As I know, the Fourier transform has the below property which is called time-differentiation:

$$ \frac{dx(t)}{dt}\leftrightarrow j\omega X(j\omega) $$

and the Fourier transform of the cosine and the sine signals are $\pi(\delta(\omega-\omega_0)+ \delta(\omega+\omega_0))$ and $\frac {\pi} {j}(\delta(\omega-\omega_0)- \delta(\omega+\omega_0))$, respectively.

The derivative of $\sin(t)$ is $\cos(t)$. However, the Fourier transform of $\cos(t)$ cannot be obtained from multiplying the Fourier transform of $\sin(t)$ with $j\omega$.

could any one help me.


It does work. Note that the Fourier transforms you stated in your question are not the transforms of $\cos(t)$ and $\sin(t)$, but of $\cos(\omega_0t)$ and $\sin(\omega_0t)$.

Let's Assume we know the transform of $\sin(\omega_0t)$:


Now we take the derivative of $\sin(\omega_0t)$:


Using $(1)$ and the differentiation property of the Fourier transform, the Fourier transform of $(2)$ must be

$$\mathcal{F}\{\omega_0\cos(\omega_0t)\}=j\omega\cdot \frac{\pi}{j}[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]\tag{3}$$

Noting that for any function $f(\omega)$ that is continuous at $\omega=\omega_0$ we have


and, consequently, we can rewrite $(3)$ as

$$\begin{align}\mathcal{F}\{\omega_0\cos(\omega_0t)\}&= \pi[\omega_0\delta(\omega-\omega_0)-(-\omega_0)\delta(\omega+\omega_0)]\\&=\omega_0\pi[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]\end{align}\tag{5}$$

as expected.

| improve this answer | |
  • $\begingroup$ Thanks a lot Matt. I have forgotten to use the sifting property of the Delta Dirac signal. $\endgroup$ – AllEs Aug 22 '16 at 6:37
  • $\begingroup$ @AllEs: You're welcome! If the answer was helpful please accept it by clicking on the check mark to its left. $\endgroup$ – Matt L. Aug 22 '16 at 8:34
  • $\begingroup$ I've done. If I should do any other work, please inform me. $\endgroup$ – AllEs Aug 22 '16 at 9:48

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