Exercise 4.2Z: About the Sampling Theorem

Harmonic oscillations of different phase

The sampling theorem states that the sampling frequency $f_{\rm A} = 1/T_{\rm A}$ must be at least twice as large as the largest frequency $f_\text {N, max}$ contained in the source signal $q(t)$:

$$f_{\rm A} \ge 2 \cdot f_{\rm N,\hspace{0.05cm}max}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} T_{\rm A} \le \frac{1}{2 \cdot f_{\rm N, \hspace{0.05cm}max}}\hspace{0.05cm}.$$

If this condition is met, then at the receiver the message signal can be passed through a rectangular (ideal) low-pass filter with frequency response

$$H(f) = \left\{ \begin{array}{l} 1 \\ 1/2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < f_{\rm G},} \\ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| = f_{\rm G},} \\ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| > f_{\rm G}} \\ \end{array}$$

can be completely reconstructed, that is, it is then $v(t) = q(t)$.

The cutoff frequency $f_{\rm G}$ is to be chosen equal to half the sampling frequency.
The equal sign is generally valid only if the spectrum $Q(f)$ does not contain a discrete spectral line at frequency $f_\text {N, max}$.

In this exercise, three different source signals are considered, each of which can be expressed as a harmonic oscillation

$$q(t) = A \cdot \cos (2 \pi \cdot f_{\rm N} \cdot t - \varphi)$$

with amplitude $A = 1\ \rm V$ and frequency $f_{\rm N}= 5 \ \rm kHz$. For the spectral function $Q(f)$ of all represented time signals generally holds:

$$Q(f) = \frac{A}{2} \cdot \delta (f- f_{\rm N}) \cdot {\rm e}^{-{\rm j}\hspace{0.04cm}\cdot \hspace{0.04cm}\varphi}+ \frac{A}{2} \cdot \delta (f+ f_{\rm N}) \cdot {\rm e}^{+{\rm j}\hspace{0.04cm}\cdot \hspace{0.04cm}\varphi}\hspace{0.05cm}.$$

The oscillations sketched in the graph differ only by the phase $φ$:

$φ_1 = 0$ ⇒ cosine signal $q_1(t)$,
$φ_2 = π/2 \ (= 90^\circ)$ ⇒ sinusoidal signal $q_2(t)$,
$φ_3 = π/4 \ (= 45^\circ)$ ⇒ signal $q_3(t)$.

Hints:

The exercise belongs to the chapter "Pulse Code Modulation".
Reference is made in particular to the page "Sampling and Signal Reconstruction".
The sampled source signal is denoted by $q_{\rm A}(t)$ and its spectral function by $Q_{\rm A}(f)$.
Sampling is always performed at $ν \cdot T_{\rm A}$.

Questions

Solution

(1) All statements are true:

Spectral function of sampled signal

The sampling theorem is satisfied by $f_{\rm A} = 11 \ \rm kHz > 2 \cdot 5 \ \rm kHz$ so that complete signal reconstruction is always possible.
The spectrum $Q_{\rm A}(f)$ results from $Q(f)$ by periodic continuation at the respective frequency spacing $f_{\rm A}$, which is generally illustrated in the graph for the spectral function $Q_3(f)$ .
By a rectangle–low-pass with $f_{\rm G} = f_{\rm A}/2 = 5.5 \ \rm kHz$ the original spectrum $Q(f)$ is obtained.

The shift by

$f_{\rm A} = 11 \ \rm kHz$ yields the lines at $+6 \ \rm kHz$ and $+16 \ \rm kHz$,
$-f_{\rm A} = -11 \ \rm kHz$ yields the lines at $-6 \ \rm kHz$ and $-16 \ \rm kHz$,
$2 - f_{\rm A} = 22 \ \rm kHz$ yields the lines at $+17 \ \rm kHz$ and $+27 \ \rm kHz$,
$-2 - f_{\rm A}= -22 \ \rm kHz$ yields the lines at $-17 \ \rm kHz$, $-27 \ \rm kHz$.

