2
$\begingroup$

I want to make sure that my units are all correct. I've been using the formula

$$ \frac{P_S}{P_N} = \frac{E_S}{N_0} \frac{R_S}{B} $$

The dimensions that I'm confident of are

  • $P_S$, $P_N$ : Watts (Joules/sec)
  • $N_0$ : Joules/(sec*Hz)
  • $B$ : Hz

The last two are then (I think)

  • $R_S$ : symbols/sec
  • $E_S$ : Joules/symbol

If those are all correct, and given that

$$ \frac{E_S}{N_0} = \frac{P_S}{P_N} \frac{B}{R_S} $$

it seems like the units of $\frac{E_S}{N_0}$ are

$$ \frac{\text{Hz}}{\text{symbols/sec}} = \frac{1}{\text{symbol}} $$

I was under the impression that it should be dimensionless. Is this correct?

$\endgroup$

1 Answer 1

4
$\begingroup$

The "symbols per second" (or, more strictly, the baud) unit is a bit weird. It is really a unit that is used for human convenience and not for mathematical rigor. Just like "samples per second", bauds can be interpreted as hertz and still make sense, since they measure the "number of times something happens per second".

So, in your calculation, the units of $B/R_S$ are $\text{Hz}/\text{Hz}$, making it unitless.

I know this is somewhat unsatisfying (at least, it is to me), but this is the best explanation I can give.

(All your unit definitions at the top of the question are correct, BTW).

$\endgroup$
2
  • $\begingroup$ That makes sense. So I guess $E_S$ is the energy in a given symbol, and the units of $E_S$ are just Joules (the power times the duration). $\endgroup$
    – Trey
    Apr 9, 2018 at 23:00
  • $\begingroup$ Yes, you can think of it like that -- the unit "symbols" helps us keep track of what we're measuring. $\endgroup$
    – MBaz
    Apr 10, 2018 at 0:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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