Aufgaben:Exercise 1.3: Rectangular Functions for Transmitter and Receiver - Revision history

Guenter at 07:07, 20 May 2022

2022-05-20T07:07:38Z

Guenter at 07:06, 20 May 2022

2022-05-20T07:06:06Z

Guenter at 07:05, 20 May 2022

2022-05-20T07:05:04Z

Hwang at 15:06, 9 May 2022

2022-05-09T15:06:00Z

Hwang at 15:04, 9 May 2022

2022-05-09T15:04:19Z

Hwang at 15:00, 9 May 2022

2022-05-09T15:00:26Z

Hwang at 14:58, 9 May 2022

2022-05-09T14:58:37Z

Hwang at 14:57, 9 May 2022

2022-05-09T14:57:47Z

Hwang at 14:56, 9 May 2022

2022-05-09T14:56:16Z

Guenter at 13:08, 30 April 2022

2022-04-30T13:08:34Z

@@ Line 96: / Line 96: @@
 '''4.'''&nbsp; Since&nbsp; '''System B'''&nbsp; uses the same receiver filter as&nbsp; '''System A''',&nbsp; the same detection noise power&nbsp; $&sigma;_{d}^2 = 1 \ \rm W$&nbsp; is also obtained.
-*However,&nbsp; the basic detection pulse '''KORREKTUR: basic transmitter pulse''' is now no longer triangular,&nbsp; but has a more pointed shape.&nbsp; At time $t = 0$ applies:
+*However,&nbsp; the basic detection pulse is now no longer triangular,&nbsp; but has a more pointed shape.&nbsp; At time $t = 0$ applies:
 :$$g_d (t = 0)  = \frac{1}{T} \cdot  \int_{ - T/2}^{
 + T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot

@@ Line 114: / Line 114: @@
-'''5.'''&nbsp; For the rectangular basic transmission pulse and the triangular impulse response &nbsp; &rArr; &nbsp; '''System C''', <br>the same basic detection pulse '''KORREKTUR: basic transmitter pulse''' is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.
+'''5.'''&nbsp; For the rectangular basic transmission pulse and the triangular impulse response &nbsp; &rArr; &nbsp; '''System C''', <br>the same basic detection pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.
 *Therefore,&nbsp; as in&nbsp; '''System B''':
 :$$g_d (t = 0)  =  \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm

@@ Line 30: / Line 30: @@
 <quiz display=simple>
-{Calculate for &nbsp;$\text{System A}$&nbsp; the basic detection pulse '''KORREKTUR: basic transmitter pulse''' &nbsp;$g_{d}(t) =  g_{ s}(t) \star h_{\rm E}(t)$.&nbsp; What value &nbsp;$g_0 = g_{d}(t=0)$&nbsp; results at time &nbsp;$t = 0$?
+{Calculate for &nbsp;$\text{System A}$&nbsp; the basic detection pulse &nbsp;$g_{d}(t) =  g_{ s}(t) \star h_{\rm E}(t)$.&nbsp; What value &nbsp;$g_0 = g_{d}(t=0)$&nbsp; results at time &nbsp;$t = 0$?
 |type="{}"}
 $g_0 \hspace{0.28cm} = \ $ { 6 3% } $\ \rm W^{1/2}$

@@ Line 30: / Line 30: @@
 <quiz display=simple>
-{Calculate for &nbsp;$\text{System A}$&nbsp; the basic transmitter pulse &nbsp;$g_{d}(t) =  g_{ s}(t) \star h_{\rm E}(t)$.&nbsp; What value &nbsp;$g_0 = g_{d}(t=0)$&nbsp; results at time &nbsp;$t = 0$?
+{Calculate for &nbsp;$\text{System A}$&nbsp; the basic detection pulse '''KORREKTUR: basic transmitter pulse''' &nbsp;$g_{d}(t) =  g_{ s}(t) \star h_{\rm E}(t)$.&nbsp; What value &nbsp;$g_0 = g_{d}(t=0)$&nbsp; results at time &nbsp;$t = 0$?
 |type="{}"}
 $g_0 \hspace{0.28cm} = \ $ { 6 3% } $\ \rm W^{1/2}$
@@ Line 96: / Line 96: @@
 '''4.'''&nbsp; Since&nbsp; '''System B'''&nbsp; uses the same receiver filter as&nbsp; '''System A''',&nbsp; the same detection noise power&nbsp; $&sigma;_{d}^2 = 1 \ \rm W$&nbsp; is also obtained.
-*However,&nbsp; the basic transmitter pulse is now no longer triangular,&nbsp; but has a more pointed shape.&nbsp; At time $t = 0$ applies:
+*However,&nbsp; the basic detection pulse '''KORREKTUR: basic transmitter pulse''' is now no longer triangular,&nbsp; but has a more pointed shape.&nbsp; At time $t = 0$ applies:
 :$$g_d (t = 0)  = \frac{1}{T} \cdot  \int_{ - T/2}^{
 + T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot
@@ Line 114: / Line 114: @@
-'''5.'''&nbsp; For the rectangular basic transmission pulse and the triangular impulse response &nbsp; &rArr; &nbsp; '''System C''', <br>the same basic transmitter pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.
+'''5.'''&nbsp; For the rectangular basic transmission pulse and the triangular impulse response &nbsp; &rArr; &nbsp; '''System C''', <br>the same basic detection pulse '''KORREKTUR: basic transmitter pulse''' is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.
 *Therefore,&nbsp; as in&nbsp; '''System B''':
 :$$g_d (t = 0)  =  \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm

