Difference between revisions of "Aufgaben:Exercise 5.6: Filter Dimensioning"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Creation_of_Predefined_ACF_Properties |
}} | }} | ||
− | [[File:P_ID566__Sto_A_5_6.png|right|frame| | + | [[File:P_ID566__Sto_A_5_6.png|right|frame|Desired ACF $\varphi_y(k \cdot T_{\rm A})$]] |
− | + | A discrete-time random variable $\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle$ with the outlined auto-correlation function $\rm (ACF)$ is to be generated using a digital filter. | |
− | + | Let the discrete-time Gaussian input values $x_\nu$ be characterized in each case by | |
− | * | + | *the mean value $m_x = 0$, |
− | * | + | *the standard deviation $\sigma_x = 1$. |
− | + | Notes: | |
− | + | *The exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Creation_of_Predefined_ACF_Properties|Creation of Predefined ACF Properties]]. | |
− | + | *Reference is also made to the chapter [[Theory_of_Stochastic_Signals/Auto-Correlation_Function_(ACF)|Auto-Correlation Function]]. | |
− | + | *Let all ACF values $\varphi_y(k \cdot T_{\rm A})$ with index $|k| \gt 2$ be zero. | |
− | |||
− | * | ||
− | * | ||
− | * | ||
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Which of the following statements are true? |
|type="[]"} | |type="[]"} | ||
− | - | + | - A first-order recursive filter is suitable. |
− | - | + | - A first-order non-recursive filter is suitable. |
− | + | + | + A second order non-recursive filter is suitable. |
− | - | + | - The output values $y_\nu$ are triangularly distributed. |
− | + | + | + The output values $y_\nu$ are mean-free $(m_y = 0)$. |
− | { | + | {Give the equations for determining the coefficients $a_0$, $a_1$ and $a_2$. Replace the three variables with $u = a_1^2$ and $w = (a_0 + a_2)^2$. Determine $u$ and $w$. <br> <u>Note:</u> There is only one reasonable solution. |
|type="{}"} | |type="{}"} | ||
$u \ = \ $ { 0.25 3% } | $u \ = \ $ { 0.25 3% } | ||
Line 43: | Line 39: | ||
− | { | + | {Determine the filter coefficients $a_0$, $a_1$ and $a_2$. Enter the following quotients: |
|type="{}"} | |type="{}"} | ||
$a_1/a_0 \ = \ $ { -0.515--0.485 } | $a_1/a_0 \ = \ $ { -0.515--0.485 } | ||
Line 49: | Line 45: | ||
− | { | + | {How many different sets of parameters $(I)$ lead to the desired ACF? |
|type="{}"} | |type="{}"} | ||
$I \ = \ $ { 2 } | $I \ = \ $ { 2 } | ||
Line 56: | Line 52: | ||
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' <u>Solutions 3 and 5</u> are correct: |
− | * | + | *A recursive filter would always cause an infinitely extended impulse response $h(t)$ and thus also an infinitely extended auto-correlation function $\rm (ACF)$. |
− | * | + | *Therefore, a non-recursive filter structure must be chosen here. The specified ACF requires the order $M= 2$. |
− | * | + | *Since the input values are Gaussian distributed and mean-free, this also applies to the output values. |
− | * | + | *When filtering stochastic signals, the following always applies: '''Gaussian remains Gaussian and non-Gaussian never becomes (exactly) Gaussian'''. |
− | '''(2)''' | + | '''(2)''' The system of equations is: |
:$$k = 2\text{:}\quad a_0 \cdot a_2 = 1.$$ | :$$k = 2\text{:}\quad a_0 \cdot a_2 = 1.$$ | ||
:$$k = 1\text{:}\quad a_0 \cdot a_1 + a_1 \cdot a_2 = - 1\quad \Rightarrow \quad \sqrt {u \cdot w} = - 1\quad \Rightarrow \quad u \cdot w = 1.$$ | :$$k = 1\text{:}\quad a_0 \cdot a_1 + a_1 \cdot a_2 = - 1\quad \Rightarrow \quad \sqrt {u \cdot w} = - 1\quad \Rightarrow \quad u \cdot w = 1.$$ | ||
:$$k = 0\text{:}\quad a_0 ^2 + a_1 ^2 + a_2 ^2 = 2.25\quad \;\;\, \Rightarrow \quad u + w = 2.25 + 2a_0 \cdot a_2 = 4.25.$$ | :$$k = 0\text{:}\quad a_0 ^2 + a_1 ^2 + a_2 ^2 = 2.25\quad \;\;\, \Rightarrow \quad u + w = 2.25 + 2a_0 \cdot a_2 = 4.25.$$ | ||
− | + | The two equations with respect to $u$ and $w$ has two solutions: | |
− | *$u = 4, \ w = 0.25$: | + | *$u = 4, \ w = 0.25$: Because of the condition $a_2 = 1/a_0$ (see first equation), $a_0$ and $a_2$ have the same sign. |
− | * | + | * Moreover, at least one of the two coefficients is greater than/equal to $1$. |
− | * | + | *Thus the condition $a_0+a_2= \sqrt{w} = 0.5$ cannot be fulfilled. |
− | * | + | *Therefore, the correct solution is $\underline{u = 0.25}, \ \underline{w = 4}$. |
− | '''(3)''' | + | '''(3)''' The result of '''(2)''' means that $a_1 = \pm \sqrt{0.25} = \pm 0.5$. |
