Difference between revisions of "Aufgaben:Exercise 2.1: ACF and PSD with Coding"

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{{quiz-Header|Buchseite=Digitalsignalübertragung/Grundlagen der codierten Übertragung
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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Basics_of_Coded_Transmission
 
}}
 
}}
  
  
  
[[File:P_ID1308__Dig_A_2_1.png|right|frame|Leistungsdichtespektrum bei Codierung]]
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[[File:P_ID1308__Dig_A_2_1.png|right|frame|Power-spectral density with coding]]
Wir betrachten das Digitalsignal
+
We consider the digital signal  $s(t)$,   using the following descriptive quantities:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g_s ( t - \nu \cdot T) \hspace{0.05cm},$$
+
*$a_{\nu}$  are the amplitude coefficients,
wobei wir folgende Beschreibungsgrößen verwenden:
+
*$g_{s}(t)$  indicates the basic transmission pulse,
*$a_{\nu}$ sind die Amplitudenkoeffizienten,
+
*$T$  is the symbol duration  (spacing of the pulses).
*$g_{s}(t)$ gibt den Sendegrundimpuls an,
 
*$T$ ist die Symboldauer (Abstand der Impulse).
 
  
  
Zur Charakterisierung der spektralen Eigenschaften, die sich aufgrund der Codierung und der Impulsformung ergeben, verwendet man unter anderem
+
Then holds:
*die Autokorrelationsfunktion (AKF)
+
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g_s ( t - \nu \cdot T) \hspace{0.05cm}.$$
 +
 
 +
To characterize the spectral properties resulting from the coding and pulse shaping,  one uses,  among other things
 +
*the auto-correlation function  $\rm (ACF)$
 
:$$\varphi_s(\tau) = \sum_{\lambda = -\infty}^{+\infty}{1}/{T} \cdot \varphi_a(\lambda)\cdot \varphi^{^{\bullet}}_{gs}(\tau - \lambda \cdot T)\hspace{0.05cm},$$
 
:$$\varphi_s(\tau) = \sum_{\lambda = -\infty}^{+\infty}{1}/{T} \cdot \varphi_a(\lambda)\cdot \varphi^{^{\bullet}}_{gs}(\tau - \lambda \cdot T)\hspace{0.05cm},$$
*das Leistungsdichtespektrum (LDS)
+
*the power-spectral density   $\rm (PSD)$
 
:$${\it \Phi}_s(f) = {1}/{T} \cdot {\it \Phi}_a(f) \cdot {\it \Phi}^{^{\bullet}}_{gs}(f) \hspace{0.05cm}.$$
 
:$${\it \Phi}_s(f) = {1}/{T} \cdot {\it \Phi}_a(f) \cdot {\it \Phi}^{^{\bullet}}_{gs}(f) \hspace{0.05cm}.$$
  
Hierbei bezeichnet $\varphi_{a}(\lambda)$ die diskrete Autokorrelationsfunktion der Amplitudenkoeffizienten, die mit der spektralen Leistungsdichte $\Phi_{a}(f)$ über die Fouriertransformation zusammenhängt. Für diese gilt somit:
+
Here,  $\varphi_{a}(\lambda)$  denotes the discrete ACF of the amplitude coefficients related to the power-spectral density  ${\it \Phi}_{a}(f)$  via the Fourier transform.  Thus,  for this holds:
 
:$${\it \Phi}_a(f) = \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} \hspace{0.05cm}.$$
 
:$${\it \Phi}_a(f) = \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} \hspace{0.05cm}.$$
Weiterhin sind in obigen Gleichungen die Energie–AKF und das Energiespektrum verwendet:
+
Furthermore,  the energy ACF and energy spectrum are used in above equations:
 
:$$\varphi^{^{\bullet}}_{gs}(\tau) = \int_{-\infty}^{+\infty} g_s ( t ) \cdot g_s ( t + \tau)\,{\rm d} t \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm} {\it \Phi}^{^{\bullet}}_{gs}(f) = |G_s(f)|^2 \hspace{0.05cm}.$$
 
