Difference between revisions of "Applets:Complementary Gaussian Error Functions"
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| − | {{ | + | {{LntAppletLinkEnDe|qfunction_en|qfunction}} |
| − | == | + | |
| + | |||
| + | ==Applet Description== | ||
| + | <br> | ||
| + | This applet allows the calculation and graphical representation of the (complementary) Gaussian error functions ${\rm Q}(x)$ and $1/2\cdot {\rm erfc}(x)$, which are of great importance for error probability calculation. | ||
| + | *Both the abscissa and the function value can be represented either linearly or logarithmically. | ||
| + | *For both functions an upper bound $\rm (UB)$ and a lower bound $\rm (LB)$ are given. | ||
| + | |||
| + | ==Theoretical Background== | ||
| + | <br> | ||
| + | In the study of digital transmission systems, it is often necessary to determine the probability that a (zero mean) Gaussian distributed random variable $x$ with variance $σ^2$ exceeds a given value $x_0$. For this probability holds: | ||
| + | :$${\rm Pr}(x > x_0)={\rm Q}(\frac{x_0}{\sigma}) = 1/2 \cdot {\rm erfc}(\frac{x_0}{\sqrt{2} \cdot \sigma}).$$ | ||
| + | |||
| + | ===The function ${\rm Q}(x )$=== | ||
| + | |||
| + | The function ${\rm Q}(x)$ is called the '''complementary Gaussian error integral'''. The following calculation rule applies: | ||
| + | :$${\rm Q}(x ) = \frac{1}{\sqrt{2\pi}}\int_{x}^{ +\infty}\hspace{-0.2cm}{\rm e}^{-u^{2}/\hspace{0.05cm} 2}\,{\rm d} u .$$ | ||
| + | *This integral cannot be solved analytically and must be taken from tables if one does not have this applet available. | ||
| + | *Specially for larger $x$ values (i.e., for small error probabilities), the bounds given below provide a useful estimate for ${\rm Q}(x)$, which can also be calculated without tables. | ||
| + | *An upper bound $\rm (UB)$ of this function is: | ||
| + | :$${\rm Q}_{\rm UB}(x )=\text{Upper Bound }\big [{\rm Q}(x ) \big ] = \frac{ 1}{\sqrt{2\pi}\cdot x}\cdot {\rm e}^{- x^{2}/\hspace{0.05cm}2} > {\rm Q}(x).$$ | ||
| + | *Correspondingly, for the lower bound $\rm (LB)$: | ||
| + | :$${\rm Q}_{\rm LB}(x )=\text{Lower Bound }\big [{\rm Q}(x ) \big ] =\frac{1-1/x^2}{\sqrt{2\pi}\cdot x}\cdot {\rm e}^{-x^ 2/\hspace{0.05cm}2} ={\rm Q}_{\rm UB}(x ) \cdot (1-1/x^2)< {\rm Q}(x).$$ | ||
| + | |||
| + | However, in many program libraries, the function ${\rm Q}(x )$ cannot be found. | ||
<br> | <br> | ||
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| + | ===The function $1/2 \cdot {\rm erfc}(x )$=== | ||
| + | |||
| + | On the other hand, in almost all program libraries, you can find the '''complementary Gaussian error function''': | ||
| + | :$${\rm erfc}(x) = \frac{2}{\sqrt{\pi}}\int_{x}^{ +\infty}\hspace{-0.2cm}{\rm e}^{-u^{2}}\,{\rm d} u ,$$ | ||
| + | which is related to ${\rm Q}(x)$ as follows: ${\rm Q}(x)=1/2\cdot {\rm erfc}(x/{\sqrt{2}}).$ | ||
| + | *Since in almost all applications this function is used with the factor $1/2$, in this applet exactly this function was realized: | ||
| + | :$$1/2 \cdot{\rm erfc}(x) = \frac{1}{\sqrt{\pi}}\int_{x}^{ +\infty}\hspace{-0.2cm}{\rm e}^{-u^{2}}\,{\rm d} u .$$ | ||
