Aufgaben:Exercise 4.3: Natural and Discrete Sampling - Revision history

Guenter at 15:13, 8 April 2022

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Guenter at 14:45, 8 April 2022

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Guenter at 13:54, 8 April 2022

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Noah at 16:29, 6 April 2022

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Noah at 21:15, 4 April 2022

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Noah at 18:57, 19 March 2022

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Guenter at 13:51, 16 March 2022

2022-03-16T13:51:56Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.3: Natürliche und diskrete Abtastung to Aufgaben:Exercise 4.3: Natural and Discrete Sampling

2022-03-16T13:43:35Z

Guenter moved page Aufgabe 4.3: Natürliche und diskrete Abtastung to Exercise 4.3: Natural and Discrete Sampling

Guenter at 16:17, 9 November 2021

2021-11-09T16:17:22Z

Javier: Text replacement - "„" to """

2021-05-28T14:34:39Z

Text replacement - "„" to """

@@ Line 104: / Line 104: @@
 *The highest&nbsp; signal frequency&nbsp; $(f = f_{\rm A}/2)$&nbsp; is attenuated the most here:
 :$$V(f = f_{\rm A}/2) = Q( f_{\rm A}/2) \cdot {\rm sinc}( \frac{T_{\rm R}}{2 \cdot T_{\rm A}})=
-  Q( f_{\rm A}/2) \cdot {\rm sinc}(\frac{\sin(\pi/4)}{\pi/4})\approx 0.9 \cdot Q( f_{\rm A}/2) \hspace{0.05cm}.$$
+  Q( f_{\rm A}/2) \cdot \frac{\sin(\pi/4)}{\pi/4}\approx 0.9 \cdot Q( f_{\rm A}/2) \hspace{0.05cm}.$$
 {{ML-Fuß}}