:$$\varphi^{^{\bullet}}_{gs}(\tau) = \int_{-\infty}^{+\infty} g_s ( t ) \cdot g_s ( t + \tau)\,{\rm d} t \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm} {\it \Phi}^{^{\bullet}}_{gs}(f) = |G_s(f)|^2 \hspace{0.05cm}.$$
In der vorliegenden Aufgabe soll für die spektrale Leistungsdichte der Amplitudenkoeffizienten folgender Funktionsverlauf angenommen werden (siehe Grafik):
+
 
 +
 
 +
In the present exercise,  the following function is to be assumed for the power-spectral density of the amplitude coefficients  (see graph):
 
:$${\it \Phi}_a(f) = {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\hspace{0.05cm}.$$
 
:$${\it \Phi}_a(f) = {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\hspace{0.05cm}.$$
Für den Sendegrundimpuls werden folgende Annahmen getroffen:
+
The following assumptions are made for the basic transmission pulse:
*n der Teilfrage (2) sei $g_{s}(t)$ ein NRZ–Rechteckimpuls, so dass eine dreieckförmige Energie–AKF vorliegt, die auf den Bereich $|\tau| ≤ T$ beschränkt ist. Das Maximum ist dabei
+
*In question  '''(2)''',  let  $g_{s}(t)$  be an NRZ rectangular pulse,  so that there is a triangular energy ACF confined to the range  $|\tau| ≤ T$.  The maximum value here is
 
:$$\varphi^{^{\bullet}}_{gs}(\tau = 0) = s_0^2 \cdot T \hspace{0.05cm}.$$
 
:$$\varphi^{^{\bullet}}_{gs}(\tau = 0) = s_0^2 \cdot T \hspace{0.05cm}.$$
*Für die Teilaufgabe (3) soll von einer Wurzel–Nyquist–Charakteristik mit Rolloff–Faktor $r = 0$ ausgegangen werden. In diesem Fall gilt:
+
*For question  '''(3)''',  assume a root-Nyquist characteristic with rolloff factor  $r = 0$.  In this case holds:
:$$|G_s(f)|^2 = \left\{ \begin{array}{c} s_0^2 \cdot T^2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array} \begin{array}{*{20}c} |f| < {1}/({2T}) \hspace{0.05cm}, \\ |f| > {1}/({2T}) \hspace{0.05cm}.\\ \end{array}$$
+
:$$|G_s(f)|^2 = \left\{ \begin{array}{c} s_0^2 \cdot T^2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}} \\ {\rm{for}} \\ \end{array} \begin{array}{*{20}c} |f| < {1}/({2T}) \hspace{0.05cm}, \\ |f| > {1}/({2T}) \hspace{0.05cm}.\\ \end{array}$$
*Für numerische Berechnungen ist stets $s_{0}^{2} = 10 \ \rm mW$ zu verwenden.
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*For numerical calculations,&nbsp; use always &nbsp;$s_{0}^{2} = 10 \ \rm mW$.&nbsp;
 +
 
  
  
''Hinweis:''
 
  
Die Aufgabe gehört zum [[Digitalsignalübertragung/Grundlagen_der_codierten_Übertragung|Grundlagen der codierten Übertragung]] des vorliegenden Buches. Berücksichtigen Sie, dass die Sendeleistung $P_{\rm S}$ gleich der AKF $\varphi_{s}(\tau)$ an der Stelle $\tau = 0$ ist, aber auch als Integral über das LDS $\Phi_{s}(f)$ berechnet werden kann.
+
Notes:
 +
*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Basics_of_Coded_Transmission|"Basics of Coded Transmission"]].
 +
 +
*Consider that the transmit power &nbsp;$P_{\rm S}$&nbsp; is equal to the ACF &nbsp;$\varphi_{s}(\tau)$&nbsp; at the point &nbsp;$\tau = 0$,&nbsp; but can also be calculated as an integral over the PSD &nbsp;${\it \Phi}_{s}(f)$.&nbsp;
  
===Fragebogen===
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===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
  