@@ Line 96: / Line 96: @@
 '''4.'''&nbsp; Since&nbsp; '''System B'''&nbsp; uses the same receiver filter as&nbsp; '''System A''',&nbsp; the same detection noise power&nbsp; $&sigma;_{d}^2 = 1 \ \rm W$&nbsp; is also obtained.
-*However,&nbsp; the basic detection pulse is now no longer triangular,&nbsp; but has a more pointed shape.&nbsp; At time $t = 0$ applies:
+*However,&nbsp; the basic transmitter pulse is now no longer triangular,&nbsp; but has a more pointed shape.&nbsp; At time $t = 0$ applies:
 :$$g_d (t = 0)  = \frac{1}{T} \cdot  \int_{ - T/2}^{
 + T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot
@@ Line 114: / Line 114: @@
-'''5.'''&nbsp; For the rectangular basic transmission pulse and the triangular impulse response &nbsp; &rArr; &nbsp; '''System C''', <br>the same basic detection pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.
+'''5.'''&nbsp; For the rectangular basic transmission pulse and the triangular impulse response &nbsp; &rArr; &nbsp; '''System C''', <br>the same basic transmitter pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.
 *Therefore,&nbsp; as in&nbsp; '''System B''':
 :$$g_d (t = 0)  =  \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm

@@ Line 63: / Line 63: @@
 \cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\rm
 W}}}\hspace{0.05cm}.$$
-There is no intersymbol interfering because for&nbsp; $| t |\ge T$&nbsp; the detection pulse is'&nbsp; $g_{d}(t) = 0$.
+There is no intersymbol interfering because for&nbsp; $| t |\ge T$&nbsp; the detection pulse is&nbsp; $g_{d}(t) = 0$.

@@ Line 58: / Line 58: @@
 ===Solution===
 {{ML-Kopf}}
-'''1.'''&nbsp; For '''System A''','&nbsp;  the convolution of the two equal-width rectangular functions&nbsp; $g_{s}(t)$&nbsp; and&nbsp; $h_{\rm E}(t)$&nbsp; leads to a triangular detection pulse with the maximum at&nbsp; $t = 0$:
+'''1.'''&nbsp; For '''System A''',&nbsp;  the convolution of the two equal-width rectangular functions&nbsp; $g_{s}(t)$&nbsp; and&nbsp; $h_{\rm E}(t)$&nbsp; leads to a triangular detection pulse with the maximum at&nbsp; $t = 0$:
 :$$g_d (t = 0)  =  \int_{ - T/2}^{
 + T/2} { g_s(t) \cdot h_{\rm E}( t )} \hspace{0.1cm}{\rm{d}}t =s_0

@@ Line 19: / Line 19: @@
 *The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission|"Error Probability for Baseband Transmission"]].
 *You can use the interactive applet &nbsp;[[Applets:Complementary_Gaussian_Error_Functions|"Complementary Gaussian Error Functions"]]&nbsp; to determine error probabilities.
-*Consider &nbsp;[[Theory_of_Stochastic_Signals/Power-Spectral_Density#Wiener-Khintchine_Theorem|Wiener-Khintchine's theorem]]&nbsp; when calculating the detection noise power:
+*Consider &nbsp;[[Theory_of_Stochastic_Signals/Power-Spectral_Density#Wiener-Khintchine_Theorem|"Wiener-Khintchine's theorem"]]&nbsp; when calculating the detection noise power:
 :$$ \sigma _d ^2  = \frac{N_0 }{2} \cdot \int_{ - \infty }^{
 + \infty } {\left| {H_{\rm E}( f )} \right|^2

