Exercise 4.2: Low-Pass for Signal Reconstruction

Examples of continuous and discrete spectra

We consider in this exercise two different source signals $q_{\rm con}(t)$ and $q_{\rm dis}(t)$ whose magnitude spectra $|Q_{\rm con}(f)|$ and $|Q_{\rm dis}(f)|$ are plotted. The highest frequency occurring in the signals is in each case $4 \rm kHz$.

Nothing more is known of the spectral function $Q_{\rm con}(f)$ than that it is a continuous spectrum, where:

$$Q_{\rm con}(|f| \le 4\,{\rm kHz}) \ne 0 \hspace{0.05cm}.$$

The spectrum $Q_{\rm dis}(f)$ contains spectral lines at $±1 \ \rm kHz$, $±2 \ \rm kHz$, $±3 \ \rm kHz$ and $±4 \ \rm kHz$. Thus:

$$q_{\rm dis}(t) = \sum_{i=1}^{4}C_i \cdot \cos (2 \pi \cdot f_i \cdot t - \varphi_i),$$

Amplitude values: $C_1 = 1.0 \ \rm V$, $C_2 = 1.8 \ \rm V$, $C_3 = 0.8 \ \rm V$, $C_4 = 0.4 \ \rm V.$

The phase values $φ_1$, $φ_2$ and $φ_3$ are respectively in the range $±180^\circ$ and it holds $φ_4 = 90^\circ$.

The signals are each sampled at frequency $f_{\rm A}$ and immediately fed to an ideal rectangular low-pass filter with cutoff frequency $f_{\rm G}$ This scenario applies, for example, to

the interference-free pulse amplitude modulation $\rm (PAM)$ and
the interference-free pulse code modulation $\rm (PCM)$ at infinitely large quantization stage number $M$.

The output signal of the (rectangular) low-pass filter is called the sink signal $v(t)$ and for the error signal:

$$ε(t) = v(t) - q(t).$$

This is different from zero only if the parameters of the sampling $($sampling frequency $f_{\rm A})$ and/or the signal reconstruction $($cutoff frequency $f_{\rm G})$ are not dimensioned in the best possible way.

Hints:

The exercise belongs to the chapter "Pulse Code Modulation".
Reference is made in particular to the page "Sampling and Signal Reconstruction".

Questions

Solution

(1) Only the first statement is correct:

Sampling $q_{\rm dis}(t)$ with sampling frequency $f_{\rm A} = 8 \ \rm kHz$ leads to an irreversible error, since $Q_{\rm dis}(f)$ involves a discrete spectral component (Dirac line) at $f_4 = 4\ \rm kHz$ and the phase value is $φ_4 ≠ 0$.
With the phase value $φ_4 = 90^\circ$ $(4 \ \rm kHz$ sinusoidal component$)$ given here holds $ε_{\rm dis}(t) = v_{\rm dis}(t) - q_{\rm dis}(t) = -0. 4 \ \rm V - \sin(2π \cdot f_4 \cdot t)$. See also solution to Exercise 4.2Z.
On the other hand, the signal $q_{\rm con}(t)$ with the continuous spectrum $Q_{\rm con}(f)$ can also then be measured with a rectangular low-pass filter $($with cutoff frequency $f_{\rm G} = 4\ \rm kHz)$ be completely reconstructed if sampling frequency $f_{\rm A} = 8\ \rm kHz$ was used. For all frequencies not equal to $f_4$ the sampling theorem is satisfied.
The contribution of the $f_4$ component to the total spectrum $Q_{\rm con}(f)$ is only vanishingly small ⇒ ${\rm Pr}(f_4) → 0$ as long as the spectrum has no Dirac line at $f_4$.

(2) Only the proposed solution 1 is correct:

With $f_{\rm A} = 10\ \rm kHz$ the sampling theorem is satisfied in both cases.
With $f_{\rm G} = f_{\rm A} /2$ both error signals $ε_{\rm con}(t)$ and $ε_{\rm dis}(t)$ are identically zero.
In addition, the signal reconstruction also works as long as $f_{\rm G} > 4 \ \rm kHz$ and $f_{\rm G} < 6 \ \rm kHz$ holds.

(3) The correct solution here is suggested solution 2:

With $f_{\rm G} = 3.5 \ \rm kHz$ the low-pass incorrectly removes the $4\ \rm kHz$ component, that is, then holds:

$$ v_{\rm dis}(t) = q_{\rm dis}(t) - 0.4\,{\rm V} \cdot \sin (2 \pi \cdot f_{\rm 4} \cdot t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \varepsilon_{\rm dis}(t) = - 0.4\,{\rm V} \cdot \sin (2 \pi \cdot f_{\rm 4} \cdot t)\hspace{0.05cm}.$$

Signal reconstruction with too large cutoff frequency

(4) The correct solution here is suggested solution 3:

Sampling with $f_{\rm A} = 10\ \rm kHz$ yields the periodic spectrum sketched on the right.
The low-pass with $f_{\rm G} = 6.5 \ \rm kHz$ removes all discrete frequency components with $|f| ≥ 7\ \rm kHz$, but not the $6\ \rm kHz$ component.

The error signal $ε_{\rm dis}(t) = v_{\rm dis}(t) - q_{\rm dis}(t)$ is then a harmonic oscillation with

the frequency $f_6 = f_{\rm A} - f_4 = 6\ \rm kHz$,
the amplitude $A_4$ of the $f_4$ component,
the phase $φ_{-4} = -φ_4$ of the $Q(f)$ component at $f = -f_4$.

	The signal $q_{\rm con}(t)$ can be completely reconstructed: $ε_{\rm con}(t) = 0$.
	The signal $q_{\rm dis}(t)$ can be completely reconstructed: $ε_{\rm dis}(t) = 0$.

	The signal $q_{\rm dis}(t)$ can be completely reconstructed: $ε_{\rm dis}(t) = 0$.
	$ε_{\rm dis}(t)$ is a harmonic oscillation with $4 \rm kHz$.
	$ε_{\rm dis}(t)$ is a harmonic oscillation with $6 \rm kHz$.