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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 $ \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 $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010