@@ Line 58: / Line 58: @@
 ===Solution===
 {{ML-Kopf}}
-'''1.'''&nbsp; For '''System A''', the convolution of the two equal-width rectangular functions $g_{s}(t)$ and $h_{\rm E}(t)$ leads to a triangular basic detection pulse with the maximum at $t = 0$:
+'''1.'''&nbsp; For '''System A''','&nbsp;  the convolution of the two equal-width rectangular functions&nbsp; $g_{s}(t)$&nbsp; and&nbsp; $h_{\rm E}(t)$&nbsp; leads to a triangular detection pulse with the maximum at&nbsp; $t = 0$:
 :$$g_d (t = 0)  =  \int_{ - T/2}^{
 + T/2} { g_s(t) \cdot h_{\rm E}( t )} \hspace{0.1cm}{\rm{d}}t =s_0
 \cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\rm
 W}}}\hspace{0.05cm}.$$
-There is no intersymbol interfering because for $| t |\ge T$ the detection pulse is $g_{d}(t) = 0$.
+There is no intersymbol interfering because for&nbsp; $| t |\ge T$&nbsp; the detection pulse is'&nbsp; $g_{d}(t) = 0$.
-'''2.'''&nbsp; The variance of the noise component of the detection signal &ndash; referred to here as the detection noise power &ndash; can be calculated in both the time and frequency domains.
+'''2.'''&nbsp; The variance of the noise component of the detection signal &ndash; referred to as the&nbsp; "detection noise power" &ndash; can be calculated in both the time and frequency domains.
-*For the present rectangular waveform, calculation in the time domain yields faster results:
+*For the present rectangular waveform,&nbsp; calculation in the time domain yields faster results:
 :$$\sigma _d ^2  \ = \ \frac{N_0 }{2} \cdot \int_{ -
 \infty }^{ + \infty } {\left| {h_{\rm E}( t )} \right|^2
@@ Line 76: / Line 76: @@
 \,\,{\rm W/Hz}}{2 \cdot 10^{-5} \,\,{\rm s}} \hspace{0.1cm}\underline {= 1\,{\rm
 W}}\hspace{0.05cm}.$$
-*The frequency domain calculation would be as follows with $H_{\rm E}(f) = {\rm si}(&pi;fT)$:
+*The frequency domain calculation would be as follows with&nbsp; $H_{\rm E}(f) = {\rm sinc}(fT)$:
 :$$\sigma _d ^2  = \frac{N_0 }{2} \cdot \int_{ - \infty }^{
 + \infty } {\left| {H_{\rm E}( f )} \right|^2
 \hspace{0.1cm}{\rm{d}}f} =  \frac{N_0 }{2}  \cdot \int_{-
-\infty }^{ \infty } {\rm si}^2(\pi f T)\hspace{0.1cm}{\rm{d}}f =
+\infty }^{ \infty } {\rm sinc}^2(f T)\hspace{0.1cm}{\rm{d}}f =
 \frac{N_0 }{2T} \hspace{0.05cm}.$$
-'''3.'''&nbsp; Due to the time-limited pulse shape (this means: no intersymbol interfering!), the bipolar approach assumed here yields:
+'''3.'''&nbsp; Due to the time-limited pulse shape&nbsp; (this means:&nbsp; no intersymbol interfering!),&nbsp; the bipolar approach assumed here yields:
 :$$p_{\rm B} =   {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {\rm Q} \left( \frac{ 6 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)
   = {\rm Q}(6) \hspace{0.1cm}\underline {= 0.987 \cdot 10^{-9}} \hspace{0.05cm}.$$
-'''System A''' represents the matched filter realization of the optimal binary receiver, so the following equations would also be applicable:
+*'''System A'''&nbsp; represents the matched filter realization of the optimal binary receiver,&nbsp; so the following equations would also be applicable:
 :$$E_{\rm B} = s_0^2 \cdot T = 36\, {\rm W} \cdot 10^{-5} {\rm s}\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right)
@@ Line 95: / Line 95: @@
-'''4.'''&nbsp; Since '''System B''' uses the same receiver filter as '''System A''', the same detection noise power $&sigma;_{d}^2 = 1 \ \rm W$ is also obtained.
+'''4.'''&nbsp; Since&nbsp; '''System B'''&nbsp; uses the same receiver filter as&nbsp; '''System A''',&nbsp; the same detection noise power&nbsp; $&sigma;_{d}^2 = 1 \ \rm W$&nbsp; is also obtained.
