Fourier Transform Sin & Cos

Question

I would like to calculate the fourier transform of cos(wt) by taking the derivative of the fourier transform of sin(wt).

We know that $\boldsymbol{F}(\dot{x})=jw\boldsymbol{F}(x)$ -->

$\boldsymbol{F}(cos(w_{o}t))=\boldsymbol{F}(\dot{\frac{1}{w_{o}}sin(w_{o}t))}=\frac{1}{w_{o}}\boldsymbol{F}(\dot{sin(w_{o}t)})=\frac{1}{w_{o}}jw\boldsymbol{F}(sin(w_{o}t))=\\\frac{1}{w_{o}}jw*\frac{1}{2j}(\delta (w-w_{o})-\delta (w+w_{o}))=\frac{w}{2w_{o}}(\delta (w-w_{o})-\delta (w+w_{o}))$

since $\delta (w-w_{o})-\delta (w+w_{o})$ is non-zero only at $w=\pm w_{o}$ ; $\boldsymbol{F}(cos(w_{o}t))$ turns out to be

$\boldsymbol{F}(cos(w_{o}t)) = \frac{1}{2}(\delta (w-w_{o})-\delta (w+w_{o}))$

However we know that $\boldsymbol{F}(cos(w_{o}t)) = \frac{1}{2}(\delta (w-w_{o})+\delta (w+w_{o}))$

I could not figure out the mistake I am doing in my derivation. Can you please point out my mistake?

Thanks and regards,

Answer 1

We have

$$\begin{align} \mathscr{F}(\cos(\omega_0 t))&=\frac{1}{\omega_0}\mathscr{F}\left(\frac{d\sin(\omega_0 t)}{dt}\right)\\\\ &=-i\frac{\omega}{\omega_0} \mathscr{F}(\sin(\omega_0 t))\\\\ &=\left(-i\frac{\omega}{\omega_0} \right)\frac{1}{2i}(\delta(\omega +\omega_0)-\delta(\omega -\omega_0))\\\\ &=\frac{1}{2}\left(-\frac{\omega}{\omega_0}\delta(\omega +\omega_0)+\frac{\omega}{\omega_0}\delta(\omega -\omega_0)\right) \tag 1\\\\ &=\frac12 (\delta(\omega+\omega_0)+\delta(\omega-\omega_0)) \tag 2 \end{align}$$

as was to be shown!

In going from $(1)$ to $(2)$, we exploited the property that $f(x)\delta(x-x_0)=f(x_0)\delta(x-x_0)$.

Applied here, we note that for the term $-\frac{\omega}{\omega_0}\delta(\omega +\omega_0)$, the Dirac Delta is "active" when $\omega=-\omega_0$. Therefore, at $\omega=-\omega_0$, $-\frac{\omega}{\omega_0}=+1$.

Answer 2

you got it right until the end, in fact $\frac{1}{2}(\delta (w-w_{o})+\delta (w+w_{o}))=\frac{w}{2w_{o}}(\delta (w-w_{o})-\delta (w+w_{o}))$