Difference between revisions of "Aufgaben:Exercise 1.3: Rectangular Functions for Transmitter and Receiver"

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[[File: P_ID1267__Dig_A_1_3.png|right|frame|Three different system concepts]]
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[[File: P_ID1267__Dig_A_1_3.png|right|frame|Three different system configurations]]
 
We consider here three variants of a binary bipolar AWGN transmission system which differ with respect to the basic transmission pulse  $g_{s}(t)$  as well as the impulse response  $h_{\rm E}(t)$  of the receiver filter:
 
We consider here three variants of a binary bipolar AWGN transmission system which differ with respect to the basic transmission pulse  $g_{s}(t)$  as well as the impulse response  $h_{\rm E}(t)$  of the receiver filter:
*For  $\text{System A}$,  both  $g_{s}(t)$  and  $h_{\rm E}(t)$  are rectangular, only the pulse heights  $(s_{\rm 0}$  and  $1/T)$  are different.
+
*For  $\text{System A}$,  both  $g_{s}(t)$  and  $h_{\rm E}(t)$  are rectangular,  only the pulse heights  $(s_{\rm 0}$  and  $1/T)$  are different.
* $\text{System B}$  differs from  $\text{System A}$  by having a triangular-shaped basic transmission pulse with  $g_{s}(t=0) = s_{\rm 0}$.
+
*$\text{System B}$  differs from  $\text{System A}$  by having a triangular-shaped basic transmission pulse with  $g_{s}(t=0) = s_{\rm 0}$.
* $\text{System C}$  has the same basic transmission pulse as  $\text{System A}$, while the impulse response is triangular with  $h_{\rm E}(t=0) = 1/T$.   
+
*$\text{System C}$  has the same rectangular basic transmission pulse as  $\text{System A}$,  while the impulse response is triangular with  $h_{\rm E}(t=0) = 1/T$.   
  
  
The absolute width of the rectangular and triangular functions considered here is  $T = 10 \ \rm µ s$ each. The bit rate is  $R = 100 \ \rm kbit/s$. The other system parameters are given as follows:
+
The absolute width of the rectangular and triangular functions considered here is  $T = 10 \ \rm µ s$  each.  The bit rate is  $R = 100 \ \rm kbit/s$.  The other system parameters are given as follows:
 
:$$s_0 = 6 \,\,\sqrt{W}\hspace{0.05cm},\hspace{0.3cm}  N_{\rm 0} = 2 \cdot 10^{-5} \,\,{\rm W/Hz}\hspace{0.05cm}.$$
 
:$$s_0 = 6 \,\,\sqrt{W}\hspace{0.05cm},\hspace{0.3cm}  N_{\rm 0} = 2 \cdot 10^{-5} \,\,{\rm W/Hz}\hspace{0.05cm}.$$
  
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''Notes:''
+
Notes:  
*The exercise belongs to the chapter   [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission|Error Probability for Baseband Transmission]].
+
*The exercise belongs to the chapter   [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission|"Error Probability for Baseband Transmission"]].
+
*You can use the interactive applet  [[Applets:Complementary_Gaussian_Error_Functions|"Complementary Gaussian Error Functions"]]  to determine error probabilities.
*You can use the interactive applet  [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]]  to determine error probabilities.
+
*Consider  [[Theory_of_Stochastic_Signals/Power-Spectral_Density#Wiener-Khintchine_Theorem|"Wiener-Khintchine's theorem"]]  when calculating the detection noise power:
*Consider  [[Theory_of_Stochastic_Signals/Power-Spectral_Density#Wiener-Khintchine_Theorem|Wiener-Chintchine's theorem]] when calculating the detection noise power:
 
 
:$$ \sigma _d ^2  = \frac{N_0 }{2} \cdot \int_{ - \infty }^{
 
:$$ \sigma _d ^2  = \frac{N_0 }{2} \cdot \int_{ - \infty }^{
 
+ \infty } {\left| {H_{\rm E}( f )} \right|^2
 
+ \infty } {\left| {H_{\rm E}( f )} \right|^2
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<quiz display=simple>
 
