Difference between revisions of "Aufgaben:Exercise 5.1: Gaussian ACF and Gaussian Low-Pass"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Stochastic_System_Theory |
}} | }} | ||
− | [[File:P_ID487__Sto_A_5_1.png|right|frame| | + | [[File:P_ID487__Sto_A_5_1.png|right|frame|Gaussian ACF at the filter input and output]] |
− | + | At the input of a low-pass filter with frequency response $H(f)$, there is a Gaussian distributed mean-free noise signal $x(t)$ with the following auto-correlation function $\rm (ACF)$: | |
− | :$${\it \ | + | |
+ | :$${\it \varphi}_{x}(\tau) = \sigma_x^2 \cdot {\rm e}^{- \pi (\tau | ||
/{\rm \nabla} \tau_x)^2}.$$ | /{\rm \nabla} \tau_x)^2}.$$ | ||
− | + | This ACF is shown in the accompanying diagram above. | |
− | + | Let the filter be Gaussian with the DC gain $H_0$ and the equivalent bandwidth $\Delta f$. Thus, for the frequency response, it can be written: | |
:$$H(f) = H_{\rm 0} \cdot{\rm e}^{- \pi (f/ {\rm \Delta} f)^2}.$$ | :$$H(f) = H_{\rm 0} \cdot{\rm e}^{- \pi (f/ {\rm \Delta} f)^2}.$$ | ||
− | + | In the course of this task, the two filter parameters $H_0$ and $\Delta f$ are to be dimensioned so that the output signal $y(t)$ has an ACF corresponding to the diagram below. | |
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− | + | Notes: | |
− | + | *The exercise belongs to the chapter [[Theory_of_Stochastic_Signals/Stochastic_System_Theory|Stochastic System Theory]]. | |
− | * | + | *Reference is also made to the chapter [[Theory_of_Stochastic_Signals/Auto-Correlation_Function_(ACF)|Auto-Correlation Function]]. |
− | * | ||
− | * | + | *Consider the following Fourier correspondence: |
:$${\rm e}^{- \pi (f/{\rm \Delta} f)^2} \hspace{0.15cm} | :$${\rm e}^{- \pi (f/{\rm \Delta} f)^2} \hspace{0.15cm} | ||
\bullet\!\!-\!\!\!-\!\!\!\hspace{0.03cm}\circ \hspace{0.15cm}{\rm \Delta} f \cdot | \bullet\!\!-\!\!\!-\!\!\!\hspace{0.03cm}\circ \hspace{0.15cm}{\rm \Delta} f \cdot | ||
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− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {What is the standard deviation of the filter input signal? |
|type="{}"} | |type="{}"} | ||
$\sigma_x \ = \ $ { 0.2 3% } $\ \rm V$ | $\sigma_x \ = \ $ { 0.2 3% } $\ \rm V$ | ||
− | { | + | {From the sketched ACF, also determine the equivalent ACF duration $\nabla\tau_x$ of the input signal. How can this be determined in general? |
|type="{}"} | |type="{}"} | ||
− | $\nabla\tau_x \ = \ $ { 1 3% } $\ µ s$ | + | $\nabla\tau_x \ = \ $ { 1 3% } $\ \rm µ s$ |
− | { | + | {What is the power-spectral density ${\it Φ}_x(f)$ of the input signal? What is the PSD value at $f= 0$? |
|type="{}"} | |type="{}"} | ||
${\it Φ}_x(f=0) \ = \ $ { 40 3% } $\ \cdot 10^{-9}\ \rm V^2/Hz$ | ${\it Φ}_x(f=0) \ = \ $ { 40 3% } $\ \cdot 10^{-9}\ \rm V^2/Hz$ | ||
− | { | + | {Calculate the PSD ${\it Φ}_y(f)$ at the filter output in general as a function of $\sigma_x$, $\nabla \tau_x$, $H_0$ and $\Delta f$. Which statements are true? |
|type="[]"} | |type="[]"} | ||
− | + | + | + The PSD ${\it Φ}_y(f)$ is also Gaussian. |
