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I know that in general it transforms the signal from the time to the frequency field but these specific cases seem pretty demanding. Do I calculate each part separately and then just leave them with convolution between them? Or do I have to calculate any integrals?

\begin{align} &\frac{1}{t^2}\cdot\cos(2πt)\cdot\cos(2πt)\\ &8\cos(20πt)\cdot\cos(40\pi{t})\cdot\cos(80\pi{t}) \end{align}

For example for the second one will the result be

$$\bigg(4\big[\delta(f-10)+\delta(f+10)\big]\bigg)\star \bigg(\frac 12\big[\delta(f-20)+\delta(f+20)\big]\bigg)\star\bigg(\frac 12\big[\delta(f-40)+\delta(f+40)\big]\bigg)$$

Is that correct?

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  • $\begingroup$ I would just use software to find the transforms. Is there any reason you need the mathematical expressions for those transforms? $\endgroup$
    – MBaz
    Jan 26, 2021 at 14:15
  • $\begingroup$ Well i am studying for a lesson at the university and i am confused about which method i should use to obtain the singal in the frequency field. $\endgroup$
    – Sarah
    Jan 26, 2021 at 15:00
  • $\begingroup$ In that case, then yes, you can find the FT of $x(t)y(t)$ as $X(f) \ast Y(f)$, if that is easier than integrating $x(t)y(t)$ directly. $\endgroup$
    – MBaz
    Jan 26, 2021 at 15:04
  • $\begingroup$ @Sarah, with the Fourier transform the good thing is that there are so many properties that are ready to use if understood properly. Armed with these, you can save yourself a lot of time. Please also see my answer below. $\endgroup$
    – Gilles
    Jan 28, 2021 at 7:30

1 Answer 1

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HINT:

It is easy to see that things can be simplified using the trigonometric product-to-sum identity in Equation $(1)$ below: $$ \cos(\alpha)\cos(\beta) = \frac 12\big[\cos(\alpha+\beta) + \cos(\alpha-\beta)\big]\tag{1} $$

  • In the first example $$ \frac{1}{t^2}\cdot\cos(2πt)\cdot\cos(2πt) = \frac 1{2t^2}\big(\cos(4\pi t) + 1\big) $$ From there you would want to visit the cosine modulation frequency-shift and differentiation properties of the Fourier transform, here in $(2)$ and $(3)$ respectively. \begin{align} \mathcal F\big\{x(t)\cos(2\pi f_0 t)\big\} &= \frac 12\big[X(f - f_0) + X(f + f_0)\big]\tag{2}\\ \mathcal F\left\{\frac{d^n x(t)}{dt^n}\right\}&= \left(j2\pi f\right)^nX(f)\tag{3} \end{align}

  • In the second example, using Equation $(1)$ you then have \begin{align} 8\cos(20\pi t)\cos(40\pi t)\cos(80\pi t) & = 8\bigg(\frac 12\big(\cos(60\pi t) + \cos(20\pi t)\big)\cos(80\pi t)\bigg)\\ & = 4\big(\cos(60\pi t) + \cos(20\pi t)\big)\cos(80\pi t)\\ & = 2\big(\cos(140\pi t) + \cos(20\pi t)\big)\\ &\quad + 2\big(\cos(100\pi t) + \cos(60\pi t)\big)\\ \end{align} With this you simply have sums and you don't have to think of convolutions, you have individual sinusoids at frequencies $10\ \rm Hz$, $30\ \rm Hz$, $50\ \rm Hz$, and $70\ \rm Hz$.

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