The derivative of $\sin(\omega_o t)$ is $\cos(\omega_o t)$.
The Fourier transform of $\sin(\omega_o t)$ is $\frac{\pi}{j}[\delta(\omega-\omega_o) - \delta(\omega+\omega_o)]$.
Differentiation in the time domain is equivalent to multiplying the transform by $j\omega$.
The transform of $\cos(\omega_o t)$ is $\pi[\delta(\omega-\omega_o) + \delta(\omega+\omega_o)]$.
What I don't understand is how multiplying the transform of $\sin(\omega_o t)$ by $j\omega$ gives you the transform of $\cos(\omega_o t)$. I see how the $j$'s will cancel out, but how does the sign of that impulse get flipped?
Answer
I assume you mean the derivative with respect to $t$. In that case, the derivative of $\sin(\omega_0t)$ is not $\cos(\omega_0t)$ but $\omega_0\cos(\omega_0t)$. And luckily, this is also obtained via the Fourier transform relation you mentioned in your question:
$$\begin{align}\mathcal{F}\left\{\frac{d}{dt}\sin(\omega_0t)\right\}&=j\omega\cdot \mathcal{F}\left\{\sin(\omega_0t)\right\}\\&=\pi\omega[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]\\&=\pi[\omega_0\delta(\omega-\omega_0)-(-\omega_0)\delta(\omega+\omega_0)]\\&=\pi\omega_0[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]\\&=\omega_0\mathcal{F}\{\cos(\omega_0t)\}\end{align}$$
where I've used the fact that $f(\omega)\delta(\omega-\omega_0)=f(\omega_0)\delta(\omega-\omega_0)$ for any function $f(\omega)$ that is continuous at $\omega=\omega_0$. Consequently you have $\omega\delta(\omega+\omega_0)=-\omega_0\delta(\omega+\omega_0)$.
No comments:
Post a Comment