Difference between revisions of "Aufgaben:Exercise 1.2: Signal Classification"
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- <math>x_3(t)</math>. | - <math>x_3(t)</math>. | ||
− | { | + | {Calculate the energy $R = 1\ Ω$ related to the unit resistance <math>E_2</math> of the signal <math>x_2(t)</math>. <br>What is the power <math>P_2</math> of this signal? |
|type="{}"} | |type="{}"} | ||
<math>E_2 \ = \ </math>{ 0.5 5% } $\ \cdot 10^{-3}\,\text{V}^2\text{s}$ | <math>E_2 \ = \ </math>{ 0.5 5% } $\ \cdot 10^{-3}\,\text{V}^2\text{s}$ | ||
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{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' The <u>solutions 1 and 3</u> are applicable: | '''(1)''' The <u>solutions 1 and 3</u> 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 <math>x_2(t)</math> and <math>x_3(t)</math> can take any values between $0$ and $1\,\text{V}$ they are therefore continuous in value. |
− | * | + | *On the other hand, with the signal <math>x_1(t)</math> only the two signal values $0$ and $1\,\text{V}$ are possible; a discrete-valued signal is present. |
− | '''(2)''' | + | '''(2)''' Correct are the <u>solutions 1 and 2</u>: |
− | * | + | *A signal is called causal if for times $t < 0$ it does not exist or is identically zero. This applies to the signals <math>x_1(t)</math> and <math>x_2(t)</math>. |
− | * | + | *In contrast, <math>x_3(t)</math> belongs to the class of non-causal signals. |
− | '''(3)''' | + | '''(3)''' According to the general definition: |
::<math>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> | ::<math>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>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> | ::<math>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)''' | + | '''(4)''' Correct are the <u>solutions 2 and 3</u>: |
− | * | + | *As already calculated in the last subtask, <math>x_2(t)</math> has a finite energy: |
− | :$$E_2= 0.5 \cdot 10^{-3}\hspace{0.1cm} {\rm V^2s}. | + | :$$E_2= 0.5 \cdot 10^{-3}\hspace{0.1cm} {\rm V^2s}. $$ |
− | * | + | *The energy of the signal <math>x_3(t)</math> is twice as large, since now the time domain $t < 0$ makes the same contribution as the time domain $t > 0$. So |
:$$E_3= 10^{-3}\hspace{0.1cm} {\rm V^2s}.$$ | :$$E_3= 10^{-3}\hspace{0.1cm} {\rm V^2s}.$$ | ||
− | * | + | *At signal <math>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>x_1(t)</math> in half the time $(t < 0)$ is identical to zero. |
− | * | + | * The signal <math>x_1(t)</math> is accordingly <u>power limited</u>. |
{{ML-Fuß}} | {{ML-Fuß}} |
Revision as of 19:04, 13 August 2020
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
(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.