@@ Line 68: / Line 68: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The spectrum of the square pulse&nbsp; $g_{\rm R}(t)$&nbsp; with amplitude&nbsp; $1$&nbsp; and duration&nbsp; $T_{\rm R}$&nbsp; is:
+'''(1)'''&nbsp; The spectrum of the rectangular pulse&nbsp; $g_{\rm R}(t)$&nbsp; with amplitude&nbsp; $1$&nbsp; and duration&nbsp; $T_{\rm R}$&nbsp; is:
-:$$ G_{\rm R}(f) = T_{\rm R} \cdot {\rm si}(\pi f T_{\rm R}) \hspace{0.3cm} {\rm with}\hspace{0.3cm} {\rm si}(x) = \sin(x)/x \hspace{0.3cm}
+:$$ G_{\rm R}(f) = T_{\rm R} \cdot {\rm sinc}(f T_{\rm R}) \hspace{0.3cm} {\rm with}\hspace{0.3cm} {\rm sinc}(x) = \sin(\pi x)/(\pi x)\hspace{0.3cm}
-\rightarrow \hspace{0.3cm} \frac{G_{\rm R}(f)}{T_{\rm A}} = \frac{T_{\rm R}}{T_{\rm A}} \cdot {\rm si}(\pi f T_{\rm R})$$
+\Rightarrow \hspace{0.3cm} \frac{G_{\rm R}(f)}{T_{\rm A}} = \frac{T_{\rm R}}{T_{\rm A}} \cdot {\rm sinc}(f T_{\rm R})\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{G_{\rm R}(f = 0)}{T_{\rm A}} =\frac{T_{\rm R}}{T_{\rm A}}\hspace{0.15cm}\underline { = 0.5} \hspace{0.05cm}.$$
+'''(2)'''&nbsp; The correct solution is the&nbsp; <u>second suggested solution</u>:
-'''(2)'''&nbsp; The correct solution is the <u>second suggested solution</u>:
+*From the given equation in the time domain,&nbsp; the convolution theorem gives:
 :$$q_{\rm A}(t) = \left [ \frac{1}{T_{\rm A}} \cdot p_{\rm \delta}(t) \star g_{\rm R}(t)\right ]\cdot q(t) \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}Q_{\rm A}(f) = \left [ \frac{1}{T_{\rm A}}\cdot P_{\rm \delta}(f) \cdot G_{\rm R}(f) \right ] \star Q(f) = \left [ P_{\rm \delta}(f) \cdot \frac{G_{\rm R}(f)}{{T_{\rm A}}} \right ] \star Q(f) \hspace{0.05cm}.$$
-*The first proposed solution is valid only for ideal sampling&nbsp; (with a Dirac comb)&nbsp; and the last one for discrete sampling.
+*The first proposed solution is valid only for ideal sampling &nbsp; (with a Dirac comb) &nbsp; and the last one for discrete sampling.
-'''(3)'''&nbsp; The answer is <u>YES</u>:
+'''(3)'''&nbsp; The answer is&nbsp; <u>YES</u>:
-* Starting from the result of the subtask&nbsp; '''(2)'''&nbsp; using the spectral function of the Dirac comb, we obtain.
+* Starting from the result of the subtask&nbsp; '''(2)'''&nbsp; using the spectral function of the Dirac comb,&nbsp; we obtain.
 :$$Q_{\rm A}(f) = \left [ P_{\rm \delta}(f) \cdot \frac{G_{\rm R}(f)}{{T_{\rm A}}} \right ] \star Q(f)= \left [ \frac{G_{\rm R}(f)}{{T_{\rm A}}} \cdot \sum_{\mu = -\infty}^{+\infty} \delta(f - \mu \cdot f_{\rm A})\right ] \star Q(f) \hspace{0.05cm}.$$
-*When the sampling theorem is satisfied and the low-pass filter is correct, of the infinite convolution products, only the convolution product with&nbsp; $μ = 0$&nbsp; lie in the passband.
+*When the sampling theorem is satisfied and the low-pass filter is correct:&nbsp; <br>From the infinite convolution products&nbsp; only the convolution product with&nbsp; $μ = 0$&nbsp; lie in the passband.
