Aufgaben:Exercise 4.2Z: About the Sampling Theorem - Revision history

Guenter at 10:28, 8 April 2022

2022-04-08T10:28:12Z

Guenter at 16:34, 7 April 2022

2022-04-07T16:34:51Z

Noah at 21:05, 4 April 2022

2022-04-04T21:05:36Z

Noah at 16:20, 19 March 2022

2022-03-19T16:20:29Z

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Guenter at 13:50, 16 March 2022

2022-03-16T13:50:42Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.2Z: Zum Abtasttheorem to Aufgaben:Exercise 4.2Z: About the Sampling Theorem

2022-03-16T13:42:32Z

Guenter moved page Aufgabe 4.2Z: Zum Abtasttheorem to Exercise 4.2Z: About the Sampling Theorem

Javier: Text replacement - "Category:Aufgaben zu Modulationsverfahren" to "Category:Modulation Methods: Exercises"

2021-03-23T13:01:39Z

Text replacement - "Category:Aufgaben zu Modulationsverfahren" to "Category:Modulation Methods: Exercises"

Javier: Text replacement - "Signal_Representation/Zeitdiskrete_Signaldarstellung" to "Signal_Representation/Time_Discrete_Signal_Representation"

2020-09-01T13:26:30Z

Text replacement - "Signal_Representation/Zeitdiskrete_Signaldarstellung" to "Signal_Representation/Time_Discrete_Signal_Representation"

Javier: Text replacement - "[[Modulationsverfahren" to "[[Modulation_Methods"

2020-07-09T12:59:40Z

Text replacement - "[[Modulationsverfahren" to "[[Modulation_Methods"

Javier: Text replacement - "[[Signaldarstellung/" to "[[Signal_Representation/"

2020-07-09T09:39:45Z

Text replacement - "[[Signaldarstellung/" to "[[Signal_Representation/"