<quiz display=simple>
  
{Calculate for &nbsp;$\text{System A}$&nbsp; the basic detection pulse &nbsp;$g_{d}(t) =  g_{ s}(t) \star h_{\rm E}(t)$. <br>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="{}"}
 
|type="{}"}
 
$g_0 \hspace{0.28cm} = \ $ { 6 3% } $\ \rm W^{1/2}$
 
$g_0 \hspace{0.28cm} = \ $ { 6 3% } $\ \rm W^{1/2}$
  
{From this, calculate the detection noise power &nbsp;$σ_{d}^2$.
+
{From this,&nbsp; calculate the detection noise power&nbsp; (variance) &nbsp;$σ_{d}^2$.
 
|type="{}"}
 
|type="{}"}
 
$σ_{d}^{\hspace{0.02cm}2} \hspace{0.2cm} = \ $ { 1 3% } $\ \rm W$
 
$σ_{d}^{\hspace{0.02cm}2} \hspace{0.2cm} = \ $ { 1 3% } $\ \rm W$
  
{Thus, what is the bit error probability &nbsp;$p_{\rm B}$&nbsp; for &nbsp;$\text{System A}$?
+
{Thus,&nbsp; what is the bit error probability &nbsp;$p_{\rm B}$&nbsp; for &nbsp;$\text{System A}$?
 
|type="{}"}
 
|type="{}"}
 
$p_{\rm B} \hspace{0.2cm} = \ $ { 0.987 10% } $\ \cdot 10^{-9}$
 
$p_{\rm B} \hspace{0.2cm} = \ $ { 0.987 10% } $\ \cdot 10^{-9}$
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===Solution===
 
===Solution===
 
{{ML-Kopf}}
 
{{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}^{
 
:$$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
 
+ 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
 
\cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\rm
 
W}}}\hspace{0.05cm}.$$
 
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_{ -
 
:$$\sigma _d ^2  \ = \ \frac{N_0 }{2} \cdot \int_{ -
 
\infty }^{ + \infty } {\left| {h_{\rm E}( t )} \right|^2
 
\infty }^{ + \infty } {\left| {h_{\rm E}( t )} \right|^2
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\,\,{\rm W/Hz}}{2 \cdot 10^{-5} \,\,{\rm s}} \hspace{0.1cm}\underline {= 1\,{\rm
 
\,\,{\rm W/Hz}}{2 \cdot 10^{-5} \,\,{\rm s}} \hspace{0.1cm}\underline {= 1\,{\rm
 
W}}\hspace{0.05cm}.$$
 
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 }^{
 
:$$\sigma _d ^2  = \frac{N_0 }{2} \cdot \int_{ - \infty }^{
 
+ \infty } {\left| {H_{\rm E}( f )} \right|^2
 
+ \infty } {\left| {H_{\rm E}( f )} \right|^2
 
\hspace{0.1cm}{\rm{d}}f} =  \frac{N_0 }{2}  \cdot \int_{-
 
\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}.$$
 
\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)
 
:$$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}.$$
 
  = {\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}
 
:$$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)
 
\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right)
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'''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}^{
 
:$$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
 
+ 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
 
\frac{s_0 }{2}  \cdot T = \frac{s_0 }{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm
 
W}}\hspace{0.05cm}.$$
 
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)
 
:$$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}.$$
 
  = {\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
 
:$$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
 
  d}t =  2\cdot s_0^2 \cdot  \int ^{+T/2} _{0} \left( 1- \frac{2t}{T}\right)^2\,{\rm
Line 112: Line 111:
 
  ={\rm Q} \left( \sqrt{12}\right)={\rm Q}(3.464) \approx 3 \cdot 10^{-4}
 
  ={\rm Q} \left( \sqrt{12}\right)={\rm Q}(3.464) \approx 3 \cdot 10^{-4}
 
\hspace{0.05cm}.$$
 
\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
 
:$$g_d (t = 0)  =  \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm
 
W}}\hspace{0.05cm}.$$
 
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
 
:$$\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}}.$$
 
  d}t = \frac{N_0}{6T}\hspace{0.1cm}\underline { = 0.333 \,{\rm W}}.$$
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:$$p_{\rm B} =    {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{0.577 \,\sqrt{\rm W}}\right)
 
