Linear and Time Invariant Systems/Properties of Coaxial Cables - Revision history

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@@ Line 175: / Line 175: @@
 ==Special features of coaxial cable systems==
 <br>
-Assuming binary transmission with non-return-to-zero&nbsp; (NRZ)&nbsp; rectangular pulses&nbsp; $($symbol duration $T)$&nbsp; and a coaxial transmission channel,&nbsp; the following system model is obtained.&nbsp; In particular, it should be noted:
+[[File:EN_LZI_4_2_S5_v2.png |right|frame| Binary transmission system with coaxial cable]]
-[[File:EN_LZI_4_2_S5_v2.png |right|frame| Binary transmission system with coaxial cable<br><br><br>]]
+*and a coaxial transmission channel,&nbsp;
-*In a simulation, the phase delay time of the coaxial cable is conveniently left out of consideration.&nbsp; Then the basic receiver pulse  &nbsp;$g_r(t)$&nbsp; (with ${\rm a}_{\rm \star}$&nbsp;in Neper)&nbsp; is approximated by
 :$$g_r(t) \approx s_0 \cdot T \cdot h_{\rm K}(t)  =  \frac {s_0 \cdot {\rm a}_{\rm \star}/\pi}{ \sqrt{2    \cdot(t/T)^3}}\cdot {\rm e}^{  -{{\rm a}_{\rm \star}^2}/( {2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}t/T}) } \hspace{0.05cm}.$$
-*Because of the good shielding of coaxial cables against other impairments,&nbsp; the&nbsp; [[Aufgaben:Exercise_1.3Z:_Thermal_Noise|$\text{thermal noise}$]]&nbsp;  is the dominant stochastic perturbation.&nbsp; In this case,&nbsp; the signal &nbsp;$n(t)$&nbsp; is Gaussian and white.&nbsp; It can described by the (two-sided) noise power density &nbsp;$N_0/2$.
-*By far the largest noise component arises in the input stage of the receiver, so that it is expedient to add the noise signal &nbsp;$n(t)$&nbsp; at the&nbsp; interface&nbsp; "cable ⇒ receiver".&nbsp;
+$(2)$&nbsp; Because of the good shielding of coaxial cables against other impairments,&nbsp; the&nbsp; [[Aufgaben:Exercise_1.3Z:_Thermal_Noise|&raquo;thermal noise&laquo;]]&nbsp;  is the dominant stochastic perturbation.&nbsp; In this case,&nbsp; the signal &nbsp;$n(t)$&nbsp; is Gaussian and white.&nbsp; It can described by the&nbsp; $($two-sided$)$&nbsp; noise power density &nbsp;$N_0/2$.
-*This noise addition point is also useful because the frequency response &nbsp;$H_{\rm K}(f)$&nbsp; decisively attenuates all noise accumulated along the cable.
-*Then the received signal is with the amplitude coefficients &nbsp;$a_{\nu}$:
+$(3)$&nbsp; By far the largest noise component arises in the input stage of the receiver,&nbsp; so that it is expedient to add the noise signal &nbsp;$n(t)$&nbsp; at the&nbsp; interface&nbsp; &raquo;cable ⇒ receiver&laquo;.&nbsp; This noise addition point is also useful because the frequency response &nbsp;$H_{\rm K}(f)$&nbsp; decisively attenuates all noise accumulated along the cable.