@@ Line 71: / Line 71: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>All statements</u> are true:
+'''(1)'''&nbsp; <u>All statements</u>&nbsp; are true:
-[[File:P_ID1611__Mod_Z_4_2a.png|P_ID1611__Mod_Z_4_2a.png|right|frame|Spectral function of sampled signal]]
+[[File:P_ID1611__Mod_Z_4_2a.png|P_ID1611__Mod_Z_4_2a.png|right|frame|Spectral function of the sampled signal]]
-*The sampling theorem is satisfied by&nbsp; $f_{\rm A} = 11 \ \rm kHz > 2 \cdot 5 \ \rm kHz$&nbsp; so that complete signal reconstruction is always possible.
+*The sampling theorem is satisfied by&nbsp; $f_{\rm A} = 11 \ \rm kHz > 2 \cdot 5 \ \rm kHz$&nbsp; so that a complete signal reconstruction is always possible.
-*The spectrum&nbsp; $Q_{\rm A}(f)$&nbsp; results from&nbsp; $Q(f)$&nbsp; by periodic continuation at the respective frequency spacing&nbsp; $f_{\rm A}$, which is generally illustrated in the graph for the spectral function&nbsp; $Q_3(f)$&nbsp;.
+*The spectrum&nbsp; $Q_{\rm A}(f)$&nbsp; results from&nbsp; $Q(f)$&nbsp; by periodic continuation at the respective frequency spacing&nbsp; $f_{\rm A}$,&nbsp; which is generally illustrated in the graph.
-*By a rectangle&ndash;low-pass with&nbsp; $f_{\rm G} = f_{\rm A}/2 = 5.5 \ \rm kHz$&nbsp; the original spectrum&nbsp; $Q(f)$ is obtained.
+*By a rectangular low-pass with&nbsp; $f_{\rm G} = f_{\rm A}/2 = 5.5 \ \rm kHz$&nbsp; the original spectrum&nbsp; $Q(f)$ is obtained.
 The shift by
-* $f_{\rm A} = 11 \ \rm kHz$&nbsp; yields the lines at&nbsp; $+6 \ \rm kHz$ and $+16 \ \rm kHz$,
+* $f_{\rm A} = 11 \ \rm kHz$&nbsp; yields the lines at&nbsp; $+6 \ \rm kHz$&nbsp; and&nbsp; $+16 \ \rm kHz$,
-* $-f_{\rm A} = -11 \ \rm kHz$&nbsp; yields the lines at&nbsp; $-6 \ \rm kHz$ and $-16 \ \rm kHz$,
+* $-f_{\rm A} = -11 \ \rm kHz$&nbsp; yields the lines at&nbsp; $-6 \ \rm kHz$&nbsp; and&nbsp; $-16 \ \rm kHz$,
-* $2 - f_{\rm A} = 22 \ \rm kHz$&nbsp; yields the lines at&nbsp; $+17 \ \rm kHz$ and $+27 \ \rm kHz$,
+* $2 - f_{\rm A} = 22 \ \rm kHz$&nbsp; yields the lines at&nbsp; $+17 \ \rm kHz$&nbsp; and&nbsp; $+27 \ \rm kHz$,
 * $-2 - f_{\rm A}= -22 \ \rm kHz$&nbsp; yields the lines at&nbsp; $-17 \ \rm kHz$, $-27 \ \rm kHz$.
@@ Line 91: / Line 91: @@
-'''(3)'''&nbsp; The correct solution is <u>suggestion 2</u>:
+'''(3)'''&nbsp; The correct solution is&nbsp; <u>suggestion 2</u>:
-*For the cosinusoidal signal, according to the next graph with&nbsp; $f_{\rm A} = 10 \rm kHz$&nbsp; the spectrum&nbsp; $Q_{\rm A}(f)$:&nbsp; &nbsp; All spectral lines are real.
+[[File:P_ID1612__Mod_Z_4_2c.png|P_ID1612__Mod_Z_4_2c.png|right|frame|Spectral function of the sampled cosine signal]]
 *The periodization of&nbsp; $Q(f)$&nbsp; with&nbsp; $f_{\rm A} = 10 \rm kHz$&nbsp; leads to a Dirac comb with spectral lines at&nbsp; $±f_{\rm N}$,&nbsp; $±f_{\rm N}± f_{\rm A}$,&nbsp; $±f_{\rm N}± 2f_{\rm A}$, . ..
-*Through the superpositions, all Dirac functions have weight&nbsp; $A$, while the two spectral lines of&nbsp; $Q(f)$&nbsp; are weighted only by&nbsp; $A/2$&nbsp; each.
+*For the cosinusoidal signal,&nbsp; according to this graph with&nbsp; $f_{\rm A} = 10 \rm \ kHz$:&nbsp;  All spectral lines of&nbsp; $Q_{\rm A}(f)$:&nbsp; are real.
-*Because&nbsp; $H(f = f_{\rm N}) = H(f = f_{\rm G}) = 0.5$&nbsp; the spectrum&nbsp; $V_1(f)$&nbsp; after the low-pass is identical to&nbsp; $Q_1(f)$&nbsp; and accordingly&nbsp; $v_1(t) = q_1(t)$.
