Exercise 2.5: Distortion and Equalization

Trapezoidal spectrum (top) and
associated impulse response (bottom)

A communication system with input $x(t)$ and output $y(t)$, which is fully described by the trapezoidal frequency response $H(f)$ according to the top graph, is considered. Using the roll-off factor $r = 0.5$ and the equivalent bandwidth $\Delta f = 16 \ \rm kHz$ the corresponding impulse response, which is computable by applying the inverse Fourier transform, is:

$$h(t) = \Delta f \cdot {\rm si}(\pi \cdot \Delta f \cdot t )\cdot {\rm si}(\pi \cdot r \cdot \Delta f \cdot t ) .$$

The available input signals are:

The sum of two harmonic oscillations:

$$x_1(t) = {1\, \rm V} \cdot \cos(\omega_1 \cdot t) + {1\, \rm V} \cdot \sin(\omega_2 \cdot t).$$

Here, the following holds: $\omega_1 = 2\pi \cdot 2000 \ {\rm 1/s}$ and $\omega_2 \gt \omega_1$.

A periodic triangular signal:

$$x_2(t) = \frac{8\, \rm V}{\pi^2} \cdot \big[\cos(\omega_0 t) + {1}/{9} \cdot \cos(3\omega_0 t) + {1}/{25} \cdot \cos(5\omega_0 t) + \hspace{0.05cm}\text{...}\big].$$

It should be noted that the fundamental frequency is $f_0 = 2 \ \rm kHz$ or $3\ \rm kHz$ beträgt. At time $t = 0$ the signal value in both cases is $1 \ \rm V$.

A rectangular pulse $x_3(t)$ with amplitude $A = 1 \ \rm V$ and duration $T = 1 \ \rm ms$. Since its spectrum $X_3(f)$ extends to infinity, $H(f)$ always results in linear distortions here.

From subtask (6) onwards, it shall be attempted to eliminate the distortions possibly generated by $H(f)$ by means of a downstream equaliser with

frequency response $H_{\rm E}(f)$,
input signal $y(t)$, and
output signal $z(t)$.

Please note:

The task belongs to the chapter Linear Distortions.
In particular, reference is made to the page Entzerrungsverfahren.
The term "overall distortion" used in the formulation of the questions refers to the input signal $x(t)$ and the output signal $z(t)$.

Questions

Solution

(1) Proposed solutions 1 and 3 are correct:

A linear system is already implicitly assumed by specifying a frequency response so that nonlinear distortions cannot occur.
Since $H(f)$ is purely real, phase distortions can also be ruled out.

(2) Proposed solutions 1 and 2 are correct:

The output signal is $y_1(t) = x_1(t)$.
Thus, the system is not only distortion-free but can also be termed ideal for this application.

(3) Proposed solution 3 is correct:

In this case, the following is obtained for the output signal:

$$y_1(t)= 1\,{\rm V}\cdot \cos(2 \pi \cdot f_1 \cdot t) + {1}/{4}\cdot 1\,{\rm V}\cdot \sin(2 \pi \cdot f_2 \cdot t).$$

While the component at $f_1$ is transmitted unchanged, the sinusoidal component at $f_2$ is attenuated and one-quarter of the original sinusoidal component.
So, there are attenuation distortions.

(4) The output signal $y_2(t)$ has the following form taking into account the fundamental frequency $f_0 = 3 \ \rm kHz$ :

$$y_2(t)= \frac{8\,{\rm V}}{\pi^2} \left( \cos(\omega_0 t) + \frac{3}{8}\cdot \frac{1}{9} \cdot \cos(3\omega_0 t)\right) .$$

The factor $3/8$ describes $H(f = 9 \ \rm kHz)$. All other spectral components at $15 \ \rm kHz$, $21 \ \rm kHz$, etc. are suppressed by the system.

The strongest deviations between $x_2(t)$ and $y_2(t)$ will occur at the triangle peaks since the missing high frequencies have the strongest effect here. For example, for the time $\underline{t= 0}$ one obtains:

$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 + {3}/{72}\right)= 0.844\,{\rm V} \hspace{0.3cm}\Rightarrow\hspace{0.3cm} \varepsilon_{\rm max} = |y_2(t=0)- x_2(t=0)| \hspace{0.15cm}\underline{= 0.156\,{\rm V}}.$$

(5) With the fundamental frequency $f_0 = 2 \ \rm kHz$ and the transmission values $H(3f_0) = 0.75$, $H(5f_0) = 0.25$, $H(7f_0) = 0$ the following is obtained:

$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 + \frac{3}{4}\cdot \frac{1}{9} + \frac{1}{4} \cdot\frac{1}{25}\right)= 0.886\,{\rm V}\hspace{0.5cm} \Rightarrow \hspace{0.5cm}\varepsilon_{\rm max} \hspace{0.15cm}\underline{= 0.114\,{\rm V}}.$$

(6) In the range up to $4 \ \rm kHz$, $H_{\rm E}(f) = H(f) = 1$ is to be set. In contrast, in the range from $4 \ \rm kHz$ to $12 \ \rm kHz$ the following holds:

$$H_{\rm E}(f)= \frac{1}{H(f)} = \frac{1}{1.5 \cdot \big[1 - f/(12\,{\rm kHz})\big]} \hspace{0.5cm} \Rightarrow \hspace{0.5cm} H_{\rm E}(f = 10\,{\rm kHz})\hspace{0.15cm}\underline{= 4} .$$

Here, the denominator expression describes the equation of the straight line of the frequency roll-off.

(7) Proposed solution 1 is correct:

Both $x_2(t)$ and $x_3(t)$ also contain spectral components at frequencies greater than $12 \ \rm kHz$.
If these have been truncated by $H(f)$ ⇒ Bandbegrenzung, so können sie durch den Entzerrer nicht mehr rekonstruiert werden.
Das heißt, dass nur das Signal $x_1(t)$ durch $H_{\rm E}(f)$ wieder hergestellt werden kann, allerdings nur dann, wenn $f_2 < 12 \ \rm kHz$ gilt:

$$z_1(t)= \underline{1} \cdot 1\,{\rm V}\cdot \cos(2 \pi \cdot f_1 \cdot t) + \underline{4} \cdot \frac{1}{4}\cdot 1\,{\rm V}\cdot \sin(2 \pi \cdot f_2 \cdot t).$$

Die jeweils ersten (unterstrichenen) Faktoren geben jeweils die Verstärkungswerte von $H_{\rm E}(f)$ an.

	Nonlinear distortions.
	Attenuation distortions.
	Phase distortions.

	It acts like an ideal system.
	It acts like a distortion-free system.
	It can be seen that the system at hand is a distorting system.

	For signal $x_1(t)$ with $f_2 = 10 \ \rm kHz$,
	for signal $x_2(t)$,
	for signal $x_3(t)$.