| − | + | *Once again, an upper and lower bound can be specified for this function: | |
| − | + | :$$\text{Upper Bound }\big [1/2 \cdot{\rm erfc}(x) \big ] = \frac{ 1}{\sqrt{\pi}\cdot 2x}\cdot {\rm e}^{- x^{2}} ,$$ | |
| − | + | :$$\text{Lower Bound }\big [1/2 \cdot{\rm erfc}(x) \big ] = \frac{ {1-1/(2x^2)}}{\sqrt{\pi}\cdot 2x}\cdot {\rm e}^{- x^{2}} .$$ | |
| + | ===When which function offers advantages?=== | ||
| − | + | {{GreyBox|TEXT= | |
| − | + | $\text{Example 1:}$ We consider binary baseband transmission. Here, the bit error probability $p_{\rm B} = {\rm Q}({s_0}/{\sigma_d})$, where the useful signal can take the values $\pm s_0$ and the noise root-mean-square value $\sigma_d$ . | |
| − | |||
| − | |||
| + | It is assumed that tables are available listing the argument of the two Gaussian error functions at distance $0.1$. With $s_0/\sigma_d = 4$ one obtains for the bit error probability according to the function ${\rm Q}(x )$: | ||
| + | :$$p_{\rm B} = {\rm Q} (4) \approx 0.317 \cdot 10^{-4}\hspace{0.05cm}.$$ | ||
| + | According to the second equation, we get: | ||
| + | :$$p_{\rm B} = {1}/{2} \cdot {\rm erfc} ( {4}/{\sqrt{2} })= {1}/{2} \cdot {\rm erfc} ( 2.828)\approx {1}/{2} \cdot {\rm erfc} ( 2.8)= 0.375 \cdot 10^{-4}\hspace{0.05cm}.$$ | ||
| + | *The first value is more correct. In the second method of calculation, one must round or – even better – interpolate, which is very difficult due to the strong nonlinearity of this function.<br> | ||
| + | *Accordingly, with the given numerical values, ${\rm Q}(x )$ is more suitable. However, outside of exercise examples $s_0/\sigma_d$ will usually have a "curvilinear" value. In this case, of course, ${\rm Q}(x)$ offers no advantage over $1/2 \cdot{\rm erfc}(x)$. }} | ||
| − | == | + | |
| + | {{GreyBox|TEXT= | ||
| + | $\text{Example 2:}$ | ||
| + | With the energy per bit $(E_{\rm B})$ and the noise power density $(N_0)$ the bit error probability of ''Binary Phase Shift Keying'' (BPSK) is: | ||
| + | :$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2 E_{\rm B} }/{N_0} }\right ) = {1}/{2} \cdot { \rm erfc} \left ( \sqrt{ {E_{\rm B} }/{N_0} }\right ) \hspace{0.05cm}.$$ | ||
| + | For the numerical values $E_{\rm B} = 16 \rm mWs$ and $N_0 = 1 \rm mW/Hz$ we obtain: | ||
| + | :$$p_{\rm B} = {\rm Q} \left (4 \cdot \sqrt{ 2} \right ) = {1}/{2} \cdot {\rm erfc} \left ( 4\right ) \hspace{0.05cm}.$$ | ||
| + | *The first way leads to the result $p_{\rm B} = {\rm Q} (5.657) \approx {\rm Q} (5.7) = 0.6 \cdot 10^{-8}\hspace{0.01cm}$, while $1/2 \cdot{\rm erfc}(x)$ yields the more correct value $p_{\rm B} \approx 0.771 \cdot 10^{-8}$ here. | ||
| + | *As in the first example, however, you can see: The functions ${\rm Q}(x)$ and $1/2 \cdot{\rm erfc}(x)$ are basically equally well suited. | ||
| + | *Advantages or disadvantages of one or the other function arise only for concrete numerical values.}} | ||
<br> | <br> | ||
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| − | === | + | ==Exercises== |
| − | + | ||
| − | + | * First select the number $(1, 2, \text{...})$ of the exercise. The number $0$ corresponds to a "Reset": Same setting as at program start. | |