+*The periodization of&nbsp; $Q(f)$&nbsp; with&nbsp; $f_{\rm A} = 10 \rm \ kHz$&nbsp; leads to a Dirac comb with spectral lines at&nbsp; $±f_{\rm N}$,&nbsp; $±f_{\rm N}± f_{\rm A}$,&nbsp; $±f_{\rm N}± 2f_{\rm A}$, . ..
 *In the time domain, the signal reconstruction can be thought of as follows: &nbsp; The samples of&nbsp; $q_1(t)$&nbsp; lie exactly at the signal maxima and minima. &nbsp;
 *The lowpass filter forms the cosine signal with correct amplitude, frequency and phase.
+<br clear=all>
+[[File:P_ID1613__Mod_Z_4_2d.png|P_ID1613__Mod_Z_4_2d.png|right|frame|Sampled sine signal]]
-[[File:P_ID1612__Mod_Z_4_2c.png|P_ID1612__Mod_Z_4_2c.png|center|frame|Spectral function of the sampled cosine signal]]
+'''(4)'''&nbsp; Correct is&nbsp; <u>suggested solution 2</u>:
+*All sampled values of&nbsp; $q_2(t)$&nbsp; now lie exactly at the zero crossings of the sinusoidal signal,&nbsp; which means that here&nbsp; $q_{\rm A}(t) \equiv 0$&nbsp; holds.&nbsp; However,&nbsp; this naturally also gives&nbsp; $v_2(t) \equiv 0$.
+*In the spectral domain,&nbsp; the result can be derived using the graph for subtask&nbsp; '''(1)'''.&nbsp; <br>⇒ &nbsp;$Q(f)$&nbsp; is purely imaginary and the imaginary parts at&nbsp; $±f_{\rm N}$&nbsp; have different signs. &nbsp;
-[[File:P_ID1613__Mod_Z_4_2d.png|P_ID1613__Mod_Z_4_2d.png|right|frame|Abgetastetes Sinussignal]]
+*Thus,&nbsp; one positive and one negative part cancel each other in periodization &nbsp; <br>⇒ &nbsp; $Q_{\rm A}(f) \equiv 0$ &nbsp; ⇒ &nbsp; $V_2(f) \equiv 0$.
 <br clear=all>
 [[File:P_ID1614__Mod_Z_4_2e.png|P_ID1614__Mod_Z_4_2e.png|right|frame|Sampled harmonic oscillation with phase&nbsp; $φ_3 = π/4$]]
-'''(5)'''&nbsp; <u>None of the given solutions</u> is correct:
+'''(5)'''&nbsp; <u>None of the given solutions</u>&nbsp; is correct:
-*If in the graph for the subtask&nbsp; '''(1)'''&nbsp; the sampling frequency&nbsp; $f_{\rm A} = 11 \ \rm kHz$&nbsp; is replaced by&nbsp; $f_{\rm A} = 10 \ \rm kHz$, the real parts add up, but the imaginary parts cancel out.
+*If in the graph for the subtask&nbsp; '''(1)'''&nbsp; the sampling frequency&nbsp; $f_{\rm A} = 11 \ \rm kHz$&nbsp; is replaced by&nbsp; $f_{\rm A} = 10 \ \rm kHz$,&nbsp; the real parts add up,&nbsp; but the imaginary parts cancel out.
-*This means that now&nbsp; $Q_{\rm A}(f)$&nbsp; and&nbsp; $V_3(f)$&nbsp; are real spectra.&nbsp; This further means that&nbsp;
+*This means that now&nbsp; $Q_{\rm A}(f)$&nbsp; and&nbsp; $V_3(f)$&nbsp; are real spectra.&nbsp; This further means:
 *The phase information is lost&nbsp; $(φ = 0)$&nbsp; and the output signal&nbsp; $v_3(t)$&nbsp; is a cosine signal.
-*The signals&nbsp; $q_3(t)$&nbsp; and&nbsp; $v_3(t)$&nbsp; thus differ in both amplitude and phase. Only the frequency remains the same.
+*$q_3(t)$&nbsp; and&nbsp; $v_3(t)$&nbsp; thus differ in both amplitude and phase.&nbsp; Only the frequency remains the same.
 The graph shows
-*turquoise the signal $q_3(t)$&nbsp; and its samples (circles) and
+*turquoise the signal $q_3(t)$&nbsp; and its samples&nbsp; (circles),&nbsp; and
 *red dashed the output signal&nbsp; $v_3(t)$&nbsp; of the low-pass.
-You can see that the lowpass gives exactly the result you would probably choose if you were to draw a curve through the samples (circles).
+You can see that the low-pass gives exactly the result you would probably choose if you were to draw a curve through the samples&nbsp; (circles).