 :$$r(t) = \sum_{\nu = - \infty}^{+ \infty}a_{\nu}\cdot g_r(t - \nu \cdot T)+ n(t) \hspace{0.05cm} .$$

@@ Line 144: / Line 144: @@
 {{BlaueBox|TEXT=
-$\text{Conclusion:}$&nbsp; When simulating and optimizing transmission systems,&nbsp;
+$\text{Conclusion:}$&nbsp; When simulating and optimizing transmission systems,&nbsp; <br>'''one usually omits the linear phase term''' &nbsp;$b_1 = β_1 · f$,&nbsp; since this results exclusively in a&nbsp; $($often not disturbing$)$&nbsp; phase delay,&nbsp; '''but no signal distortions'''.}}
 ==Basic receiver pulse==
 <br>
-With the basic transmission pulse &nbsp;$g_s(t)$ &nbsp; &rArr; &nbsp; "basic  pulse of the transmitted signal&nbsp; $s(t)$"&nbsp; and the impulse response &nbsp;$h_{\rm K}(t)$&nbsp; of the channel,&nbsp; the result for the basic receiver pulse  &nbsp;$g_r(t)$ &nbsp; &rArr; &nbsp; "basic  pulse of the received signal&nbsp; $r(t)$"&nbsp; is:
+With the&nbsp; &raquo;basic transmission pulse&laquo; &nbsp;$g_s(t)$ &nbsp; &rArr; &nbsp; &raquo;basic  pulse of the transmitted signal&laquo;&nbsp; $s(t)$&nbsp; and the&nbsp; &raquo;impulse response &nbsp;$h_{\rm K}(t)$&nbsp; of the channel,&nbsp; the result for the&nbsp; &raquo;basic receiver pulse&laquo; &nbsp;$g_r(t)$ &nbsp; &rArr; &nbsp; &raquo;basic  pulse of the received signal&laquo;&nbsp; $r(t)$&nbsp; is:
 :$$g_r(t) = g_s(t) \star h_{\rm K}(t)\hspace{0.05cm}.$$
-If a non-return-to-zero (NRZ) rectangular pulse &nbsp;$g_s(t)$&nbsp; with amplitude &nbsp;$s_0$&nbsp; and duration &nbsp;$Δt_s = T$ is used at the transmitter,&nbsp; the following results for the basic pulse at the receiver input:
+If a non-return-to-zero&nbsp; $\rm (NRZ)$&nbsp; rectangular pulse &nbsp;$g_s(t)$&nbsp; with amplitude &nbsp;$s_0$&nbsp; and duration &nbsp;$Δt_s = T$&nbsp; is used at the transmitter,&nbsp; the following results for the basic pulse at the receiver input:
 :$$g_r(t) = 2 s_0 \cdot \left [ {\rm Q} \left (\frac {{\rm a}_{\rm \star}/\sqrt {\pi}}{ \sqrt{ (t/T - 0.5)}}\right  ) -
   {\rm Q} \left (\frac {{\rm a}_{\rm \star}/\sqrt {\pi}}{ \sqrt{ (t/T + 0.5)}}\right  ) \right ]\hspace{0.05cm}.$$
-Here &nbsp;$\rm a_∗$&nbsp; denotes the characteristic cable attenuation&nbsp; (in Neper)&nbsp; and  &nbsp;${\rm Q}(x)$&nbsp; the&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|$\text{complementary Gaussian error function}$]].
+Here &nbsp;$\rm a_∗$&nbsp; denotes the&nbsp; &raquo;characteristic cable attenuation&laquo;&nbsp; $($in Neper$)$&nbsp; and  &nbsp;${\rm Q}(x)$&nbsp; the&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|&raquo;complementary Gaussian error function&laquo;]].
 {{GraueBox|TEXT=
 $\text{Example 2:}$&nbsp;
-The figure shows for the characteristic cable attenuations &nbsp;$\rm a_∗ = 40 \ \rm dB$, ... ,&nbsp; $100 \ \rm dB$&nbsp; (smaller&nbsp; $\rm a_∗$&ndash;values &nbsp; are not relevant for practice)
+The figure shows for the characteristic cable attenuations &nbsp;$\rm a_∗ = 40 \ \rm dB$, ... ,&nbsp; $100 \ \rm dB$&nbsp; $($smaller&nbsp; $\rm a_∗$&ndash;values &nbsp; are not relevant for practice$)$
 *the normalized coaxial cable impulse response &nbsp;$T · h_{\rm K}(t)$ &nbsp;  &rArr; &nbsp; solid curves,
-*the basic receiver pulse&nbsp; ("rectangular response")&nbsp; $g_r(t)$&nbsp; normalized to the transmission amplitude &nbsp;$s_0$ &nbsp;  &rArr; &nbsp; dotted line.