{Welche diskreten AKF–Werte $\varphi_{a}(\lambda)$ der Amplitudenkoeffizienten ergeben sich? Geben Sie die Zahlenwerte für $\lambda = 0$, $\lambda = 1$ und $\lambda = 2$ ein.
+
{What are the discrete ACF values &nbsp;$\varphi_{a}(\lambda)$&nbsp; of the amplitude coefficients? Enter the numerical values for &nbsp;$\lambda = 0$, &nbsp;$\lambda = 1$&nbsp; and &nbsp;$\lambda = 2$.&nbsp;
 
|type="{}"}
 
|type="{}"}
 
$\varphi_{a}(\lambda = 0)  \ = \ $ { 0.5 3% }
 
$\varphi_{a}(\lambda = 0)  \ = \ $ { 0.5 3% }
$\varphi_{a}(\lambda = 1) \ = \ $ { 0 3% }
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$\varphi_{a}(\lambda = 1) \ = \ $ { 0. }
$\varphi_{a}(\lambda = 2) \ = \ $ { -0.2575--02425 }
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$\varphi_{a}(\lambda = 2) \ = \ $ { -0.2575--0.2425 }
  
{Welche Sendeleistung ergibt sich mit dem NRZ–Sendegrundimpuls?
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{What is the transmit power with the &nbsp;<u>NRZ basic transmission pulse</u>?
 
|type="{}"}
 
|type="{}"}
NRZ–Rechteck: $P_{\rm S} \ = \ $ { 5 3% } $ \ \rm mW$
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$P_{\rm S} \ = \ $ { 5 3% } $ \ \rm mW$
  
{Wie groß ist die Sendeleistung bei Wurzel–Nyquist–Charakteristik $(r = 0)$?
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{What is the transmit power with &nbsp;<u>root-Nyquist characteristic</u> &nbsp;$(r = 0)$?
 
|type="{}"}
 
|type="{}"}
Wurzel–Nyquist: $P_{\rm S} \ = \ $ { 5 3% } $ \ \rm mW$
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$P_{\rm S} \ = \ $ { 5 3% } $ \ \rm mW$
  
  
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</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp;
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'''(1)'''&nbsp; Since&nbsp; ${\it \Phi}_{a}(f)$&nbsp; as a power-spectral density is always real &nbsp;(plus even and positive,&nbsp; but that does not matter here)&nbsp; and the ACF values&nbsp; $\varphi_{a}(\lambda)$&nbsp; are symmetric about&nbsp; $\lambda = 0$,&nbsp; the given equation can be transformed as follows:
'''(2)'''&nbsp;
+
:$${\it \Phi}_a(f) = \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} = \varphi_a(0) + \sum_{\lambda = 1}^{\infty}2 \cdot \varphi_a(\lambda)\cdot\cos ( 2 \pi f \hspace{0.02cm} \lambda T) \hspace{0.05cm}.$$
'''(3)'''&nbsp;
+
*By comparison with the sketched function
'''(4)'''&nbsp;
+
:$${\it \Phi}_a(f) = {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\hspace{0.05cm}.$$
'''(5)'''&nbsp;
+
:one obtains:
'''(6)'''&nbsp;
+
:$${\it \varphi}_a(\lambda = 0)\hspace{0.15cm}\underline { = 0.5}, \hspace{0.2cm} {\it \varphi}_a(\lambda = 2) = {\it \varphi}_a(\lambda = -2) \hspace{0.15cm}\underline {= -0.25} \hspace{0.05cm}.$$
 +
*All other ACF values result to zero,&nbsp; so also&nbsp; $\varphi_{a}(\lambda = ±1)\hspace{0.15cm}\underline {=0}$.
 +
 
 +
 
 +
 
 +
'''(2)'''&nbsp; For the rectangular NRZ basic pulse,&nbsp; due to the limitation of the energy ACF to the range $|\tau| ≤ T$,&nbsp; we obtain:
 +
:$$P_{\rm S} = \varphi_s(\tau = 0) = \frac{1}{T} \cdot \varphi_a(\lambda = 0)\cdot \varphi^{^{\bullet}}_{gs}(\tau = 0)= \frac{1}{T} \cdot \frac{1}{2} \cdot s_0^2 \cdot T = \frac{s_0^2}{2} \hspace{0.15cm}\underline {= 5\,\,{\rm mW}}\hspace{0.05cm}.$$
 +
 