| − | :$${\rm | + | *A task description is displayed. The parameter values are adjusted. Solution after pressing "Show solution". <br> |
| − | + | ||
| − | * | + | |
| − | + | {{BlueBox|TEXT= | |
| − | + | '''(1)''' Find the values of the function ${\rm Q}(x)$ for $x=1$, $x=2$, $x=4$ and $x=6$. Interpret the graphs for linear and logarithmic ordinates.}} | |
| − | + | ||
| − | + | *The applet returns the values ${\rm Q}(1)=1.5866 \cdot 10^{-1}$, ${\rm Q}(2)=2. 275 \cdot 10^{-2}$, ${\rm Q}(4)=3.1671 \cdot 10^{-5}$ and ${\rm Q}(6)=9.8659 \cdot 10^{-10}$. | |
| + | *With linear ordinate, the values for $x>3$ are indistinguishable from the zero line. More interesting is the plot with logarithmic ordinate. | ||
| + | |||
| + | |||
| + | {{BlueBox|TEXT= | ||
| + | '''(2)''' Evaluate the two bounds ${\rm UB}(x )=\text{Upper Bound }\big [{\rm Q}(x ) \big ]$ and ${\rm LB}(x )=\text{Lower Bound }\big [{\rm Q}(x ) \big ]$ for the ${\rm Q}$ function. }} | ||
| + | |||
| + | *For $x \ge 2$ the upper bound is only slightly above ${\rm Q}(x)$ and the lower bound is only slightly below ${\rm Q}(x)$. | ||
| + | *For example: ${\rm Q}(x=4)=3.1671 \cdot 10^{-5}$ ⇒ ${\rm LB}(x=4)=3.1366 \cdot 10^{-5}$, ${\rm UB}(x=4)=3.3458 \cdot 10^{-5}$. | ||
| + | *The upper bound has greater significance for assessing a communications system than "LB", since this corresponds to a "worst case" consideration. | ||
| + | |||
| + | |||
| + | {{BlueBox|TEXT= | ||
| + | '''(3)''' Try to use the app to determine ${\rm Q}(x=2 \cdot \sqrt{2} \approx 2.828)$ as accurately as possible despite the quantization of the input parameter. }} | ||
| + | *The program returns for $x=2.8$ the too large result $2.5551 \cdot 10^{-3}$ and for $x=2.85$ the result $2.186 \cdot 10^{-3}$. The exact value lies in between. | ||
| + | *But it also holds: ${\rm Q}(x=2 \cdot \sqrt{2})=0.5 \cdot {\rm erfc}(x=2)$. This gives the exact value ${\rm Q}(x=2 \cdot \sqrt{2})=2.3389 \cdot 10^{-3}$. | ||
| + | |||
| + | |||
| + | {{BlueBox|TEXT= | ||
| + | '''(4)''' Find the values of the function $0.5 \cdot {\rm erfc}(x)$ for $x=1$, $x=2$, $x=3$ and $x=4$. Interpret the exact results and the bounds.}} | ||
| + | |||
| + | *The applet returns: $0.5 \cdot {\rm erfc}(1)=7.865 \cdot 10^{-2}$, $0.5 \cdot {\rm erfc}(2)=2. 3389 \cdot 10^{-3}$, $0.5 \cdot {\rm erfc}(3)=1.1045 \cdot 10^{-5}$ and $0.5 \cdot {\rm erfc}(4)=7.7086 \cdot 10^{-9}$. | ||
| + | *All the above statements about ${\rm Q}(x)$ with respect to suitable representation type and upper and lower bounds also apply to the function $0.5 \cdot {\rm erfc}(x)$. | ||
| + | |||
| + | |||
| + | {{BlueBox|TEXT= | ||
| + | '''(5)''' The results of '''(4)''' are now to be converted for the case of a logarithmic abscissa. The conversion is done according to $\rho\big[{\rm dB}\big ] = 20 \cdot \lg(x)$. }} | ||
| + | |||
| + | * The linear abscissa value $x=1$ leads to the logarithmic abscissa value $\rho=0\ \rm dB$ ⇒ $0. 5 \cdot {\rm erfc}(\rho=0\ {\rm dB})={0.5 \cdot \rm erfc}(x=1)=7.865 \cdot 10^{-2}$. | ||
| + | *Similarly $0.5 \cdot {\rm erfc}(\rho=6.021\ {\rm dB}) =0.5 \cdot {\rm erfc}(x=2)=2. 3389 \cdot 10^{-3}$, $0.5 \cdot {\rm erfc}(\rho=9.542\ {\rm dB})=0.5 \cdot {\rm erfc}(3)=1.1045 \cdot 10^{-5}$, | ||
