To avoid distortion we preprocess $x$ with a low pass filter, i.e., we transform $x$ into a signal $x_{f_s}$ with spectrum $X_{f_s} = \ccalF(x_{f_s})$ where $X_{f_s}(f) = X(f)\sqcap_{f_s}(f)$. This can be implemented as a convolution in the time domain,
\begin{equation} x_{f_s} = x * f_s \mbox{sinc}(\pi f_s t)\ . \end{equation}

The signal $x_{f_s}$ has bandwith $f_s$ and can be sampled without aliasing. In other words, frequency components between $-f_s/2$ and $f_s/2$ are retained without distortion.

# Reconstruction with arbitrary pulse trains

While it is mathematically possible to reconstruct $x(t)$ from $x_\delta(t)$, it is physically implausible to generate a Dirac train. However, we can approximate $\delta(t)$ with narrow pulses $p(t)$. In problem $\textbf{(1.3)}$, you are asked to derive the conditions on $p(t)$ that minimize signal distortion.

To start, the spectra of the actual modulated train pulse using $\delta(t)$ and the approximated modulated train pulse using $p(t)$ can be denoted

\begin{align} X_\delta(f) &= \sum_{k=-\infty}^{\infty} X(f-kf_s), X_p(f) &= P(f)X_\delta(f) = P(f)\sum_{k=-\infty}^{\infty}X(f-kf_s), \end{align}

where $P(f)$ is the spectrum of $p(t)$. The reconstruction method described in $\textbf{(1.1)}$ recovers the signal by low pass filtering the sampled signal in the frequency domain. Using this approach, the spectrum $\tilde{X}_\delta (f)$ of the reconstructed signal using $X_\delta(f)$ and the spectrum $\tilde{X}_p(f)$ of the reconstructed signal using $X_p(f)$ can be computed as

\begin{align} \tilde{X}_\delta(f) &= \sqcap_{f_s}(f)\sum_{k=-\infty}^{\infty} X(f-kf_s) = \sqcap_{f_s}(f)X(f), \tilde{X}_p(f) &= \sqcap_{f_s}(f)P(f)\sum_{k=-\infty}^{\infty} X(f-kf_s) = P(f)\sqcap_{f_s}(f)X(f), \end{align}

where we use the fact that the low pass filter eliminates all frequencies outside of $[-f_s/2,f_s/2]$. Notice that $\tilde{X}_\delta(f)$ coincides with $\tilde{X}_p(f)$, i.e., no distortion results from using $p(t)$ to approximate $\delta(t)$, when the spectrum of $p(t)$ satisfies

\begin{equation} P(f) = 1,\mbox{ for all } f \in [-f_s/2,f_s/2]. \end{equation}

In particular, a pulse satisfying this property is $p(t) = f_s\mbox{sinc}(\pi f_s t)$ with $P(f) = \sqcap_{f_s}(f)$. Interestingly, this pulse is not that narrow.

# Subsampling

In problem $\textbf{(2.1)}$, you are asked to derive a theorem relating the spectrum of the discrete time signal $x$ and its subsampled version $x_\delta$. Starting from equation (11) of the handout, note that it is equivalent to write

\begin{equation} x_\delta(n) = x(n) \sum_{m=-\infty}^{\infty} \delta\left(n – m \frac{\tau}{T_s}\right) \end{equation}

which is $x(n)$ multiplied with a train of discrete deltas with spacing time $\tau$ and sampling time $T_s$. Suppose the sampling time is $\tau$; the train of discrete deltas becomes $\sum_{m=-\infty}^{\infty} \delta(n-m)$ which is a discrete time constant, and its DTFT is a Dirac train with spacing $\nu = 1/\tau$ (see slides 30-32 in the sampling lecture). The following equation in the DTFT derivation will be used later,

\begin{equation} \tau \sum_{n=-\infty}^{\infty} e^{-j2\pi f n \tau} = \sum_{k=-\infty}^{\infty} \delta(f-k\nu)\ . \end{equation}

Back to our setting where the sampling time is $T_s$, the train of discrete deltas is a discrete signal with values either $0$ or $1$, depending on whether $n$ is a multiple of $\tau/T_s$. Its DTFT can be computed as

\begin{equation} X_c(f) = T_s \sum_{n=-\infty}^{\infty} \sum_{m=-\infty}^{\infty} \delta(n – m\tau/T_s) e^{-j 2 \pi f n T_s}\ . \end{equation}

We may define $n’ = n T_s/\tau$ and utilize the result from (9) in evaluating (10),

\begin{align} X_c(f) = T_s \sum_{n’=-\infty}^{\infty} \sum_{m=-\infty}^{\infty} \delta(n’ – m) e^{-j2 \pi f n’ \tau} &= T_s \sum_{n’=-\infty}^{\infty} e^{-j2 \pi f n’ \tau} &= \frac{T_s}{\tau} \sum_{k=-\infty}^{\infty} \delta(f-k\nu)\ . \end{align}

The multiplication in the time domain in (8) implies that the spectrum $X_\delta(f)$ of $x_\delta(n)$ can be written as the convolution of $X(f)$ and the spectrum of the train of discrete deltas. Therefore, we may write

\begin{equation} X_\delta(f) = X(f) * \frac{T_s}{\tau} \sum_{k=-\infty}^{\infty} \delta(f-k\nu) = \frac{T_s}{\tau} \sum_{k=0}^{\tau/T_s – 1} X\left(f-\frac{k}{\tau}\right)\ . \end{equation}

In the above derivation, we encountered two entities that appear to be similar — the train of discrete deltas $\sum_{m=-\infty}^{\infty} \delta(n – m \tau/T_s)$ and the Dirac train $\sum_{k=-\infty}^{\infty} \delta(f – k \nu)$ — and here we emphasize their difference. The train of discrete deltas $x(n) = \sum_{m=-\infty}^{\infty} \delta(n – m \tau/T_s)$ is a discrete signal and $x(n)=1$ whenever $n$ is a multiple of $\tau/T_s$ and zero otherwise. The Dirac train $\sum_{k=-\infty}^{\infty} \delta(f – k \nu)$, on the other hand, is a continuous signal and for each $k \in \mbZ$ there is a Dirac delta function (which is a continuous signal itself) centered at $k \nu$. When $\tau = T_s$, $x(n) = \sum_{m=-\infty}^{\infty} \delta(n – m \tau/T_s) = \sum_{m=-\infty}^{\infty} \delta(n-m)$ is also called a discrete time constant.

This spectrum periodization result is verified in problems $\textbf{(2.2)}$ and $\textbf{(2.3)}$. The gaussian pulse sampled at $f_s = 40$kHz with duration $T=2$ and parameters $\mu=1$ and $\sigma=0.1$ is shown on the top of Figure 2. Its subsampled version, with subsampling frequency $\nu = 4$kHz, is shown on the bottom of the same figure.