Difference between revisions of "Theory of Stochastic Signals/Moments of a Discrete Random Variable"
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{{GraueBox|TEXT= | {{GraueBox|TEXT= | ||
$\text{Example 2:}$ | $\text{Example 2:}$ | ||
− | A binary signal $x(t)$ | + | A binary signal $x(t)$ with the amplitude values |
− | *$1\hspace{0.03cm}\rm V$ $($ | + | *$1\hspace{0.03cm}\rm V$ $($for the symbol $\rm L)$, |
− | *$3\hspace{0.03cm}\rm V$ $($ | + | *$3\hspace{0.03cm}\rm V$ $($for the symbol $\rm H)$ |
− | + | and the probabilities of occurrence $p_{\rm L} = 0.2$ and $p_{\rm H} = 0.8$ , respectively, has the total signal power | |
:$$P_{\rm Gesamt} = 0.2 \cdot (1\,{\rm V})^2+ 0.8 \cdot (3\,{\rm V})^2 = 7.4 \hspace{0.05cm}{\rm V}^2,$$ | :$$P_{\rm Gesamt} = 0.2 \cdot (1\,{\rm V})^2+ 0.8 \cdot (3\,{\rm V})^2 = 7.4 \hspace{0.05cm}{\rm V}^2,$$ | ||
− | + | if one assumes the reference resistance $R = 1 \hspace{0.05cm} Ω$ . | |
− | + | With the DC component $m_1 = 2.6 \hspace{0.05cm}\rm V$ $($see [[Theory_of_Stochastic_Signals/Momente_einer_diskreten_Zufallsgröße#Linear_mean_-_DC_component|$\text{example 1})$]] it follows for | |
− | * | + | *the alternating power (variance) $P_{\rm W} = σ^2 = 7.4 \hspace{0.05cm}{\rm V}^2 - \big [2.6 \hspace{0.05cm}\rm V\big ]^2 = 0.64\hspace{0.05cm} {\rm V}^2$, |
− | * | + | *the rms value $s_{\rm eff} = σ = 0.8 \hspace{0.05cm} \rm V$. |
− | :::'' | + | :::''Parenthesis'': With other reference resistance ⇒ $R \ne 1 \hspace{0.1cm} Ω$ , not all these calculations apply. For example, with $R = 50 \hspace{0.1cm} Ω$ , the power $P_{\rm Gesamt} $, , the alternating power $P_{\rm W}$ , and the rms value $s_{\rm eff}$ have the following values: |
− | ::::$$P_{\rm | + | ::::$$P_{\rm Total} \hspace{-0.05cm}= \hspace{-0.05cm} \frac{m_2}{R} \hspace{-0.05cm}= \hspace{-0.05cm} \frac{7.4\,{\rm V}^2}{50\,{\rm \Omega} } \hspace{-0.05cm}= \hspace{-0.05cm}0.148\,{\rm W},\hspace{0.5cm} |
P_{\rm W} \hspace{-0.05cm} = \hspace{-0.05cm} \frac{\sigma^2}{R} \hspace{-0.05cm}= \hspace{-0.05cm}12.8\,{\rm mW} \hspace{0.05cm},\hspace{0.5cm} | P_{\rm W} \hspace{-0.05cm} = \hspace{-0.05cm} \frac{\sigma^2}{R} \hspace{-0.05cm}= \hspace{-0.05cm}12.8\,{\rm mW} \hspace{0.05cm},\hspace{0.5cm} | ||
s_{\rm eff} \hspace{-0.05cm} = \hspace{-0.05cm}\sqrt{R \cdot P_{\rm W} } \hspace{-0.05cm}= \hspace{-0.05cm} \sigma \hspace{-0.05cm}= \hspace{-0.05cm} 0.8\,{\rm V}.$$ | s_{\rm eff} \hspace{-0.05cm} = \hspace{-0.05cm}\sqrt{R \cdot P_{\rm W} } \hspace{-0.05cm}= \hspace{-0.05cm} \sigma \hspace{-0.05cm}= \hspace{-0.05cm} 0.8\,{\rm V}.$$ | ||
− | + | The same variance and rms value $s_{\rm eff}$ are obtained for amplitudes $0\hspace{0.05cm}\rm V$ $($for symbol $\rm L)$ and $2\hspace{0.05cm}\rm V$ $($for symbol $\rm H)$ , provided that the probabilities of occurrence $p_{\rm L} = 0.2$ and $p_{\rm H} = 0.8$ remain the same. Only the DC component and the total power change: | |
:$$m_1 = 1.6 \hspace{0.05cm}{\rm V}, \hspace{0.5cm}P_{\rm Gesamt} = P_{\rm W} + {m_1}^2 = 3.2 \hspace{0.05cm}{\rm V}^2.$$ | :$$m_1 = 1.6 \hspace{0.05cm}{\rm V}, \hspace{0.5cm}P_{\rm Gesamt} = P_{\rm W} + {m_1}^2 = 3.2 \hspace{0.05cm}{\rm V}^2.$$ | ||
}} | }} | ||
− | == | + | ==Exercises for the chapter== |
<br> | <br> | ||
− | [[Aufgaben: | + | [[Aufgaben:Exercise_2.2:_Multi-Level_Signals|Exercise 2.2: Multi-Level Signals]] |
− | [[Aufgaben: | + | [[Aufgaben:Exercise_2.2Z:_Discrete_Random_Variables|Exercise 2.2Z: Discrete Random Variables]] |
{{Display}} | {{Display}} |
Revision as of 19:56, 3 December 2021
Contents
Calculation as ensemble average or time average
The probabilities and the relative frequencies provide extensive information about a discrete random variable. Reduced information is obtained by the so-called moments $m_k$, where $k$ represents a natural number.