 +
 
 +
'''(3)'''&nbsp; For rectangular spectral function,&nbsp; it is more convenient to calculate the transmit power by integration over the power-spectral density:
 +
:$$P_{\rm S}  = \ \int_{-1/(2T)}^{+1/(2T)} {\it \Phi}_s(f) \,{\rm d} f = \frac{1}{T} \cdot \int_{-1/(2T)}^{+1/(2T)} {\it \Phi}_a(f) \cdot {\it \Phi}^{^{\bullet}}_{gs}(f) \,{\rm d} f$$
 +
:$$\Rightarrow\hspace{0.3cm}P_{\rm S} = \ \frac{1}{T} \cdot \left [ s_0^2 \cdot T^2 \right ] \cdot \int_{-1/(2T)}^{+1/(2T)} \left( {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\right ) \,{\rm d} f\hspace{0.05cm} = {s_0^2}/{2}\hspace{0.15cm}\underline { = 5\,\,{\rm mW}} .$$
 +
*Here it is considered that the energy PSD&nbsp; $|G_{s}(f)|^{2}$&nbsp; is given as constant&nbsp; (within the integration interval)&nbsp; and thus can be drawn in front of the integral.
 +
*In spite of a completely different signal form&nbsp; $s(t)$,&nbsp; the same transmit power results here,&nbsp; since the integral yields the value $1/(2T)$.
 +
*It should be noted that this simple calculation is only possible for the rolloff factor $r = 0$.
 +
 
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
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[[Category:Aufgaben zu Digitalsignalübertragung|^2.1 Codierte Übertragung - Grundlagen^]]
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[[Category:Digital Signal Transmission: Exercises|^2.1 Basics of Coded Transmission^]]

Latest revision as of 15:23, 23 May 2022



Power-spectral density with coding

We consider the digital signal  $s(t)$,  using the following descriptive quantities:

  • $a_{\nu}$  are the amplitude coefficients,
  • $g_{s}(t)$  indicates the basic transmission pulse,
  • $T$  is the symbol duration  (spacing of the pulses).


Then holds:

$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g_s ( t - \nu \cdot T) \hspace{0.05cm}.$$

To characterize the spectral properties resulting from the coding and pulse shaping,  one uses,  among other things

  • the auto-correlation function  $\rm (ACF)$
$$\varphi_s(\tau) = \sum_{\lambda = -\infty}^{+\infty}{1}/{T} \cdot \varphi_a(\lambda)\cdot \varphi^{^{\bullet}}_{gs}(\tau - \lambda \cdot T)\hspace{0.05cm},$$
  • the power-spectral density  $\rm (PSD)$
$${\it \Phi}_s(f) = {1}/{T} \cdot {\it \Phi}_a(f) \cdot {\it \Phi}^{^{\bullet}}_{gs}(f) \hspace{0.05cm}.$$

Here,  $\varphi_{a}(\lambda)$  denotes the discrete ACF of the amplitude coefficients related to the power-spectral density  ${\it \Phi}_{a}(f)$  via the Fourier transform.  Thus,  for this holds:

$${\it \Phi}_a(f) = \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} \hspace{0.05cm}.$$

Furthermore,  the energy ACF and energy spectrum are used in above equations:

$$\varphi^{^{\bullet}}_{gs}(\tau) = \int_{-\infty}^{+\infty} g_s ( t ) \cdot g_s ( t + \tau)\,{\rm d} t \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm} {\it \Phi}^{^{\bullet}}_{gs}(f) = |G_s(f)|^2 \hspace{0.05cm}.$$


In the present exercise,  the following function is to be assumed for the power-spectral density of the amplitude coefficients  (see graph):

$${\it \Phi}_a(f) = {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\hspace{0.05cm}.$$

The following assumptions are made for the basic transmission pulse:

  • In question  (2),  let  $g_{s}(t)$  be an NRZ rectangular pulse,  so that there is a triangular energy ACF confined to the range  $|\tau| ≤ T$.  The maximum value here is
$$\varphi^{^{\bullet}}_{gs}(\tau = 0) = s_0^2 \cdot T \hspace{0.05cm}.$$
  • For question  (3),  assume a root-Nyquist characteristic with rolloff factor  $r = 0$.  In this case holds:
$$|G_s(f)|^2 = \left\{ \begin{array}{c} s_0^2 \cdot T^2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}} \\ {\rm{for}} \\ \end{array} \begin{array}{*{20}c} |f| < {1}/({2T}) \hspace{0.05cm}, \\ |f| > {1}/({2T}) \hspace{0.05cm}.\\ \end{array}$$
  • For numerical calculations,  use always  $s_{0}^{2} = 10 \ \rm mW$. 