-*Taking into account the gain factor&nbsp; $T_{\rm A}/T_{\rm R}$&nbsp; we thus obtain for the spectrum at the filter output:
+*Taking into account the gain factor&nbsp; $T_{\rm A}/T_{\rm R}$,&nbsp; we thus obtain for the spectrum at the filter output:
-:$$V(f) = \frac{T_{\rm A}}{T_{\rm R}} \cdot \left [ \frac{G_{\rm R}(f = 0)}{{T_{\rm A}} \cdot \delta(f )\right ] \star Q(f)= Q(f) \hspace{0.05cm}.$$
+:$$V(f) = \frac{T_{\rm A}}{T_{\rm R}} \cdot \left [ \frac{G_{\rm R}(f = 0)}{{T_{\rm A}}} \cdot \delta(f )\right ] \star Q(f)= Q(f) \hspace{0.05cm}.$$
-'''(4)'''&nbsp; The<u> last suggested solution</u> is correct.
+'''(4)'''&nbsp; The&nbsp; <u>last suggested solution</u>&nbsp; is correct.
-*Shifting the factor&nbsp; $1/T_{\rm A}$&nbsp; to the rectangular pulse, we obtain with discrete sampling using the convolution theorem:
+*Shifting the factor&nbsp; $1/T_{\rm A}$&nbsp; to the rectangular pulse,&nbsp; we obtain with discrete sampling using the convolution theorem:
 :$$ q_{\rm A}(t) = \big [ p_{\rm \delta}(t)\cdot q(t) \big ] \star \frac{g_{\rm R}(t)}{T_{\rm A}}\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}Q_{\rm A}(f)= \big [ P_{\rm \delta}(f)\star Q(f) \big ] \cdot \frac{G_{\rm R}(f)}{T_{\rm A}}\hspace{0.05cm}.$$
-'''(5)'''&nbsp; The answer is <u>NO</u>:
+'''(5)'''&nbsp; The answer is&nbsp; <u>NO</u>:
 *The weighting function&nbsp; $G_{\rm R}(f)$&nbsp; now involves the inner kernel&nbsp; $(μ = 0)$&nbsp; of the convolution product.
-*All other terms&nbsp; $(μ ≠ 0)$&nbsp; are eliminated by the low-pass filter.&nbsp; One obtains here in the relevant range&nbsp; $|f| < f_{\rm A}/2$:
+*All other terms&nbsp; $(μ ≠ 0)$&nbsp; are eliminated by the low-pass filter.&nbsp;
-:$$V(f) = \frac{T_{\rm A}}{T_{\rm R}} \cdot \frac{G_{\rm R}(f )}{{T_{\rm A}}} \cdot Q(f) = 2 \cdot 0.5 \cdot {\rm si}(\pi f T_{\rm R})\cdot Q(f) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}V(f) = Q(f) \cdot {\rm si}(\pi f T_{\rm R})\hspace{0.05cm}.$$
+*One obtains here in the relevant range&nbsp; $|f| < f_{\rm A}/2$:
-*If no additional equalization is provided here, the higher frequencies are attenuated according to the&nbsp; $\rm si$ function.
+:$$V(f) = \frac{T_{\rm A}}{T_{\rm R}} \cdot \frac{G_{\rm R}(f )}{{T_{\rm A}}} \cdot Q(f) = 2 \cdot 0.5 \cdot {\rm sinc}(f T_{\rm R})\cdot Q(f) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}V(f) = Q(f) \cdot {\rm sinc}(f T_{\rm R})\hspace{0.05cm}.$$
 *The highest&nbsp; signal frequency&nbsp; $(f = f_{\rm A}/2)$&nbsp; is attenuated the most here:
-:$$V(f = \frac{f_{\rm A}}{2}) = Q( \frac{f_{\rm A}}{2}) \cdot {\rm si}(\pi \cdot \frac{T_{\rm R}}{2 \cdot T_{\rm A}})=
+:$$V(f = f_{\rm A}/2) = Q( f_{\rm A}/2) \cdot {\rm sinc}( \frac{T_{\rm R}}{2 \cdot T_{\rm A}})=
-  Q( \frac{f_{\rm A}}{2}) \cdot {\rm si}(\pi \cdot \frac{\sin(\pi/4)}{\pi/4})\approx 0.9 \cdot Q( \frac{f_{\rm A}}{2}) \hspace{0.05cm}.$$
+  Q( f_{\rm A}/2) \cdot {\rm sinc}(\frac{\sin(\pi/4)}{\pi/4})\approx 0.9 \cdot Q( f_{\rm A}/2) \hspace{0.05cm}.$$
 {{ML-Fuß}}