@@ Line 4: / Line 4: @@
 [[File:P_ID1610__Mod_Z_4_2.png|right|frame|Harmonic oscillations of different phase]]
-The&nbsp; [[Signal_Representation/Discrete-Time_Signal_Representation#Sampling_theorem|sampling theorem]]&nbsp; states that the sampling frequency&nbsp; $f_{\rm A} = 1/T_{\rm A}$&nbsp; must be at least twice as large as the largest frequency contained in the source signal&nbsp; $q(t)$&nbsp; $f_\text {N, max}$:
+The&nbsp; [[Signal_Representation/Discrete-Time_Signal_Representation#Sampling_theorem|sampling theorem]]&nbsp; states that the sampling frequency&nbsp; $f_{\rm A} = 1/T_{\rm A}$&nbsp; must be at least twice as large as the largest frequency&nbsp; $f_\text {N, max}$&nbsp; contained in the source signal&nbsp; $q(t)$:
 :$$f_{\rm A} \ge 2 \cdot f_{\rm N,\hspace{0.05cm}max}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} T_{\rm A} \le \frac{1}{2 \cdot f_{\rm N, \hspace{0.05cm}max}}\hspace{0.05cm}.$$
-If this condition is met, then at the receiver the message signal can be passed through a rectangular (ideal) low-pass filter with frequency response
+If this condition is met,&nbsp; then at the receiver the message signal can be passed through a rectangular&nbsp; (ideal)&nbsp; low-pass filter with frequency response
-:$$H(f) = \left\{ \begin{array}{l} 1 \\ 1/2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{for}} \\{\rm{for}} \\{\rm{for}} \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < f_{\rm G},} \ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| = f_{\rm G},} \ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| > f_{\rm G}} \end{array}$$
+:$$H(f) = \left\{ \begin{array}{l} 1 \\ 1/2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < f_{\rm G},} \\ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| = f_{\rm G},} \\ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| > f_{\rm G}} \\ \end{array}$$
 can be completely reconstructed, that is, it is then&nbsp; $v(t) = q(t)$.
 *The cutoff frequency&nbsp; $f_{\rm G}$&nbsp; is to be chosen equal to half the sampling frequency.
-*The equal sign is generally valid only if the spectrum&nbsp; $Q(f)$&nbsp; does not contain a discrete spectral line at frequency&nbsp; $f_\text {N, max}$&nbsp;.
+*The equal sign is generally valid only if the spectrum&nbsp; $Q(f)$&nbsp; does not contain a discrete spectral line at frequency&nbsp; $f_\text {N, max}$.
-In this exercise, three different source signals are considered, each of which can be expressed as a harmonic oscillation
+In this exercise,&nbsp; three different source signals are considered,&nbsp; each of which can be expressed as a harmonic oscillation
 :$$q(t) = A \cdot \cos (2 \pi \cdot f_{\rm N} \cdot t - \varphi)$$
-with amplitude&nbsp; $A = 1\ \rm V$&nbsp; and frequency&nbsp; $f_{\rm N}= 5 \ \rm kHz$&nbsp; For the spectral function&nbsp; $Q(f)$&nbsp; of all represented time signals generally holds:
