Difference between revisions of "Aufgaben:Exercise 2.7: Coherence Bandwidth"
m (Text replacement - "power spectral density" to "power-spectral density") |
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}} |
− | [[File:P_ID2172__Mob_A_2_7.png|right|frame| | + | [[File:P_ID2172__Mob_A_2_7.png|right|frame|Delay power-spectral density and <br>frequency correlation function]] |
− | For the power spectral density | + | For the delay power-spectral density, we assume an exponential behavior. With ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$ we have |
:$${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$ | :$${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$ | ||
− | The constant $\tau_0$ can be determined from the tangent in the point $\tau = 0$ according to the upper graph. Note that ${\it \Phi}_{\rm V}(\tau)$ has | + | The constant $\tau_0$ can be determined from the tangent in the point $\tau = 0$ according to the upper graph. Note that ${\it \Phi}_{\rm V}(\tau)$ has unit $[1/\rm s]$ . Furthermore, |
− | * The probability density function (PDF) $f_{\rm V}(\tau)$ has the same form as ${\it \Phi}_{\rm V}(\tau)$, but is normalized to area $1$ . | + | * The probability density function $\rm (PDF)$ $f_{\rm V}(\tau)$ has the same form as ${\it \Phi}_{\rm V}(\tau)$, but is normalized to area $1$ . |
− | * The <b>average excess delay</b> or <b>mean excess delay</b> $m_{\rm V}$ is equal to the linear expectation $E\big [\tau \big]$ and can be determined from the PDF | + | * The <b>average excess delay</b> or <b>mean excess delay</b> $m_{\rm V}$ is equal to the linear expectation $E\big [\tau \big]$ and can be determined from the PDF $f_{\rm V}(\tau)$ . |
− | * The <b>multipath spread</b> or <b>delay spread</b> $\sigma_{\rm V}$ gives the standard deviation | + | * The <b>multipath spread</b> or <b>delay spread</b> $\sigma_{\rm V}$ gives the standard deviation of the random variable $\tau$ . In the theory part we also use the term $T_{\rm V}$ for this. |
− | * The displayed frequency correlation function $\varphi_{\rm F}(\ | + | * The displayed frequency correlation function $\varphi_{\rm F}(\Delta f)$ can be calculated as the Fourier transform of the delay power-spectral density ${\it \Phi}_{\rm V}(\tau)$ : |
:$$\varphi_{\rm F}(\Delta f) | :$$\varphi_{\rm F}(\Delta f) | ||
\hspace{0.2cm} {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$ | \hspace{0.2cm} {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$ | ||
− | * The <b>coherence bandwidth</b> $B_{\rm K}$ is the value of | + | * The <b>coherence bandwidth</b> $B_{\rm K}$ is the value of $\Delta f$ at which the frequency correlation function $\varphi_{\rm F}(\Delta f)$ has dropped to half in absolute value. |
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''Notes:'' | ''Notes:'' | ||
− | * This | + | * This exercise belongs to the chapter [[Mobile_Communications/The_GWSSUS_Channel_Model|The GWSSUS Channel Model]]. |
− | * This | + | * This exercise requires knowledge of [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected values and moments]] from the book "Theory of Stochastic Signals". |
* In addition, the following Fourier transform is given: | * In addition, the following Fourier transform is given: | ||
:$$x(t) = \left\{ \begin{array}{c} {\rm e}^{- \lambda \hspace{0.05cm}\cdot \hspace{0.05cm} t}\\ | :$$x(t) = \left\{ \begin{array}{c} {\rm e}^{- \lambda \hspace{0.05cm}\cdot \hspace{0.05cm} t}\\ | ||