 One recognizes the following from this diagram:
-*With &nbsp;$\rm a_∗ = 40 \ \rm dB$,&nbsp; the normalized rectangular response  &nbsp;$g_r(t)/s_0$&nbsp; is slightly&nbsp; (about a factor of $0.95)$&nbsp; smaller at the peak than the normalized impulse response &nbsp;$T · h_{\rm K}(t)$.&nbsp; Here is a small difference between impulse response and basic receiver pulse .
+#With &nbsp;$\rm a_∗ = 40 \ \rm dB$,&nbsp; the normalized basic receiver pulse  &nbsp;$g_r(t)/s_0$&nbsp; is at the peak slightly&nbsp; $($about a factor of $0.95)$&nbsp; smaller than the normalized impulse response &nbsp;$T · h_{\rm K}(t)$.&nbsp; Here is a small difference between impulse response and basic receiver pulse .
-*In contrast,&nbsp; for the case &nbsp;$a_∗ ≥ 60 \ \rm dB$,&nbsp; the rectangular response and the impulse response are indistinguishable within the drawing accuracy.
+#In contrast,&nbsp; for  &nbsp;$a_∗ ≥ 60 \ \rm dB$,&nbsp; the basic receiver pulse and the impulse response are indistinguishable within the drawing accuracy.
-*For a return-to-zero&nbsp; (RZ)&nbsp; pulse,&nbsp; the above equation for the basic receiver pulse  would still need to be multiplied by the duty cycle &nbsp;$Δt_s/T$&nbsp;.&nbsp; In this case &nbsp;$g_r(t)/s_0$&nbsp; is smaller than &nbsp;$T · h_{\rm K}(t)$&nbsp; by at least this factor.
+#For a return-to-zero&nbsp; $\rm (RZ)$&nbsp; pulse,&nbsp; the above equation for &nbsp;$g_r(t)$&nbsp;  would still need to be multiplied by the factor &nbsp;$Δt_s/T$.&nbsp; In this case &nbsp;$g_r(t)/s_0$&nbsp; is smaller than &nbsp;$T · h_{\rm K}(t)$&nbsp; by at least this factor.
-*The equation modified in this way is also a good approximation for other basic transmission pulses as long as &nbsp;$\rm a_∗≥ 60 \ \rm dB$&nbsp; is sufficiently large.&nbsp; $Δt_s$&nbsp; then indicates the&nbsp;[[Signal_Representation/Special_Cases_of_Pulses#Gaussian_pulse|$\text{equivalent pulse duration}$]]&nbsp; of &nbsp;$g_r(t)$.}}
+#The equation modified in this way is also a good approximation for other basic transmission pulses as long as &nbsp;$\rm a_∗≥ 60 \ \rm dB$&nbsp; is sufficiently large.&nbsp; $Δt_s$&nbsp; then indicates the&nbsp;[[Signal_Representation/Special_Cases_of_Pulses#Gaussian_pulse|&raquo;equivalent pulse duration&laquo;]]&nbsp; of &nbsp;$g_r(t)$.}}

@@ Line 85: / Line 85: @@
 ==Impulse response of a coaxial cable==
 <br>
-To calculate the impulse response,&nbsp; the first two components of the five attenuation  components of the complex propagation function (per unit length) can be neglected&nbsp; (the reasoning can be found in the previous section).&nbsp; So we start from the following equation:
+To calculate the impulse response,&nbsp; the first two of the in total five&nbsp;   components of the complex propagation function&nbsp; $($per unit length$)$&nbsp; can be neglected&nbsp; $($the reasoning can be found in the previous section$)$.&nbsp; So we start from the following equation:
 :$$\gamma(f)  = \alpha_0 + \alpha_1 \cdot f +  {\rm j} \cdot \beta_1 \cdot f  +\alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot \beta_2 \cdot \sqrt {f} \approx    {\rm j} \cdot \beta_1 \cdot f  +\alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot \beta_2 \cdot \sqrt {f} \hspace{0.05cm}.$$
 Considering
-*the cable length &nbsp;$l$,
+#the cable length &nbsp;$l$,
-*the characteristic cable attenuation &nbsp;$\rm a_∗$&nbsp; and
+#the characteristic cable attenuation &nbsp;$\rm a_∗$,&nbsp; and
-*that &nbsp;$α_2$&nbsp; (in&nbsp;  "Np")&nbsp; and &nbsp;$β_2$&nbsp; (in&nbsp; "rad")&nbsp; are numerically equal,
+#that &nbsp;$α_2$&nbsp; $($in&nbsp;  &raquo;Np&laquo;$)$&nbsp; and &nbsp;$β_2$&nbsp; (in&nbsp; &raquo;rad&laquo;$)$&nbsp; are numerically equal,