@@ Line 4: / Line 4: @@
 [[File:EN_Mod_A_4_3.png|right|frame|For natural and discrete sampling]]
-Ideal sampling can be described in the time domain by multiplying the analog source signal &nbsp;$q(t)$&nbsp; by a&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation#Diracpulse_in_Time_and_in_Frequency_Domain|Dirac comb]]&nbsp; $p_δ(t)$&nbsp; :
+Ideal sampling can be described in time domain by multiplying the analog source signal &nbsp;$q(t)$&nbsp; by a&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation#Diracpulse_in_Time_and_in_Frequency_Domain|Dirac comb]]&nbsp; $p_δ(t)$&nbsp; :
 :$$ q_{\rm A}(t) = p_{\delta}(t) \cdot q(t) \hspace{0.05cm}.$$
-Dirac combs- infinitely narrow and infinitely high - and accordingly also the Dirac comb &nbsp;$p_δ(t)$&nbsp; cannot be realized in practice, however.&nbsp; Here we must assume instead the square pulse &nbsp;$p_{\rm R}(t)$&nbsp; where the following relation holds:
+Dirac impulses&nbsp; &ndash; infinitely narrow and infinitely high &ndash;&nbsp; and accordingly also the&nbsp; "Dirac comb" &nbsp;$p_δ(t)$&nbsp; cannot be realized in practice,&nbsp; however.&nbsp;
-:$$ p_{\rm R}(t) = \left [ \frac{1}{T_{\rm A}} \cdot p_{\rm \delta}(t) \right ]\star g_{\rm R}(t)\hspace{0.9cm}\text{with}\hspace{0.9cm}
-  g_{\rm R}(t) = \left\{ \begin{array}{l} 1 \\ 1/2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{for}} \\{\rm{for}} \\{\rm{for}} \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} < T_{\rm R}/2\hspace{0.05cm}, \ {\hspace{0.04cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} = T_{\rm R}/2\hspace{0.05cm}, \ {\hspace{0.005cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} > T_{\rm R}/2\hspace{0.05cm}. \hspace{0.05cm} \end{array}$$
+Here we must assume instead the&nbsp; "rectangular pulse comb" &nbsp;$p_{\rm R}(t)$&nbsp; where the following relation holds:
 The duration &nbsp;$T_{\rm R}$&nbsp; of a rectangular pulse &nbsp;$g_{\rm R}(t)$&nbsp; should be (significantly) smaller than the distance $T_{\rm A}$ of two samples.
-In the diagram this ratio is chosen very large with&nbsp; $T_{\rm R}/T_{\rm A} = 0.5$&nbsp; to make the difference between&nbsp; "natural sampling"&nbsp; and&nbsp; "discrete sampling"&nbsp; especially clear:
+In the diagram this ratio is chosen with&nbsp; $T_{\rm R}/T_{\rm A} = 0.5$&nbsp;  very large to make the difference between&nbsp; "natural sampling"&nbsp; and&nbsp; "discrete sampling"&nbsp; especially clear:
-* In natural sampling, the sampled signal &nbsp;$q_{\rm A}(t)$&nbsp; is equal to the product of square pulse &nbsp;$p_{\rm R}(t)$&nbsp; and analog source signal &nbsp;$q(t)$:
+* In natural sampling,&nbsp; the sampled signal &nbsp;$q_{\rm A}(t)$&nbsp; is equal to the product of the rectangular pulse comb &nbsp;$p_{\rm R}(t)$&nbsp; and the analog source signal &nbsp;$q(t)$:
 :$$q_{\rm A}(t) = p_{\rm R}(t) \cdot q(t) = \left [ \frac{1}{T_{\rm A}} \cdot p_{\rm \delta}(t) \star g_{\rm R}(t)\right ]\cdot q(t)\hspace{0.05cm}.$$
-* In contrast, the corresponding equation for discrete sampling is:
+* In contrast,&nbsp; the corresponding equation for discrete sampling is:
 :$$ q_{\rm A}(t) = \left [ \frac{1}{T_{\rm A}} \cdot p_{\rm \delta}(t) \cdot q(t)\right ]\star g_{\rm R}(t)\hspace{0.05cm}.$$
-In the graph, these signals are sketched in blue&nbsp; (natural sampling)&nbsp; and green&nbsp; (discrete sampling)&nbsp; respectively.
+In the graph,&nbsp; these signals are sketched in blue&nbsp; (natural sampling)&nbsp; and green&nbsp; (discrete sampling)&nbsp; respectively.
 Hints:
-*The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Pulse_Code_Modulation|Puls 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#Natural_and_discrete_sampling|Natural and discrete sampling]].
+*Reference is made in particular to the page&nbsp; [[Modulation_Methods/Pulse_Code_Modulation#Natural_and_discrete_sampling|"Natural and discrete sampling"]].
 *The sampled source signal is denoted by&nbsp; $q_{\rm A}(t)$&nbsp; and its spectral function by&nbsp; $Q_{\rm A}(f)$.
 * Sampling is always performed at&nbsp; $ν \cdot T_{\rm A}$.
@@ Line 42: / Line 38: @@
 <quiz display=simple>
-{Let &nbsp;$T_{\rm R}/T_{\rm A} = 0.5$.&nbsp; For this, give the normalized spectrum &nbsp;$G_{\rm R}(f)/T_{\rm A}$&nbsp; What spectral value occurs at &nbsp;$f = 0$&nbsp;?
+{Let &nbsp;$T_{\rm R}/T_{\rm A} = 0.5$.&nbsp; For this,&nbsp; give the normalized spectrum &nbsp;$G_{\rm R}(f)/T_{\rm A}$.&nbsp; What spectral value occurs at &nbsp;$f = 0$&nbsp;?
 |type="{}"}
 $G_{\rm R}(f=0)/T_{\rm A} \ = \ $ { 0.5 3% }
@@ Line 52: / Line 48: @@
 - It holds &nbsp;$Q_{\rm A}(f) = \big[P_{\rm δ}(f) ∗ Q(f)\big] \cdot (G_{\rm R}(f)/T_{\rm A})$.
-{ For natural sampling, is the specified low-pass suitable for interpolation?
+{ For natural sampling:&nbsp; Is the specified low-pass suitable for interpolation?
 |type="()"}
 + Yes.
@@ Line 63: / Line 59: @@
 + It holds &nbsp;$Q_{\rm A}(f) = \big[P_{\rm δ}(f) ∗ Q(f)\big] \cdot (G_{\rm R}(f)/T_{\rm A})$.
-{For discrete sampling, is the specified low-pass suitable for interpolation?
+{For discrete sampling:&nbsp; Is the specified low-pass suitable for interpolation?
 |type="()"}
 - Yes.

