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?


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 $\omega\delta(\omega+\omega_0)=-\omega_0\delta(\omega+\omega_0)$.

  • $\begingroup$ Forgot about that chain rule, and understand the rest too, thanks! $\endgroup$ – CMDoolittle Oct 24 '16 at 18:44

So, the delta function satisfies: $$\int_{-\infty}^{\infty} f(x) \delta(x - a)\, \mathrm{d}x = f(a)$$ Now, suppose we substitute $f(x) = x$ $$\int_{-\infty}^{\infty} f(x) \delta(x - a)\, \mathrm{d}x = a = \int_{-\infty}^{\infty} a \delta(x - a)\, \mathrm{d}x$$ This means that multiplying by $\omega$ is the same as multiplying by a constant. If you substitute in your equation you'll get the result. Keep in mind that the derivative of your sine function will be multiplied by a constant term.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.