Latest revision as of 10:57, 15 September 2010

How to obtain the CTFT of a rect in terms of f in hertz (from the formula in terms of $\omega$ )

Recall:

$x(t)=\left\{\begin{array}{ll}1, & \text{ if }|t|<T,\\ 0, & \text{else.}\end{array} \right.$

$\mathcal{X}(\omega)=\frac{2 \sin \left( T \omega \right)}{\omega}$

To obtain X(f), use the substitution

$\omega= 2 \pi f$ .

More specifically

$X(f)=\mathcal{X}(2\pi f)=\frac{\sin \left(2\pi Tf \right)}{\pi f} \$

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-{|
+=How to obtain the CTFT of a rect in terms of f in hertz (from the formula in terms of <math>\omega</math>) =
-| align="left" style="padding-left: 0em;" | CTFT of a rect
-|-
+Recall:
-| <math> X(f)=\mathcal{X}(2\pi f)=\frac{\sin \left(2\pi Tf \right)}{\pi f}  \ </math>
-|-
+<math>x(t)=\left\{\begin{array}{ll}1, &  \text{ if }|t|<T,\\ 0, & \text{else.}\end{array} \right.</math>
-|}
+<math>\mathcal{X}(\omega)=\frac{2 \sin \left( T \omega \right)}{\omega}</math>
+To obtain X(f), use the substitution
+<math>\omega= 2 \pi f </math>.
+More specifically
+<math> X(f)=\mathcal{X}(2\pi f)=\frac{\sin \left(2\pi Tf \right)}{\pi f}  \ </math>
+----
+[[ECE438_HW1_Solution|Back to Table]]