# Exercise 1.2: Signal Classification

From LNTwww

Three signal curves are shown on the right:

- The blue signal \(x_1(t)\) is switched on at time $t = 0$ and has the value $1\,\text{V}\,$ for $t > 0$ .
- The red signal \(x_2(t)\equiv 0\) 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:

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

- Correspondingly, the following applies to the green signal \(x_3(t)\) for all $t$:

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

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

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

*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 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 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 it 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 \(x_1(t)\) therefore is
__power–limited__.