Exercise 1.2: Signal Classification

Three signal curves are shown on the right:

The blue signal [math]\displaystyle{ x_1(t) }[/math] is switched on at time $t = 0$ and has the value $1\,\text{V}\,$ for $t > 0$ .
The red signal [math]\displaystyle{ x_2(t)\equiv 0 }[/math] for $t < 0$ and jumps to $1\,\text{V}$ at $t = 0$ . It then decreases with the time constant $1\,\text{ms}$ .
For $t > 0$ the following applies:

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

Correspondingly, the following applies to the green signal [math]\displaystyle{ x_3(t) }[/math] for all $t$:

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

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

deterministic or stochastic,
causal or non–causal,
energy-limited or power-limited,
continuous-valued or discrete-valued,
continuous in time or discrete in time.

Note:

This exercise belongs to the chapter Signal Classification.

Questions

Solution

(1) The solutions 1 and 3 are applicable:

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

(2) Correct are the solutions 1 and 2:

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

(3) According to the general definition:

[math]\displaystyle{ 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. }[/math]

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

[math]\displaystyle{ 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}. }[/math]

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, [math]\displaystyle{ x_2(t) }[/math] has a finite energy:

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

The energy of the signal [math]\displaystyle{ x_3(t) }[/math] 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 [math]\displaystyle{ x_1(t) }[/math] 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 [math]\displaystyle{ x_1(t) }[/math] in half the time $(t < 0)$ is identical to zero.
The signal [math]\displaystyle{ x_1(t) }[/math] therefore is power–limited.

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

	[math]\displaystyle{ x_1(t) }[/math],
	[math]\displaystyle{ x_2(t) }[/math],
	[math]\displaystyle{ x_3(t) }[/math].