HW 05: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
Fonggr (talk | contribs)
Fonggr (talk | contribs)
Line 60: Line 60:
|-
|-
|
|
|<math>=\sum_{-\infty}^{\infty}\alpha_n \delta_{\frac{n}{T}-f} </math>
|<math>=\sum_{-\infty}^{\infty}\alpha_n \delta\left(\frac{n}{T}-f\right) </math>
|-
|
|<math>=\alpha_{fT}\,\!</math>
|}
|}
Is the last problem done correctly?

Revision as of 21:25, 23 November 2008

Find the following Fourier Transforms

  • F[ejω0t]
  • F[cos⁡ω0t]
  • F[∑−∞∞αnej2πnt/T]
  • F[sin⁡ω0t]

Solutions

F[ejω0t] =∫−∞∞ejω0te−jωtdt
=∫−∞∞ej(ω0−ω)tdt
=2π[12π∫−∞∞ej(ω0−ω)tdt]
=2πδ(ω0−ω)
F[cos⁡ω0t] =∫−∞∞ejω0t+e−jω0t2e−jωtdt
=12∫−∞∞(ejω0t+e−jω0t)e−jωtdt
=12∫−∞∞[ej(ω0−ω)t+e−j(ω0+ω)t]dt
=π[12π∫−∞∞(ej(ω0−ω)t+e−j(ω0+ω)t)dt]
=πδ(ω0−ω)+πδ(ω0+ω)
F[sin⁡ω0t] =∫−∞∞ejω0t−e−jω0t2je−jωtdt
=12j∫−∞∞(ejω0t−e−jω0t)e−jωtdt
=12j∫−∞∞(ej(ω0−ω)t−e−j(ω0+ω)t)dt
=πj[12π∫−∞∞(ej(ω0−ω)t−e−j(ω0+ω)t)dt]
=−jπδ(ω0−ω)+jπδ(ω0+ω)
F[∑−∞∞αnej2πnt/T] =∫−∞∞(∑−∞∞αnej2πnt/T)e−jωtdt
=∑−∞∞αn(∫−∞∞ej2πnt/Te−j2πftdt)
=∑−∞∞αn(∫−∞∞ej2πt(nT−f)dt)
=∑−∞∞αnδ(nT−f)