10/02 - Fourier Series: Difference between revisions

Latest revision as of 23:12, 4 December 2008

Fourier Series (as compared to vectors)

If a function is periodic, $x (t) = x (t + T)$ , and it meets the Dirichlet conditions, then we can write it as $x (t) = \sum_{n = - \infty}^{\infty} α_{n} e^{j 2 π n t / T}$

Dirichlet conditions
- x(t) must have a finite number of extrema in any given interval
- x(t) must have a finite number of discontinuities in any given interval
- x(t) must be absolutely integrable over a period
- x(t) must be bounded

Like vectors we can change to a new basis function by taking the inner product of $x (t)$ with the mth basis function. Don't forget that the inner product of two vectors requires a complex conjugate. $\vec{a} \cdot \vec{b} = \sum_{i = 1}^{n} a_{i} \cdot b_{i}^{*}$

$x (t) \cdot e^{j 2 π m t / T}$	$= \int_{- T / 2}^{T / 2} x (t) e^{- j 2 π m t / T} d t$
	$= \int_{- T / 2}^{T / 2} (\sum_{n = - \infty}^{\infty} α_{n} e^{- j 2 π n t / T}) e^{j 2 π m t / T} d t$

	$= \sum_{n = - \infty}^{\infty} α_{n} \int_{- T / 2}^{T / 2} e^{j 2 π n t / T} e^{- j 2 π m t / T} d t$
	$= \sum_{n = - \infty}^{\infty} α_{n} \int_{- T / 2}^{T / 2} e^{j 2 π (n - m) t / T} d t$
	$= \sum_{n = - \infty}^{\infty} α_{n} {\begin{matrix} T, & n = m \\ 0, & n \neq m \end{matrix}}$
	$= \sum_{n = - \infty}^{\infty} α_{n} T δ (n - m)$
	$= T \sum_{n = - \infty}^{\infty} α_{n} δ (n - m)$
	$= T α_{m}$

Sticking this back into the top equation gives

$α_{m} = \frac{1}{T} \int_{- T / 2}^{T / 2} x (t) e^{- j 2 π m t / T} d t$

Notes

The integral range reflects the period T as defined at the top of the page. The current range carries over to the Fourier series better than going from 0 to T.

When switching the order of integrals and summations, you can "cavalierly" switch the order as long as there aren't summations/integrals to infinity.

To see ${\begin{matrix} T, & n = m \\ 0, & n \neq m \end{matrix}}$ remember Euler's identities for sine and cosine. Since we are integrating the sine and cosine waves over a single period (assuming $m \neq n$ ), the value integrates to 0.

\cos (x) = \frac{e^{i x} + e^{- i x}}{2}

\sin (x) = \frac{e^{i x} - e^{- i x}}{2 i}

@@ Line 7: / Line 7: @@
 **x(t) must be bounded
-Like vectors we can change to a new basis function by taking the inner product of <math>x(t)\,\!</math> with the mth basis function.
+Like vectors we can change to a new basis function by taking the inner product of <math>x(t)\,\!</math> with the mth basis function. Don't forget that the inner product of two vectors requires a complex conjugate. <math> \vec a \cdot \vec b = \sum_{i=1}^n a_i \cdot b_i^* </math>
+{| border="0" cellpadding="0" cellspacing="0"
+|-
+|<math>x(t) \cdot e^{j2\pi mt/T}</math>
+|<math>=\int_{-T/2}^{T/2} x(t) e^{-j2\pi mt/T}\, dt</math>
+|-
+|
+|<math>=\int_{-T/2}^{T/2}\left (\sum_{n=-\infty}^\infty \alpha_n e^{-j2\pi nt/T} \right ) e^{j2\pi mt/T}\, dt</math>
+|-
+|
+|-
+|
+|<math>=\sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2}e^{j2\pi nt/T}  e^{-j2\pi mt/T}\, dt</math>
+|-
+|
+|<math>=\sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2}e^{j2\pi (n-m)t/T}\, dt</math>
+|-
+|
+|<math>=\sum_{n=-\infty}^\infty \alpha_n \begin{Bmatrix} T, & n=m \\ 0, & n \ne m \end{Bmatrix} </math>
+|-
+|
+|<math>=\sum_{n=-\infty}^\infty \alpha_n T \delta(n-m) </math>
+|-
+|
+|<math>=T\sum_{n=-\infty}^\infty \alpha_n \delta(n-m) </math>
+|-
+|
+|<math>=T \alpha_m \,\!</math>
+|}
+Sticking this back into the top equation gives
+<math> \alpha_m = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{-j2\pi mt/T}\, dt</math>
+==Notes==
+*The integral range reflects the period T as defined at the top of the page. The current range carries over to the Fourier series better than going from 0 to T.
+*When switching the order of integrals and summations, you can "cavalierly" switch the order as long as there aren't summations/integrals to infinity.
+*To see <math> \begin{Bmatrix} T, & n=m \\ 0, & n \ne m \end{Bmatrix} </math> remember Euler's identities for sine and cosine. Since we are integrating the sine and cosine waves over a single period (assuming <math> m \ne n </math>), the value integrates to 0.
+:<math>\cos(x) = \frac{e^{ix} + e^{-ix}}{2} </math>
+:<math>\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}</math>

10/02 - Fourier Series: Difference between revisions

Latest revision as of 23:12, 4 December 2008

Fourier Series (as compared to vectors)

Notes

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