0 \end{array} \right.\quad | 0 \end{array} \right.\quad | ||
− | \begin{array}{*{1}c} \hspace{-0.35cm} {\rm | + | \begin{array}{*{1}c} \hspace{-0.35cm} {\rm for} \hspace{0.15cm} t \ge 0 |
− | \\ \hspace{-0.35cm} {\rm | + | \\ \hspace{-0.35cm} {\rm for} \hspace{0.15cm} t < 0 \\ \end{array} |
\hspace{0.4cm} {\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.4cm} X(f) = \frac{1}{\lambda + {\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$ | \hspace{0.4cm} {\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.4cm} X(f) = \frac{1}{\lambda + {\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$ | ||
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===Questionnaire=== | ===Questionnaire=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | {What is the probability density $f_{\rm V}(\tau)$ of the delay? | + | {What is the probability density function $f_{\rm V}(\tau)$ of the delay time? |
− | |type=" | + | |type="()"} |
- $f_{\rm V}(\tau) = {\rm e}^{-\tau/\tau_0}$. | - $f_{\rm V}(\tau) = {\rm e}^{-\tau/\tau_0}$. | ||
+ $f_{\rm V}(\tau) = 1/\tau_0 \cdot {\rm e}^{-\tau/\tau_0}$, | + $f_{\rm V}(\tau) = 1/\tau_0 \cdot {\rm e}^{-\tau/\tau_0}$, | ||
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$m_{\rm V} \ = \ ${ 1 3% } $\ \rm µ s$ | $m_{\rm V} \ = \ ${ 1 3% } $\ \rm µ s$ | ||
− | {Which value results for the multipath | + | {Which value results for the multipath spread with $\tau_0 = 1 \ \ \rm µ s$? |
|type="{}"} | |type="{}"} | ||
$\sigma_{\rm V} \ = \ ${ 1 3% } $\ \rm µ s$ | $\sigma_{\rm V} \ = \ ${ 1 3% } $\ \rm µ s$ | ||
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' Let ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$. The integral of the power spectral density | + | '''(1)''' Let ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$. The integral of the delay power-spectral density gives |
:$$\int_{0}^{+\infty} {\it \Phi}_{\rm V}(\tau) \hspace{0.15cm}{\rm d} \tau = | :$$\int_{0}^{+\infty} {\it \Phi}_{\rm V}(\tau) \hspace{0.15cm}{\rm d} \tau = | ||
{\it \Phi}_{\rm 0} \cdot \int_{0}^{+\infty} {\rm e}^{-\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau = | {\it \Phi}_{\rm 0} \cdot \int_{0}^{+\infty} {\rm e}^{-\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau = | ||
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\hspace{0.05cm}. $$ | \hspace{0.05cm}. $$ | ||
− | * | + | *Then the probability density function is |
:$$f_{\rm V}(\tau) = \frac{{\it \Phi}_{\rm V}(\tau) }{{\it \Phi}_{\rm 0} \cdot \tau_0}= \frac{1}{\tau_0} \cdot {\rm e}^{-\tau / \tau_0} | :$$f_{\rm V}(\tau) = \frac{{\it \Phi}_{\rm V}(\tau) }{{\it \Phi}_{\rm 0} \cdot \tau_0}= \frac{1}{\tau_0} \cdot {\rm e}^{-\tau / \tau_0} | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | + | *<u>Solution 2</u> is correct. | |
− | *<u>Solution 2</u> is | ||
− | |||
− | '''(2)''' The $k$-th moment of an [[ | + | '''(2)''' The $k$-th moment of an [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#One-sided_exponential_distribution|one-sided exponential random variable]] is $m_k = k! \cdot \tau_0^k$. |
− | *With $k = 1$, this results in the linear mean value $m_1 = m_{\rm V}$: | + | *With $k = 1$, this results in the linear mean value $m_1 = m_{\rm V}$: |
:$$m_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}} | :$$m_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}} | ||
\hspace{0.05cm}. $$ | \hspace{0.05cm}. $$ | ||
− | '''(3)''' According to the [[ | + | |
