10/09 - Fourier Transform: Difference between revisions

Revision as of 14:28, 17 November 2008

$\left\langle e^{j2\pi nt/T}\mid e^{j2\pi mt/T}\right\rangle$	$=\int _{-\infty }^{\infty }e^{j2\pi nt/T}e^{-j2\pi mt/T}\,dt$
	$=\int _{-\infty }^{\infty }e^{j2\pi (n-m)t/T}\,dt$
	$=\int _{-T/2}^{T/2}e^{j2\pi (n-m)t/T}\,dt$	Assuming the function is perodic with the period T
	$=T\delta _{m,n}\,\!$

If we let $T\rightarrow \infty$

${\frac {1}{T}}$	$\rightarrow df$
${\frac {n}{T}}$	$\rightarrow f$	Remember $f={\frac {2\pi n}{T}}\,\!$
$T\,\!$	$\rightarrow \infty$
$\sum _{n=-\infty }^{\infty }{\frac {1}{T}}$	$\rightarrow \int _{-\infty }^{\infty }()df$

$F[x(t)]\,\!$	$=X(f)\,\!$	$=\int _{-\infty }^{\infty }x(t)e^{-j2\pi ft}dt$	$=\left\langle x(t)\mid e^{j2\pi ft}\right\rangle _{t}$
$F^{-1}[x(t)]\,\!$	$=x(t)\,\!$	$=\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df$	$=\left\langle X(f)\mid e^{-j2\pi ft}\right\rangle _{f}$

$F^{-1}[F[x(t)]]\,\!$	$=\int _{-\infty }^{\infty }\left[\int _{-\infty }^{\infty }x(t)e^{-j2\pi ft}dt\right]e^{j2\pi ft}df$
	$=\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df$
	$=x(t)\,\!$

$\int _{-\infty }^{\infty }e^{j2\pi ft}e^{-j2\pi f\lambda }df$	$=\left\langle e^{j2\pi ft}\mid e^{j2\pi ft}\right\rangle _{f}$	$=\delta (t-\lambda )\,\!$
$\int _{-\infty }^{\infty }e^{j2\pi tf}e^{-j2\pi tf_{0}}dt$	$=\left\langle e^{j2\pi tf}\mid e^{j2\pi tf_{0}}\right\rangle _{t}$	$=\delta (f-f_{0})\,\!$

@@ Line 64: / Line 64: @@
 |
 |<math>=x(t)\,\!</math>
+|}
+{| border="0" cellpadding="0" cellspacing="0"
+|-
+|<math>\int_{-\infty}^{\infty}e^{j2\pi ft}e^{-j2\pi f\lambda}df</math>
+|<math>=\left\langle e^{j2\pi ft}\mid e^{j2\pi ft}\right\rangle_f</math>
+|<math>=\delta(t-\lambda)\,\!</math>
+|-
+|<math>\int_{-\infty}^{\infty}e^{j2\pi tf}e^{-j2\pi tf_0}dt</math>
+|<math>=\left\langle e^{j2\pi tf}\mid e^{j2\pi tf_0}\right\rangle_t</math>
+|<math>=\delta(f-f_0)\,\!</math>
 |}