Difference between revisions of "Aufgaben:Exercise 5.5Z: ACF after 1st Order Filter"
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− | [[File:P_ID565__Sto_Z_5_5_neu.png|right|frame|Non-recursive filter with DC component]] | + | [[File:P_ID565__Sto_Z_5_5_neu.png|right|frame|Non-recursive filter <br>with DC component]] |
We consider here a first order non-recursive filter $(M = 1)$. | We consider here a first order non-recursive filter $(M = 1)$. | ||
*Let the filter coefficients be $a_0 = 0.4$ and $a_1 = 0.3$. | *Let the filter coefficients be $a_0 = 0.4$ and $a_1 = 0.3$. | ||
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The individual elements of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle$ | The individual elements of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle$ | ||
* are Gaussian as well as mean-free, and | * are Gaussian as well as mean-free, and | ||
− | * have in each case the | + | * have in each case the standard deviation $\sigma_x = 1$. |
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{Which statements are true regarding the output ACF when $K = 0$? Justify your results. | {Which statements are true regarding the output ACF when $K = 0$? Justify your results. | ||
|type="[]"} | |type="[]"} | ||
− | - The ACF value $\varphi_y(0)$ indicates the | + | - The ACF value $\varphi_y(0)$ indicates the standard deviation $\sigma_y$. |
+ All ACF values $\varphi_y(k \cdot T_{\rm A})$ with $k \ge 2$ are zero. | + All ACF values $\varphi_y(k \cdot T_{\rm A})$ with $k \ge 2$ are zero. | ||
− | + The power-spectral density | + | + The power-spectral density $\rm (PSD)$ ${\it \Phi}_y(f)$ is cosinusoidal. |
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− | {What values do you need to set for $a_0$ and $a_1$ if you want the | + | {What values do you need to set for $a_0$ and $a_1$ if you want the standard deviation to be $\sigma_y = 1$ for the same ACF shape? Let $a_0 > a_1$. |
|type="{}"} | |type="{}"} | ||
$a_0 \ = \ $ { 0.8 3% } | $a_0 \ = \ $ { 0.8 3% } | ||
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− | {What is the | + | {What is the standard deviation $\sigma_y$ now? |
|type="{}"} | |type="{}"} | ||
$\sigma_y \ = \ $ { 0.5 3% } | $\sigma_y \ = \ $ { 0.5 3% } | ||
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===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' <u>Solutions 2 and 3</u> are correct: | + | '''(1)''' <u>Solutions 2 and 3</u> are correct: |
− | *The ACF value $\varphi_y(0)$ gives the variance (power) $\sigma_y^2$ and not the | + | *The ACF value $\varphi_y(0)$ gives the variance ("power") $\sigma_y^2$ and not the "standard deviation" $\sigma_y$. |
− | *Since a first-order | + | *Since a first-order non-recursive filter is present, all ACF values are $\varphi_y(k \cdot T_{\rm A})= 0$ for $|k| \ge 2$. |
*The ACF value $\varphi_y(- T_{\rm A})$ is equal to $\varphi_y(+ T_{\rm A})$. | *The ACF value $\varphi_y(- T_{\rm A})$ is equal to $\varphi_y(+ T_{\rm A})$. | ||
− | *These two ACF values result in a cosine function in the power density | + | *These two ACF values result in a cosine function in the power-spectral density, to which the DC component $\varphi_y(0)$ is added. |
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− | '''(3)''' With the previous settings, the variance is $\sigma_y^2 = 0.25$ and thus the | + | '''(3)''' With the previous settings, the variance is $\sigma_y^2 = 0.25$ and thus the standard deviation $\sigma_y = 0.5$. |
− | *Doubling the coefficients gives $\sigma_y = 1$ as desired: | + | *Doubling the coefficients gives $\sigma_y = 1$ as desired: |
:$$\hspace{0.15cm}\underline {a_0 = 0.8},\quad \hspace{0.15cm}\underline {a_1 = 0.6}.$$ | :$$\hspace{0.15cm}\underline {a_0 = 0.8},\quad \hspace{0.15cm}\underline {a_1 = 0.6}.$$ | ||
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− | '''(5)''' All ACF values are now larger by the constant value $K^2 = 0.25$ | + | '''(5)''' All ACF values are now larger by the constant value $K^2 = 0.25$. Thus |
:$$\varphi _y ( { T_{\rm A} } ) = 0.12 + 0.25 \hspace{0.15cm}\underline {= 0.37},$$ | :$$\varphi _y ( { T_{\rm A} } ) = 0.12 + 0.25 \hspace{0.15cm}\underline {= 0.37},$$ | ||
:$$\varphi _y ( { 2T_{\rm A} } ) = 0 + 0.25 \hspace{0.15cm}\underline {= 0.25}.$$ | :$$\varphi _y ( { 2T_{\rm A} } ) = 0 + 0.25 \hspace{0.15cm}\underline {= 0.25}.$$ | ||
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− | '''(6)''' The constant $K$ does not change the | + | '''(6)''' The constant $K$ does not change the standard deviation, i.e. $\sigma_y = 0.5$ is still valid. |
− | *Formally, this quantity can also be calculated as follows: | + | *Formally, this quantity can also be calculated as follows: |
:$$\sigma _y ^2 = \varphi _y ( 0 ) - \mathop {\lim }\limits_{k \to \infty } \varphi _y ( {k \cdot T_{\rm A} } ) = 0.5 - 0.25 = 0.25.$$ | :$$\sigma _y ^2 = \varphi _y ( 0 ) - \mathop {\lim }\limits_{k \to \infty } \varphi _y ( {k \cdot T_{\rm A} } ) = 0.5 - 0.25 = 0.25.$$ | ||
− | *Again, this gives $\sigma_y \hspace{0.15cm}\underline {= 0.5}$. | + | *Again, this gives $\sigma_y \hspace{0.15cm}\underline {= 0.5}$. |
{{ML-Fuß}} | {{ML-Fuß}} |
Latest revision as of 12:07, 25 February 2022
We consider here a first order non-recursive filter $(M = 1)$.
