Difference between revisions of "DTFTCosinePawling" - Rhea

• Introduction
• Sampling Above Nyquist Frequency
• Sampling Below Nyquist Frequency
• Conclusion

Sampling Above Nyquist Frequency

For this signal $ f_{s} > 2f_{m} = 2(349)Hz = 698Hz $ or else aliasing will occur. We will choose a sampling frequency of $ f_{s} = 1/T_{1} = 1000Hz $.

$ \begin{align} x_{1}(n) &= x(nT_{1}) \\ &= cos(2\pi349nT_{1}) \\ &= cos(\frac{2\pi349n}{1000}) \\ &= \frac{1}{2}(e^{\frac{-j2\pi349n}{1000}} + e^{\frac{j2\pi349n}{1000}}) \\ \\ \\ \end{align} $

$ Note\ that:\ 0 < |\pm2\pi\frac{349}{1000}| < \pi $

This means the original signal can be properly represented when sampled at $ f_{s} = 1000Hz. $

Using the discrete-time Fourier transform pair for cosine:

$ x[n] \ \ \ \ \ \ \ \ \ \ -----> \ \ \ \ X(\omega) $

$ cos(\omega_{0}n) \ \ \ \ -----> \ \ \ \ \pi\sum_{k=-\infty}^\infty (\delta(\omega -\omega_{0}+2\pi k) + \delta(\omega + \omega_{0}+2\pi k)) $

$ \begin{align} X_{1}(\omega) &= \pi \sum_{k=-\infty}^\infty(\delta(\omega - 2\pi\frac{349}{1000} +2\pi k) + \delta(\omega + 2\pi\frac{349}{1000} + 2\pi k)) \\ &= rep_{2\pi}\pi (\delta(\omega - 2\pi\frac{349}{1000}) + \delta(\omega + 2\pi\frac{349}{1000})) \\ &= rep_{2\pi}\frac{1000}{2} ((\delta(\frac{1000}{2\pi}\omega - 349) + \delta(\frac{1000}{2\pi}\omega + 349)) \\ \end{align} $

Plot of $ X_{1}(\omega) $:

Repeats every $ 2\pi $

The frequency content of the original signal lies within the -$ \pi $ to $ \pi $ band. Therefore the signal can be properly reconstructed.

Sampling Below Nyquist Frequency

For this signal $ f_{s} > 2f_{m} = 2(349)Hz = 698Hz $ or else aliasing will occur. We will choose a sampling frequency that does not satisfy this condition this time. Let's use $ f_{s} = 1/T_{2} = 500Hz $.

$ \begin{align} x_{2}(n) &= x(nT_{2}) \\ &= cos(2\pi349nT_{2}) \\ &= cos(\frac{2\pi349n}{500}) \\ &= \frac{1}{2}(e^{\frac{-j2\pi349n}{500}} + e^{\frac{j2\pi349n}{500}}) \\ \\ \\ \end{align} $

$ Note\ that:\ \pi < |\pm2\pi\frac{349}{500}| < 2\pi $

This means the original signal cannot be properly represented when sampled at $ f_{s} = 500Hz. $

Using the discrete-time Fourier transform pair for cosine again:

$ x[n] \ \ \ \ \ \ \ \ \ \ -----> \ \ \ \ X(\omega) $

$ cos(\omega_{0}n) \ \ \ \ -----> \ \ \ \ \pi\sum_{k=-\infty}^\infty (\delta(\omega -\omega_{0}+2\pi k) + \delta(\omega + \omega_{0}+2\pi k)) $

$ \begin{align} X_{1}(\omega) &= \pi \sum_{k=-\infty}^\infty(\delta(\omega - 2\pi\frac{349}{500} +2\pi k) + \delta(\omega + 2\pi\frac{349}{500} + 2\pi k)) \\ &= rep_{2\pi}\pi (\delta(\omega - 2\pi\frac{349}{500}) + \delta(\omega + 2\pi\frac{349}{500})) \\ &= rep_{2\pi}\frac{500}{2} ((\delta(\frac{500}{2\pi}\omega - 349) + \delta(\frac{500}{2\pi}\omega + 349)) \\ \end{align} $

Use periodicity property of cosine to shift from -$ 2\pi $ to $ 2\pi $ to -$ \pi $ to $ \pi $ band.

$ X_{1}(\omega) = rep_{2\pi}\frac{500}{2} ((\delta(\frac{500}{2\pi}\omega - 151) + \delta(\frac{500}{2\pi}\omega + 151)) $

Plot of $ X_{2}(\omega) $:

As you can see the original signal frequency exists outside of the -$ \pi $ to $ \pi $ band. The frequencies that exists within this band are copies of the original signal that are 'rep'ed. Therefore these frequencies do not represent the frequency content of the the original signal, resulting in aliasing.

Conclusion

When sampling above the Nyquist frequency the signal can be accurately represented but when sampling below the signal will become aliased.

Questions and Comments