:$$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}.$$
 
  \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}}
 
{{ML-Fuß}}

Latest revision as of 08:07, 20 May 2022

Three different system configurations

We consider here three variants of a binary bipolar AWGN transmission system which differ with respect to the basic transmission pulse  $g_{s}(t)$  as well as the impulse response  $h_{\rm E}(t)$  of the receiver filter:

  • For  $\text{System A}$,  both  $g_{s}(t)$  and  $h_{\rm E}(t)$  are rectangular,  only the pulse heights  $(s_{\rm 0}$  and  $1/T)$  are different.
  • $\text{System B}$  differs from  $\text{System A}$  by having a triangular-shaped basic transmission pulse with  $g_{s}(t=0) = s_{\rm 0}$.
  • $\text{System C}$  has the same rectangular basic transmission pulse as  $\text{System A}$,  while the impulse response is triangular with  $h_{\rm E}(t=0) = 1/T$. 


The absolute width of the rectangular and triangular functions considered here is  $T = 10 \ \rm µ s$  each.  The bit rate is  $R = 100 \ \rm kbit/s$.  The other system parameters are given as follows:

$$s_0 = 6 \,\,\sqrt{W}\hspace{0.05cm},\hspace{0.3cm} N_{\rm 0} = 2 \cdot 10^{-5} \,\,{\rm W/Hz}\hspace{0.05cm}.$$



Notes:

$$ \sigma _d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{ + \infty } {\left| {H_{\rm E}( f )} \right|^2 \hspace{0.1cm}{\rm{d}}f} = \frac{N_0 }{2} \cdot \int_{ - \infty }^{ + \infty } {\left| {h_{\rm E}( t )} \right|^2 \hspace{0.1cm}{\rm{d}}t}\hspace{0.05cm}.$$


Questions

1

Calculate for  $\text{System A}$  the basic detection pulse  $g_{d}(t) = g_{ s}(t) \star h_{\rm E}(t)$.  What value  $g_0 = g_{d}(t=0)$  results at time  $t = 0$?

$g_0 \hspace{0.28cm} = \ $

$\ \rm W^{1/2}$

2

From this,  calculate the detection noise power  (variance)  $σ_{d}^2$.

$σ_{d}^{\hspace{0.02cm}2} \hspace{0.2cm} = \ $

$\ \rm W$

3

Thus,  what is the bit error probability  $p_{\rm B}$  for  $\text{System A}$?

$p_{\rm B} \hspace{0.2cm} = \ $

$\ \cdot 10^{-9}$

4

Determine the corresponding quantities for  $\text{System B}$ .

$g_0 \hspace{0.28cm} = \ $

$\ \rm W^{1/2}$
$σ_{d}^{\hspace{0.02cm}2} \hspace{0.2cm} = \ $

$\ \rm W$
$p_{\rm B} \hspace{0.2cm} = \ $

$\ \cdot 10^{-2}$

5

What are the characteristics for  $\text{System C}$ ?

$g_0 \hspace{0.28cm} = \ $

$\ \rm W^{1/2}$
$σ_{d}^{\hspace{0.02cm}2} \hspace{0.2cm} = \ $

$\ \rm W$
$p_{\rm B} \hspace{0.2cm} = \ $

$\ \cdot 10^{-7}$


Solution

1.  For System A,  the convolution of the two equal-width rectangular functions  $g_{s}(t)$  and  $h_{\rm E}(t)$  leads to a triangular detection pulse with the maximum at  $t = 0$:

$$g_d (t = 0) = \int_{ - T/2}^{ + T/2} { g_s(t) \cdot h_{\rm E}( t )} \hspace{0.1cm}{\rm{d}}t =s_0 \cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\rm W}}}\hspace{0.05cm}.$$

There is no intersymbol interfering because for  $| t |\ge T$  the detection pulse is  $g_{d}(t) = 0$.


2.  The variance of the noise component of the detection signal – referred to as the  "detection noise power" – can be calculated in both the time and frequency domains.