− | - | + | - The smaller $\Delta f$ is, the wider ${\it Φ}_y(f)$. |
− | + $H_0$ | + | + $H_0$ only affects the height, but not the width of ${\it Φ}_y(f)$. |
− | { | + | {How large must the equivalent filter bandwidth $\Delta f$ be chosen so that $\nabla \tau_y = 3 \ \rm µ s$ holds for the equivalent ACF duration? |
|type="{}"} | |type="{}"} | ||
$\Delta f \ = \ $ { 0.5 3% } $\ \rm MHz$ | $\Delta f \ = \ $ { 0.5 3% } $\ \rm MHz$ | ||
− | { | + | {How large must one select the DC signal transfer factor $H_0$ so that the condition $\sigma_y = \sigma_x$ is fulfilled? |
|type="{}"} | |type="{}"} | ||
$H_0 \ = \ $ { 1.732 3% } | $H_0 \ = \ $ { 1.732 3% } | ||
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' The variance is equal to the ACF value at $\tau = 0$, so $\sigma_x^2 = 0.04 \ \rm V^2$. |
+ | *From this follows $\sigma_x\hspace{0.15cm}\underline {= 0.2 \ \rm V}$. | ||
+ | |||
+ | |||
+ | '''(2)''' The equivalent ACF duration can be determined via the rectangle of equal area. | ||
+ | *According to the sketch, we obtain $\nabla \tau_x\hspace{0.15cm}\underline {= 1 \ \rm µ s}$. | ||
− | |||
− | '''(3)''' | + | '''(3)''' The PSD is the Fourier transform of the ACF. |
− | * | + | *With the given Fourier correspondence holds: |
:$${\it \Phi}_{x}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot | :$${\it \Phi}_{x}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot | ||
{\rm e}^{- \pi ({\rm \nabla} \tau_x \hspace{0.03cm}\cdot \hspace{0.03cm}f)^2} .$$ | {\rm e}^{- \pi ({\rm \nabla} \tau_x \hspace{0.03cm}\cdot \hspace{0.03cm}f)^2} .$$ | ||
− | * | + | *At frequency $f = 0$, we obtain: |
:$${\it \Phi}_{x}(f = 0) = \sigma_x^2 \cdot {\rm \nabla} \tau_x = | :$${\it \Phi}_{x}(f = 0) = \sigma_x^2 \cdot {\rm \nabla} \tau_x = | ||
\rm 0.04 \hspace{0.1cm} V^2 \cdot 10^{-6} \hspace{0.1cm} s \hspace{0.15cm} \underline{= 40 | \rm 0.04 \hspace{0.1cm} V^2 \cdot 10^{-6} \hspace{0.1cm} s \hspace{0.15cm} \underline{= 40 | ||
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− | '''(4)''' | + | |
− | * | + | '''(4)''' <u>Solutions 1 and 3</u> are correct: |
+ | *In general, ${\it \Phi}_{y}(f) = {\it \Phi}_{x}(f) \cdot |H(f)|^2$. It follows: | ||
:$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot | :$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot | ||
{\rm e}^{- \pi ({\rm \nabla} \tau_x \cdot f)^2}\cdot H_{\rm 0}^2 | {\rm e}^{- \pi ({\rm \nabla} \tau_x \cdot f)^2}\cdot H_{\rm 0}^2 | ||
\cdot{\rm e}^{- 2 \pi (f/ {\rm \Delta} f)^2} .$$ | \cdot{\rm e}^{- 2 \pi (f/ {\rm \Delta} f)^2} .$$ | ||
− | * | + | *By combining the two exponential functions, we obtain: |
:$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot H_0^2 \cdot | :$${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot H_0^2 \cdot | ||
− | {\rm e}^{- \pi\cdot ({\rm \nabla} \tau_x^2 + 2/ | + | {\rm e}^{- \pi\cdot ({\rm \nabla} \tau_x^2 + 2/\Delta f^2 ) \hspace{0.1cm}\cdot f^2}.$$ |
− | * | + | *Also ${\it \Phi}_{y}(f)$ is Gaussian and never wider than ${\it \Phi}_{x}(f)$. For $f \to \infty$, the approximation ${\it \Phi}_{y}(f) \approx {\it \Phi}_{x}(f)$ holds. |