(2) The sampling distance is equal to the reciprocal of the sampling frequency:

$$ T_{\rm A} = {1}/{f_{\rm A} }\hspace{0.15cm}\underline { = 0.1\,{\rm ms}} \hspace{0.05cm}.$$

(3) The correct solution is suggestion 2:

For the cosinusoidal signal, according to the next graph with $f_{\rm A} = 10 \rm kHz$ the spectrum $Q_{\rm A}(f)$: All spectral lines are real.
The periodization of $Q(f)$ with $f_{\rm A} = 10 \rm kHz$ leads to a Dirac comb with spectral lines at $±f_{\rm N}$, $±f_{\rm N}± f_{\rm A}$, $±f_{\rm N}± 2f_{\rm A}$, . ..
Through the superpositions, all Dirac functions have weight $A$, while the two spectral lines of $Q(f)$ are weighted only by $A/2$ each.
Because $H(f = f_{\rm N}) = H(f = f_{\rm G}) = 0.5$ the spectrum $V_1(f)$ after the low-pass is identical to $Q_1(f)$ and accordingly $v_1(t) = q_1(t)$.
In the time domain, the signal reconstruction can be thought of as follows: The samples of $q_1(t)$ lie exactly at the signal maxima and minima.
The lowpass filter forms the cosine signal with correct amplitude, frequency and phase.

Spectral function of the sampled cosine signal

Abgetastetes Sinussignal

(4) Correct is suggested solution 2:

All sampled values of $q_2(t)$ now lie exactly at the zero crossings of the sinusoidal signal, which means that here $q_{\rm A}(t) \equiv 0$ holds. However, this naturally also gives $v_2(t) \equiv 0$.
In the spectral domain, the result can be derived using the graph for subtask (1) . $Q(f)$ is purely imaginary and the imaginary parts at $±f_{\rm N}$ have different signs.
Thus, one positive and one negative part cancel each other in periodization ⇒ $Q_{\rm A}(f) \equiv 0$ ⇒ $V_2(f) \equiv 0$.

Sampled harmonic oscillation with phase $φ_3 = π/4$

(5) None of the given solutions is correct:

If in the graph for the subtask (1) the sampling frequency $f_{\rm A} = 11 \ \rm kHz$ is replaced by $f_{\rm A} = 10 \ \rm kHz$, the real parts add up, but the imaginary parts cancel out.
This means that now $Q_{\rm A}(f)$ and $V_3(f)$ are real spectra. This further means that
The phase information is lost $(φ = 0)$ and the output signal $v_3(t)$ is a cosine signal.
The signals $q_3(t)$ and $v_3(t)$ thus differ in both amplitude and phase. Only the frequency remains the same.

The graph shows

turquoise the signal $q_3(t)$ and its samples (circles) and
red dashed the output signal $v_3(t)$ of the low-pass.

You can see that the lowpass gives exactly the result you would probably choose if you were to draw a curve through the samples (circles).

	The sampling theorem is always satisfied.
	All signals can be reconstructed by a low-pass filter.
	It is always true: $Q_{\rm A}(f = 5 \ {\rm kHz}) = Q(f = 5 \ \rm kHz)$.

	It holds $Q_{\rm A}(f = 5 \ {\rm kHz)} = Q_1(f = 5 \ \rm kHz)$.
	A complete signal reconstruction is possible ⇒ $v_1(t) = q_1(t)$.
	The reconstructed signal is $v_1(t) \equiv 0$.

	It holds $Q_{\rm A}(f = 5 \ {\rm kHz)} = Q_2(f = 5 \ \rm kHz)$.
	A complete signal reconstruction is possible ⇒ $v_2(t) = q_2(t)$.
	The reconstructed signal is $v_2(t) \equiv 0$.

	It holds $Q_{\rm A}(f = 5 \ {\rm kHz)} = Q_3(f = 5 \ \rm kHz)$.
	A complete signal reconstruction is possible ⇒ $v_3(t) = q_3(t)$.
	The reconstructed signal is $v_3(t) \equiv 0$.