Exercise 4.1: Low-Pass and Band-Pass Signals
Three signal curves are sketched on the right, the first two having the following curve:
- $$x(t) = 10\hspace{0.05cm}{\rm V} \cdot {\rm si} ( \pi \cdot {t}/{T_x}) ,$$
- $$y(t) = 6\hspace{0.05cm}{\rm V} \cdot {\rm si}( \pi \cdot {t}/{T_y}) .$$
$T_x = 100 \,{\rm µ}\text{s}$ and $T_y = 166.67 \,{\rm µ}\text{s}$ indicate the first zero of $x(t)$ bzw. $y(t)$ respectively. The signal $d(t)$ results from the difference of the two upper signals (lower graph):
- $$d(t) = x(t)-y(t) .$$
In the subtask (4) the integral areas of the impulsive signals $x(t)$ and $d(t)$ are asked for. For these holds:
- $$F_x = \int_{- \infty}^{+\infty}\hspace{-0.4cm}x(t)\hspace{0.1cm}{\rm d}t , \hspace{0.5cm}F_d = \int_{- \infty}^{+\infty}\hspace{-0.4cm}d(t)\hspace{0.1cm}{\rm d}t .$$
On the other hand, for the corresponding signal energies with Parseval's theorem:
- $$E_x = \int_{- \infty}^{+\infty}\hspace{-0.4cm}|x(t)|^2\hspace{0.1cm}{\rm d}t = \int_{- \infty}^{+\infty}\hspace{-0.4cm}|X(f)|^2\hspace{0.1cm}{\rm d}f ,$$
- $$E_d = \int_{- \infty}^{+\infty}\hspace{-0.4cm}|d(t)|^2\hspace{0.1cm}{\rm d}t = \int_{- \infty}^{+\infty}\hspace{-0.4cm}|D(f)|^2\hspace{0.1cm}{\rm d}f .$$
Hints:
- This exercise belongs to the chapter Differences and Similarities of LP and BP Signals.
- The Fourier retransform of a rectangular spectrum $X(f)$ leads to an $\rm si$–shaped time function $x(t)$:
- $$X(f)=\left\{ {X_0 \; \rm f\ddot{u}r\; |\it f| < \rm B, \atop {\rm 0 \;\;\; \rm sonst}}\right. \;\; \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \;\;x(t) = 2 \cdot X_0 \cdot B \cdot {\rm si} ( 2\pi B t) .$$
- In this task, the function $\rm si(x) = \rm sin(x)/x = \rm sinc(x/π)$ is used.
Questions
Solution
- The absolute, two-sided bandwidth $2 \cdot B_x$ is equal to the reciprocal of the first zero. It follows that:
- $$B_x = \frac{1}{2 \cdot T_x} = \frac{1}{2 \cdot 0.1 \hspace{0.1cm}{\rm ms}}\hspace{0.15 cm}\underline{ = 5 \hspace{0.1cm}{\rm kHz}}.$$
- Since the signal value at $t = 0$ is equal to the rectangular area, the constant height is given by:
- $$X(f=0) = \frac{x(t=0)}{2 B_x} = \frac{10 \hspace{0.1cm}{\rm V}}{10 \hspace{0.1cm}{\rm kHz}} \hspace{0.15 cm}\underline{= 1 \hspace{0.1cm}{\rm mV/Hz}}.$$
(2) From $T_y = 0.167 \,\text{ms}$ we get $B_y \;\underline{= 3 \,\text{kHz}}$.
- Together with $y(t = 0) = 6\,\text{V}$ this leads to the same spectral value $Y(f = 0)\; \underline{= 1\, \text{mV/Hz}}$ as in subtaske (1).
(3) From $d(t) = x(t) - y(t)$ follows because of the linearity of the Fourier transformation: $D(f) = X(f) - Y(f).$
- The difference of the two equal square wave functions leads to a rectangular band-pass spectrum between $3 \,\text{kHz}$ and $5 \,\text{kHz}$.
- The (one-sided) bandwidth is thus $B_d \;\underline{= 2 \,\text{kHz}}$. In this frequency interval $D(f) = 1 \,\text{mV/Hz}$. Outside, i.e. also at $f = 0$, gilt $D(f)\;\underline{ = 0}$ applies.
(4) According to the fundamental laws of the Fourier transformation, the integral over the time function is equal to the spectral value at $f = 0$. Daraus folgt:
- $$F_x = X(f=0) = \frac{x(t=0)}{2 \cdot B_x} = 10^{-3} \hspace{0.1cm}{\rm V/Hz}\hspace{0.15 cm}\underline{= 0.001 \hspace{0.1cm}{\rm Vs}},$$
- $$F_d = D(f=0) \hspace{0.15 cm}\underline{= 0}.$$
⇒ For each band-pass signal, the areas of the positive signal components are equal to the areas of the negative components.
(5) In both cases, the calculation of the signal energy is easier in the frequency domain than in the time domain, because here the integration can be reduced to an area calculation of rectangles:
- $$E_x = (10^{-3} \hspace{0.1cm}{\rm V/Hz})^2 \cdot 2 \cdot 5 \hspace{0.1cm}{\rm kHz} \hspace{0.15 cm}\underline{= 0.01 \hspace{0.1cm}{\rm V^2s}},$$
- $$E_d = (10^{-3} \hspace{0.1cm}{\rm V/Hz})^2 \cdot 2 \cdot 2 \hspace{0.1cm}{\rm kHz} \hspace{0.15 cm}\underline{= 0.004 \hspace{0.1cm}{\rm V^2s}}.$$
