10/09 - Fourier Transform: Difference between revisions

Revision as of 15:54, 17 November 2008

$⟨ e^{j 2 π n t / T} ∣ e^{j 2 π m t / T} ⟩$	$= \int_{- \infty}^{\infty} e^{j 2 π n t / T} e^{- j 2 π m t / T} d t$
	$= \int_{- \infty}^{\infty} e^{j 2 π (n - m) t / T} d t$
	$= \int_{- T / 2}^{T / 2} e^{j 2 π (n - m) t / T} d t$	Assuming the function is perodic with the period T
	$= T δ_{m, n}$

If we let $T \to \infty$

$F [x (t)]$	$= X (f)$	$= \int_{- \infty}^{\infty} x (t) e^{- j 2 π f t} d t$	$= {⟨ x (t) ∣ e^{j 2 π f t} ⟩}_{t}$
$F^{- 1} [x (t)]$	$= x (t)$	$= \int_{- \infty}^{\infty} X (f) e^{j 2 π f t} d f$	$= {⟨ X (f) ∣ e^{- j 2 π f t} ⟩}_{f}$

$F^{- 1} [F [x (t)]]$	$= \int_{- \infty}^{\infty} [\int_{- \infty}^{\infty} x (λ) e^{- j 2 π f λ} d λ] e^{j 2 π f t} d f$	$= \int_{- \infty}^{\infty} X (f) e^{j 2 π f t} d f$	$= x (t)$
		$= \int_{- \infty}^{\infty} x (λ) \int_{- \infty}^{\infty} e^{j 2 π f (t - λ)} d f d λ$	$= \int_{- \infty}^{\infty} x (λ) δ (t - λ) d λ$	$= x (t)$

$\int_{- \infty}^{\infty} e^{j 2 π f t} e^{- j 2 π f λ} d f$	$= {⟨ e^{j 2 π f t} ∣ e^{j 2 π f t} ⟩}_{f}$	$= δ (t - λ)$
$\int_{- \infty}^{\infty} e^{j 2 π t f} e^{- j 2 π t f_{0}} d t$	$= {⟨ e^{j 2 π t f} ∣ e^{j 2 π t f_{0}} ⟩}_{t}$	$= δ (f - f_{0})$

@@ Line 57: / Line 57: @@
 |-
 |<math>F^{-1}[F[x(t)]]\,\!</math>
-|<math>=\int_{-\infty}^{\infty}\left [ \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}dt\right ]e^{j2\pi ft} df</math>
+|<math>=\int_{-\infty}^{\infty}\left [ \int_{-\infty}^{\infty} x(\lambda) e^{-j2\pi f\lambda}d\lambda\right ]e^{j2\pi ft} df</math>
+|<math>=\int_{-\infty}^{\infty}X(f) e^{j2\pi ft} df</math>
+|<math>=x(t)\,\!</math>
 |-
 |
-|<math>=\int_{-\infty}^{\infty}X(f) e^{j2\pi ft} df</math>
-|-
 |
+|<math>=\int_{-\infty}^{\infty}x(\lambda) \int_{-\infty}^{\infty}e^{j2\pi f(t-\lambda)} df d\lambda</math>
+|<math>=\int_{-\infty}^{\infty}x(\lambda) \delta(t-\lambda) d\lambda</math>
 |<math>=x(t)\,\!</math>
 |}