@@ Line 4: / Line 4: @@
 [[File:EN_Mod_A_4_3.png|right|frame|For natural and discrete sampling]]
-Ideal sampling can be described in the time domain by multiplying the analog source signal &nbsp;$q(t)$&nbsp; by a&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation#Diracpulse_in_Time_and_in_Frequency_Domain|Diracpulse]]&nbsp; $p_δ(t)$&nbsp; :
+Ideal sampling can be described in the time domain by multiplying the analog source signal &nbsp;$q(t)$&nbsp; by a&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation#Diracpulse_in_Time_and_in_Frequency_Domain|Dirac comb]]&nbsp; $p_δ(t)$&nbsp; :
 :$$ q_{\rm A}(t) = p_{\delta}(t) \cdot q(t) \hspace{0.05cm}.$$
-Dirac pulses - infinitely narrow and infinitely high - and accordingly also the Dirac pulse &nbsp;$p_δ(t)$&nbsp; cannot be realized in practice, however.&nbsp; Here we must assume instead the square pulse &nbsp;$p_{\rm R}(t)$&nbsp; where the following relation holds:
+Dirac combs- infinitely narrow and infinitely high - and accordingly also the Dirac comb &nbsp;$p_δ(t)$&nbsp; cannot be realized in practice, however.&nbsp; Here we must assume instead the square pulse &nbsp;$p_{\rm R}(t)$&nbsp; where the following relation holds:
 :$$ p_{\rm R}(t) = \left [ \frac{1}{T_{\rm A}} \cdot p_{\rm \delta}(t) \right ]\star g_{\rm R}(t)\hspace{0.9cm}\text{with}\hspace{0.9cm}
   g_{\rm R}(t) = \left\{ \begin{array}{l} 1 \\ 1/2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{for}} \\{\rm{for}} \\{\rm{for}} \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} < T_{\rm R}/2\hspace{0.05cm}, \ {\hspace{0.04cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} = T_{\rm R}/2\hspace{0.05cm}, \ {\hspace{0.005cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} > T_{\rm R}/2\hspace{0.05cm}. \hspace{0.05cm} \end{array}$$
@@ Line 20: / Line 20: @@
 For signal reconstruction, a rectangular low-pass filter&nbsp; $H(f)$&nbsp; with cutoff frequency&nbsp; $f_{\rm G} = f_{\rm A}/2$&nbsp; and gain&nbsp; $T_{\rm A}/T_{\rm R}$&nbsp; is used in the passband:
-:$$H(f) = \left\{ \begin{array}{l} T_{\rm A}/T_{\rm R} \ 0 \end{array} \right.\quad \begin{array}{*{5}c}{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < f_{\rm A}/2}\hspace{0.05cm}, \ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| > f_{\rm A}/2}\hspace{0.05cm}. \end{array}$$
+:$$H(f) = \left\{ \begin{array}{l} T_{\rm A}/T_{\rm R} \ 0 \end{array} \right.\quad \begin{array}{*{5}c}{\rm{for}} \\{\rm{for}} \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < f_{\rm A}/2}\hspace{0.05cm}, \ {\hspace{0.04cm}\left| \hspace{0.005cm} f\hspace{0.05cm} \right| > f_{\rm A}/2}\hspace{0.05cm}. \end{array}$$
@@ Line 34: / Line 34: @@
 *Reference is made in particular to the page&nbsp; [[Modulation_Methods/Pulse_Code_Modulation#Natural_and_discrete_sampling|Natural and discrete sampling]].
 *The sampled source signal is denoted by&nbsp; $q_{\rm A}(t)$&nbsp; and its spectral function by&nbsp; $Q_{\rm A}(f)$.
-* Sampling is always performed at&nbsp; $ν - T_{\rm A}$.
+* Sampling is always performed at&nbsp; $ν \cdot T_{\rm A}$.
@@ Line 49: / Line 49: @@
 |type="[]"}
 - It holds &nbsp;$Q_{\rm A}(f) = P_{\rm δ}(f) ∗ Q(f)$.
-+ It holds &nbsp;$Q_{\rm A}(f) = \big[{\rm δ}(f) - (G_{\rm R}(f)/T_{\rm A})\big] ∗ Q(f)$.
++ It holds &nbsp;$Q_{\rm A}(f) = \big[{\rm δ}(f) \cdot (G_{\rm R}(f)/T_{\rm A})\big] ∗ Q(f)$.
-- It holds &nbsp;$Q_{\rm A}(f) = \big[P_{\rm δ}(f) ∗ Q(f)\big] - (G_{\rm R}(f)/T_{\rm A})$.
+- It holds &nbsp;$Q_{\rm A}(f) = \big[P_{\rm δ}(f) ∗ Q(f)\big] \cdot (G_{\rm R}(f)/T_{\rm A})$.
 { For natural sampling, is the specified low-pass suitable for interpolation?
@@ Line 60: / Line 60: @@
 |type="[]"}
 - It holds &nbsp;$Q_{\rm A}(f) = P_{\rm δ}(f) ∗ Q(f)$.
-- It holds &nbsp;$Q_{\rm A}(f) = \big[{\rm δ}(f) - (G_{\rm R}(f)/T_{\rm A})\big] ∗ Q(f)$.
+- It holds &nbsp;$Q_{\rm A}(f) = \big[{\rm δ}(f) \cdot (G_{\rm R}(f)/T_{\rm A})\big] ∗ Q(f)$.
-+ It holds &nbsp;$Q_{\rm A}(f) = \big[P_{\rm δ}(f) ∗ Q(f)\big] - (G_{\rm R}(f)/T_{\rm A})$.
++ It holds &nbsp;$Q_{\rm A}(f) = \big[P_{\rm δ}(f) ∗ Q(f)\big] \cdot (G_{\rm R}(f)/T_{\rm A})$.
 {For discrete sampling, is the specified low-pass suitable for interpolation?
@@ Line 83: / Line 83: @@
 :$$q_{\rm A}(t) = \left [ \frac{1}{T_{\rm A}} \cdot p_{\rm \delta}(t) \star g_{\rm R}(t)\right ]\cdot q(t) \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}Q_{\rm A}(f) = \left [ \frac{1}{T_{\rm A}}\cdot P_{\rm \delta}(f) \cdot G_{\rm R}(f) \right ] \star Q(f) = \left [ P_{\rm \delta}(f) \cdot \frac{G_{\rm R}(f)}{{T_{\rm A}}} \right ] \star Q(f) \hspace{0.05cm}.$$
-*The first proposed solution is valid only for ideal sampling&nbsp; (with a Dirac pulse)&nbsp; and the last one for discrete sampling.
+*The first proposed solution is valid only for ideal sampling&nbsp; (with a Dirac comb)&nbsp; and the last one for discrete sampling.
@@ Line 89: / Line 89: @@
 '''(3)'''&nbsp; The answer is <u>YES</u>:
-* Starting from the result of the subtask&nbsp; '''(2)'''&nbsp; using the spectral function of the Dirac pulse, we obtain.
+* Starting from the result of the subtask&nbsp; '''(2)'''&nbsp; using the spectral function of the Dirac comb, we obtain.
 :$$Q_{\rm A}(f) = \left [ P_{\rm \delta}(f) \cdot \frac{G_{\rm R}(f)}{{T_{\rm A}}} \right ] \star Q(f)= \left [ \frac{G_{\rm R}(f)}{{T_{\rm A}}} \cdot \sum_{\mu = -\infty}^{+\infty} \delta(f - \mu \cdot f_{\rm A})\right ] \star Q(f) \hspace{0.05cm}.$$
 *When the sampling theorem is satisfied and the low-pass filter is correct, of the infinite convolution products, only the convolution product with&nbsp; $μ = 0$&nbsp; lie in the passband.