Notes:

  • Consider that the transmit power  $P_{\rm S}$  is equal to the ACF  $\varphi_{s}(\tau)$  at the point  $\tau = 0$,  but can also be calculated as an integral over the PSD  ${\it \Phi}_{s}(f)$. 

Questions

1

What are the discrete ACF values  $\varphi_{a}(\lambda)$  of the amplitude coefficients? Enter the numerical values for  $\lambda = 0$,  $\lambda = 1$  and  $\lambda = 2$. 

$\varphi_{a}(\lambda = 0) \ = \ $

$\varphi_{a}(\lambda = 1) \ = \ $

$\varphi_{a}(\lambda = 2) \ = \ $

2

What is the transmit power with the  NRZ basic transmission pulse?

$P_{\rm S} \ = \ $

$ \ \rm mW$

3

What is the transmit power with  root-Nyquist characteristic  $(r = 0)$?

$P_{\rm S} \ = \ $

$ \ \rm mW$


Solution

(1)  Since  ${\it \Phi}_{a}(f)$  as a power-spectral density is always real  (plus even and positive,  but that does not matter here)  and the ACF values  $\varphi_{a}(\lambda)$  are symmetric about  $\lambda = 0$,  the given equation can be transformed as follows:

$${\it \Phi}_a(f) = \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} = \varphi_a(0) + \sum_{\lambda = 1}^{\infty}2 \cdot \varphi_a(\lambda)\cdot\cos ( 2 \pi f \hspace{0.02cm} \lambda T) \hspace{0.05cm}.$$
  • By comparison with the sketched function
$${\it \Phi}_a(f) = {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\hspace{0.05cm}.$$
one obtains:
$${\it \varphi}_a(\lambda = 0)\hspace{0.15cm}\underline { = 0.5}, \hspace{0.2cm} {\it \varphi}_a(\lambda = 2) = {\it \varphi}_a(\lambda = -2) \hspace{0.15cm}\underline {= -0.25} \hspace{0.05cm}.$$
  • All other ACF values result to zero,  so also  $\varphi_{a}(\lambda = ±1)\hspace{0.15cm}\underline {=0}$.


(2)  For the rectangular NRZ basic pulse,  due to the limitation of the energy ACF to the range $|\tau| ≤ T$,  we obtain:

$$P_{\rm S} = \varphi_s(\tau = 0) = \frac{1}{T} \cdot \varphi_a(\lambda = 0)\cdot \varphi^{^{\bullet}}_{gs}(\tau = 0)= \frac{1}{T} \cdot \frac{1}{2} \cdot s_0^2 \cdot T = \frac{s_0^2}{2} \hspace{0.15cm}\underline {= 5\,\,{\rm mW}}\hspace{0.05cm}.$$


(3)  For rectangular spectral function,  it is more convenient to calculate the transmit power by integration over the power-spectral density:

$$P_{\rm S} = \ \int_{-1/(2T)}^{+1/(2T)} {\it \Phi}_s(f) \,{\rm d} f = \frac{1}{T} \cdot \int_{-1/(2T)}^{+1/(2T)} {\it \Phi}_a(f) \cdot {\it \Phi}^{^{\bullet}}_{gs}(f) \,{\rm d} f$$
$$\Rightarrow\hspace{0.3cm}P_{\rm S} = \ \frac{1}{T} \cdot \left [ s_0^2 \cdot T^2 \right ] \cdot \int_{-1/(2T)}^{+1/(2T)} \left( {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\right ) \,{\rm d} f\hspace{0.05cm} = {s_0^2}/{2}\hspace{0.15cm}\underline { = 5\,\,{\rm mW}} .$$
  • Here it is considered that the energy PSD  $|G_{s}(f)|^{2}$  is given as constant  (within the integration interval)  and thus can be drawn in front of the integral.
  • In spite of a completely different signal form  $s(t)$,  the same transmit power results here,  since the integral yields the value $1/(2T)$.
  • It should be noted that this simple calculation is only possible for the rolloff factor $r = 0$.