Difference between revisions of "DTFTCosinePawling" - Rhea

Lets look at a pure tone frequency F4 = 349Hz

We will represent this tone as a cosine signal, $ cos*(2\pi349t) $

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) $: