Aufgaben:Exercise 3.4: GMSK Modulation - Revision history

Guenter at 15:26, 11 January 2023

2023-01-11T15:26:22Z

Guenter at 15:13, 11 January 2023

2023-01-11T15:13:38Z

Guenter at 15:05, 11 January 2023

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Guenter at 14:08, 11 January 2023

2023-01-11T14:08:59Z

Guenter at 16:39, 3 January 2023

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Guenter at 14:08, 2 January 2023

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Guenter at 16:36, 31 December 2022

2022-12-31T16:36:08Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.4: GMSK–Modulation to Aufgaben:Exercise 3.4: GMSK Modulation

2022-12-31T16:27:25Z

Guenter moved page Aufgabe 3.4: GMSK–Modulation to Exercise 3.4: GMSK Modulation

Noah at 20:42, 30 December 2022

2022-12-30T20:42:12Z

Noah at 20:41, 30 December 2022

2022-12-30T20:41:51Z

@@ Line 92: / Line 92: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; If all amplitude coefficients&nbsp; $a_{\nu}=+1$,&nbsp; then&nbsp; $q_{\rm R}(t) = 1$&nbsp; is a constant.
-*The Gaussian low-pass therefore has no effect and $q_{\rm G}(t) \equiv 1$.
+*The Gaussian low-pass therefore has no effect and&nbsp; $q_{\rm G}(t) \equiv 1$.
 *The maximum frequency is therefore
 :$${\rm Max}\hspace{0.05cm}[f_{\rm A}(t)] = f_{\rm T} + \Delta f_{\rm A} \hspace{0.15cm} \underline{= 900.068\,{\rm MHz}} \hspace{0.05cm}.$$
-*The minimum of the instantaneous frequency results when all amplitude coefficients are negative:
+*The minimum of the instantaneous frequency results when all coefficients are negative:
 :$${\rm Min}\hspace{0.05cm}[f_{\rm A}(t)] = f_{\rm T} - \Delta f_{\rm A} \hspace{0.15cm} \underline { = 899.932\,{\rm MHz}} \hspace{0.05cm}$$
 *In this case,&nbsp; $q_{\rm R}(t) = q_{\rm G}(t) = -1$.
@@ Line 104: / Line 104: @@
-'''(2)'''&nbsp; The frequency at which the logarithmized power transfer function is $3 \ \rm dB$ smaller than $f = 0$ is called the $3\hspace{0.05cm}\rm dB$ cutoff frequency.
+'''(2)'''&nbsp; The frequency at which the logarithmized power transfer function is&nbsp; $3 \ \rm dB$&nbsp; smaller than at&nbsp; $f = 0$&nbsp; is&nbsp; called the&nbsp; "$3\hspace{0.05cm}\rm dB$"&nbsp; cutoff frequency.
 *This can also be expressed as follows:
 :$$\frac {|H(f = f_{\rm 3\hspace{0.05cm}dB})|}{|H(f = 0)|}= \frac{1}{\sqrt{2}} \hspace{0.05cm}.$$
-*In particular, for the Gaussian lowpass because of $H(f = 0) = 1$:
+*In particular,&nbsp; for the Gaussian low-pass because of&nbsp; $H(f = 0) = 1$:
 :$$H(f = f_{\rm 3dB})= {\rm e}^{-\pi\cdot ({f_{\rm 3dB}}/{2 f_{\rm G}})^2} = \frac{1}{\sqrt{2}}\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}(\frac{f_{\rm 3dB}}{2 f_{\rm G}})^2 = \frac{{\rm ln}\hspace{0.1cm}\sqrt{2}}{\pi} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm G} = \sqrt{\frac{\pi}{4 \cdot {\rm ln}\hspace{0.1cm}\sqrt{2}}}\cdot f_{\rm 3dB}\hspace{0.05cm}.$$
-*The numerical evaluation leads to $f_{\rm G} \approx 1.5 \cdot f_{\rm 3\hspace{0.05cm}dB}$.
+*The numerical evaluation leads to&nbsp; $f_{\rm G} \approx 1.5 \cdot f_{\rm 3\hspace{0.05cm}dB}$.
 *From $f_{\rm 3\hspace{0.05cm}dB} \cdot T = 0.3$ thus follows $f_{\rm G} \cdot T \hspace{0.15cm}\underline{\approx 0.45}$.
-'''(3)'''&nbsp; The frequency pulse is obtained by convolution of square wave function $g_{\rm R}(t)$ and impulse response $h_{\rm G}(t)$:
+'''(3)'''&nbsp; The frequency pulse is obtained by convolution of the rectangular function&nbsp; $g_{\rm R}(t)$&nbsp; and the impulse response&nbsp; $h_{\rm G}(t)$:
 :$$g(t) = g_{\rm R} (t) \star h_{\rm G}(t) = 2 f_{\rm G} \cdot \int \limits^{t + T/2} _{t - T/2} {\rm e}^{-\pi\cdot (2 f_{\rm G}\cdot \tau)^2}\,{\rm d}\tau \hspace{0.05cm}.$$
-*With the substitution $u^{2 } = 8π \cdot f_{\rm G}^{2} \cdot \tau^{2}$ and the function $\phi (x)$ can also be written for this:
+*With the substitution &nbsp; $u^{2 } = 8π \cdot f_{\rm G}^{2} \cdot \tau^{2}$ &nbsp; and the function &nbsp; $\phi (x)$&nbsp; can also be written:
 :$$g(t) \ = \ \frac {1}{\sqrt{2 \pi}} \cdot \int \limits^{2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t + T/2)} _{2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t - T/2)} {\rm e}^{-u^2/2}\,{\rm d}u = \phi(2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t + T/2))- \phi(2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t - T/2)) \hspace{0.05cm}.$$
-*For the time $t = 0$, taking into account $\phi (-x) = 1 - \phi (x)$ and $f_{\rm G} \cdot T = 0.45$:
+*For time&nbsp; $t = 0$,&nbsp; taking into account &nbsp; $\phi (-x) = 1 - \phi (x)$ &nbsp; and &nbsp; $f_{\rm G} \cdot T = 0.45$:
 :$$g(t = 0) \ = \ \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)- \phi(-\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)= 2 \cdot \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)-1 \approx 2 \cdot \phi(1. 12)-1 \hspace{0.15cm} \underline {= 0.737} \hspace{0.05cm}.$$
@@ Line 129: / Line 130: @@
-'''(4)'''&nbsp; With $a_{3} = +1$, $q_{\rm G}(t = 3 T) = 1$ would result. Thus, due to linearity:
+'''(4)'''&nbsp; With&nbsp; $a_{3} = +1$ &nbsp; &rArr; &nbsp; $q_{\rm G}(t = 3 T) = 1$&nbsp; would result. Thus,&nbsp; due to linearity:
 :$$q_{\rm G}(t = 3 T ) = 1 - 2 \cdot g(t = 0)= 1 - 2 \cdot 0.737 \hspace{0.15cm} \underline {= -0.474} \hspace{0.05cm}.$$
-'''(5)'''&nbsp; Using the result of subtask '''(3)''' and $f_{\rm G} \cdot T = 0.45$ we obtain:
+'''(5)'''&nbsp; Using the result of subtask&nbsp; '''(3)'''&nbsp; and&nbsp; $f_{\rm G} \cdot T = 0.45$&nbsp; we obtain:
 :$$g(t = T) \ = \ \phi(3 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot T)- \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T) \approx \ \phi(3.36)-\phi(1.12) = 0.999 - 0.868 \hspace{0.15cm} \underline { = 0.131} \hspace{0.05cm}.$$
-'''(6)'''&nbsp; In the alternating sequence, for symmetry reasons, the amounts $|q_{\rm G}(\nu \cdot T)|$ are all the same for all multiples of the bit duration $T$.
+'''(6)'''&nbsp; In the alternating sequence,&nbsp; for symmetry reasons, the magnitudes &nbsp; $|q_{\rm G}(\nu \cdot T)|$ &nbsp; are all the same for all multiples of the bit duration&nbsp; $T$.
-*All intermediate values at $t \neq \nu - T$ are smaller.
+*All intermediate values at&nbsp; $t \neq \nu \cdot T$&nbsp; are smaller.
 *Considering $g(t ≥ 2T) \approx 0$, every single pulse value $g(0)$ is reduced by the preceding pulse with $g(t = T)$, additionally by the following one with $g(t = -T)$.
-*Thus, pulse interference results and one obtains:
+*Considering&nbsp; $g(t ≥ 2T) \approx 0$,&nbsp; every single pulse value&nbsp; $g(0)$&nbsp; is reduced by the preceding pulse with&nbsp; $g(t = T)$,&nbsp; additionally by the following one with&nbsp; $g(t = -T)$.
 :$${\rm Max} \hspace{0.08cm}q_{\rm G}(t) = g(0) - 2 \cdot g(T) = 0.737 - 2 \cdot 0.131 \hspace{0.15cm} \underline {= 0.475 }\hspace{0.05cm}.$$

@@ Line 64: / Line 64: @@
 |type="{}"}
 ${\rm Max} \ \big [f_{\rm A}(t) \big ] \ = \ $ { 900.068 3% }
 {Which system theoretic cutoff frequency of the Gaussian low-pass results from the requirement&nbsp; $f_{\rm 3\hspace{0.05cm} dB} \cdot T = 0.3$?
@@ Line 90: / Line 91: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; If all amplitude coefficients $a_{\nu}$ are equal to $+1$, $q_{\rm R}(t) = 1$ is a constant.
+'''(1)'''&nbsp; If all amplitude coefficients&nbsp; $a_{\nu}=+1$,&nbsp; then&nbsp; $q_{\rm R}(t) = 1$&nbsp; is a constant.
-*The Gaussian low-pass therefore has no effect and $q_{\rm G}(t) = 1$ is obtained.
+*The Gaussian low-pass therefore has no effect and $q_{\rm G}(t) \equiv 1$.
 *The maximum frequency is therefore
 :$${\rm Max}\hspace{0.05cm}[f_{\rm A}(t)] = f_{\rm T} + \Delta f_{\rm A} \hspace{0.15cm} \underline{= 900.068\,{\rm MHz}} \hspace{0.05cm}.$$
@@ Line 97: / Line 99: @@
 *The minimum of the instantaneous frequency results when all amplitude coefficients are negative:
 :$${\rm Min}\hspace{0.05cm}[f_{\rm A}(t)] = f_{\rm T} - \Delta f_{\rm A} \hspace{0.15cm} \underline { = 899.932\,{\rm MHz}} \hspace{0.05cm}$$
-*In this case, $q_{\rm R}(t) = q_{\rm G}(t) = -1$.
+*In this case,&nbsp; $q_{\rm R}(t) = q_{\rm G}(t) = -1$.

@@ Line 73: / Line 73: @@
 $g(t = 0) \ = \ $ { 0.737 3% }
-{What is the &nbsp; $q_{\rm G}(t = 3T)$&nbsp; value  if all coefficients except&nbsp; $a_{3} = -1$&nbsp; continue&nbsp; $a_{\nu \neq 3} = +1$&nbsp; What is the amplitude here&nbsp; $f_{\rm A}(t = 3T)$?
+{What is the &nbsp; $q_{\rm G}(t = 3T)$&nbsp; value  if all coefficients except&nbsp; $a_{3} = -1$&nbsp; continue&nbsp; $a_{\nu \neq 3} = +1$&nbsp; What is here the instantaneous frequency&nbsp; $f_{\rm A}(t = 3T)$?
 |type="{}"}
 $q_{\rm G}(t = 3T) \ = \ $ { -0.48822--0.45978 }
@@ Line 81: / Line 81: @@
 $ g(t = ±T) \ = \ $ { 0.131 3% }
-{What is the maximum amplitude of&nbsp; $q_{\rm G}(t)$&nbsp; with alternating coefficients? Consider that&nbsp; $g(t ≥ 2 T) \approx 0$&nbsp; is.
+{What is the maximum&nbsp; $q_{\rm G}(t)$&nbsp; value with alternating coefficients? Consider that&nbsp; $g(t ≥ 2 T) \approx 0$.
 |type="{}"}
 ${\rm Max} \ |q_{\rm G}(t)| \ = \ $ { 0.475 3% }

