Exercise 1.2: Signal Classification

predetermined characteristics

Three signal curves are shown on the Right:

The blue signal $x_1(t)$ is switched on at time $t = 0$ and has at $t > 0$ the value $1\,\text{V}$.
The blue signal $x_2(t)$ is for $t < 0$ equals zero, jumps at $t = 0$ to $1\,\text{V}$ and then falls down with the time constant $1\,\text{ms}$ . For $t > 0$ the following applies:

\[x_2(t) = 1\,\text{V} \cdot {\rm e}^{- {t}/(1\,\text{ms})}.\]

Correspondingly, the signal shown in green applies to all times $t$:

\[x_3(t) = 1\,\text{V} \cdot {\rm e}^{- {|\hspace{0.05cm}t\hspace{0.05cm}|}/(1\,\text{ms})}.\]

You will now classify these three signals according to the following criteria:

deterministic or stochastic,
causal or acausal,
energy limited or power limited,
value-continuous or value-discrete,
time-continuous or time-discrete.

Notes:

This exercise belongs to the chapter Klassifizierung von Signalen.

Questions

Solutions

Solution

(1) The solutions 1 and 3 are applicable:

All signals can be described completely in analytical form; therefore they are also deterministic.
All signals are also clearly defined for all times $t$ not only at certain times. Therefore, they are always time-continuous signals.
The signal amplitudes of $x_2(t)$ and $x_3(t)$ can take any values between $0$ and $1\,\text{V}$ they are therefore continuous in value.
On the other hand, with the signal $x_1(t)$ only the two signal values $0$ and $1\,\text{V}$ are possible; a discrete-valued signal is present.

(2) Correct are the solutions 1 and 2:

A signal is called causal if for times $t < 0$ it does not exist or is identically zero. This applies to the signals $x_1(t)$ and $x_2(t)$.
In contrast, $x_3(t)$ belongs to the class of non-causal signals.

(3) According to the general definition:

\[E_2=\lim_{T_{\rm M}\to\infty}\int^{T_{\rm M}/2}_{-T_{\rm M}/2}x^2_2(t)\,\hspace{0.1cm}{\rm d}t.\]

In this case, the lower integration limit is zero and the upper integration limit $+\infty$. You get:

\[E_2=\int^\infty_0 (1{\rm V})^2\cdot{\rm e}^{-2t/(1\rm ms)}\,\hspace{0.1cm}{\rm d}t = 5 \cdot 10^{-4}\hspace{0.1cm} \rm V^2s \hspace{0.15cm}\underline{= 0.5 \cdot 10^{-3}\hspace{0.1cm} \rm V^2s}. \]

With finite energy, the associated power is always negligible. From this follows $P_2\hspace{0.15cm}\underline{ = 0}$.

(4) Correct are the solutions 2 and 3:

As already calculated in the last subtask, $x_2(t)$ has a finite energy:

$$E_2= 0.5 \cdot 10^{-3}\hspace{0.1cm} {\rm V^2s}. $$

The energy of the signal $x_3(t)$ is twice as large, since now the time domain $t < 0$ makes the same contribution as the time domain $t > 0$. So

$$E_3= 10^{-3}\hspace{0.1cm} {\rm V^2s}.$$

At signal $x_1(t)$ the energy integral diverges: $E_1 \rightarrow \infty$. This signal has a finite power ⇒ $P_1= 0.5 \hspace{0.1cm} {\rm V}^2$.
The result also takes into account that the signal $x_1(t)$ in half the time $(t < 0)$ is identical to zero.
The signal $x_1(t)$ is accordingly power limited.

	All signals considered here are deterministic.
	All signals considered here are of stochastic nature.
	The signals are always continuous in time.
	They are always signals of continuous value.

	\(x_1(t)\),
	\(x_2(t)\),
	\(x_3(t)\).

	\(x_1(t)\),
	\(x_2(t)\),
	\(x_3(t)\).