− | *This yields $m_2 = 2 \cdot \tau_0^2$, and therefore | + | '''(3)''' According to the [[Theory_of_Stochastic_Signals/Erwartungswerte_und_Momente#Some common central moments| Steiner's Theorem]], the variance of any random variable is $\sigma^2 = m_2 \, -m_1^2$. |
+ | *This yields $m_2 = 2 \cdot \tau_0^2$, and therefore | ||
:$$\sigma_{\rm V}^2 = m_2 - m_1^2 = 2 \cdot \tau_0^2 - (\tau_0)^2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} | :$$\sigma_{\rm V}^2 = m_2 - m_1^2 = 2 \cdot \tau_0^2 - (\tau_0)^2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} | ||
\sigma_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}} | \sigma_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}} | ||
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− | '''(4)''' ${\it \Phi}_{\rm V}(\tau)$ is identical to $x(t)$ in the given Fourier transform pair if $t$ is replaced by $\tau$ and $\lambda$ by $1/\tau_0$. | + | '''(4)''' ${\it \Phi}_{\rm V}(\tau)$ is identical to $x(t)$ in the given Fourier transform pair if $t$ is replaced by $\tau$ and $\lambda$ by $1/\tau_0$. |
− | *Thus, $\varphi_{\rm F}(\ | + | *Thus, $\varphi_{\rm F}(\Delta f)$ is equal to $X(f)$ with the substitution $f → \Delta f$: |
:$$\varphi_{\rm F}(\Delta f) = \frac{1}{1/\tau_0 + {\rm j} \cdot 2\pi \Delta f} | :$$\varphi_{\rm F}(\Delta f) = \frac{1}{1/\tau_0 + {\rm j} \cdot 2\pi \Delta f} | ||
= \frac{\tau_0}{1 + {\rm j} \cdot 2\pi \cdot \tau_0 \cdot \Delta f}\hspace{0.05cm}.$$ | = \frac{\tau_0}{1 + {\rm j} \cdot 2\pi \cdot \tau_0 \cdot \Delta f}\hspace{0.05cm}.$$ | ||
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B_{\rm K}= \frac{\sqrt{3}}{2\pi \cdot \tau_0} \approx \frac{0.276}{ \tau_0}\hspace{0.05cm}. $$ | B_{\rm K}= \frac{\sqrt{3}}{2\pi \cdot \tau_0} \approx \frac{0.276}{ \tau_0}\hspace{0.05cm}. $$ | ||
− | *With $\tau_0 = 1 \ \ \rm µ s$, the coherence bandwidth is $B_{\rm K} \ \underline {= 276 \ \ \rm kHz}$. | + | *With $\tau_0 = 1 \ \ \rm µ s$, the coherence bandwidth is $B_{\rm K} \ \underline {= 276 \ \ \rm kHz}$. |
{{ML-Fuß}} | {{ML-Fuß}} | ||
− | [[Category: | + | [[Category:Mobile Communications: Exercises|^2.3 The GWSSUS Channel Model^]] |
Latest revision as of 12:41, 17 February 2022
For the delay power-spectral density, we assume an exponential behavior. With ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$ we have
- $${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$
The constant $\tau_0$ can be determined from the tangent in the point $\tau = 0$ according to the upper graph. Note that ${\it \Phi}_{\rm V}(\tau)$ has unit $[1/\rm s]$ . Furthermore,
- The probability density function $\rm (PDF)$ $f_{\rm V}(\tau)$ has the same form as ${\it \Phi}_{\rm V}(\tau)$, but is normalized to area $1$ .
- The average excess delay or mean excess delay $m_{\rm V}$ is equal to the linear expectation $E\big [\tau \big]$ and can be determined from the PDF $f_{\rm V}(\tau)$ .
- The multipath spread or delay spread $\sigma_{\rm V}$ gives the standard deviation of the random variable $\tau$ . In the theory part we also use the term $T_{\rm V}$ for this.
- The displayed frequency correlation function $\varphi_{\rm F}(\Delta f)$ can be calculated as the Fourier transform of the delay power-spectral density ${\it \Phi}_{\rm V}(\tau)$ :
- $$\varphi_{\rm F}(\Delta f) \hspace{0.2cm} {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$
- The coherence bandwidth $B_{\rm K}$ is the value of $\Delta f$ at which the frequency correlation function $\varphi_{\rm F}(\Delta f)$ has dropped to half in absolute value.