-thus applies to the frequency response of the coaxial cable:
+thus applies to the&nbsp; &raquo;frequency response of the coaxial cable&laquo;:
 :$$H_{\rm K}(f)    = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1 f} \cdot {\rm e}^{-{\rm a}_{\rm \star}\hspace{0.05cm}\cdot \hspace{0.05cm} \sqrt{2f/R} }\cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}{\rm a}_{\rm \star}\hspace{0.05cm}\cdot \hspace{0.02cm} \sqrt{2f/R}}= {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1 f} \cdot {\rm e}^{-2{\rm a}_{\rm \star}\hspace{0.03cm}\cdot \hspace{0.03cm} \sqrt{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}f/R}} \hspace{0.05cm}.$$
 The following abbreviations are used here:
 :$$b_1\hspace{0.1cm}{(\rm in }\hspace{0.15cm}{\rm rad)}= \beta_1 \cdot l \hspace{0.05cm}, \hspace{0.8cm} {\rm a}_{\rm \star}\hspace{0.1cm}{(\rm in }\hspace{0.15cm}{\rm Np)}= \alpha_2 \cdot \sqrt {R/2} \cdot l \hspace{0.05cm}.$$
-The time domain display is obtained by applying the&nbsp; [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_second_Fourier_integral|$\text{Fourier inverse transform}$]]&nbsp; and the&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|$\text{Convolution theorem}$]]:
+The time domain representation is obtained by applying the&nbsp; [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_second_Fourier_integral|&raquo;inverse Fourier transform&laquo;]]&nbsp; and the&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|&raquo;convolution theorem&laquo;]]:
 :$$h_{\rm K}(t)  = \mathcal{F}^{-1} \left \{ {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1
   f}\right \} \star\mathcal{F}^{-1} \left \{ {\rm e}^{-2{\rm a}_{\rm \star}\hspace{0.03cm}\cdot \hspace{0.03cm} \sqrt{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}f/R} }\right \} \hspace{0.05cm}.$$
-To be considered here:
+It must be taken into account here:
-*The first term yields the Dirac delta function &nbsp;$δ(t - τ_{\rm P})$ shifted by the phase delay &nbsp;$τ_{\rm P} = b_1/2π$&nbsp;.
+*The first term yields the Dirac delta function &nbsp;$δ(t - τ_{\rm P})$&nbsp; shifted by the phase delay &nbsp;$τ_{\rm P} = b_1/2π$&nbsp;.
 *The second term can be given analytically closed.&nbsp; We write &nbsp;$h_{\rm K}(t + τ_P)$,&nbsp; so that the phase delay &nbsp;$τ_{\rm P}$&nbsp; need not be considered further.
 :$$h_{\rm K}(t + \tau_{\rm P})  = \frac {{\rm a}_{\rm \star}}{\pi \cdot \sqrt{2 \cdot R  \cdot t^3}}\cdot {\rm exp} \left [ -\frac {{\rm a}_{\rm \star}^2}{ {2\pi \cdot R\cdot t}} \right ]\hspace{0.05cm},\hspace{0.2cm}{\rm a}_{\rm \star}\hspace{0.15cm}{\rm in\hspace{0.15cm} Np}\hspace{0.05cm}.$$
-*Since the bit rate &nbsp;$R$&nbsp; also  has already been considered in the definition of the characteristic cable attenuation &nbsp;$\rm a_∗$&nbsp; this equation can be easily represented with the normalized time &nbsp;$t\hspace{0.05cm}' = t/T$&nbsp;:
+*Since the bit rate &nbsp;$R$&nbsp; also  has already been considered in the&nbsp;$\rm a_∗$&nbsp; definition;&nbsp;   this equation can be easily represented with the normalized time &nbsp;$t\hspace{0.05cm}' = t/T$&nbsp;:
 :$$h_{\rm K}(t\hspace{0.05cm}' + \tau_{\rm P}\hspace{0.05cm} ')  = \frac {1}{T} \cdot \frac {{\rm a}_{\rm \star}}{\pi \cdot \sqrt{2   \cdot t\hspace{0.05cm}'\hspace{0.05cm}^3}}\cdot {\rm exp} \left [ -\frac {{\rm a}_{\rm \star}^2}{ {2\pi \cdot t\hspace{0.05cm}'}} \right ]\hspace{0.05cm},\hspace{0.2cm}{\rm a}_{\rm \star}\hspace{0.15cm}{\rm in\hspace{0.15cm} Np}\hspace{0.05cm}.$$
-:Here&nbsp; $T = 1/R$&nbsp; denotes the symbol duration of a binary system and it holds &nbsp;$τ_{\rm P} \hspace{0.05cm}' = τ_{\rm P}/T$.
