The figure below shows Fourier spectrum of a signal $g(t)$ figure

Sketch the spectrum of the signal $2g(t)\cos^2(100\pi t)$. Show value in sketch.

  • $\begingroup$ I assume "100pit" is supposed to mean $100\pi t$? Writing "100pit" is a horrible way to express that - you don't absolutely have to use MathJab for simple formulas such as this, but you at least have to make it legible by adding spaces... $\endgroup$ – fgp May 7 '14 at 12:11
  • $\begingroup$ If you split the $\cos^2$ into a sum of fourier components (double angle formula and Euler's formula), you'll find it just stacks a couple of copies of the spectrum at different frequency shifts. This is exactly how AM radio works. $\endgroup$ – orion May 7 '14 at 12:22

You are probably supposed to use that $$ \mathcal{F}\left(g\cdot h\right) = \left(\mathcal{F}g\right) * \left(\mathcal{F}g\right) \text{,} $$ where $*$ denotes the convolution of $\mathcal{F}g$ and $\mathcal{F}h$, defined by $$ (F * G)(\omega) = \int_{-\infty}^\infty F(\lambda)G(\omega - \lambda) \,d\lambda \text{.} $$

Note, however, that you need to turn to distributions to do this computation, because the fourier transform of $h(t) = \cos^2(100\pi t)$ isn't a function. For shifted $\delta-$distributions, i.e. "functions" $\delta_{x_0}$ that are zero everywhere except at some $x_0$, but with "$\int_{-\infty}^\infty \delta(x) \,dx = 1$", you have that $$ (F \cdot \delta_{\omega_0})(\omega) = F(\omega - \omega_0) \text{.} $$

Since $\cos^2(x) = \frac{1}{2} + \frac{1}{2}\cos(2x)$, the distributional fourier transform of $h(t) = 2\cos^2(100\pi t)$ is $$ H = \mathcal{F}\left(\cos^2(100\pi t)\right) = \delta_0 + \tfrac{1}{2}\delta_{200\pi} + \tfrac{1}{2}\delta_{-200\pi} \text{,} $$ and therefore $$\begin{eqnarray} \mathcal{F}\left(g \cdot h\right) &=& F * \left(\delta_0 + \delta_{200\pi}\right) = (F * \delta_0) + \tfrac{1}{2}(F * \delta_{200\pi}) + \tfrac{1}{2}(F * \delta_{-200\pi}) \\ &=& F(\omega) + \tfrac{1}{2}F(\omega - 200\pi) + \tfrac{1}{2}F(\omega + 200\pi) \text{.} \end{eqnarray}$$

The spectrum of $2g(t)\cos^2(100\pi t)$ is the the original spectrum of $g$, plus two copies of that spectrum shifted by $200\pi$ and $-200\pi$ and scaled by $\frac{1}{2}$.

As orion already points out in the comments, this is the basic principle behind AM modulation. You multiply your signal with a carrier, and because of the above the spectrum of the resulting signal contains a version of the original signal shifted by the frequency of the carrier.

Note that the above assume that the definition of the fourier transform $F = \mathcal{F}(f)$ is such that the reverse transform is given by $$ f(t) = \int_{-\infty}^\infty F(\omega)e^{i\omega t} \,d\omega \text{.} $$ There are slightly different ways to define the fourier transform (the reverse transform can have an additional factor, or have $-i\omega t$ in place of $i\omega t$, or something like that), and for these you have to adjust things slightly above.

| cite | improve this answer | |
  • 1
    $\begingroup$ Don't forget that $\cos (200\pi t)$ contains two fourier components - negative and positive frequency. $\endgroup$ – orion May 7 '14 at 12:37
  • $\begingroup$ @orion Yup, realized that right after posting. Fixed now. Thanks. $\endgroup$ – fgp May 7 '14 at 12:41
  • $\begingroup$ Smart to use $\cos^2$ identity. I would have done it the long way with $G(\omega) \ast \left( \delta( \omega - 100\pi ) + \delta( \omega + 100\pi ) \right) \ast \left( \delta( \omega - 100\pi ) + \delta( \omega + 100\pi ) \right)$. You'd get the same thing this way, but with more legwork. $\endgroup$ – AnonSubmitter85 May 7 '14 at 14:01
  • $\begingroup$ @AnonSubmitter85 Yup ;-) Some EE background helps here - the fact that squaring a $\sin$ or $\cos$ doubles the frequency and adds an offset is a frequently used fact there. $\endgroup$ – fgp May 7 '14 at 14:55

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.