  • For the present rectangular waveform,  calculation in the time domain yields faster results:
$$\sigma _d ^2 \ = \ \frac{N_0 }{2} \cdot \int_{ - \infty }^{ + \infty } {\left| {h_{\rm E}( t )} \right|^2 \hspace{0.1cm}{\rm{d}}t} =\frac{N_0 }{2} \cdot \int_{ - T/2 }^{ + T/2 } {\left| {h_{\rm E}( t )} \right|^2 \hspace{0.1cm}{\rm{d}}t} = \ \frac{N_0 }{2} \cdot\frac{1 }{T^2} \cdot T = \frac{N_0 }{2T} = \frac{2 \cdot 10^{-5} \,\,{\rm W/Hz}}{2 \cdot 10^{-5} \,\,{\rm s}} \hspace{0.1cm}\underline {= 1\,{\rm W}}\hspace{0.05cm}.$$
  • The frequency domain calculation would be as follows with  $H_{\rm E}(f) = {\rm sinc}(fT)$:
$$\sigma _d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{ + \infty } {\left| {H_{\rm E}( f )} \right|^2 \hspace{0.1cm}{\rm{d}}f} = \frac{N_0 }{2} \cdot \int_{- \infty }^{ \infty } {\rm sinc}^2(f T)\hspace{0.1cm}{\rm{d}}f = \frac{N_0 }{2T} \hspace{0.05cm}.$$


3.  Due to the time-limited pulse shape  (this means:  no intersymbol interfering!),  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:
$$E_{\rm B} = s_0^2 \cdot T = 36\, {\rm W} \cdot 10^{-5} {\rm s}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right) ={\rm Q} \left( \sqrt{\frac{2 \cdot 36 \cdot 10^{-5}\,\, {\rm Ws}}{2 \cdot 10^{-5} \,\, {\rm Ws}}}\right)={\rm Q}(6) \hspace{0.05cm}.$$


4.  Since  System B  uses the same receiver filter as  System A,  the same detection noise power  $σ_{d}^2 = 1 \ \rm W$  is also obtained.

  • However,  the basic detection pulse is now no longer triangular,  but has a more pointed shape.  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:
$$p_{\rm B} = {\rm Q} \left( \frac{g_d (t = 0)}{\sigma_d}\right)= {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right) = {\rm Q}(3) \hspace{0.1cm}\underline {= 0.135 \cdot 10^{-2}} \hspace{0.05cm}.$$
  • On the other hand,  the following calculation is  not  applicable here:
$$E_{\rm B} = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm d}t = 2\cdot s_0^2 \cdot \int ^{+T/2} _{0} \left( 1- \frac{2t}{T}\right)^2\,{\rm d}t = \frac{s_0^2 \cdot T }{3} = 12 \cdot 10^{-5} \,{\rm Ws}$$
$$\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right) ={\rm Q} \left( \sqrt{12}\right)={\rm Q}(3.464) \approx 3 \cdot 10^{-4} \hspace{0.05cm}.$$
  • One would thus compute a bit error probability that is too low,  since the implicit assumption of a matched filter does not hold.


5.  For the rectangular basic transmission pulse and the triangular impulse response   ⇒   System C,
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:
$$g_d (t = 0) = \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm W}}\hspace{0.05cm}.$$
  • In contrast,  the detection noise power is now smaller than in systems  A  and  B:
$$\sigma _d ^2 = \frac{N_0}{2} \cdot \frac{1}{T^2} \cdot \int^{+T/2} _{-T/2} \left( 1- \frac{2t}{T}\right)^2\,{\rm d}t = \frac{N_0}{6T}\hspace{0.1cm}\underline { = 0.333 \,{\rm W}}.$$
  • This now gives us for the bit error probability:
$$p_{\rm B} = {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{0.577 \,\sqrt{\rm W}}\right) \approx {\rm Q}(5.2)\hspace{0.1cm}\underline { \approx 10^{-7} } \hspace{0.05cm}.$$
  • 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 improvement over subtask  (4)  is due to the higher signal energy.