+with amplitude&nbsp; $A = 1\ \rm V$&nbsp; and frequency&nbsp; $f_{\rm N}= 5 \ \rm kHz$.&nbsp; For the spectral function&nbsp; $Q(f)$&nbsp; of all represented time signals generally holds:
 :$$Q(f) = \frac{A}{2} \cdot \delta (f- f_{\rm N}) \cdot {\rm e}^{-{\rm j}\hspace{0.04cm}\cdot \hspace{0.04cm}\varphi}+ \frac{A}{2} \cdot \delta (f+ f_{\rm N}) \cdot {\rm e}^{+{\rm j}\hspace{0.04cm}\cdot \hspace{0.04cm}\varphi}\hspace{0.05cm}.$$
 The oscillations sketched in the graph differ only by the phase&nbsp; $φ$:
-* $φ_1 = 0$ &nbsp; ⇒ &nbsp; Cosine signal $q_1(t)$,
+* $φ_1 = 0$ &nbsp; ⇒ &nbsp; cosine signal&nbsp; $q_1(t)$,
-* $φ_2 = π/2 \ (= 90^\circ)$ &nbsp; ⇒ &nbsp; sinusoidal signal $q_2(t)$,
+* $φ_2 = π/2 \ (= 90^\circ)$ &nbsp; ⇒ &nbsp; sinusoidal signal&nbsp; $q_2(t)$,
-* $φ_3 = π/4 \ (= 45^\circ)$ &nbsp; ⇒ &nbsp; signal $q_3(t)$.
+* $φ_3 = π/4 \ (= 45^\circ)$ &nbsp; ⇒ &nbsp; signal&nbsp; $q_3(t)$.
@@ Line 28: / Line 28: @@
 Hints:
-*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Pulse_Code_Modulation|Pulse Code Modulation]].
+*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Pulse_Code_Modulation|"Pulse Code Modulation"]].
-*Reference is made in particular to the page&nbsp; [[Modulation_Methods/Pulse_Code_Modulation#Sampling_and_signal_reconstruction|Sampling and Signal Reconstruction]].
+*Reference is made in particular to the page&nbsp; [[Modulation_Methods/Pulse_Code_Modulation#Sampling_and_signal_reconstruction|"Sampling and Signal Reconstruction"]].
 *The sampled source signal is denoted by&nbsp; $q_{\rm A}(t)$&nbsp; and its spectral function by&nbsp; $Q_{\rm A}(f)$.&nbsp;
-*Sampling is always performed at&nbsp; $ν - T_{\rm A}$.
+*Sampling is always performed at&nbsp; $ν \cdot T_{\rm A}$.
@@ Line 41: / Line 41: @@
 + The sampling theorem is always satisfied.
 + All signals can be reconstructed by a low-pass filter.
-+ It is always true &nbsp;$Q_{\rm A}(f = 5 \ {\rm kHz}) = Q(f = 5 \ \rm kHz)$.
++ It is always true: &nbsp;$Q_{\rm A}(f = 5 \ {\rm kHz}) = Q(f = 5 \ \rm kHz)$.
@@ Line 50: / Line 50: @@
 {Which statements are valid for the signal &nbsp;$q_1(t)$&nbsp; and &nbsp;$f_{\rm A} = 10\ \rm kHz$?
 |type="[]"}
-- It is &nbsp;$Q_{\rm A}(f = 5 \ {\rm kHz)} = Q_1(f = 5 \ \rm kHz)$.
+- It holds &nbsp;$Q_{\rm A}(f = 5 \ {\rm kHz)} = Q_1(f = 5 \ \rm kHz)$.
 + A complete signal reconstruction is possible &nbsp; ⇒ &nbsp; $v_1(t) = q_1(t)$.
 - The reconstructed signal is &nbsp;$v_1(t) \equiv 0$.
@@ Line 56: / Line 56: @@
 {What statements hold for the signal &nbsp;$q_2(t)$&nbsp; and &nbsp;$f_{\rm A} = 10\ \rm kHz$?
 |type="[]"}
-- It is &nbsp;$Q_{\rm A}(f = 5 \ {\rm kHz)} = Q_2(f = 5 \ \rm kHz)$.
+- It holds &nbsp;$Q_{\rm A}(f = 5 \ {\rm kHz)} = Q_2(f = 5 \ \rm kHz)$.
 - A complete signal reconstruction is possible &nbsp; ⇒ &nbsp; $v_2(t) = q_2(t)$.
 + The reconstructed signal is &nbsp;$v_2(t) \equiv 0$.

