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| − | CTFT of a cosine | + | =How to obtain the CTFT of a cosine in terms of f in hertz (from the formula in terms of <math>\omega</math>) = |
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| + | Recall: | ||
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| + | <math>x(t)= \cos(\omega_0 t)</math> | ||
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| + | <math>\mathcal{X}(\omega)=\pi \left[\delta (\omega - \omega_0) + \delta (\omega + \omega_0)\right] </math> | ||
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| + | To obtain X(f), use the substitution | ||
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| + | <math>\omega= 2 \pi f </math>. | ||
| + | |||
| + | More specifically | ||
<math> | <math> | ||
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<math>Since\ k\delta (kt)=\delta (t),\forall k\ne 0</math> | <math>Since\ k\delta (kt)=\delta (t),\forall k\ne 0</math> | ||
| + | |||
| + | ---- | ||
| + | [[ECE438_HW1_Solution|Back to Table]] | ||
Latest revision as of 11:50, 15 September 2010
How to obtain the CTFT of a cosine in terms of f in hertz (from the formula in terms of $ \omega $)
Recall:
$ x(t)= \cos(\omega_0 t) $
$ \mathcal{X}(\omega)=\pi \left[\delta (\omega - \omega_0) + \delta (\omega + \omega_0)\right] $
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ \begin{align} X(f) &=\mathcal{X}(2\pi f) \\ &=\pi \left[\delta (2\pi f- \omega_0) + \delta (2\pi f+ \omega_0)\right] \\ &=\frac{1}{2} \left[\delta (f - \frac{\omega_0}{2\pi}) + \delta (f + \frac{\omega_0}{2\pi})\right] \end{align} $
$ Since\ k\delta (kt)=\delta (t),\forall k\ne 0 $
