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I was trying to solve fourier transform of $cos(2\pi f_0 t)u(t)$ where $u(t)$ is a unit step function,

$$ u(t) = \begin{cases} 1 \text{ for t$\ge$ 0}\\ 0 \text{ for $t< 0$}\\ \end{cases} $$

As far i have done that $$ \mathscr{F}\{cos(2\pi f_0 t)u(t)\} = \mathscr{F}\{cos(2\pi f_0 t)\} * \mathscr{F}\{u(t)\}\\ = \frac{1}{2}[\delta(f-f_0)+\delta(f+f_0)]*[\frac{1}{2}\delta(f)+\frac{1}{2j\pi f}] $$ I know that $\delta(f-f_0)*x(t) = x(f-f_0)$, so $$ \frac{1}{2}[\delta(f-f_0)+\delta(f+f_0)]*[\frac{1}{2}\delta(f)+\frac{1}{2j\pi f}]\\ = \frac{1}{4j\pi (f-f_0)}+\frac{1}{4j\pi (f+f_0)}+\frac{1}{2}[\delta(f-f_0)+\delta(f+f_0)]*[\frac{1}{2}\delta(f)] $$ but what is the value of $\frac{1}{2}[\delta(f-f_0)+\delta(f+f_0)]*[\frac{1}{2}\delta(f)] = ?$

Thanks in advance.

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1 Answer 1

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For every distribution $u$ one has $u(f) * \delta(f) = u(f)$ so, $$\delta(f \pm f_0) * \delta(f) = \delta(f \pm f_0).$$

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