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Bei_A_3_4.png|right|frame|Various signals of the GMSK modulation '''Korrektur''']]
+[[File:EN_Bei_A_3_4.png|right|frame|Various signals of the GMSK modulation]]
-The modulation method used for GSM is known as Gaussian Minimum Shift Keying (GMSK). This is a type of&nbsp; '' Frequency Shift Keying''&nbsp; (FSK) with continuous phase adjustment&nbsp; $(\rm CP-FSK)$, in which
+The modulation method used for GSM is known as&nbsp; "Gaussian Minimum Shift Keying"&nbsp; $\rm (GMSK)$.&nbsp; This is a type of&nbsp; "Frequency Shift Keying"&nbsp; $\rm (FSK)$&nbsp; with continuous phase adjustment&nbsp; $(\text{CP-FSK)}$,&nbsp; in which
-*the modulation index is smallest possible to still satisfy the orthogonality condition&nbsp; <br> &nbsp; &nbsp; $h = 0.5$ &nbsp; &rArr; &nbsp;  ''Minimum Shift Keying'',
+*the modulation index is smallest possible to still satisfy the orthogonality condition&nbsp; <br> &nbsp; &nbsp; $h = 0.5$ &nbsp; &rArr; &nbsp;  "Minimum Shift Keying",
 *a Gaussian low-pass filter with impulse response&nbsp; $h_{\rm G}(t)$&nbsp; is introduced before the FSK modulator, <br>to further save bandwidth.
-The graphic illustrates the facts:
+The graph illustrates the facts:
-*The digital message is represented by the amplitude coefficients&nbsp; $a_{\nu} ∈ ±1$&nbsp; to which a Dirac pulse is applied. Note that the plotted (red) sequence is assumed for the subtask '''(3)'''.
+*The digital message is represented by the amplitude coefficients&nbsp; $a_{\nu} ∈ ±1$&nbsp; to which a Dirac delta comb is applied.&nbsp; Note that the plotted&nbsp; (red)&nbsp; sequence is assumed for subtask&nbsp; '''(3)'''.
-*Let the rectangular pulse be dimensionless, symmetric and have the GSM bit duration $T_{\rm B} = T$:
+*Let the rectangular pulse be dimensionless,&nbsp; symmetric and have the GSM bit duration $T_{\rm B} = T$:
 :$$g_{\rm R}(t) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c} {\rm{{\rm{for}}}} \\ {\rm{{\rm{for}}}} \\ \end{array}\begin{array}{*{5}c} |\hspace{0.05cm} t \hspace{0.05cm}| < T/2 \hspace{0.05cm}, \\ |\hspace{0.05cm} t \hspace{0.05cm}| > T/2 \hspace{0.05cm}. \\ \end{array}$$
-:This gives the following for the rectangular wave signal:
+*This gives the following for the rectangular source signal:
 :$$q_{\rm R} (t) = q_{\rm \delta} (t) \star g_{\rm R}(t) = \sum_{\nu} a_{\nu}\cdot g_{\rm R}(t - \nu \cdot T)\hspace{0.05cm}.$$
-*The Gaussian low pass is given by its frequency response or impulse response:
+*The Gaussian low-pass is given by its frequency response and its impulse response:
 :$$H_{\rm G}(f) = {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}[f/(2 f_{\rm G})]^2} \hspace{0.2cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.2cm} h_{\rm G}(t) = 2 f_{\rm G} \cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(2 f_{\rm G}\cdot t)^2}\hspace{0.05cm},$$
-:where the system theoretic cutoff frequency&nbsp; $f_{\rm G}$&nbsp; is used. However, in the GSM specification, the&nbsp; $3 \hspace{0.05cm}\rm dB$ cutoff frequency is given as&nbsp; $f_{\rm 3\hspace{0.05cm} dB} = 0.3/T$&nbsp;. From this&nbsp; $f_{\rm G}$&nbsp; can be calculated directly.
+:where the system theoretic cutoff frequency&nbsp; $f_{\rm G}$&nbsp; is used.
 *The signal after the Gaussian low-pass is thus:
 :$$q_{\rm G} (t) = q_{\rm R} (t) \star h_{\rm G}(t) = \sum_{\nu} a_{\nu}\cdot g(t - \nu \cdot T)\hspace{0.05cm}.$$
