Thread: Fourier trigonometric series - amplitude and phase spectrum

1. Fourier trigonometric series - amplitude and phase spectrum

Hello,

I don't understand why is amplitude spectrum of Fourier series defined as:

$c_0 = |a_0|$
$c_n = \sqrt{a_n^{2} + b_n^{2}}$

and why is phase spectrum:

$Arg(a_n - b_n)$

Can someone give me an understandable easy explanation.
Thanks.

2. Re: Fourier trigonometric series - amplitude and phase spectrum

Try writing
$a_n \cos{ (nt)} + b_n \sin{ (nt) }= c_n \cos {(nt+\alpha)}$ and solving for $c_n$ and $\alpha$.

Hint: use $\cos{ (A+B) }= \cos{ (A)}\cos{(B)}-\sin{(A)}\sin{(B)}$ on the right hand side.

3. Re: Fourier trigonometric series - amplitude and phase spectrum

Ok, I got this:

$c_n = \frac{a_n(cos(nt)+b_n(nt)}{cos(nt)cos(\alpha)-sin(nt)(sin(\alpha)}$

How to get from this to Pythagorean theorem?

4. Re: Fourier trigonometric series - amplitude and phase spectrum

No, we equate coefficients:
\begin{align*} c_n \cos{(nt-\alpha_n)} &= c_n \big( \cos{(\alpha_n)}\cos{(nt)} + \sin{(\alpha_n)}\sin{(nt)} \big) \\ &= c_n \cos{(\alpha_n)}\cos{(nt)} + c_n \sin{(\alpha_n)}\sin{(nt)} = a_n \cos{(nt)} + b_n \sin{(nt)} \end{align*}

And so we can say that
$a_n = c_n \cos{(\alpha_n)} \quad b_n = c_n \sin{(\alpha_n)}$

Thus,
$a_n^2 + b_n^2 = c_n^2 \quad \tan{(\alpha_n)} = \frac{b_n}{a_n}$

5. Re: Fourier trigonometric series - amplitude and phase spectrum

Hey thank you.

I have another question. In this equation the left hand side is the definition of Fourier trigonometric series (Infinite sum of cos and sin terms). How did you get to the right hand side?

$a_n \cos{ (nt)} + b_n \sin{ (nt) }= c_n \cos {(nt+\alpha)}$

Well, I have another question, but let's stick with one at the time.

6. Re: Fourier trigonometric series - amplitude and phase spectrum

It's a trigonometric identity.

7. Re: Fourier trigonometric series - amplitude and phase spectrum

Which trigonometric identity? I didn't find it on wikipedia or any other site.

And another question: Why are there two these series? (Exponential Fourier series and Trigonometric)

8. Re: Fourier trigonometric series - amplitude and phase spectrum

No one knows which trigonometric identity is this? What about other question?

9. Re: Fourier trigonometric series - amplitude and phase spectrum

The "trigonometric identity" in question is cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)

Taking A= nt, $B= \alpha$, $cos(nt+\alpha)= cos(nt)cos(\alpha)- sin(nt)sin(\alpha)$

So $a_ncos(nt)+ b_nsin(nt)= c_ncos(nt+ \alpha)$ with $a_n= c_ncos(\alpha)$ and $b_n= c_nsin(\alpha)$