@@ Line 8: / Line 8: @@
 Dirac pulses - infinitely narrow and infinitely high - and accordingly also the Dirac pulse &nbsp;$p_δ(t)$&nbsp; cannot be realized in practice, however.&nbsp; Here we must assume instead the square pulse &nbsp;$p_{\rm R}(t)$&nbsp; where the following relation holds:
 :$$ p_{\rm R}(t) = \left [ \frac{1}{T_{\rm A}} \cdot p_{\rm \delta}(t) \right ]\star g_{\rm R}(t)\hspace{0.9cm}\text{with}\hspace{0.9cm}
-  g_{\rm R}(t) = \left\{ \begin{array}{l} 1 \ 1/2 \ 0 \ \end{array} \right.\quad \begin{array}{*{5}c}{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \\{\rm{f\ddot{u}r}} \end{array}\begin{array}{*{10}c} {\hspace{0.04cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} < T_{\rm R}/2\hspace{0.05cm}, \ {\hspace{0.04cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} = T_{\rm R}/2\hspace{0.05cm}, \ {\hspace{0.005cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} > T_{\rm R}/2\hspace{0.05cm}. \hspace{0.05cm} \end{array}$$
+  g_{\rm R}(t) = \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{0.06cm} t \hspace{0.05cm} \right|} < T_{\rm R}/2\hspace{0.05cm}, \ {\hspace{0.04cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} = T_{\rm R}/2\hspace{0.05cm}, \ {\hspace{0.005cm}\left|\hspace{0.06cm} t \hspace{0.05cm} \right|} > T_{\rm R}/2\hspace{0.05cm}. \hspace{0.05cm} \end{array}$$
 The duration &nbsp;$T_{\rm R}$&nbsp; of a rectangular pulse &nbsp;$g_{\rm R}(t)$&nbsp; should be (significantly) smaller than the distance $T_{\rm A}$ of two samples.