- Let the filter coefficients be $a_0 = 0.4$ and $a_1 = 0.3$.
- A constant $K$ is added at the filter output, which is to be set to zero up to and including subtask (3).
The individual elements of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle$
- are Gaussian as well as mean-free, and
- have in each case the standard deviation $\sigma_x = 1$.
Notes:
- The exercise belongs to the chapter Creation of Predefined ACF Properties.
- Reference is also made to the chapters Auto-Correlation Function and Power-Spectral Density.
Questions
Solution
(1) Solutions 2 and 3 are correct:
- The ACF value $\varphi_y(0)$ gives the variance ("power") $\sigma_y^2$ and not the "standard deviation" $\sigma_y$.
- Since a first-order non-recursive filter is present, all ACF values are $\varphi_y(k \cdot T_{\rm A})= 0$ for $|k| \ge 2$.
- The ACF value $\varphi_y(- T_{\rm A})$ is equal to $\varphi_y(+ T_{\rm A})$.
- These two ACF values result in a cosine function in the power-spectral density, to which the DC component $\varphi_y(0)$ is added.
(2) The general equation with $M = 1$ for $k \in \{0, \ 1\}$ is:
- $$\varphi _y ( {k \cdot T_{\rm A} } ) = \sigma _x ^2 \cdot \sum\limits_{\mu = 0}^{M - k} {a_\mu \cdot a_{\mu + k} } .$$
- From this we obtain with $\sigma_x = 1$:
- $$\varphi _y( 0 ) = a_0 ^2 + a_1 ^2 = 0.4^2 + 0.3^2 \hspace{0.15cm}\underline { = 0.25},$$
- $$\varphi _y ( { T_{\rm A} } ) = a_0 \cdot a_1 = 0.4 \cdot 0.3 \hspace{0.15cm}\underline {= 0.12}.$$
(3) With the previous settings, the variance is $\sigma_y^2 = 0.25$ and thus the standard deviation $\sigma_y = 0.5$.
- Doubling the coefficients gives $\sigma_y = 1$ as desired:
- $$\hspace{0.15cm}\underline {a_0 = 0.8},\quad \hspace{0.15cm}\underline {a_1 = 0.6}.$$
(4) The constant $K$ raises the total ACF by $K^2$. Using the result from (2), it follows:
- $$K^2 = 0.5 - 0.25 = 0.25\quad \Rightarrow \quad \hspace{0.15cm}\underline {K = 0.5}.$$
(5) All ACF values are now larger by the constant value $K^2 = 0.25$. Thus
- $$\varphi _y ( { T_{\rm A} } ) = 0.12 + 0.25 \hspace{0.15cm}\underline {= 0.37},$$
- $$\varphi _y ( { 2T_{\rm A} } ) = 0 + 0.25 \hspace{0.15cm}\underline {= 0.25}.$$
(6) The constant $K$ does not change the standard deviation, i.e. $\sigma_y = 0.5$ is still valid.
- Formally, this quantity can also be calculated as follows:
- $$\sigma _y ^2 = \varphi _y ( 0 ) - \mathop {\lim }\limits_{k \to \infty } \varphi _y ( {k \cdot T_{\rm A} } ) = 0.5 - 0.25 = 0.25.$$
- Again, this gives $\sigma_y \hspace{0.15cm}\underline {= 0.5}$.