| + | *$0.5 \cdot {\rm erfc}(\rho=12.041\ {\rm dB})= 0.5 \cdot {\rm erfc}(4)=7.7086 \cdot 10^{-9}$. | ||
| + | *As per right diagram: $0.5 \cdot {\rm erfc}(\rho=6\ {\rm dB}) =2.3883 \cdot 10^{-3}$, $0.5 \cdot {\rm erfc}(\rho=9. 5\ {\rm dB}) =1.2109 \cdot 10^{-5}$, $0.5 \cdot {\rm erfc}(\rho=12\ {\rm dB}) =9.006 \cdot 10^{-9}$. | ||
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| + | |||
| + | {{BlueBox|TEXT= | ||
| + | '''(6)''' Find ${\rm Q}(\rho=0\ {\rm dB})$, ${\rm Q}(\rho=5\ {\rm dB})$ and ${\rm Q}(\rho=10\ {\rm dB})$, and establish the relationship between linear and logarithmic abscissa.}} | ||
| + | *The program returns for logarithmic abscissa ${\rm Q}(\rho=0\ {\rm dB})=1. 5866 \cdot 10^{-1}$, ${\rm Q}(\rho=5\ {\rm dB})=3.7679 \cdot 10^{-2}$, ${\rm Q}(\rho=10\ {\rm dB})=7.827 \cdot 10^{-4}$. | ||
| + | *The conversion is done according to the equation $x=10^{\hspace{0.05cm}0.05\hspace{0.05cm} \cdot\hspace{0.05cm} \rho[{\rm dB}]}$. For $\rho=0\ {\rm dB}$ we get $x=1$ ⇒ ${\rm Q}(\rho=0\ {\rm dB})={\rm Q}(x=1) =1.5866 \cdot 10^{-1}$. | ||
| + | *For $\rho=5\ {\rm dB}$ we get $x=1.1778$ ⇒ ${\rm Q}(\rho=5\ {\rm dB})={\rm Q}(x=1. 778) =3.7679 \cdot 10^{-2}$. From the left diagram: ${\rm Q}(x=1.8) =3.593 \cdot 10^{-2}$. | ||
| + | *For $\rho=10\ {\rm dB}$ we get $x=3.162$ ⇒ ${\rm Q}(\rho=10\ {\rm dB})={\rm Q}(x=3. 162) =7.827 \cdot 10^{-4}$. After "quantization": ${\rm Q}(x=3.15) =8.1635 \cdot 10^{-4}$. | ||
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| − | == | + | ==Applet Manual== |
<br> | <br> | ||
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| − | + | [[File:Qfunction bedienung.png|right|550px]] | |
| − | + | '''(A)''' Equations used in the example ${\rm Q}(x)$ | |
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| − | + | '''(B)''' Selection option for ${\rm Q}(x)$ or ${\rm 0.5 \cdot erfc}(x)$ | |
| − | |||
| − | '''( | ||
| − | '''( | + | '''(C)''' Bounds ${\rm LB}$ and ${\rm UB}$ are drawn |
| − | '''( | + | '''(D)''' Selection whether abscissa is linear $\rm (lin)$ or logarithmic $\rm (log)$ |
| − | '''( | + | '''(E)''' Select whether ordinate is linear $\rm (lin)$ or logarithmic $\rm (log)$ |
| − | '''( | + | '''(F)''' Numerical output using the example ${\rm Q}(x)$ with linear abscissa |
| − | '''( | + | '''(G)''' Slider input of abscissa value $x$ for linear abscissa |
| − | '''( | + | '''(H)''' Slider input of abscissa value $\rho \ \rm [dB]$ for logarithmic abscissa |
| − | '''( | + | '''(I)''' Graphical output of function ${\rm Q}(x)$ – here: linear abscissa |
| − | $\ | + | '''(J)''' Graph output of function ${\rm 0.5 \cdot erfc}(x)$ – here: linear abscissa |
| − | + | '''(K)''' Variation possibility for the graphical representations | |
| − | $\hspace{1.5cm}$ | + | $\hspace{1.5cm}$"$+$" (zoom in), |
| − | $\hspace{1.5cm}$ | + | $\hspace{1.5cm}$"$-$" (zoom out) |
| − | + | $\hspace{1.5cm}$ "$\rm o$" (Reset) | |
| − | + | $\hspace{1.5cm}$ "$\leftarrow$" (Move left), etc. | |
<br clear=all> | <br clear=all> | ||
| − | <br> | + | ==About the Authors== |
| − | * | + | <br> |