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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 13:43, 16 March 2022
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@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1615__Mod_A_4_3.png|right|frame|Zur natürlichen und diskreten Abtastung]]
+[[File:EN_Mod_A_4_3.png|right|frame|Zur natürlichen und diskreten Abtastung]]
 Die ideale Abtastung lässt sich im Zeitbereich durch Multiplikation des analogen Quellensignals &nbsp;$q(t)$&nbsp; mit einem&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation#Diracpuls_im_Zeit-_und_im_Frequenzbereich|Diracpuls]]&nbsp; $p_δ(t)$&nbsp; beschreiben:
 :$$ q_{\rm A}(t) = p_{\delta}(t) \cdot q(t) \hspace{0.05cm}.$$

@@ Line 11: / Line 11: @@
 Die Dauer &nbsp;$T_{\rm R}$&nbsp; eines Rechteckimpulses &nbsp;$g_{\rm R}(t)$&nbsp; sollte dabei (deutlich) kleiner sein als der Abstand $T_{\rm A}$ zweier Abtastwerte.
-In der Grafik ist dieses Verhältnis mit&nbsp; $T_{\rm R}/T_{\rm A} = 0.5$&nbsp; sehr groß gewählt, um den Unterschied zwischen der&nbsp; &bdquo;natürlichen Abtastung"&nbsp; und der&nbsp; &bdquo;diskreten Abtastung"&nbsp; besonders deutlich werden zu lassen:
+In der Grafik ist dieses Verhältnis mit&nbsp; $T_{\rm R}/T_{\rm A} = 0.5$&nbsp; sehr groß gewählt, um den Unterschied zwischen der&nbsp; "natürlichen Abtastung"&nbsp; und der&nbsp; "diskreten Abtastung"&nbsp; besonders deutlich werden zu lassen:
 * Bei natürlicher Abtastung ist das abgetastete Signal &nbsp;$q_{\rm A}(t)$&nbsp; gleich dem Produkt aus Rechteckpuls &nbsp;$p_{\rm R}(t)$&nbsp; und analogem Quellensignal &nbsp;$q(t)$:
 :$$q_{\rm A}(t) = p_{\rm R}(t) \cdot q(t) = \left [ \frac{1}{T_{\rm A}} \cdot p_{\rm \delta}(t) \star g_{\rm R}(t)\right ]\cdot q(t)\hspace{0.05cm}.$$