− | * | + | *As $\Delta f$ gets smaller, ${\it \Phi}_{y}(f)$ gets narrower (so the second statement is false). |
− | *$H_0$ | + | *$H_0$ actually affects only the PSD height, but not the width of the PSD. |
− | '''(5)''' | + | '''(5)''' Analogous to task '''(1)''', it can be written for the PSD of the output signal $y(t)$: |
:$${\it \Phi}_{y}(f) = \sigma_y^2 \cdot {\rm \nabla} \tau_y \cdot | :$${\it \Phi}_{y}(f) = \sigma_y^2 \cdot {\rm \nabla} \tau_y \cdot | ||
{\rm e}^{- \pi \cdot {\rm \nabla} \tau_y^2 \cdot f^2 }.$$ | {\rm e}^{- \pi \cdot {\rm \nabla} \tau_y^2 \cdot f^2 }.$$ | ||
− | * | + | *By comparing with the result from '''(4)''' we get: |
:$${{\rm \nabla} \tau_y^2} = {{\rm \nabla} \tau_x^2} + \frac {2}{{\rm | :$${{\rm \nabla} \tau_y^2} = {{\rm \nabla} \tau_x^2} + \frac {2}{{\rm | ||
\Delta} f^2}.$$ | \Delta} f^2}.$$ | ||
− | * | + | *Solving the equation for $\Delta f$ and considering the values $\nabla \tau_x {= 1 \ \rm µ s}$ as well as $\nabla \tau_y {= 3 \ \rm µ s}$, it follows: |
:$${\rm \Delta} f = \sqrt{\frac{2}{{\rm \nabla} \tau_y^2 - {\rm | :$${\rm \Delta} f = \sqrt{\frac{2}{{\rm \nabla} \tau_y^2 - {\rm | ||
\nabla} \tau_x^2}} = \sqrt{\frac{2}{9 - 1}} \hspace{0.1cm}\rm MHz | \nabla} \tau_x^2}} = \sqrt{\frac{2}{9 - 1}} \hspace{0.1cm}\rm MHz | ||
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− | '''(6)''' | + | |
− | * | + | '''(6)''' The condition $\sigma_y = \sigma_x$ is equivalent to $\varphi_y(\tau = 0)= \varphi_x(\tau = 0)$. |
− | * | + | *Moreover, since $\nabla \tau_y = 3 \cdot \nabla \tau_x$ is given, therefore ${\it \Phi}_{y}(f= 0) = 3 \cdot {\it \Phi}_{x}(f= 0)$ must also hold. |
+ | *From this we obtain: | ||
:$$H_{\rm 0} = \sqrt{\frac{\it \Phi_y (f \rm = 0)}{\it \Phi_x (f = \rm 0)}} = \sqrt | :$$H_{\rm 0} = \sqrt{\frac{\it \Phi_y (f \rm = 0)}{\it \Phi_x (f = \rm 0)}} = \sqrt | ||
{3}\hspace{0.15cm} \underline{=1.732}.$$ | {3}\hspace{0.15cm} \underline{=1.732}.$$ | ||
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− | [[Category: | + | [[Category:Theory of Stochastic Signals: Exercises|^5.1 Stochastic Systems Theory^]] |
Latest revision as of 12:11, 17 February 2022
At the input of a low-pass filter with frequency response $H(f)$, there is a Gaussian distributed mean-free noise signal $x(t)$ with the following auto-correlation function $\rm (ACF)$:
- $${\it \varphi}_{x}(\tau) = \sigma_x^2 \cdot {\rm e}^{- \pi (\tau /{\rm \nabla} \tau_x)^2}.$$
This ACF is shown in the accompanying diagram above.
Let the filter be Gaussian with the DC gain $H_0$ and the equivalent bandwidth $\Delta f$. Thus, for the frequency response, it can be written:
- $$H(f) = H_{\rm 0} \cdot{\rm e}^{- \pi (f/ {\rm \Delta} f)^2}.$$
In the course of this task, the two filter parameters $H_0$ and $\Delta f$ are to be dimensioned so that the output signal $y(t)$ has an ACF corresponding to the diagram below.
Notes:
- The exercise belongs to the chapter Stochastic System Theory.
- Reference is also made to the chapter Auto-Correlation Function.
