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| − | + | =How to obtain the CTFT of a impulse train in terms of f in hertz (from the formula in terms of <math>\omega</math>) = | |
| − | + | ||
| − | + | Recall: | |
| − | + | ||
| − | + | <math>x(t)=\sum^{\infty}_{n=-\infty} \delta(t-nT) </math> | |
| − | + | ||
| − | | | + | <math>\mathcal{X}(\omega)=\frac{2\pi}{T}\sum^{\infty}_{k=-\infty}\delta(w-\frac{2\pi k}{T}) </math> |
| + | |||
| + | To obtain X(f), use the substitution | ||
| + | |||
| + | <math>\omega= 2 \pi f </math>. | ||
| + | |||
| + | More specifically | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | X(f) &=\mathcal{X}(2\pi f) \\ | ||
| + | &=\frac{2\pi}{T}\sum^{\infty}_{k=-\infty}\delta(2\pi f-\frac{2\pi k}{T}) \\ | ||
| + | &=\frac{1}{T}\sum^{\infty}_{k=-\infty}\delta(f-\frac{k}{T}) | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | <math>Since\ k\delta (kt)=\delta (t),\forall k\ne 0</math> | ||
| + | |||
| + | ---- | ||
| + | [[ECE438_HW1_Solution|Back to Table]] | ||
Latest revision as of 11:06, 15 September 2010
How to obtain the CTFT of a impulse train in terms of f in hertz (from the formula in terms of $ \omega $)
Recall:
$ x(t)=\sum^{\infty}_{n=-\infty} \delta(t-nT) $
$ \mathcal{X}(\omega)=\frac{2\pi}{T}\sum^{\infty}_{k=-\infty}\delta(w-\frac{2\pi k}{T}) $
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ \begin{align} X(f) &=\mathcal{X}(2\pi f) \\ &=\frac{2\pi}{T}\sum^{\infty}_{k=-\infty}\delta(2\pi f-\frac{2\pi k}{T}) \\ &=\frac{1}{T}\sum^{\infty}_{k=-\infty}\delta(f-\frac{k}{T}) \end{align} $
$ Since\ k\delta (kt)=\delta (t),\forall k\ne 0 $
