Question

- Question about Fourier Formulae.

Answer 1

In the formulae below, Ao is essentially 1/2pi.

Answer 2

Is it also possible to call Ao = 1/pi, and just divide Ao/2 in the equation attached instead?

Answer 3

It is possible to rename the coefficient and remember to divide by 2
but in general A0 is not 1/2pi
the integral of f(x) is the area under the curve of f(x) for "one cycle"
that area is dependent on what f(x) is

Answer 4

Ah ok.

Dividing ao by 2 is not necessary when using half range?

Ie:\[\frac{ 4 }{ T } \int\limits_{0}^{T/2} f(t)\]

Answer 5

that won't work (in general)

The first term in a fourier series represents the "average" value of f(x) over one cycle
If you integrated e.g. sin(x) for a full cycle, the area will sum to zero (half above the x-axis, half below). If you integrate just 1/2 of a cycle, you will not get zero.

Answer 6

Ok. I think I understand, can i just ask you one last thing to clarify Phi? I appreciate your help.

Answer 7

?

Answer 8

My lecturer's general equation is:

\[\frac{ ao }{ 2 } + \sum_{n=1}^{\inf} ancos (\frac{ 2n \pi t }{ T }) + bnsin (\frac{ 2n \pi t }{ T })\]

Answer 9

Where ao:
\[ao = \frac{ 2 }{ T } \int\limits_{-T/2}^{T/2} f(t) dt\]

Answer 10

So if I work out Ao for a full range, i need to then divide it by 2 when putting it back into the general formula

Answer 11

if you use the general form
\[ \frac{ a_0 }{ 2 } + \sum_{n=1}^{\infty} a_n \cos \left(\frac{ 2n \pi t }{ T }\right) + b_n\sin \left(\frac{ 2n \pi t }{ T }\right) \]

then all the "a" coeffients use the same formula
\[ a_n= \frac{2}{T} \int_{-\frac{T}{2}}^\frac{T}{2} f(x) \cos \left(\frac{ 2n \pi t }{ T }\right)\ dt \]
for all n, including 0
of course, cos(0) is 1, so a0 simplifies to your version.

Answer 12

Yeah.

It's just, if I've gotten ao = 1.

I've often left it 1, but when stating the final general form, I need to make it 1/2, right?

Answer 13

yes, in the series, you use a0/2

Answer 14

Ok. It's nice to meet someone with knowledge of this!

Answer 15

Got a big engineering exam tomorrow. :)

Answer 16

good luck!