+:Here&nbsp; $T = 1/R$&nbsp; denotes&nbsp; &raquo;the symbol duration of a binary system&laquo;&nbsp; and it holds &nbsp;$τ_{\rm P} \hspace{0.05cm}' = τ_{\rm P}/T$.
 {{GraueBox|TEXT=
@@ Line 115: / Line 116: @@
 The results of this section are illustrated by the following graph as an example.
 [[File:EN_LZI_4_2_S3_neu.png|right|frame| Impulse response of a coaxial cable with &nbsp;$\rm a_∗ = 60 \ dB$]]
-*The normalized impulse response &nbsp;$T · h_{\rm K}(t)$&nbsp; of a coaxial cable with  &nbsp;$\rm a_∗ = 60 \ dB  \ \ (6.9\  Np)$ is shown.
+#The normalized impulse response &nbsp;$T · h_{\rm K}(t)$&nbsp; of a coaxial cable with  &nbsp;$\rm a_∗ = 60 \ dB  \ \ (6.9\  Np)$ is shown.
-*The attenuation coefficients &nbsp;$α_0$&nbsp; and &nbsp;$α_1$&nbsp; can thus be neglected,&nbsp; as shown in the last section.
+#The attenuation coefficients &nbsp;$α_0$&nbsp; and &nbsp;$α_1$&nbsp; can thus be neglected,&nbsp; as shown in the last section.
-*For the left graph, the parameter &nbsp;$β_1 = 0$&nbsp; was also set.
+#For the left graph, the parameter &nbsp;$β_1 = 0$&nbsp; was also set.
-Because of the parameterization by the coefficient &nbsp;$a_∗$&nbsp; suitable for this purpose  and the normalization of the time to the symbol duration &nbsp;$T$&nbsp; the left curve is equally valid for systems with small or normal coaxial cable,&nbsp; different lengths and different bit rates,&nbsp; for example for a
+Because of the parameterization by the coefficient &nbsp;$\rm a_∗$,&nbsp; and the time normalization to the symbol duration &nbsp;$T$&nbsp; the left curve is equally valid
-*normal coaxial cable&nbsp; $\text{2.6/9.5 mm}$,&nbsp; bit rate &nbsp;$R = 140 \ \rm Mbit/s$,&nbsp; cable length &nbsp;$l = 3 \ \rm km$  &nbsp; ⇒ &nbsp;  system&nbsp; $\rm A$,
+#for systems with small or normal coaxial cable,&nbsp;
-*small coaxial cable&nbsp; $\text{1.2/4.4 mm mm}$,&nbsp; bit rate &nbsp;$R = 35 \ \rm Mbit/s$, cable length &nbsp;$l = 2.8 \ \rm km$  &nbsp; ⇒ &nbsp;  system&nbsp; $\rm B$.
