Todai Entrance Exam: Subject 2019 – Problem 5

This \delta is Kronecker delta. The equation f_{s}(t) = f(t)\sum_{i=-\infty}^{\infty} \delta(t - it_{s}) can be interpreted as: If there exists a i \in \mathbb{Z} that t = it_{s} then we return f_{s}(t) = f(t), otherwise we return f_{s}(t) = 0.

(1) According the given formula at the end of the problem:

    \[ \delta_{s}(t) = \sum_{k=-\infty}^{\infty} \frac{e^{-j\frac{2\pi kt}{t_{s}}}}{t_{s}}\int_{0}^{t_{s}} \delta_{s}(t)e^{-j\frac{2\pi kt}{t_{s}}}dt = \sum_{k=-\infty}^{\infty} \frac{e^{-j\frac{2\pi kt}{t_{s}}}}{t_{s}}(1 + e^{-j\frac{2\pi kt_{s}}{t_{s}}}) \]

(2) Ok, this is Discrete Fourier Transform:

    \[ \Delta_{s}(w) = \int_{-\infty}^{\infty} \delta_{s}(t)e^{-jwt}dt = \sum_{k=-\infty}^{\infty} e^{-jwkt_{s}} = \sum_{k=-\infty}^{\infty} e^{-\frac{jwk2\pi}{w_{s}}} \]

(3)

By Modulation Property of the Fourier Transform, we have:

    \[ F_{s}(w) = \mathcal{F}\{ f(t)\delta_{s}(t) \} = \frac{1}{2\pi}F(w) * \Delta_{s}(w) = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\tau)\Delta_{s}(w - \tau)d\tau \]

I don’t know much calculus but I think I can apply the inverse Fourier Transform formula here:

    \[ F_{s}(w) = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\tau)\sum_{k=-\infty}^{\infty} e^{-\frac{j(w - \tau)k2\pi}{w_{s}}}d\tau = \sum_{k=-\infty}^{\infty}e^{-\frac{jwk2\pi}{w_{s}}}\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\tau)e^{\frac{j\tau k2\pi}{w_{s}}}d\tau = \sum_{k=-\infty}^{\infty} e^{-\frac{jwk2\pi}{w_{s}}}f\left (\frac{k2\pi}{w_{s}} \right ) \]

    \[ F_{s}(w) = \sum_{k=-\infty}^{\infty} e^{-\frac{jwk2\pi}{w_{s}}}f\left (kt_{s} \right ) \]

(4) Thanks to this and this, basically, aliasing is the phenomenon when the continuous function is not bandwidth limited to less than the Nyquist critical frequency (f_{c} = \frac{1}{2t_{s}} \Leftrightarrow w_{c} = \frac{w_{s}}{2}), the power spectral density that lies outside of the range [-f_{c}, f_{c}] is aliased into the range.

We have:

    \[ F(w) = \int_{-\infty}^{\infty} f(t)e^{-jwt}dt \]

and

    \[ F_{s}(w) = \sum_{k=-\infty}^{\infty}e^{-\frac{jwk2\pi}{w_{s}}} f\left (\frac{k2\pi}{w_{s}} \right ) \]

It seems like F_{s}(w) is a discrete version of F(w) with t being approximated by kt_{s} = \frac{k2\pi}{w_{s}}. So the larger w_{s} is (the bigger the sampling rate is), the more F_{s}(w) approximates F(w).

It is not so easy to see that:

    \[ e^{-\frac{j(w+nw_{s})k2\pi}{w_{s}}} = e^{-\frac{jwk2\pi}{w_{s}}}e^{-\frac{jnw_{s}k2\pi}{w_{s}}} = e^{-\frac{jwk2\pi}{w_{s}}}e^{-jnk2\pi} =e^{-\frac{jwk2\pi}{w_{s}}} \]

So:

    \[ F_{s}(w) = F_{s}(w + nw_{s}) \]

For any w_{1} \notin [-w_{c}, w_{c}], there exists a w_{2} \in [-w_{c}, w_{c}] such that F_{s}(w_{1}) = F_{s}(w_{2}). Therefore, if f(t) is not bandwith limited to less than w_{c}, power spectral density that lies outside of the range [-w_{c}, w_{c}] is aliased into the range.

To avoid aliasing, w_{s} must be larger than twice of the largest non-zero angular frequency of f(t).


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