@@ Line 7: / Line 7: @@
 :$$f_{\rm A} \ge 2 \cdot f_{\rm N,\hspace{0.05cm}max}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} T_{\rm A} \le \frac{1}{2 \cdot f_{\rm N, \hspace{0.05cm}max}}\hspace{0.05cm}.$$
 If this condition is met, then at the receiver the message signal can be passed through a rectangular (ideal) low-pass filter with frequency response
-:$$H(f) = \left\{ \begin{array}{l} 1 \ 1/2 \ 0 \end{array} \right.\quad \begin{array}{*{5}c}{\rm{for}} \\{\rm{for}} \\{\rm{for}} \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < f_{\rm G},} \ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| = f_{\rm G},} \ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| > f_{\rm G}} \end{array}$$
+:$$H(f) = \left\{ \begin{array}{l} 1 \\ 1/2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{for}} \\{\rm{for}} \\{\rm{for}} \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < f_{\rm G},} \ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| = f_{\rm G},} \ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| > f_{\rm G}} \end{array}$$
 can be completely reconstructed, that is, it is then&nbsp; $v(t) = q(t)$.
 *The cutoff frequency&nbsp; $f_{\rm G}$&nbsp; is to be chosen equal to half the sampling frequency.
@@ Line 20: / Line 20: @@
 * $φ_1 = 0$ &nbsp; ⇒ &nbsp; Cosine signal $q_1(t)$,
 * $φ_2 = π/2 \ (= 90^\circ)$ &nbsp; ⇒ &nbsp; sinusoidal signal $q_2(t)$,
-* $φ_3 = π/4 \ (= 45^\circ)$ &nbsp; ⇒ &nbsp; signal $q_3(t)$.terscheiden sich allein durch die Phase&nbsp; $φ$:
+* $φ_3 = π/4 \ (= 45^\circ)$ &nbsp; ⇒ &nbsp; signal $q_3(t)$.
@@ Line 78: / Line 73: @@
 '''(1)'''&nbsp; <u>All statements</u> are true:
 [[File:P_ID1611__Mod_Z_4_2a.png|P_ID1611__Mod_Z_4_2a.png|right|frame|Spectral function of sampled signal]]
-*The sampling theorem is satisfied by&nbsp; $f_{\rm A} = 11 \ \rm kHz > 2 - 5 \ \rm kHz$&nbsp; so that complete signal reconstruction is always possible.
+*The sampling theorem is satisfied by&nbsp; $f_{\rm A} = 11 \ \rm kHz > 2 \cdot 5 \ \rm kHz$&nbsp; so that complete signal reconstruction is always possible.
 *The spectrum&nbsp; $Q_{\rm A}(f)$&nbsp; results from&nbsp; $Q(f)$&nbsp; by periodic continuation at the respective frequency spacing&nbsp; $f_{\rm A}$, which is generally illustrated in the graph for the spectral function&nbsp; $Q_3(f)$&nbsp;.
 *By a rectangle&ndash;low-pass with&nbsp; $f_{\rm G} = f_{\rm A}/2 = 5.5 \ \rm kHz$&nbsp; the original spectrum&nbsp; $Q(f)$ is obtained.
@@ Line 98: / Line 93: @@
 '''(3)'''&nbsp; The correct solution is <u>suggestion 2</u>:
 *For the cosinusoidal signal, according to the next graph with&nbsp; $f_{\rm A} = 10 \rm kHz$&nbsp; the spectrum&nbsp; $Q_{\rm A}(f)$:&nbsp; &nbsp; All spectral lines are real.
-*The periodization of&nbsp; $Q(f)$&nbsp; with&nbsp; $f_{\rm A} = 10 \rm kHz$&nbsp; leads to a Dirac pulse with spectral lines at&nbsp; $±f_{\rm N}$,&nbsp; $±f_{\rm N}± f_{\rm A}$,&nbsp; $±f_{\rm N}± 2f_{\rm A}$, . ..
+*The periodization of&nbsp; $Q(f)$&nbsp; with&nbsp; $f_{\rm A} = 10 \rm kHz$&nbsp; leads to a Dirac comb with spectral lines at&nbsp; $±f_{\rm N}$,&nbsp; $±f_{\rm N}± f_{\rm A}$,&nbsp; $±f_{\rm N}± 2f_{\rm A}$, . ..
 *Through the superpositions, all Dirac functions have weight&nbsp; $A$, while the two spectral lines of&nbsp; $Q(f)$&nbsp; are weighted only by&nbsp; $A/2$&nbsp; each.
 *Because&nbsp; $H(f = f_{\rm N}) = H(f = f_{\rm G}) = 0.5$&nbsp; the spectrum&nbsp; $V_1(f)$&nbsp; after the low-pass is identical to&nbsp; $Q_1(f)$&nbsp; and accordingly&nbsp; $v_1(t) = q_1(t)$.