| − | * | + | This interactive calculation tool was designed and implemented at the [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering] at the [https://www.tum.de/en Technical University of Munich]. |
| + | *The first version was created in 2007 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Thomas_Gro.C3.9Fer_.28Diplomarbeit_LB_2006.2C_danach_freie_Mitarbeit_bis_2010.29|Thomas Großer]] as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]). | ||
| + | *In 2018 the program was redesigned by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Xiaohan_Liu_.28Bachelorarbeit_2018.29|Xiaohan Liu]] as part of her bachelor thesis (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5". | ||
| + | *Last revision and English version 2021 by [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]] in the context of a working student activity. | ||
| + | The conversion of this applet to HTML 5 was financially supported by [https://www.ei.tum.de/studium/studienzuschuesse/ "Studienzuschüsse"] (Faculty EI of the TU Munich). We thank. | ||
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| − | {{ | + | ==Once again: Open Applet in new Tab== |
| + | <br> | ||
| + | {{LntAppletLinkEnDe|qfunction_en|qfunction}} | ||
Latest revision as of 18:49, 16 April 2023
Open Applet in new Tab Deutsche Version Öffnen
Contents
Applet Description
This applet allows the calculation and graphical representation of the (complementary) Gaussian error functions ${\rm Q}(x)$ and $1/2\cdot {\rm erfc}(x)$, which are of great importance for error probability calculation.
- Both the abscissa and the function value can be represented either linearly or logarithmically.
- For both functions an upper bound $\rm (UB)$ and a lower bound $\rm (LB)$ are given.
Theoretical Background
In the study of digital transmission systems, it is often necessary to determine the probability that a (zero mean) Gaussian distributed random variable $x$ with variance $σ^2$ exceeds a given value $x_0$. For this probability holds:
- $${\rm Pr}(x > x_0)={\rm Q}(\frac{x_0}{\sigma}) = 1/2 \cdot {\rm erfc}(\frac{x_0}{\sqrt{2} \cdot \sigma}).$$
The function ${\rm Q}(x )$
The function ${\rm Q}(x)$ is called the complementary Gaussian error integral. The following calculation rule applies:
- $${\rm Q}(x ) = \frac{1}{\sqrt{2\pi}}\int_{x}^{ +\infty}\hspace{-0.2cm}{\rm e}^{-u^{2}/\hspace{0.05cm} 2}\,{\rm d} u .$$
- This integral cannot be solved analytically and must be taken from tables if one does not have this applet available.
- Specially for larger $x$ values (i.e., for small error probabilities), the bounds given below provide a useful estimate for ${\rm Q}(x)$, which can also be calculated without tables.
- An upper bound $\rm (UB)$ of this function is:
- $${\rm Q}_{\rm UB}(x )=\text{Upper Bound }\big [{\rm Q}(x ) \big ] = \frac{ 1}{\sqrt{2\pi}\cdot x}\cdot {\rm e}^{- x^{2}/\hspace{0.05cm}2} > {\rm Q}(x).$$
- Correspondingly, for the lower bound $\rm (LB)$:
- $${\rm Q}_{\rm LB}(x )=\text{Lower Bound }\big [{\rm Q}(x ) \big ] =\frac{1-1/x^2}{\sqrt{2\pi}\cdot x}\cdot {\rm e}^{-x^ 2/\hspace{0.05cm}2} ={\rm Q}_{\rm UB}(x ) \cdot (1-1/x^2)< {\rm Q}(x).$$
However, in many program libraries, the function ${\rm Q}(x )$ cannot be found.
