1. Relationship between continuous and discrete signals.
Task
Let the signal \(v_0(t)\) be sampled at the sampling frequency \(f_s\).
 Under which condition is the information of \(v_o(t)\) fully preserved after sampling?
 Explain the meaning of 'aliasing'.
 How can the original continuous signal \(v_0(t)\) be reconstructed from its sampled version?
 Consider the signal \(v_0(t)=\mbox{cos}(2\pi\cdot 12000\cdot t)\) sampled at \(f_s = 16000 \). Sketch and interpret the FOURIER spectrum.
Amount and difficulty
 Working time: approx. 20 minutes
 Difficulty: easy
Solution

The Sampling theorem has to be fulfilled.
If a continuous signal \(v_o(t)\) has a bandlimited FOURIER transform \(V_o(j\omega)\), that is, \(V_o(j\omega) = 0\) for \(\omega \geq 2\pi f_c\), then \(v_o(t)\) can be uniquely reconstructed without errors from equally spaced samples \(v_o(nT_s), \infty <n < \infty \), if \begin{equation*} f_s > 2\cdot f_c \end{equation*} where \(f_s = \frac{1}{T_s}\) is the sampling frequency and \(f_c\) is the cutoff frequency.
 If the signal \(v_o(t)\) is sampled below the Nyquist frequency (\(f_s=2f_c\)), then distortions due to the spectral fold over (aliasing) occur.
 The original signal can be reconstructed by applying a lowpass filter whose cutoff frequency \(\omega_{\text{LP}}\) lies between \(\omega_c<\omega_{\text{LP}}<\omega_s  \omega_c\).
 The following approach was used: \begin{equation*}V_o(j\omega) = \mathfrak{F}\{v_o(t)\} = \mathfrak{F}\{\cos(2\pi \omega t)\} = \pi \left[ \delta_0 (\omega \omega_0) + \delta_0 (\omega +\omega_0) \right] \end{equation*} Due to the repetition of the spectra caused by the sampling, the FOURIER spectrum is given by: \begin{align*} V_s(j\omega) &= \sum \limits_{\mu = \infty}^{\infty} V_o\left( j[\omega  \mu \omega_s] \right)\\ V_s(j\omega) &= \pi \sum \limits_{\mu = \infty}^{\infty} \left[ \delta_0 (\omega \omega_0 \mu \omega_s) + \delta_0 (\omega +\omega_0 \mu \omega_s) \right] \end{align*} The following result can be sketched and it can be concluded that the sampling theorem is not fulfilled.
2. Sampling of periodic signals.
Task
Let the signal \(v_0(t)\) be sampled at the sampling frequency \(f_s\).
 Consider the Tperiodic signal \(v_o(t)\) sampled at \(f_s=1/T_s\). Determine \(\alpha = T/T_s\) for which the discrete (sampled) signal \(v(n)\) is periodic.
 Assume the signal \(v_o(t) = \sin(\omega_0t)\), where \(\omega_0 = \frac{2\pi}{T}\), is sampled with \(T_s=T/\alpha\). Show whether or not \(v(n)\) is periodic and determine its period \(K\) (if possible) for each of the following cases:
 \(\alpha = 5,\)
 \(\alpha = 5.5,\)
 \(\alpha = \frac{16}{3},\)
 \(\alpha = \pi,\)
 \(\alpha = 1. \)
Amount and difficulty
 Working time: approx. 25 minutes
 Difficulty: easy
Solution
 The Signal \(v_o(t)\) is a continuous Tperiodic signal, defined by \begin{equation*} v_o(t) = v_o(t+\mu T) \ \ \ ,\ \ \mu\in \mathbb{Z}, \ \ T\in \mathbb{R}^+ \text{ .} \end{equation*} The sequence \(v(n)\) is defined by: \begin{equation*} v(n) = v_o(nT_s)=v_o(nT_s+\mu T) = v_o(n T_s + \mu \alpha T_s) = v_o((n+\mu \alpha)T_s) = v(n+\mu \alpha) \end{equation*}
The sequence is periodic if one value \(\mu\) exists for which \(K = \mu\cdot \alpha \in \mathbb{N}\) (natural, nonzero number) holds true.
The period \(K\) is given by
\begin{equation} K = \min\limits_{\mu} \left\{ \mu \cdot \alpha \  \ \mu \cdot \alpha \in \mathbb{N} \right\} \end{equation} and it is depended on \(\alpha\). \begin{align*} 1) \ & \alpha \in \mathbb{N} : && \mu_{\text{min}} = 1 \ \rightarrow \ K=\alpha & \Longrightarrow v(n) \text{ is periodic}\\ 2) \ & \alpha \in \mathbb{Q}^+ : && \alpha = \frac{m}{n} \ \text{ where } m,n \in \mathbb{N} & \\ \ & && \mu_{\text{min}} = n \ \rightarrow \ K=m & \Longrightarrow v(n) \text{ is periodic}\\ 3) \ & \alpha \in \mathbb{R}^+ \setminus \mathbb{Q}^+ : && \text{there is no } \mu \in \mathbb{Z} \text{ for which} & \\ \ & && \mu \cdot \alpha \in \mathbb{N} & \Longrightarrow v(n) \text{ is nonperiodic} \end{align*}  The period \(K\) can be found by utilizing equation (1). \begin{align*} (i) \ & \alpha = 5: && \mu \cdot \alpha = 5, \ \mu_{\text{min}} = 1, \ K=5 & \Longrightarrow v(n) \text{ is periodic}\\ (ii) \ & \alpha = 5.5:&& \mu \cdot \alpha = 11, \ \mu_{\text{min}} = 2, \ K=11 & \Longrightarrow v(n) \text{ is periodic}\\ (iii) \ & \alpha = \frac{16}{3}: && \mu \cdot \alpha = 16, \ \mu_{\text{min}} = 3, \ K=16 & \Longrightarrow v(n) \text{ is periodic}\\ (iv) \ & \alpha = \pi: && \text{there is no } \mu \cdot \alpha \in \mathbb{N} \text{ with } \mu \in \mathbb{Z} & \Longrightarrow v(n) \text{ is nonperiodic}\\ (v) \ & \alpha = 1: && \mu \cdot \alpha = 1, \ \mu_{\text{min}} = 1, \ K=1& \Longrightarrow v(n) \text{ is periodic,} \\ & && & \text{but the sampling theorem is not fulfilled}\\ \end{align*}