$\text{Two alternative ways of calculation:}$
Under the Ergodicity implied here, there are two different calculation possibilities for the moment $k$-th order:
- the ensemble averaging' or expected value formation ⇒ averaging over all possible values $\{ z_\mu\}$ with the index $\mu = 1 , \hspace{0.1cm}\text{ ...} \hspace{0.1cm} , M$:
- $$m_k = {\rm E} \big[z^k \big] = \sum_{\mu = 1}^{M}p_\mu \cdot z_\mu^k \hspace{2cm} \rm with \hspace{0.1cm} {\rm E\big[\text{ ...} \big]\hspace{-0.1cm}:} \hspace{0.3cm} \rm expected\hspace{0.1cm}value ;$$
- the time averaging over the random sequence $\langle z_ν\rangle$ with the index $ν = 1 , \hspace{0.1cm}\text{ ...} \hspace{0.1cm} , N$:
- $$m_k=\overline{z_\nu^k}=\hspace{0.01cm}\lim_{N\to\infty}\frac{1}{N}\sum_{\nu=\rm 1}^{\it N}z_\nu^k\hspace{1.7cm}\rm with\hspace{0.1cm}horizontal\hspace{0.1cm}line\hspace{-0.1cm}:\hspace{0.1cm}time\hspace{0.1cm}average.$$
Note:
- Both types of calculations lead to the same asymptotic result for sufficiently large values of $N$ .
- For finite $N$ , a comparable error results as when the probability is approximated by the relative frequency.
Linear mean - DC component
$\text{Definition:}$ With $k = 1$ we obtain from the general equation for moments the linear mean:
- $$m_1 =\sum_{\mu=1}^{M}p_\mu\cdot z_\mu =\lim_{N\to\infty}\frac{1}{N}\sum_{\nu=1}^{N}z_\nu.$$
- The left part of this equation describes the ensemble averaging (over all possible values),
- while the right equation gives the determination as time average.
- In the context of signals, this quantity is also referred to as the DC component
$\text{Example 1:}$ A binary signal $x(t)$ with the two possible amplitude values.
- $1\hspace{0.03cm}\rm V$ $($for the symbol $\rm L)$,
- $3\hspace{0.03cm}\rm V$ $($for the symbol $\rm H)$
as well as the occurrence probabilities $p_{\rm L} = 0.2$ respectively $p_{\rm H} = 0.8$ has the linear mean (DC)
- $$m_1 = 0.2 \cdot 1\,{\rm V}+ 0.8 \cdot 3\,{\rm V}= 2.6 \,{\rm V}. $$
This is drawn as a red line in the graph.
If we determine this parameter by time averaging over the displayed $N = 12$ signal values, we obtain a slightly smaller value:
- $$m_1\hspace{0.01cm}' = 4/12 \cdot 1\,{\rm V}+ 8/12 \cdot 3\,{\rm V}= 2.33 \,{\rm V}. $$
Here, the probabilities of occurrence $p_{\rm L} = 0.2$ and $p_{\rm H} = 0.8$ were replaced by the corresponding frequencies $h_{\rm L} = 4/12$ and $h_{\rm H} = 8/12$ respectively. The relative error due to insufficient sequence length $N$ is greater than $10\%$ in the example.
Note about our (admittedly somewhat unusual) nomenclature:.
We denote binary symbols here as in circuit theory with $\rm L$ (Low) and $\rm H$ (High) to avoid confusion.
- In coding theory, it is useful to map $\{ \text{L, H}\}$ to $\{0, 1\}$ to take advantage of the possibilities of modulo algebra.
- In contrast, to describe modulation with bipolar (antipodal) signals, one better chooses the mapping $\{ \text{L, H}\}$ ⇔ $ \{-1, +1\}$.