-*However, the basic detection pulse is now no longer triangular, but has a more pointed shape. At time $t = 0$ applies:
+*However,&nbsp; the basic detection pulse is now no longer triangular,&nbsp; but has a more pointed shape.&nbsp; At time $t = 0$ applies:
 :$$g_d (t = 0)  = \frac{1}{T} \cdot  \int_{ - T/2}^{
 + T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot
 \frac{s_0 }{2}  \cdot T = \frac{s_0 }{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm
 W}}\hspace{0.05cm}.$$
-*'''System B''' is also free of intersymbol interfering. Therefore, one obtains for the bit error probability:
+*'''System B'''&nbsp; is also free of intersymbol interfering.&nbsp; Therefore,&nbsp; one obtains for the bit error probability:
 :$$p_{\rm B} =   {\rm Q} \left( \frac{g_d (t = 0)}{\sigma_d}\right)= {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)
   = {\rm Q}(3) \hspace{0.1cm}\underline {= 0.135 \cdot 10^{-2}} \hspace{0.05cm}.$$
-*On the other hand, the following calculation is not applicable here:
+*On the other hand,&nbsp; the following calculation is&nbsp; '''not'''&nbsp; applicable here:
 :$$E_{\rm B} =    \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm
   d}t =  2\cdot s_0^2 \cdot  \int ^{+T/2} _{0} \left( 1- \frac{2t}{T}\right)^2\,{\rm
@@ Line 111: / Line 111: @@
   ={\rm Q} \left( \sqrt{12}\right)={\rm Q}(3.464) \approx 3 \cdot 10^{-4}
 \hspace{0.05cm}.$$
-*One would thus compute a bit error probability that is too low, since the implicit assumption of a matched filter does not hold.
+*One would thus compute a bit error probability that is too low,&nbsp; since the implicit assumption of a matched filter does not hold.
-'''5.'''&nbsp; For rectangular basic transmission pulse and triangular impulse response &nbsp; &rArr; &nbsp; '''System C''', the same basic detection pulse is obtained as for triangular $g_{\rm s}(t)$ and rectangular $h_{\rm E}(t)$.
+'''5.'''&nbsp; For the rectangular basic transmission pulse and the triangular impulse response &nbsp; &rArr; &nbsp; '''System C''', <br>the same basic detection pulse is obtained as for the triangular $g_{\rm s}(t)$ and the rectangular $h_{\rm E}(t)$.
-*Therefore, as in '''System B''':
+*Therefore,&nbsp; as in&nbsp; '''System B''':
 :$$g_d (t = 0)  =  \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm
 W}}\hspace{0.05cm}.$$
-*In contrast, the detection noise power is now smaller than in systems A and B:
+*In contrast,&nbsp; the detection noise power is now smaller than in systems&nbsp; '''A'''&nbsp; and&nbsp; '''B''':
 :$$\sigma _d ^2 =  \frac{N_0}{2}  \cdot \frac{1}{T^2} \cdot \int^{+T/2} _{-T/2} \left( 1- \frac{2t}{T}\right)^2\,{\rm
   d}t = \frac{N_0}{6T}\hspace{0.1cm}\underline { = 0.333 \,{\rm W}}.$$
@@ Line 124: / Line 124: @@
 :$$p_{\rm B} =    {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{0.577 \,\sqrt{\rm W}}\right)
   \approx {\rm Q}(5.2)\hspace{0.1cm}\underline { \approx  10^{-7} } \hspace{0.05cm}.$$
-*The apparent increase in error probability by a factor of about $100$ compared to subtask '''(3)''' is due to the severe mismatch compared to the matched filter.
+*The apparent increase in error probability by a factor of about&nbsp; $100$&nbsp; compared to subtask&nbsp; '''(3)'''&nbsp; is due to the severe mismatch compared to the matched filter.
-*The improvement over subtask '''(4)''' is due to the higher signal energy.
+*The improvement over subtask&nbsp; '''(4)'''&nbsp; is due to the higher signal energy.
 {{ML-Fuß}}