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What is the value of $$a/b/c,$$

for $a,b,c>0$?

Is it $ac/b$ or $a/bc$?

I get confused about this when I was studying wireless communications. The unit of the spectral efficiency is defined as $bit/second/Hz$.

  • if it is $ac/b$ then the unit of spectral efficiency will be $bit Hz/second$ which is $bit/second^2$; and
  • if it is $a/bc$ then I will have $bit/secondHz$ so it is simply $bit$.

Hence, I guess

$$a/b/c=a/bc.$$

But that's not true, isn't it?

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  • $\begingroup$ It is usually $a/bc$ AFAIK. But maybe it means $a/(b/c)=ac/b$. :-/ $\endgroup$ – Simply Beautiful Art Jan 1 '17 at 18:26
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    $\begingroup$ That's usually taken to mean $(a/b)/c=a/(bc)$. For spectral efficiency in particular, wikipedia makes the explicit mention that the unit is in fact (bit/s)/Hz. $\endgroup$ – dxiv Jan 1 '17 at 18:27
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    $\begingroup$ We can guess from the name of the unit, spectral efficiency, that the answer will be $bit/second^2$, not $bit$. $\endgroup$ – Théophile Jan 1 '17 at 18:30
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    $\begingroup$ Putting in my two cents as a comment since it's not appropriate as an answer and the question has otherwise been sufficiently addressed anyway. I disagree with the answers that say $a/b/c$ is ambiguous because standard order of operations says it's $(a/b)/c$. Also, the notation $a/bc$ to represent $\frac a{bc}$ isn't good form because, strictly speaking, order of operations says $$a/bc = a/b \cdot c = \frac ab \cdot c = \frac{ac}b$$ $\endgroup$ – tilper Jan 2 '17 at 2:12
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Answering the question you really seemed to want to ask.

Yes. Bits per second per Hertz comes to just bits. But, it needs to interpreted. That unit is used when you are using a channel with a given bandwith $W$ (measure in Hertz, so for example you may have access to a band of width $W=5$ MHz). Then if you use that channel with efficiency of $k$ bits/second/Hertz for a time slice of length $T$ (seconds), you expect to transmit $kTW$ bits. The Hertz's and seconds cancel, and this gives a pure number of bits. As you would expect: if you double your time you double the number of transmitted bits, if you double your bandwidth you again double the number of bits you can transmit.

For details about the definition of spectral efficiency you probably want to ask at DSP.SE. I will add that often spectral efficiency is measure in bpcu (= bits per channel use). I'm not a telcomm engineer, so I won't say whether you include things like FEC code rate into spectral efficiency, or whether you just take into account the modulation scheme (such as QPSK-modulation = 2 bpcu, 16-QAM = 4 bpcu et cetera).

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That depends on where you choose to put the big fractions because $\frac{\frac ab}{c}\not=\frac{a}{\frac bc}$, the term $a/b/c$ is ambiguous.

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    $\begingroup$ it is really $$\frac{a}{bc}$$ if the variables are non zero $\endgroup$ – Dr. Sonnhard Graubner Jan 1 '17 at 18:28
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    $\begingroup$ We can't tell without the parenthesis. $\endgroup$ – FuzzyPixelz Jan 1 '17 at 18:29
  • $\begingroup$ aha why? try it with Wolfram alpha $\endgroup$ – Dr. Sonnhard Graubner Jan 1 '17 at 18:31
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It's a bit ambiguous without parentheses, but usually I would perform equivalent operations from left to right. So it would be

$$(a/b)/c = \frac{a/b}{c} = \frac{a}{b}\frac{1}{c} = \frac{a}{bc}$$

In this case you're dealing with physical units, so think about what spectral efficiency is trying to say. It's supposed to be a measure of how fast information can be transmitted with a given frequency range. It's the information "velocity" for a given bandwidth. If my communication system can transfer $100 bit/s$ with a $5Hz$ bandwidth, we would expect the spectral efficiency to be $20$ of some unit. In other words we're dividing

$$\frac{100bit/s}{5Hz}= 20\frac{bit/s}{Hz}$$

So the unit is (bit/s)/Hz.

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