10/09 - Fourier Transform: Difference between revisions

Latest revision as of 12:49, 4 December 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
	$=T\delta _{m,n}\,\!$

Fourier Transform

Remember from 10/02 - Fourier Series

$\alpha _{n}={\frac {1}{T}}\int _{-T/2}^{T/2}x(t)e^{-j\,2\,\pi \,n\,t/T}\,dt$
$x(t)=x(t+T)=\sum _{n=-\infty }^{\infty }\alpha _{n}e^{j\,2\pi \,n/T}$

If we let $T\rightarrow \infty$

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

Definitions

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

Examples

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

$F^{-1}[F[x(t)]]\,\!$	$=\int _{-\infty }^{\infty }\left[\int _{-\infty }^{\infty }x(\lambda )e^{-j2\pi f\lambda }d\lambda \right]e^{j2\pi ft}df$	$=\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df$	$=x(t)\,\!$
		$=\int _{-\infty }^{\infty }x(\lambda )\int _{-\infty }^{\infty }e^{j2\pi f(t-\lambda )}dfd\lambda$	$=\int _{-\infty }^{\infty }x(\lambda )\delta (t-\lambda )d\lambda$	$=x(t)\,\!$
	$=\int _{-\infty }^{\infty }\left[\int _{-\infty }^{\infty }x(\lambda )e^{-j\omega \lambda }d\lambda \right]e^{j\omega t}{\frac {1}{2\pi }}d\omega$	$=\int _{-\infty }^{\infty }x(\lambda )\left[{\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{j(t-\omega )\lambda }d\omega \right]d\lambda$	$=\int _{-\infty }^{\infty }x(\lambda )\delta (t-\omega )d\lambda$	$=x(t)\,\!$

Sifting property of the delta function

The dirac delta function is defined as any function, denoted as $\delta (t-u)\,\!$ , that works for all variables that makes the following equation true: $x(t)=\int _{-\infty }^{\infty }x(u)\delta (t-u)du$

When dealing with $\omega \,\!$ , it behaves slightly different than dealing with $f\,\!$ . When dealing with $=\int _{-\infty }^{\infty }x(\lambda )\left[{\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{j(t-\omega )\lambda }d\omega \right]d\lambda$ , note that the delta function is ${\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{j(t-\omega )\lambda }d\omega$ . The ${\frac {1}{2\pi }}$ is tacked onto the front. Thus, when dealing with $\omega \,\!$ , you will often need to multiply it by $2\pi \,\!$ to cancel out the ${\frac {1}{2\pi }}$ .

More properties of the delta function

$\delta (a\,t)={\frac {1}{\left|a\right|}}$

$\delta (\omega )=\delta (2\pi f)={\frac {1}{2\pi }}\delta (f)$

$\int _{-\infty }^{\infty }\delta (a\,t)\,dt$	$=\int _{-\infty }^{\infty }\delta (u\,t)\,{\frac {du}{\left\|a\right\|}}$	Let $a\,t=u$ and $du=a\,dt$
	$={\frac {1}{\left\|a\right\|}}$

@@ Line 20: / Line 20: @@
 ==Fourier Transform==
 Remember from [[10/02 - Fourier Series]]
-*<math> \alpha_m = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{-j2\pi mt/T}\, dt</math>
+*<math> \alpha_n = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{-j\,2\,\pi \,n\,t/T}\, dt</math>
-*<math>x(t) = x(t+T) = \sum_{n=-\infty}^\infty \alpha_m e^{j2\pi m/T}</math>
+*<math>x(t) = x(t+T) = \sum_{n=-\infty}^\infty \alpha_n e^{j\,2\pi \,n/T}</math>
 If we let <math> T \rightarrow \infty</math>
@@ Line 46: / Line 46: @@
 |<math>=X(f)\,\!</math>
 |<math>=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}dt</math>
-|<math>=\left \langle x(t) \mid e^{j2\pi ft}\right \rangle</math>
+|<math>=\left \langle x(t) \mid e^{j2\pi ft}\right \rangle_t</math>
 |-
-|<math>F^{-1}[x(t)]\,\!</math>
+|<math>F^{-1}[X(f)]\,\!</math>
 |<math>=x(t)\,\!</math>
-|<math>=\int_{-\infty}^{\infty} X(f) e^{j2\pi ft}dt</math>
+|<math>=\int_{-\infty}^{\infty} X(f) e^{j2\pi ft}df</math>
-|<math>=\left \langle X(f) \mid e^{-j2\pi ft}\right \rangle</math>
+|<math>=\left \langle X(f) \mid e^{-j2\pi ft}\right \rangle_f</math>
 |}
 ==Examples==
+{| 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>
+|}
+{| border="0" cellpadding="0" cellspacing="0"
+|-
+|<math>F^{-1}[F[x(t)]]\,\!</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(\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>
+|-
+|
+|<math>=\int_{-\infty}^{\infty}\left [ \int_{-\infty}^{\infty} x(\lambda) e^{-j\omega\lambda}d\lambda\right ]e^{j\omega t} \frac{1}{2\pi}d\omega </math>
+|<math>=\int_{-\infty}^{\infty}x(\lambda) \left [ \frac{1}{2\pi} \int_{-\infty}^{\infty}  e^{j(t-\omega) \lambda}d\omega\right ] d\lambda</math>
+|<math>=\int_{-\infty}^{\infty}x(\lambda) \delta(t-\omega) d\lambda</math>
+|<math>=x(t)\,\!</math>
+|}
+===Sifting property of the delta function===
+The dirac delta function is defined as any function, denoted as <math> \delta(t-u)\,\!</math>, that works for all variables that makes the following equation true:
+<math>x(t)=\int_{-\infty}^{\infty} x(u)\delta(t-u) du</math>
+*When dealing with <math>\omega\,\!</math>, it behaves slightly different than dealing with <math>f\,\!</math>. When dealing with <math>=\int_{-\infty}^{\infty}x(\lambda) \left [ \frac{1}{2\pi} \int_{-\infty}^{\infty}  e^{j(t-\omega) \lambda}d\omega\right ] d\lambda</math>, note that the delta function is <math>\frac{1}{2\pi} \int_{-\infty}^{\infty}  e^{j(t-\omega) \lambda}d\omega</math>. The <math>\frac{1}{2\pi}</math> is tacked onto the front. Thus, when dealing with <math>\omega\,\!</math>, you will often need to multiply it by <math>2\pi\,\!</math> to cancel out the <math>\frac{1}{2\pi}</math>.
+===More properties of the delta function===
+<math>\delta(a\,t) = \frac{1}{\left | a \right |}</math>
+<math>\delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f)</math>
+{| border="0" cellpadding="0" cellspacing="0"
+|-
+|<math>\int_{-\infty}^{\infty}\delta(a\,t)\,dt</math>
+|<math>=\int_{-\infty}^{\infty}\delta(u\,t)\,\frac{du}{\left|a\right|}</math>
+|Let <math>a\,t=u</math> and <math>du=a\,dt</math>
+|-
+|
+|<math>=\frac{1}{\left|a\right|}</math>
+|}

10/09 - Fourier Transform: Difference between revisions

Latest revision as of 12:49, 4 December 2008

Contents

Fourier Transform

Definitions

Examples

Sifting property of the delta function

More properties of the delta function

Navigation menu

10/09 - Fourier Transform: Difference between revisions

Latest revision as of 12:49, 4 December 2008

Fourier Transform

Definitions

Examples

Sifting property of the delta function

More properties of the delta function

Navigation menu

Search