The function $1/2 \cdot {\rm erfc}(x )$
On the other hand, in almost all program libraries, you can find the complementary Gaussian error function:
- $${\rm erfc}(x) = \frac{2}{\sqrt{\pi}}\int_{x}^{ +\infty}\hspace{-0.2cm}{\rm e}^{-u^{2}}\,{\rm d} u ,$$
which is related to ${\rm Q}(x)$ as follows: ${\rm Q}(x)=1/2\cdot {\rm erfc}(x/{\sqrt{2}}).$
- Since in almost all applications this function is used with the factor $1/2$, in this applet exactly this function was realized:
- $$1/2 \cdot{\rm erfc}(x) = \frac{1}{\sqrt{\pi}}\int_{x}^{ +\infty}\hspace{-0.2cm}{\rm e}^{-u^{2}}\,{\rm d} u .$$
- Once again, an upper and lower bound can be specified for this function:
- $$\text{Upper Bound }\big [1/2 \cdot{\rm erfc}(x) \big ] = \frac{ 1}{\sqrt{\pi}\cdot 2x}\cdot {\rm e}^{- x^{2}} ,$$
- $$\text{Lower Bound }\big [1/2 \cdot{\rm erfc}(x) \big ] = \frac{ {1-1/(2x^2)}}{\sqrt{\pi}\cdot 2x}\cdot {\rm e}^{- x^{2}} .$$
When which function offers advantages?
$\text{Example 1:}$ We consider binary baseband transmission. Here, the bit error probability $p_{\rm B} = {\rm Q}({s_0}/{\sigma_d})$, where the useful signal can take the values $\pm s_0$ and the noise root-mean-square value $\sigma_d$ .
It is assumed that tables are available listing the argument of the two Gaussian error functions at distance $0.1$. With $s_0/\sigma_d = 4$ one obtains for the bit error probability according to the function ${\rm Q}(x )$:
- $$p_{\rm B} = {\rm Q} (4) \approx 0.317 \cdot 10^{-4}\hspace{0.05cm}.$$
According to the second equation, we get:
- $$p_{\rm B} = {1}/{2} \cdot {\rm erfc} ( {4}/{\sqrt{2} })= {1}/{2} \cdot {\rm erfc} ( 2.828)\approx {1}/{2} \cdot {\rm erfc} ( 2.8)= 0.375 \cdot 10^{-4}\hspace{0.05cm}.$$
- The first value is more correct. In the second method of calculation, one must round or – even better – interpolate, which is very difficult due to the strong nonlinearity of this function.
- Accordingly, with the given numerical values, ${\rm Q}(x )$ is more suitable. However, outside of exercise examples $s_0/\sigma_d$ will usually have a "curvilinear" value. In this case, of course, ${\rm Q}(x)$ offers no advantage over $1/2 \cdot{\rm erfc}(x)$.
$\text{Example 2:}$ With the energy per bit $(E_{\rm B})$ and the noise power density $(N_0)$ the bit error probability of Binary Phase Shift Keying (BPSK) is:
- $$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2 E_{\rm B} }/{N_0} }\right ) = {1}/{2} \cdot { \rm erfc} \left ( \sqrt{ {E_{\rm B} }/{N_0} }\right ) \hspace{0.05cm}.$$
For the numerical values $E_{\rm B} = 16 \rm mWs$ and $N_0 = 1 \rm mW/Hz$ we obtain:
- $$p_{\rm B} = {\rm Q} \left (4 \cdot \sqrt{ 2} \right ) = {1}/{2} \cdot {\rm erfc} \left ( 4\right ) \hspace{0.05cm}.$$
- The first way leads to the result $p_{\rm B} = {\rm Q} (5.657) \approx {\rm Q} (5.7) = 0.6 \cdot 10^{-8}\hspace{0.01cm}$, while $1/2 \cdot{\rm erfc}(x)$ yields the more correct value $p_{\rm B} \approx 0.771 \cdot 10^{-8}$ here.