− | * | + | *The positive value leads to the equations |
− | $$(1) \hspace{0.5cm}0.5 \cdot \left( {a_0 + a_2 } \right) = - 1\quad \Rightarrow \quad a_0 + a_2 = - 2,$$ | + | :$$(1) \hspace{0.5cm}0.5 \cdot \left( {a_0 + a_2 } \right) = - 1\quad \Rightarrow \quad a_0 + a_2 = - 2,$$ |
− | $$(2) \hspace{0.5cm}a_0 \cdot a_2 = 1.$$ | + | :$$(2) \hspace{0.5cm}a_0 \cdot a_2 = 1.$$ |
− | * | + | *From this follows $a_0=a_2=-1$. With $a_1= 0.5$, the final result is: |
:$$a_1/a_0 \hspace{0.15 cm}\underline{= -0.5}, \hspace{0.5 cm} | :$$a_1/a_0 \hspace{0.15 cm}\underline{= -0.5}, \hspace{0.5 cm} | ||
a_2/a_0 \hspace{0.15 cm}\underline{= 1}.$$ | a_2/a_0 \hspace{0.15 cm}\underline{= 1}.$$ | ||
− | * | + | *The solution $a_1= -0.5$ leads to $a_0=a_2=+1$ and thus to the same quotients. |
− | '''(4)''' | + | '''(4)''' In general, this problem has $I = 4$ equivalent solutions $($mirroring/shifting as well as the multiplication by $-1$ in each case$)$. |
− | * | + | *Since here the impulse response is symmetrical, there are however only $\underline{I = 2}$ different solutions: |
− | :$$\text{ | + | :$$\text{Solution 1:} \ \ a_0 = +1,\quad a_1 = - 0.5,\quad a_2 = +1; $$ |
− | :$$\text{ | + | :$$\text{Solution 2:} \ \ a_0 = - 1,\quad a_1 = +0.5,\quad a_2 = - 1. $$ |
{{ML-Fuß}} | {{ML-Fuß}} | ||
− | [[Category:Theory of Stochastic Signals: Exercises|^5.3 | + | [[Category:Theory of Stochastic Signals: Exercises|^5.3 Filter Matching to ACF^]] |
Latest revision as of 17:22, 17 February 2022
A discrete-time random variable $\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle$ with the outlined auto-correlation function $\rm (ACF)$ is to be generated using a digital filter.
Let the discrete-time Gaussian input values $x_\nu$ be characterized in each case by
- the mean value $m_x = 0$,
- the standard deviation $\sigma_x = 1$.
Notes:
- The exercise belongs to the chapter Creation of Predefined ACF Properties.
- Reference is also made to the chapter Auto-Correlation Function.
- Let all ACF values $\varphi_y(k \cdot T_{\rm A})$ with index $|k| \gt 2$ be zero.
Questions
Solution
(1) Solutions 3 and 5 are correct:
- A recursive filter would always cause an infinitely extended impulse response $h(t)$ and thus also an infinitely extended auto-correlation function $\rm (ACF)$.
- Therefore, a non-recursive filter structure must be chosen here. The specified ACF requires the order $M= 2$.
- Since the input values are Gaussian distributed and mean-free, this also applies to the output values.
- When filtering stochastic signals, the following always applies: Gaussian remains Gaussian and non-Gaussian never becomes (exactly) Gaussian.
(2) The system of equations is:
- $$k = 2\text{:}\quad a_0 \cdot a_2 = 1.$$
- $$k = 1\text{:}\quad a_0 \cdot a_1 + a_1 \cdot a_2 = - 1\quad \Rightarrow \quad \sqrt {u \cdot w} = - 1\quad \Rightarrow \quad u \cdot w = 1.$$
- $$k = 0\text{:}\quad a_0 ^2 + a_1 ^2 + a_2 ^2 = 2.25\quad \;\;\, \Rightarrow \quad u + w = 2.25 + 2a_0 \cdot a_2 = 4.25.$$
The two equations with respect to $u$ and $w$ has two solutions:
- $u = 4, \ w = 0.25$: Because of the condition $a_2 = 1/a_0$ (see first equation), $a_0$ and $a_2$ have the same sign.
- Moreover, at least one of the two coefficients is greater than/equal to $1$.
- Thus the condition $a_0+a_2= \sqrt{w} = 0.5$ cannot be fulfilled.
- Therefore, the correct solution is $\underline{u = 0.25}, \ \underline{w = 4}$.
(3) The result of (2) means that $a_1 = \pm \sqrt{0.25} = \pm 0.5$.
- The positive value leads to the equations
- $$(1) \hspace{0.5cm}0.5 \cdot \left( {a_0 + a_2 } \right) = - 1\quad \Rightarrow \quad a_0 + a_2 = - 2,$$
- $$(2) \hspace{0.5cm}a_0 \cdot a_2 = 1.$$
- From this follows $a_0=a_2=-1$. With $a_1= 0.5$, the final result is:
- $$a_1/a_0 \hspace{0.15 cm}\underline{= -0.5}, \hspace{0.5 cm} a_2/a_0 \hspace{0.15 cm}\underline{= 1}.$$
- The solution $a_1= -0.5$ leads to $a_0=a_2=+1$ and thus to the same quotients.
(4) In general, this problem has $I = 4$ equivalent solutions $($mirroring/shifting as well as the multiplication by $-1$ in each case$)$.
- Since here the impulse response is symmetrical, there are however only $\underline{I = 2}$ different solutions:
- $$\text{Solution 1:} \ \ a_0 = +1,\quad a_1 = - 0.5,\quad a_2 = +1; $$
- $$\text{Solution 2:} \ \ a_0 = - 1,\quad a_1 = +0.5,\quad a_2 = - 1. $$