Notes:
- This exercise belongs to the chapter The GWSSUS Channel Model.
- This exercise requires knowledge of Expected values and moments from the book "Theory of Stochastic Signals".
- In addition, the following Fourier transform is given:
- $$x(t) = \left\{ \begin{array}{c} {\rm e}^{- \lambda \hspace{0.05cm}\cdot \hspace{0.05cm} t}\\ 0 \end{array} \right.\quad \begin{array}{*{1}c} \hspace{-0.35cm} {\rm for} \hspace{0.15cm} t \ge 0 \\ \hspace{-0.35cm} {\rm for} \hspace{0.15cm} t < 0 \\ \end{array} \hspace{0.4cm} {\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.4cm} X(f) = \frac{1}{\lambda + {\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$
Questionnaire
Solution
- $$\int_{0}^{+\infty} {\it \Phi}_{\rm V}(\tau) \hspace{0.15cm}{\rm d} \tau = {\it \Phi}_{\rm 0} \cdot \int_{0}^{+\infty} {\rm e}^{-\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau = {\it \Phi}_{\rm 0} \cdot \tau_0 \hspace{0.05cm}. $$
- Then the probability density function is
- $$f_{\rm V}(\tau) = \frac{{\it \Phi}_{\rm V}(\tau) }{{\it \Phi}_{\rm 0} \cdot \tau_0}= \frac{1}{\tau_0} \cdot {\rm e}^{-\tau / \tau_0} \hspace{0.05cm}.$$
- Solution 2 is correct.
(2) The $k$-th moment of an one-sided exponential random variable is $m_k = k! \cdot \tau_0^k$.
- With $k = 1$, this results in the linear mean value $m_1 = m_{\rm V}$:
- $$m_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}} \hspace{0.05cm}. $$
(3) According to the Steiner's Theorem, the variance of any random variable is $\sigma^2 = m_2 \, -m_1^2$.
- This yields $m_2 = 2 \cdot \tau_0^2$, and therefore
- $$\sigma_{\rm V}^2 = m_2 - m_1^2 = 2 \cdot \tau_0^2 - (\tau_0)^2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \sigma_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}} \hspace{0.05cm}. $$
(4) ${\it \Phi}_{\rm V}(\tau)$ is identical to $x(t)$ in the given Fourier transform pair if $t$ is replaced by $\tau$ and $\lambda$ by $1/\tau_0$.
- Thus, $\varphi_{\rm F}(\Delta f)$ is equal to $X(f)$ with the substitution $f → \Delta f$:
- $$\varphi_{\rm F}(\Delta f) = \frac{1}{1/\tau_0 + {\rm j} \cdot 2\pi \Delta f} = \frac{\tau_0}{1 + {\rm j} \cdot 2\pi \cdot \tau_0 \cdot \Delta f}\hspace{0.05cm}.$$
- The first expression is correct.
(5) The coherence bandwidth is implicit in the following equation:
- $$|\varphi_{\rm F}(\Delta f = B_{\rm K})| \stackrel {!}{=} \frac{1}{2} \cdot |\varphi_{\rm F}(\Delta f = 0)| = \frac{\tau_0}{2}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}|\varphi_{\rm F}(\Delta f = B_{\rm K})|^2 = \frac{\tau_0^2}{1 + (2\pi \cdot \tau_0 \cdot B_{\rm K})^2} \stackrel {!}{=} \frac{\tau_0^2}{4}$$
- $$\Rightarrow \hspace{0.3cm}(2\pi \cdot \tau_0 \cdot B_{\rm K})^2 = 3 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} B_{\rm K}= \frac{\sqrt{3}}{2\pi \cdot \tau_0} \approx \frac{0.276}{ \tau_0}\hspace{0.05cm}. $$
- With $\tau_0 = 1 \ \ \rm µ s$, the coherence bandwidth is $B_{\rm K} \ \underline {= 276 \ \ \rm kHz}$.