- As in the first example, however, you can see: The functions ${\rm Q}(x)$ and $1/2 \cdot{\rm erfc}(x)$ are basically equally well suited.
- Advantages or disadvantages of one or the other function arise only for concrete numerical values.
Exercises
- First select the number $(1, 2, \text{...})$ of the exercise. The number $0$ corresponds to a "Reset": Same setting as at program start.
- A task description is displayed. The parameter values are adjusted. Solution after pressing "Show solution".
(1) Find the values of the function ${\rm Q}(x)$ for $x=1$, $x=2$, $x=4$ and $x=6$. Interpret the graphs for linear and logarithmic ordinates.
- The applet returns the values ${\rm Q}(1)=1.5866 \cdot 10^{-1}$, ${\rm Q}(2)=2. 275 \cdot 10^{-2}$, ${\rm Q}(4)=3.1671 \cdot 10^{-5}$ and ${\rm Q}(6)=9.8659 \cdot 10^{-10}$.
- With linear ordinate, the values for $x>3$ are indistinguishable from the zero line. More interesting is the plot with logarithmic ordinate.
(2) Evaluate the two bounds ${\rm UB}(x )=\text{Upper Bound }\big [{\rm Q}(x ) \big ]$ and ${\rm LB}(x )=\text{Lower Bound }\big [{\rm Q}(x ) \big ]$ for the ${\rm Q}$ function.
- For $x \ge 2$ the upper bound is only slightly above ${\rm Q}(x)$ and the lower bound is only slightly below ${\rm Q}(x)$.
- For example: ${\rm Q}(x=4)=3.1671 \cdot 10^{-5}$ ⇒ ${\rm LB}(x=4)=3.1366 \cdot 10^{-5}$, ${\rm UB}(x=4)=3.3458 \cdot 10^{-5}$.
- The upper bound has greater significance for assessing a communications system than "LB", since this corresponds to a "worst case" consideration.
(3) Try to use the app to determine ${\rm Q}(x=2 \cdot \sqrt{2} \approx 2.828)$ as accurately as possible despite the quantization of the input parameter.
- The program returns for $x=2.8$ the too large result $2.5551 \cdot 10^{-3}$ and for $x=2.85$ the result $2.186 \cdot 10^{-3}$. The exact value lies in between.
- But it also holds: ${\rm Q}(x=2 \cdot \sqrt{2})=0.5 \cdot {\rm erfc}(x=2)$. This gives the exact value ${\rm Q}(x=2 \cdot \sqrt{2})=2.3389 \cdot 10^{-3}$.
(4) Find the values of the function $0.5 \cdot {\rm erfc}(x)$ for $x=1$, $x=2$, $x=3$ and $x=4$. Interpret the exact results and the bounds.
- The applet returns: $0.5 \cdot {\rm erfc}(1)=7.865 \cdot 10^{-2}$, $0.5 \cdot {\rm erfc}(2)=2. 3389 \cdot 10^{-3}$, $0.5 \cdot {\rm erfc}(3)=1.1045 \cdot 10^{-5}$ and $0.5 \cdot {\rm erfc}(4)=7.7086 \cdot 10^{-9}$.
- All the above statements about ${\rm Q}(x)$ with respect to suitable representation type and upper and lower bounds also apply to the function $0.5 \cdot {\rm erfc}(x)$.
(5) The results of (4) are now to be converted for the case of a logarithmic abscissa. The conversion is done according to $\rho\big[{\rm dB}\big ] = 20 \cdot \lg(x)$.
- The linear abscissa value $x=1$ leads to the logarithmic abscissa value $\rho=0\ \rm dB$ ⇒ $0. 5 \cdot {\rm erfc}(\rho=0\ {\rm dB})={0.5 \cdot \rm erfc}(x=1)=7.865 \cdot 10^{-2}$.
- Similarly $0.5 \cdot {\rm erfc}(\rho=6.021\ {\rm dB}) =0.5 \cdot {\rm erfc}(x=2)=2. 3389 \cdot 10^{-3}$, $0.5 \cdot {\rm erfc}(\rho=9.542\ {\rm dB})=0.5 \cdot {\rm erfc}(3)=1.1045 \cdot 10^{-5}$,
- $0.5 \cdot {\rm erfc}(\rho=12.041\ {\rm dB})= 0.5 \cdot {\rm erfc}(4)=7.7086 \cdot 10^{-9}$.
