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| − | + | =How to obtain the CTFT of a complex exponential in terms of f in hertz (from the formula in terms of <math>\omega</math>) = | |
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| − | + | To obtain X(f), use the substitution | |
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| − | + | <math>\omega= 2 \pi f </math>. | |
| − | + | ||
| − | + | More specifically | |
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| − | + | <math>a.\text{ } x(t)=e^{i\omega_0 t}</math> | |
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| + | <math>\mathcal{X}(\omega)=2\pi \delta (\omega - \omega_0)</math> | ||
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| + | <math>X(f)= \mathcal{X}(2\pi f)=2\pi \delta (2\pi f-\omega_0)</math> | ||
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| + | <math>Since\text{ } k\delta (kt)=\delta (t),\forall k\ne 0</math> | ||
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| + | <math>X(f)=\delta (f-\frac{\omega_0}{2\pi})</math> | ||
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| + | <math>b.\text{ } x(t)=e^{-at}u(t)\ </math>, where <math>a\in {\mathbb R}, a>0 </math> | ||
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| + | <math>\mathcal{X}(\omega)=\frac{1}{a+i\omega}</math> | ||
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| + | <math>X(f)= \mathcal{X}(2\pi f)=\frac{1}{a+i2\pi f}</math> | ||
| + | |||
| + | <math>c.\text{ } x(t)=te^{-at}u(t)\ </math>, where <math>a\in {\mathbb R}, a>0 </math> | ||
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| + | <math>\mathcal{X}(\omega)=\left( \frac{1}{a+i\omega}\right)^2</math> | ||
| + | |||
| + | <math>X(f)= \mathcal{X}(2\pi f)=\left( \frac{1}{a+i2\pi f}\right)^2</math> | ||
Latest revision as of 11:16, 15 September 2010
How to obtain the CTFT of a complex exponential in terms of f in hertz (from the formula in terms of $ \omega $)
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ a.\text{ } x(t)=e^{i\omega_0 t} $
$ \mathcal{X}(\omega)=2\pi \delta (\omega - \omega_0) $
$ X(f)= \mathcal{X}(2\pi f)=2\pi \delta (2\pi f-\omega_0) $
$ Since\text{ } k\delta (kt)=\delta (t),\forall k\ne 0 $
$ X(f)=\delta (f-\frac{\omega_0}{2\pi}) $
$ b.\text{ } x(t)=e^{-at}u(t)\ $, where $ a\in {\mathbb R}, a>0 $
$ \mathcal{X}(\omega)=\frac{1}{a+i\omega} $
$ X(f)= \mathcal{X}(2\pi f)=\frac{1}{a+i2\pi f} $
$ c.\text{ } x(t)=te^{-at}u(t)\ $, where $ a\in {\mathbb R}, a>0 $
$ \mathcal{X}(\omega)=\left( \frac{1}{a+i\omega}\right)^2 $
$ X(f)= \mathcal{X}(2\pi f)=\left( \frac{1}{a+i2\pi f}\right)^2 $