← Older revision		Revision as of 13:50, 16 March 2022
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−	[[Category:Modulation Methods: Exercises\|^4.1 ~~Pulscodemodulation~~^]]	+	[[Category:Modulation Methods: Exercises\|^4.1 Pulse Code Modulation^]]

← Older revision		Revision as of 13:01, 23 March 2021
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−	[[Category:~~Aufgaben zu Modulationsverfahren~~\|^4.1 Pulscodemodulation^]]	+	[[Category:Modulation Methods: Exercises\|^4.1 Pulscodemodulation^]]

← Older revision	Revision as of 13:42, 16 March 2022
(No difference)

@@ Line 4: / Line 4: @@
 [[File:P_ID1610__Mod_Z_4_2.png|right|frame|Harmonische Schwingungen unterschiedlicher Phase]]
-Das&nbsp; [[Signal_Representation/Zeitdiskrete_Signaldarstellung#Das_Abtasttheorem|Abtasttheorem]]&nbsp; besagt, dass die Abtastfrequenz&nbsp; $f_{\rm A} = 1/T_{\rm A}$&nbsp; mindestens doppelt so groß sein muss wie die größte im Quellensignal&nbsp; $q(t)$&nbsp; enthaltene Frequenz&nbsp; $f_\text {N, max}$:
+Das&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation#Das_Abtasttheorem|Abtasttheorem]]&nbsp; besagt, dass die Abtastfrequenz&nbsp; $f_{\rm A} = 1/T_{\rm A}$&nbsp; mindestens doppelt so groß sein muss wie die größte im Quellensignal&nbsp; $q(t)$&nbsp; enthaltene Frequenz&nbsp; $f_\text {N, max}$:
 :$$f_{\rm A} \ge 2 \cdot f_{\rm N,\hspace{0.05cm}max}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} T_{\rm A} \le \frac{1}{2 \cdot f_{\rm N, \hspace{0.05cm}max}}\hspace{0.05cm}.$$
 Wird diese Bedingung erfüllt, so kann beim Empfänger das Nachrichtensignal durch einen rechteckförmigen (idealen) Tiefpass mit dem Frequenzgang

@@ Line 30: / Line 30: @@
 ''Hinweise:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Modulationsverfahren/Pulscodemodulation|Pulscodemodulation]].
+*Die Aufgabe gehört zum  Kapitel&nbsp; [[Modulation_Methods/Pulscodemodulation|Pulscodemodulation]].
-*Bezug genommen wird insbesondere auf die Seite&nbsp;  [[Modulationsverfahren/Pulscodemodulation#Abtastung_und_Signalrekonstruktion|Abtastung und Signalrekonstruktion]].
+*Bezug genommen wird insbesondere auf die Seite&nbsp;  [[Modulation_Methods/Pulscodemodulation#Abtastung_und_Signalrekonstruktion|Abtastung und Signalrekonstruktion]].
 *Das abgetastete Quellensignal wird mit&nbsp; $q_{\rm A}(t)$&nbsp; bezeichnet und dessen Spektralfunktion mit&nbsp; $Q_{\rm A}(f)$.&nbsp;
 *Die Abtastung erfolgt stets bei&nbsp; $ν · T_{\rm A}$.