- As per right diagram: $0.5 \cdot {\rm erfc}(\rho=6\ {\rm dB}) =2.3883 \cdot 10^{-3}$, $0.5 \cdot {\rm erfc}(\rho=9. 5\ {\rm dB}) =1.2109 \cdot 10^{-5}$, $0.5 \cdot {\rm erfc}(\rho=12\ {\rm dB}) =9.006 \cdot 10^{-9}$.
(6) Find ${\rm Q}(\rho=0\ {\rm dB})$, ${\rm Q}(\rho=5\ {\rm dB})$ and ${\rm Q}(\rho=10\ {\rm dB})$, and establish the relationship between linear and logarithmic abscissa.
- The program returns for logarithmic abscissa ${\rm Q}(\rho=0\ {\rm dB})=1. 5866 \cdot 10^{-1}$, ${\rm Q}(\rho=5\ {\rm dB})=3.7679 \cdot 10^{-2}$, ${\rm Q}(\rho=10\ {\rm dB})=7.827 \cdot 10^{-4}$.
- The conversion is done according to the equation $x=10^{\hspace{0.05cm}0.05\hspace{0.05cm} \cdot\hspace{0.05cm} \rho[{\rm dB}]}$. For $\rho=0\ {\rm dB}$ we get $x=1$ ⇒ ${\rm Q}(\rho=0\ {\rm dB})={\rm Q}(x=1) =1.5866 \cdot 10^{-1}$.
- For $\rho=5\ {\rm dB}$ we get $x=1.1778$ ⇒ ${\rm Q}(\rho=5\ {\rm dB})={\rm Q}(x=1. 778) =3.7679 \cdot 10^{-2}$. From the left diagram: ${\rm Q}(x=1.8) =3.593 \cdot 10^{-2}$.
- For $\rho=10\ {\rm dB}$ we get $x=3.162$ ⇒ ${\rm Q}(\rho=10\ {\rm dB})={\rm Q}(x=3. 162) =7.827 \cdot 10^{-4}$. After "quantization": ${\rm Q}(x=3.15) =8.1635 \cdot 10^{-4}$.
Applet Manual
(A) Equations used in the example ${\rm Q}(x)$
(B) Selection option for ${\rm Q}(x)$ or ${\rm 0.5 \cdot erfc}(x)$
(C) Bounds ${\rm LB}$ and ${\rm UB}$ are drawn
(D) Selection whether abscissa is linear $\rm (lin)$ or logarithmic $\rm (log)$
(E) Select whether ordinate is linear $\rm (lin)$ or logarithmic $\rm (log)$
(F) Numerical output using the example ${\rm Q}(x)$ with linear abscissa
(G) Slider input of abscissa value $x$ for linear abscissa
(H) Slider input of abscissa value $\rho \ \rm [dB]$ for logarithmic abscissa
(I) Graphical output of function ${\rm Q}(x)$ – here: linear abscissa
(J) Graph output of function ${\rm 0.5 \cdot erfc}(x)$ – here: linear abscissa
(K) Variation possibility for the graphical representations
$\hspace{1.5cm}$"$+$" (zoom in),
$\hspace{1.5cm}$"$-$" (zoom out)
$\hspace{1.5cm}$ "$\rm o$" (Reset)
$\hspace{1.5cm}$ "$\leftarrow$" (Move left), etc.
About the Authors
This interactive calculation tool was designed and implemented at the Institute for Communications Engineering at the Technical University of Munich.
- The first version was created in 2007 by Thomas Großer as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: Günter Söder).
- In 2018 the program was redesigned by Xiaohan Liu as part of her bachelor thesis (Supervisor: Tasnád Kernetzky ) via "HTML5".
- Last revision and English version 2021 by Carolin Mirschina in the context of a working student activity.
The conversion of this applet to HTML 5 was financially supported by "Studienzuschüsse" (Faculty EI of the TU Munich). We thank.
