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?


2 Answers 2


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! $\endgroup$ Oct 24, 2016 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.


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