-:Here&nbsp; $g(t)$&nbsp; is called the frequency pulse. For this holds:
+*$g(t)$&nbsp; is called the&nbsp; "frequency pulse".&nbsp; For this holds:
 :$$g(t) = q_{\rm R} (t) \star h_{\rm G}(t) \hspace{0.05cm}.$$
-*With the low-pass filtered signal&nbsp; $q_{\rm G}(t)$, the carrier frequency&nbsp; $f_{\rm T}$&nbsp; and the frequency deviation&nbsp; $\delta f_{\rm A}$&nbsp; can thus be written for the instantaneous frequency at the output of the FSK modulator:
+*For the instantaneous frequency at the  FSK modulator output can thus be written
-:$$f_{\rm A}(t) = f_{\rm T} + \Delta f_{\rm A} \cdot q_{\rm G} (t)\hspace{0.05cm}.$$
+:$$f_{\rm A}(t) = f_{\rm T} + \Delta f_{\rm A} \cdot q_{\rm G} (t)\hspace{0.05cm},$$
+:*with the low-pass filtered signal&nbsp; $q_{\rm G}(t)$,
-:For your calculations, use the example values&nbsp; $f_{\rm T} = 900 \ \rm MHz$&nbsp; and&nbsp; $\Delta f_{\rm A} = 68 \ \rm kHz$.
+:*the carrier frequency&nbsp; $f_{\rm T}$,&nbsp; and
+*Reference is made in particular to the section&nbsp; [[Examples_of_Communication_Systems/Radio_Interface#Gaussian_Minimum_Shift_Keying_.28GMSK.29|"Gaussian Minimum Shift Keying"]]
-*Use the Gaussian integral to solve this exercise (see adjacent table):
+*Use the Gaussian integral to solve this exercise&nbsp; $($see adjacent table$)$:
 :$$\phi(x) =\frac {1}{\sqrt{2 \pi}} \cdot \int^{x} _{-\infty} {\rm e}^{-u^2/2}\,{\rm d}u \hspace{0.05cm}.$$
@@ Line 57: / Line 61: @@
 <quiz display=simple>
-{In what range can the instantaneous frequency&nbsp; $f_{\rm A}(t)$&nbsp; vary? What conditions must be met for this to happen?
+{In what range can the instantaneous frequency&nbsp; $f_{\rm A}(t)$&nbsp; vary?&nbsp; What conditions must be met for this to happen?
 |type="{}"}
 ${\rm Max} \ \big [f_{\rm A}(t) \big ] \ = \ $ { 900.068 3% }
@@ Line 65: / Line 69: @@
 $f_{\rm G} \cdot T \ = \ $ { 0.45 3% }
-{Calculate the frequency pulse&nbsp; $g(t)$&nbsp; using the function&nbsp; $\Phi (x)$. What is the value of&nbsp; $g(t = 0)$?
+{Calculate the frequency pulse&nbsp; $g(t)$&nbsp; using the function&nbsp; $\Phi (x)$.&nbsp; What is the &nbsp; $g(t = 0)$&nbsp; value?
 |type="{}"}
 $g(t = 0) \ = \ $ { 0.737 3% }
-{What is the value of&nbsp; $q_{\rm G}(t = 3T)$ if all coefficients except&nbsp; $a_{3} = -1$&nbsp; continue&nbsp; $a_{\nu \neq 3} = +1$&nbsp; What is the amplitude here&nbsp; $f_{\rm A}(t = 3T)$?
+{What is the &nbsp; $q_{\rm G}(t = 3T)$&nbsp; value  if all coefficients except&nbsp; $a_{3} = -1$&nbsp; continue&nbsp; $a_{\nu \neq 3} = +1$&nbsp; What is the amplitude here&nbsp; $f_{\rm A}(t = 3T)$?
 |type="{}"}
 $q_{\rm G}(t = 3T) \ = \ $ { -0.48822--0.45978 }

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1229__Bei_A_3_4.png|right|frame|Various signals of the GMSK modulation '''Korrektur''']]
+[[File:EN_Bei_A_3_4.png|right|frame|Various signals of the GMSK modulation '''Korrektur''']]
 The modulation method used for GSM is known as Gaussian Minimum Shift Keying (GMSK). This is a type of&nbsp; '' Frequency Shift Keying''&nbsp; (FSK) with continuous phase adjustment&nbsp; $(\rm CP-FSK)$, in which
 *the modulation index is smallest possible to still satisfy the orthogonality condition&nbsp; <br> &nbsp; &nbsp; $h = 0.5$ &nbsp; &rArr; &nbsp;  ''Minimum Shift Keying'',

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