined. dB are not physically defined in the same way: that's some log representation of the ratio of two intensities, but it does not state the intensities of what. \$\endgroup\$ – kriss Sep 20 '13 at 12:47
4
\$\begingroup\$

Decibels are sometimes a more convenient "unit" to work with.

The same question applies to voltage gain of an op-amp - the tendency is to state open-loop gain in decibels. Ditto closed-loop gain.

Same with filters - low pass filters(for example) have a "gain" reduction with an increase in frequency and this is usually expressed as "so many" dB per octave or decade.

Plenty of things are stated in decibels.

EDIT

The decibel is not a unit like watts, ohms, volts or amps. It's a reminder that the number preceding it is derived a certain way. A different example is scientific notation such as the number 5000 - it can be expressed as 5E3 - this doesn't mean the E3 is a unit of any type.

Same applies to the "k" in 10k\$\Omega\$ resistor - "k" is not part of the unit. It tells us that the number of ohms is 10 x 1000.

\$\endgroup\$
  • 2
    \$\begingroup\$ \$e^3\$ is about 20.1, not 1000. \$\endgroup\$ – The Photon Sep 19 '13 at 16:11
  • \$\begingroup\$ @ThePhoton, I took the liberty to edit Andy's answer to reflect what I feel certain he meant: en.wikipedia.org/wiki/Scientific_notation#E_notation \$\endgroup\$ – Alfred Centauri Sep 19 '13 at 16:34
  • \$\begingroup\$ For the record, the "E notation" is a workaround for use in programming languages that don't recognize typesetting (like superscripts). In text,, using superscripts like in "\$5\times{}10^3\$" is much preferred. \$\endgroup\$ – The Photon Sep 19 '13 at 16:38
  • \$\begingroup\$ @ThePhoton In semiconductors scientific notation is written as 1.2e6 etc. Sloppy yes but also a standard shorthand. NOT 5E3 as edited but 5e3 is correct, Just check on IEEE.org. Just like Angstrom as a unit persists, this does too. \$\endgroup\$ – placeholder Sep 19 '13 at 17:16
  • 1
    \$\begingroup\$ @rawbrawb I've only seen people do that if they have gotten confused and can no longer tell the difference between FORTRAN and English. \$\endgroup\$ – The Photon Sep 19 '13 at 18:24
4
\$\begingroup\$

As you plainly stated, decibels are used to quantify the relationship between two signals. They are relative, not absolute. Saying that a transmitter has 1dB of output is meaningless. Therefore it must be referenced to some other unit. For example, 1dBm is 1dB with respect to 1 miliwatt.

In the case of Signal to Noise ratios, the dB is the only thing that makes sense to use. Typically, a signal in RF or other applications will be much above the noise, hundreds of thousands or millions of times stronger. In that case it is simpler and shorter to write that it is 60dB above instead of 1000000 since a mistake could easily be made.

\$\endgroup\$
  • \$\begingroup\$ @ChrisStratton: You're right. Not sure why I wrote milivolt \$\endgroup\$ – Gustavo Litovsky Sep 19 '13 at 17:40
1
\$\begingroup\$

It's a particular transfer function, it really depends on the application Like in circuits analysis for op amps, we often care of the voltage signal to noise ratio So it could be V/V or A/A, or a mixture of two.

Decibels are often used to look closer at the amplitude or frequency of signal amplification and attenuation

Edit

It's a logarithmic unit, an abstract mathematic unit (not physical units)

Ohms for example is a measure of Voltage/Current, it is dimensionless.

\$\endgroup\$
  • \$\begingroup\$ Thanks for replying. Unfortunately, that does not answer my question. What I mean is a ratio must not have a unit. That's what basic mathematics teaches us. But a decibel is a unit for a ratio. Why is the rule changing in this case? Is there something special about decibels? \$\endgroup\$ – iluvthee07 Sep 19 '13 at 14:03
  • 1
    \$\begingroup\$ @iluvthee07, ratios most certainly can have a unit, e.g., feet per second. However, to express a ratio in dB, the ratio must not have units. dB is not a unit per se, while, for example, dBm is. \$\endgroup\$ – Alfred Centauri Sep 19 '13 at 14:10
  • \$\begingroup\$ @iluvthee07 edited \$\endgroup\$ – Iancovici Sep 19 '13 at 14:12
  • \$\begingroup\$ @echad: I meant a ratio of two similar quantities. As for the abstract thing, are you implying that we use it just for the sake of denoting that log is being used here and not for some other more subtle reason? \$\endgroup\$ – iluvthee07 Sep 19 '13 at 14:14
  • \$\begingroup\$ The unit tells you what you are measuring; another example of a ratio with units is specifying gain as "20 V/V", to indicate that you're referring to voltage gain and not current gain. \$\endgroup\$ – pjc50 Sep 19 '13 at 14:22
1
\$\begingroup\$

I think the problem here is that the OP is confusing units with magnitude. If I say the gain of an amplifier is 1000 or 60 dB, I am simply expressing the magnitude of the gain in 2 different ways. In either case, there are no units since gain is normally volts per volt (or amps per amp, etc.). dB's are just another way of expressing the magnitude of a number. They are very convenient for use with numbers that can be very large or very small. As already pointed out, it is much more convenient to express 0.00001 as -100 dB or 1,000,000 as 120 dB. Both expressions are simply number magnitudes. No units are involved.

\$\endgroup\$
1
\$\begingroup\$

I like to think like this to solve your ambiguity:

decibels (dB) are a "appropriate" measure of how much a quantity is larger or smaller than other. In signal to noise ratios, you are willing to know how much the power of your signal is larger than the power of the noise. If you do the math you will end up with things like (Psignal / Pnoise) = 100000 which is cumbersome. Here cames the venerable log function wich transforms it in something like:

10*log(100000) = 50dB

Its a convient and consagrated notation. Just that.

\$\endgroup\$
0
\$\begingroup\$

The way I like to think of it is that a decibel is not a unit, it's a function. (This idea is not original to me---I read it in a paper at some point, which I can't find at the moment.) Regular units like meters, seconds, and coulombs behave like irreducible constants that are multiplied by pure numbers. Even things like % and radians can be treated as multiplying constants in dimensional analysis, where %=0.01, and radian=1. But decibels are different. When someone tells you that a power ratio is equal to "3 dB", what they're really saying is that the ratio is equal to \$10^{3/10}\$, or approximately 2. So rather than writing "PR = 3 dB", we arguably should write "PR = dB(3)", where \$\mathrm{dB}(x)=10^{x/10}\$. And for the same reasons that you generally don't take exponentials and logarithms of anything except a pure number, you also don't take the dB() of anything but a pure number.

Fahrenheit and celsius degrees are similar. Neither behaves like a regular unit in dimensional analysis, they behave like functions. So "10 degC" should really be degC(10), where \$\mathrm{degC}(x)=(273.15+x)\ \mathrm{K}\$, where \$\mathrm{K}\$ is Kelvins. (Kelvin is a regular unit.) And "32 degF" should really be degF(32), where \$\mathrm{degF}(x)=5/9 \cdot (x+459.67)\ \mathrm{K}\$.

The one other wrinkle with dB is that people will often say that the "amplitude" of a signal is "x dB". What they mean is that the power of the signal is dB(x) times more than the power in some reference signal. So for instance audio engineers use "dBV" to mean the power in a signal, relative to the power in a 1 V sine wave. Since the mean power is equal to the RMS amplitude squared, that means that $$\frac{A_{rms}^2}{(1\ V)^2} = \mathrm{dB}(x)\ ,$$ which in turn implies that $$A_{rms} = 10^{x/20}\ V\ .$$

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.