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How do I should sample two continuous sinusoids of $f_1$=1/4 Hz and $f_2$=1/2 Hz at a 1 Hz rate?

Can i recover both if i bandlimit after sampling at a 1/2 Hz limit?

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I have:

  • The function $h_1(t)=\cos(2\pi\frac14 t)$ sampled by $|||(t)$ has fourier transform $H_1(f)=4||(4f)*|||(f)=2|||(2(f-1/4))$.
  • The function $h_2(t)=\cos(2\pi\frac12 t)$ sampled by $|||(t)$ has fourier transform $H_2(f)=2||(2f)*|||(f)=|||(f-1/2)$,

When I bandlimit $h_1$ with $\Pi(2f)$ I obtain $4||(4f)$. If I try to bandlimit $h_2$ by the same window, exploiting the Heaviside half maximum convention $\Pi(\pm1/2)=1/2$, I obtain $2||(2f)$.

Both sinuoids are perfectly recovered, though the Heaviside limit in the second case is not applicable in practice for dozens of reasons.

In here, I am using the pulse, sampling and sinusoidal symbols. $H$ is the Heaviside unit step function: $$ \Pi(f)=H(f+1/2)-H(f-1/2),\\ |||(t)=\sum_{t_n \in \mathbb{Z}}\delta(t-t_n)\\ ||(t)=\frac12\delta(t+1)+\frac12\delta(t-1) $$

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  • $\begingroup$ i think i would write the "Sha function" as $$ \operatorname{III}(t) \triangleq \sum\limits_{n=-\infty}^{\infty} \delta(t-n) $$ dunno how easy it is to get "Ш" as a character in $\LaTeX$. $\endgroup$ – robert bristow-johnson May 10 '18 at 19:14
  • $\begingroup$ as long as you're using "$H(\cdot)$" for transfer functions or frequency response, it might not be a good idea to use "$H()$" for the unit step function. maybe use "$\mathrm{u}(\cdot)$". "$\Pi(\cdot)$" is a good symbol for the rectangular function. still your "Sha function" symbols should maybe be changed to something more conventional. what you have is a mixture of conventions from the mathematical literature and the electrical engineering literature. i might suggest picking one convention. $\endgroup$ – robert bristow-johnson May 10 '18 at 22:17

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