# Calculating average value of a signal

The formula is $$\frac{1}{T}\int_{a}^{b}S(t) .dt$$ I understand how the integral gives us the area under the curve, but average means the sum of all over how much all is, the integral here is the sum of all $S(t)dt$ between $a$ and $b$ but shouldn't it really be just the sum of the $S(t)$'s between $a$ and $b$ divided by $T$?

• Your suggestion has the wrong units for the average value of the signal (namely, the units in your suggestion are signal amplitude per unit time, while the average value should just have units of signal amplitude). – probably_someone Sep 10 '18 at 17:09
• honestly I'm asking for an explanation and for someone to tell me what I'm missing here, thank you anyways! – user184836 Sep 10 '18 at 17:11
• @probably_someone, there's a $\rm dt$ term in the numerator and a $T$ term in the denominator, so no time units in the final result. – The Photon Sep 10 '18 at 17:11
• @ThePhoton What I was talking about was his suggestion of "just the sum of the $S(t)$s between $a$ and $b$ divided by $T$." – probably_someone Sep 10 '18 at 17:13
• Think about this: If you wanted to take a sum (rather than an integral) of S(t) between a and b, how many terms would there be in your sum? – The Photon Sep 10 '18 at 17:15

To take an average of a few discrete data points, we would add all the data points and divide by the number of data points. If you had a time interval $T$ divided into steps of size $\Delta t$, the number of data points would be $N=T/\Delta t$. Therefore, the average of measurements at those times would be $$\frac{S(t=1)+S(t=2)+...S(t=N)}{N}=\frac{S(t=1)\Delta t+S(t=2)\Delta t+...S(t=N)\Delta t}{T}$$ Now think of taking the limit that $\Delta t$ becomes infinitesimal and you get the integral.