Quadratic Mean – Variance – Dispersion
$\text{Definitions:}$
- Analogous to the linear mean, $k = 2$ is obtained for the root mean square:
- $$m_2 =\sum_{\mu=\rm 1}^{\it M}p_\mu\cdot z_\mu^2 =\lim_{N\to\infty}\frac{\rm 1}{\it N}\sum_{\nu=\rm 1}^{\it N}z_\nu^2.$$
- Together with the DC component $m_1$ , the variance $σ^2$ can be determined from this as a further parameter (Steiner's Theorem):
- $$\sigma^2=m_2-m_1^2.$$
- In statistics, the dispersion $σ$ is the square root of the variance; sometimes this quantity is also called standard deviation :
- $$\sigma=\sqrt{m_2-m_1^2}.$$
Notes on units:
- For message signals, $m_2$ indicates the (average) power power of a random signal, referenced to $1 \hspace{0.03cm} Ω$ resistance.
- If $z$ describes a voltage, $m_2$ accordingly has the unit${\rm V}^2$.
- The variance $σ^2$ of a random signal corresponds physically to the alternating power and the dispersion $σ$ to therms (root mean square) value.
- These definitions are based on the reference resistance $1 \hspace{0.03cm} Ω$ zugrunde.
The (German) learning video Momentenberechnung bei diskreten Zufallsgrößen $\Rightarrow$ Moment Calculation for Discrete Random Variables, illustrates the defined quantities using the example of a digital signal.
$\text{Example 2:}$ A binary signal $x(t)$ with the amplitude values
- $1\hspace{0.03cm}\rm V$ $($for the symbol $\rm L)$,
- $3\hspace{0.03cm}\rm V$ $($for the symbol $\rm H)$
and the probabilities of occurrence $p_{\rm L} = 0.2$ and $p_{\rm H} = 0.8$ , respectively, has the total signal power
- $$P_{\rm Gesamt} = 0.2 \cdot (1\,{\rm V})^2+ 0.8 \cdot (3\,{\rm V})^2 = 7.4 \hspace{0.05cm}{\rm V}^2,$$
if one assumes the reference resistance $R = 1 \hspace{0.05cm} Ω$ .
With the DC component $m_1 = 2.6 \hspace{0.05cm}\rm V$ $($see $\text{example 1})$ it follows for
- the alternating power (variance) $P_{\rm W} = σ^2 = 7.4 \hspace{0.05cm}{\rm V}^2 - \big [2.6 \hspace{0.05cm}\rm V\big ]^2 = 0.64\hspace{0.05cm} {\rm V}^2$,
- the rms value $s_{\rm eff} = σ = 0.8 \hspace{0.05cm} \rm V$.
- Parenthesis: With other reference resistance ⇒ $R \ne 1 \hspace{0.1cm} Ω$ , not all these calculations apply. For example, with $R = 50 \hspace{0.1cm} Ω$ , the power $P_{\rm Gesamt} $, , the alternating power $P_{\rm W}$ , and the rms value $s_{\rm eff}$ have the following values:
- $$P_{\rm Total} \hspace{-0.05cm}= \hspace{-0.05cm} \frac{m_2}{R} \hspace{-0.05cm}= \hspace{-0.05cm} \frac{7.4\,{\rm V}^2}{50\,{\rm \Omega} } \hspace{-0.05cm}= \hspace{-0.05cm}0.148\,{\rm W},\hspace{0.5cm} P_{\rm W} \hspace{-0.05cm} = \hspace{-0.05cm} \frac{\sigma^2}{R} \hspace{-0.05cm}= \hspace{-0.05cm}12.8\,{\rm mW} \hspace{0.05cm},\hspace{0.5cm} s_{\rm eff} \hspace{-0.05cm} = \hspace{-0.05cm}\sqrt{R \cdot P_{\rm W} } \hspace{-0.05cm}= \hspace{-0.05cm} \sigma \hspace{-0.05cm}= \hspace{-0.05cm} 0.8\,{\rm V}.$$
- Parenthesis: With other reference resistance ⇒ $R \ne 1 \hspace{0.1cm} Ω$ , not all these calculations apply. For example, with $R = 50 \hspace{0.1cm} Ω$ , the power $P_{\rm Gesamt} $, , the alternating power $P_{\rm W}$ , and the rms value $s_{\rm eff}$ have the following values:
The same variance and rms value $s_{\rm eff}$ are obtained for amplitudes $0\hspace{0.05cm}\rm V$ $($for symbol $\rm L)$ and $2\hspace{0.05cm}\rm V$ $($for symbol $\rm H)$ , provided that the probabilities of occurrence $p_{\rm L} = 0.2$ and $p_{\rm H} = 0.8$ remain the same. Only the DC component and the total power change:
- $$m_1 = 1.6 \hspace{0.05cm}{\rm V}, \hspace{0.5cm}P_{\rm Gesamt} = P_{\rm W} + {m_1}^2 = 3.2 \hspace{0.05cm}{\rm V}^2.$$
Exercises for the chapter
Exercise 2.2: Multi-Level Signals
Exercise 2.2Z: Discrete Random Variables