+#different lengths,
 <br clear=all>
-It can be seen that even at this moderate cable attenuation &nbsp;$\rm a_∗ = 60 \ \rm dB$&nbsp; the impulse response already extends over more than &nbsp; $200$&nbsp; symbol durations due to the skin effect &nbsp;$(α_2 = β_2 ≠ 0)$.&nbsp; Since the integral over &nbsp;$h_{\rm K}(t)$&nbsp; is equal to &nbsp;$H_{\rm K}(f = 0) = 1$,&nbsp; the maximum value becomes very small:
+It can be seen in the left&ndash;hand diagram that even at this moderate cable attenuation &nbsp;$\rm a_∗ = 60 \ \rm dB$&nbsp; the impulse response already extends over more than &nbsp; $200$&nbsp; symbol durations due to the skin effect &nbsp;$(α_2 = β_2 ≠ 0)$.&nbsp; Since the integral over &nbsp;$h_{\rm K}(t)$&nbsp; is equal to &nbsp;$H_{\rm K}(f = 0) = 1$,&nbsp; the maximum value becomes very small:
 :$${\rm Max}\big [h_{\rm K}(t)\big ] \approx 0.03.$$
-In the diagram on the right, the effects of the phase parameter &nbsp;$β_1$&nbsp; can be seen.&nbsp; Note the different time scales of the left and the right diagram:
+In the diagram on the right,&nbsp; the effects of the phase parameter &nbsp;$β_1$&nbsp; can be seen.&nbsp; Note the different time scales of the left and the right diagram:
-*For system&nbsp; $\rm A$ &nbsp;$(β_1 = 21.78 \ \rm rad/(km · MHz)$, &nbsp;$T = 7.14\ \rm  ns)$&nbsp; &nbsp;$β_1$&nbsp; leads to a phase delay of
+*For&nbsp; $\text{system A}$ &nbsp;$(β_1 = 21.78 \ \rm rad/(km · MHz)$, &nbsp;$T = 7.14\ \rm  ns)$&nbsp; &nbsp;$β_1$&nbsp; leads to a phase delay of
 :$$\tau_{\rm A}= \frac {\beta_1 \cdot l}{2\pi} =\frac {21.78\, { {\rm rad} }/{ {(\rm km \cdot MHz)} }\cdot 3\,{\rm km} }{2\pi} = 10.4\,{\rm \mu s}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\tau_{\rm A}\hspace{0.05cm}' = {\tau_{\rm A} }/{T} \approx 1457\hspace{0.05cm}.$$
-*On the other hand,&nbsp; the following can be obtained for system&nbsp; $\rm B$ &nbsp;$(β_1 = 22.18 \ \rm  rad/(km · MHz)$, &nbsp;$T = 30 \ \rm  ns)$:
+*On the other hand,&nbsp; the following can be obtained for&nbsp; $\text{system B}$ &nbsp;$(β_1 = 22.18 \ \rm  rad/(km · MHz)$, &nbsp;$T = 30 \ \rm  ns)$:
 :$$\tau_{\rm B}= \frac {\beta_1 \cdot l}{2\pi} =\frac {22.18\, { {\rm rad} }/{ {(\rm km \cdot MHz)} }\cdot 2.8\,{\rm km} }{2\pi} = 9.9\,{\rm &micro; s}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\tau_{\rm B}\hspace{0.05cm}' ={\tau_{\rm B} }/{T} \approx 330\hspace{0.05cm}.$$
@@ Line 137: / Line 144: @@
 {{BlaueBox|TEXT=
-$\text{Conclusion:}$&nbsp; When simulating and optimizing communication systems,&nbsp;
+$\text{Conclusion:}$&nbsp; When simulating and optimizing transmission systems,&nbsp;
 *one usually omits the phase term with &nbsp;$b_1 = β_1 · l$,&nbsp;
 *since this results exclusively in a&nbsp; (often not disturbing)&nbsp; phase delay,&nbsp; but no signal distortion.}}
 ==Basic receiver pulse==

@@ Line 169: / Line 169: @@
 Assuming binary transmission with non-return-to-zero&nbsp; (NRZ)&nbsp; rectangular pulses&nbsp; $($symbol duration $T)$&nbsp; and a coaxial transmission channel,&nbsp; the following system model is obtained.&nbsp; In particular, it should be noted:
-[[File:EN_LZI_4_2_S5.png |right|frame| Binary transmission system with coaxial cable '''KORREKTUR: receiver filter, noise faults'''<br><br><br>]]
