HW 05: Difference between revisions

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|<math>=2\pi\delta(\omega_0-\omega) - 2\pi\delta(\omega_0+\omega)\,\!</math>
|<math>=2\pi\delta(\omega_0-\omega) - 2\pi\delta(\omega_0+\omega)\,\!</math>
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|<math>F[\sum_{-\infty}^{\infty}\alpha_n e^{j2\pi nt/T}]</math>
|<math>=\int_{-\infty}^{\infty} \left (\sum_{-\infty}^{\infty}\alpha_n e^{j2\pi nt/T} \right )e^{-j \omega t}dt</math>
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|<math>=\sum_{-\infty}^{\infty}\alpha_n \left (\int_{-\infty}^{\infty} e^{j2\pi nt/T} e^{-j2\pi ft}dt\right )</math>
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||<math>=\sum_{-\infty}^{\infty}\alpha_n \left (\int_{-\infty}^{\infty} e^{j2\pi t (\frac{n}{T}-f)} dt\right )</math>
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*Think something went wrong here.

Revision as of 00:43, 18 November 2008

Find the following Fourier Transforms

  • F[ejω0t]
  • F[cosω0t]
  • F[αnej2πnt/T]
  • F[sinω0t]

Solutions

F[ejω0t] =ejω0tejωtdt
=ej(ω0ω)tdt
=2π[12πej(ω0ω)tdt]
=2πδ(ω0ω)
F[cosω0t] =ejω0t+ejω0t2ejωtdt
=12(ejω0t+ejω0t)2ejωtdt
=12[2ej(ω0ω)t+2ej(ω0+ω)t]dt
=2π[12π(ej(ω0ω)t+ej(ω0+ω)t)dt]
=2πδ(ω0ω)+2πδ(ω0+ω)
F[sinω0t] =ejω0tejω0t2jejωtdt
=12j(ejω0tejω0t)2jejωtdt
=(ej(ω0ω)tej(ω0+ω)t)dt
=2π[12π(ej(ω0ω)tej(ω0+ω)t)dt]
=2πδ(ω0ω)2πδ(ω0+ω)
F[αnej2πnt/T] =(αnej2πnt/T)ejωtdt
=αn(ej2πnt/Tej2πftdt)
=αn(ej2πt(nTf)dt)
  • Think something went wrong here.