- Consider the following Fourier correspondence:
- $${\rm e}^{- \pi (f/{\rm \Delta} f)^2} \hspace{0.15cm} \bullet\!\!-\!\!\!-\!\!\!\hspace{0.03cm}\circ \hspace{0.15cm}{\rm \Delta} f \cdot {\rm e}^{- \pi ({\rm \Delta} f \hspace{0.03cm} \cdot \hspace{0.03cm} t)^2}.$$
Questions
Solution
- From this follows $\sigma_x\hspace{0.15cm}\underline {= 0.2 \ \rm V}$.
(2) The equivalent ACF duration can be determined via the rectangle of equal area.
- According to the sketch, we obtain $\nabla \tau_x\hspace{0.15cm}\underline {= 1 \ \rm µ s}$.
(3) The PSD is the Fourier transform of the ACF.
- With the given Fourier correspondence holds:
- $${\it \Phi}_{x}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot {\rm e}^{- \pi ({\rm \nabla} \tau_x \hspace{0.03cm}\cdot \hspace{0.03cm}f)^2} .$$
- At frequency $f = 0$, we obtain:
- $${\it \Phi}_{x}(f = 0) = \sigma_x^2 \cdot {\rm \nabla} \tau_x = \rm 0.04 \hspace{0.1cm} V^2 \cdot 10^{-6} \hspace{0.1cm} s \hspace{0.15cm} \underline{= 40 \cdot 10^{-9} \hspace{0.1cm} V^2 / Hz}.$$
(4) Solutions 1 and 3 are correct:
- In general, ${\it \Phi}_{y}(f) = {\it \Phi}_{x}(f) \cdot |H(f)|^2$. It follows:
- $${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot {\rm e}^{- \pi ({\rm \nabla} \tau_x \cdot f)^2}\cdot H_{\rm 0}^2 \cdot{\rm e}^{- 2 \pi (f/ {\rm \Delta} f)^2} .$$
- By combining the two exponential functions, we obtain:
- $${\it \Phi}_{y}(f) = \sigma_x^2 \cdot {\rm \nabla} \tau_x \cdot H_0^2 \cdot {\rm e}^{- \pi\cdot ({\rm \nabla} \tau_x^2 + 2/\Delta f^2 ) \hspace{0.1cm}\cdot f^2}.$$
- Also ${\it \Phi}_{y}(f)$ is Gaussian and never wider than ${\it \Phi}_{x}(f)$. For $f \to \infty$, the approximation ${\it \Phi}_{y}(f) \approx {\it \Phi}_{x}(f)$ holds.
- As $\Delta f$ gets smaller, ${\it \Phi}_{y}(f)$ gets narrower (so the second statement is false).
- $H_0$ actually affects only the PSD height, but not the width of the PSD.
(5) Analogous to task (1), it can be written for the PSD of the output signal $y(t)$:
- $${\it \Phi}_{y}(f) = \sigma_y^2 \cdot {\rm \nabla} \tau_y \cdot {\rm e}^{- \pi \cdot {\rm \nabla} \tau_y^2 \cdot f^2 }.$$
- By comparing with the result from (4) we get:
- $${{\rm \nabla} \tau_y^2} = {{\rm \nabla} \tau_x^2} + \frac {2}{{\rm \Delta} f^2}.$$
- Solving the equation for $\Delta f$ and considering the values $\nabla \tau_x {= 1 \ \rm µ s}$ as well as $\nabla \tau_y {= 3 \ \rm µ s}$, it follows:
- $${\rm \Delta} f = \sqrt{\frac{2}{{\rm \nabla} \tau_y^2 - {\rm \nabla} \tau_x^2}} = \sqrt{\frac{2}{9 - 1}} \hspace{0.1cm}\rm MHz \hspace{0.15cm} \underline{= 0.5\hspace{0.1cm} MHz} .$$
(6) The condition $\sigma_y = \sigma_x$ is equivalent to $\varphi_y(\tau = 0)= \varphi_x(\tau = 0)$.
- Moreover, since $\nabla \tau_y = 3 \cdot \nabla \tau_x$ is given, therefore ${\it \Phi}_{y}(f= 0) = 3 \cdot {\it \Phi}_{x}(f= 0)$ must also hold.
- From this we obtain:
- $$H_{\rm 0} = \sqrt{\frac{\it \Phi_y (f \rm = 0)}{\it \Phi_x (f = \rm 0)}} = \sqrt {3}\hspace{0.15cm} \underline{=1.732}.$$