+[[File:EN_LZI_4_2_S5_v2.png |right|frame| Binary transmission system with coaxial cable<br><br><br>]]
 *In a simulation, the phase delay time of the coaxial cable is conveniently left out of consideration.&nbsp; Then the basic receiver pulse  &nbsp;$g_r(t)$&nbsp; (with ${\rm a}_{\rm \star}$&nbsp;in Neper)&nbsp; is approximated by

@@ Line 171: / Line 171: @@
 [[File:EN_LZI_4_2_S5.png |right|frame| Binary transmission system with coaxial cable '''KORREKTUR: receiver filter, noise faults'''<br><br><br>]]
-*In a simulation, the phase delay time of the coaxial cable is conveniently left out of consideration.&nbsp; Then the basic reception pulse '''KORREKTUR: received/receiver pulse''' &nbsp;$g_r(t)$&nbsp; (with ${\rm a}_{\rm \star}$&nbsp;in Neper)&nbsp; is approximated by
+*In a simulation, the phase delay time of the coaxial cable is conveniently left out of consideration.&nbsp; Then the basic receiver pulse  &nbsp;$g_r(t)$&nbsp; (with ${\rm a}_{\rm \star}$&nbsp;in Neper)&nbsp; is approximated by
 :$$g_r(t) \approx s_0 \cdot T \cdot h_{\rm K}(t)  =  \frac {s_0 \cdot {\rm a}_{\rm \star}/\pi}{ \sqrt{2    \cdot(t/T)^3}}\cdot {\rm e}^{  -{{\rm a}_{\rm \star}^2}/( {2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}t/T}) } \hspace{0.05cm}.$$
 *Because of the good shielding of coaxial cables against other impairments,&nbsp; the&nbsp; [[Aufgaben:Exercise_1.3Z:_Thermal_Noise|$\text{thermal noise}$]]&nbsp;  is the dominant stochastic perturbation.&nbsp; In this case,&nbsp; the signal &nbsp;$n(t)$&nbsp; is Gaussian and white.&nbsp; It can described by the (two-sided) noise power density &nbsp;$N_0/2$.

@@ Line 33: / Line 33: @@
 <br>
 The graph shows the frequency-dependent attenuation curve for the normal coaxial cable and the small coaxial cable.&nbsp; Shown on the left is the cable attenuation per unit length of the two coaxial cable types in the frequency range up to&nbsp; $\text{500 MHz}$:
-[[File:EN_LZI_4_2_S2_neu.png |right|frame| Attenuation function and characteristic attenuation of coaxial cables <br>'''Korrektur:''' alpha, Np/km]]
+[[File:EN_LZI_4_2_S2_neu.png |right|frame| Attenuation function and characteristic attenuation of coaxial cables]]
 :$${\alpha}_{\rm K}(f)  \hspace{-0.05cm} =  \alpha_0  \hspace{-0.05cm}+ \hspace{-0.05cm} \alpha_1 \cdot f  \hspace{-0.05cm}+ \hspace{-0.05cm} \alpha_2  \hspace{-0.05cm}\cdot  \hspace{-0.05cm}\sqrt {f} \hspace{0.01cm}  \hspace{0.01cm}.$$
 :$$\Rightarrow \hspace{0.3cm}{\rm a}_{\rm K}(f) =\alpha_{\rm K}(f) \cdot l $$
@@ Line 108: / Line 108: @@
 $\text{Example 1:}$&nbsp;
 The results of this section are illustrated by the following graph as an example.
-[[File:EN_LZI_4_2_S3_neu.png|right|frame| Impulse response of a coaxial cable with &nbsp;$\rm a_∗ = 60 \ dB$<br>'''Korrektur:''' Rahmen]]
+[[File:EN_LZI_4_2_S3_neu.png|right|frame| Impulse response of a coaxial cable with &nbsp;$\rm a_∗ = 60 \ dB$]]
 *The normalized impulse response &nbsp;$T · h_{\rm K}(t)$&nbsp; of a coaxial cable with  &nbsp;$\rm a_∗ = 60 \ dB  \ \ (6.9\  Np)$ is shown.
 *The attenuation coefficients &nbsp;$α_0$&nbsp; and &nbsp;$α_1$&nbsp; can thus be neglected,&nbsp; as shown in the last section.