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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.

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  • $\begingroup$ 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
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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.

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    $\begingroup$ 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

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