MHB What is the Derivation of the Average Value of a Waveform Using Calculus?

[bergausstein]

Hello.

I was having a hard time determining how the formula for the average value of the waveform can be achieved.

I've read that the ave since the waveform is symmetric is (Imin+Imax)/2. I want to know how to derive it using calculus.
Thanks.

 

[MarkFL]

The average value $M$ of a function $f(x)$ over some interval $[a,b]$ where $a<b$ is given by:

$$M=\frac{1}{b-a}\int_a^b f(x)\,dx$$

Suppose we use a sine function of the form:

$$f(x)=A\sin(x)+B$$

Now, over one period from $0\le x\le2\pi$, we see that the area above (in red) $y=B$ is equal to the area below (in green):

View attachment 6550

And so we may state (without having to actually do any calculus):

$$\int_0^{2\pi} A\sin(x)+B\,dx=2\pi B$$

We are stating in effect that the integral is equal to the area of a rectangle having base $2\pi$ and height $B$, because the area above and the area below cancel each other out, that is, they add to zero.

And so the average value of the function is given by:

$$M=\frac{1}{2\pi-0}\int_0^{2\pi} A\sin(x)+B\,dx=\frac{2\pi B}{2\pi}=B$$

Now, we know:

$$f_{\max}=B+A$$

$$f_{\min}=B-A$$

And so we could write:

$$M=B=\frac{B+A+B-A}{2}=\frac{f_{min}+f_{max}}{2}$$

 

[bergausstein]

We are stating in effect that the integral is equal to the area of a rectangle having base $2\pi$ and height $B$, because the area above and the area below cancel each other out, that is, they add to zero.

Hello MarkFL! I kind of find it difficult to understand why the area in effect is equal to the area of a rectangle? And also the part where the area above and below cancel each other? Because in your graph it has an offset of B. As I understand it, they will only cancel each other if there was no offset B. meaning the upper part of the graph is above the x-axis and the lower part is below the x-axis and if we add them together we get zero. because they have equal but opposite value. Please, if you have time, elaborate it for me. THANKS!

 

[MarkFL]

Hello MarkFL! I kind of find it difficult to understand why the area in effect is equal to the area of a rectangle? And also the part where the area above and below cancel each other? Because in your graph it has an offset of B. As I understand it, they will only cancel each other if there was no offset B. meaning the upper part of the graph is above the x-axis and the lower part is below the x-axis and if we add them together we get zero. because they have equal but opposite value. Please, if you have time, elaborate it for me. THANKS!

Well, think of it this way:

Consider first only from $x=0$ to $x=\pi$. The area $A_1$ under the curve is equal to the area of the rectangle with base $\pi$ and height B PLUS the area in red, which we'll call $C$...and so this area is:

$$A_1=B\pi+C$$

Now consider only from $x=\pi$ to $x=2\pi$. The area $A_2$ under the curve is equal to the area of the rectangle with base $\pi$ and height B MINUS the area in green, which we'll also call $C$...and so this area is:

$$A_2=B\pi-C$$

And so, we may write:

$$\int_0^{2\pi} A\sin(x)+B\,dx=A_1+A_2=(B\pi+C)+(B\pi-C)=2\pi B$$

Does this make sense?

 

[bergausstein]

Hello again! I tried to make a picture out of what you have explained. Please check if I'am on the right track here.

View attachment 6554$B\pi$ the area of rectangle in blue. $C$ is the red/green area
$B(2\pi-\pi)-C$ the area of brown part.

 

[MarkFL]

Hello again! I tried to make a picture out of what you have explained. Please check if I'am on the right track here.

$B\pi$ the area of rectangle in blue. $C$ is the red/green area
$B(2\pi-\pi)-C$ the area of brown part.

Yes...that's good. :D