Difference between revisions of "Aufgaben:Exercise 3.4Z: Trapezoid, Rectangle and Triangle"

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===Solution===
 
===Solution===
 
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'''(1)'''   Die äquivalente Impulsdauer ist  $\Delta t = t_1 + t_2 \;\underline{= 10 \,\text{ms}}$  und der Rolloff-Faktor  $r_t = 2/10 \;\underline{= 0.2}$.
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'''(1)'''   The equivalent pulse duration is  $\Delta t = t_1 + t_2 \;\underline{= 10 \,\text{ms}}$  and the rolloff factor  $r_t = 2/10 \;\underline{= 0.2}$.
  
  
  
'''(2)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 2 und 3</u>:
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'''(2)'''&nbsp; Proposed <u>solutions 2 and 3</u> are correct:
*Der Spektralwert bei&nbsp; $f = 0$&nbsp; beträgt&nbsp; $A \cdot \Delta t = 10 \,\text{mV/Hz}$.  
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*The spectral value at&nbsp; $f = 0$&nbsp; is&nbsp; $A \cdot \Delta t = 10 \,\text{mV/Hz}$.  
*Da&nbsp; ${X(f)}$&nbsp; reell ist und sowohl positive als auch negative Werte annehmen kann, sind nur die zwei Phasenwerte&nbsp; $0$&nbsp; und&nbsp; $\pi$&nbsp; möglich.
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*Since&nbsp; ${X(f)}$&nbsp; is real and can assume both positive and negative values, only the two phase values&nbsp; $0$&nbsp; und&nbsp; $\pi$&nbsp; are possible.
*Nullstellen gibt es aufgrund der ersten si-Funktion bei allen Vielfachen von&nbsp; $1/\Delta t = 100\, \text{Hz}$.  
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*Zeros exist due to the first si-function at all multiples of&nbsp; $1/\Delta t = 100\, \text{Hz}$.  
*Die zweite si-Funktion führt zu Nulldurchgängen im Abstand&nbsp; $1/(r_t \cdot \Delta t) = 500 \,\text{Hz}$. Diese fallen exakt mit den Nullstellen der ersten si-Funktion zusammen.  
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*The second si function leads to zero crossings at intervals of&nbsp; $1/(r_t \cdot \Delta t) = 500 \,\text{Hz}$. These coincide exactly with the zeros of the first si-function.  
  
  
  
'''(3)'''&nbsp;  <u>Alle Lösungsvorschläge</u> sind zutreffend:
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'''(3)'''&nbsp;  <u>All proposed solutions</u> are correct:
*Mit der äquivalenten Impulsdauer&nbsp; $\Delta t = 10 \,\text{ms}$&nbsp; und dem Rolloff-Faktor&nbsp; $r_t = 0$&nbsp; erhält man: &nbsp; $R( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ( {{\rm{\pi }} \cdot \Delta t \cdot f} ).$
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*With the equivalent pulse duration&nbsp; $\Delta t = 10 \,\text{ms}$&nbsp; and the rolloff factor&nbsp; $r_t = 0$&nbsp; one obtains: &nbsp; $R( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ( {{\rm{\pi }} \cdot \Delta t \cdot f} ).$
*Daraus folgt&nbsp; $R( f = 0) = A \cdot \Delta t = X( f = 0).$
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*It follows that&nbsp; $R( f = 0) = A \cdot \Delta t = X( f = 0).$
  
  
  
'''(4)'''&nbsp; Hier sind die <u>Lösungsvorschläge 1 und 3</u> zutreffend:
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'''(4)'''&nbsp; Proposed <u>solutions 2 and 3</u> are correct:
*Beim Dreieckimpuls ist der Rolloff-Faktor&nbsp; $r_t = 1$.  
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*For the triangular pulse, the rolloff factor is&nbsp; $r_t = 1$.  
*Die äquivalente Impulsdauer ist&nbsp; $\Delta t = 10 \,\text{ms}$. Daraus folgt &nbsp; $D( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ^2 ( {{\rm{\pi }} \cdot \Delta t \cdot f} )$&nbsp; und&nbsp; $D( f = 0) = A \cdot \Delta t  = X( f = 0)$.  
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*The equivalent pulse duration is&nbsp; $\Delta t = 10 \,\text{ms}$. It follows that &nbsp; $D( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ^2 ( {{\rm{\pi }} \cdot \Delta t \cdot f} )$&nbsp; and&nbsp; $D( f = 0) = A \cdot \Delta t  = X( f = 0)$.  
*Da&nbsp; ${D(f)}$&nbsp; nicht negativ werden kann, ist die Phase&nbsp; $[{\rm arc} \; {D(f)}]$&nbsp; stets Null. Der Phasenwert&nbsp; $\pi$&nbsp; $(180°)$&nbsp; ist also bei der Dreieckform nicht möglich.  
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*Since&nbsp; ${D(f)}$&nbsp; cannot become negative, the phase&nbsp; $[{\rm arc} \; {D(f)}]$&nbsp; is always zero. The phase value&nbsp; $\pi$&nbsp; $(180°)$&nbsp; is therefore not possible with the triangular form.  
 
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Revision as of 21:31, 23 January 2021

Trapezimpuls und dessen Grenzfälle „Rechteck” und „Dreieck”

Three different pulse shapes are considered. The pulse  ${x(t)}$  is trapezoidal. For  $| t | < t_1 = 4 \,\text{ms}$ the time course is constant equal to  ${A} = 1\, \text{V}$. Afterwards,  ${x(t)}$  drops linearly to the value zero until the time  $t_2 = 6\, \text{ms}$ . With the two derived system quantities, namely

$$\Delta t = t_1 + t_2$$
  • and the so-called roll-off factor (in the time domain)
$$r_t = \frac{t_2 - t_1 }{t_2 + t_1 }$$

is the spectral function of the trapezoidal pulse:

$$X( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits}( {{\rm \pi} \cdot \Delta t \cdot f} ) \cdot \hspace{0.1cm}{\mathop{\rm si}\nolimits}( {{\rm \pi}\cdot \Delta t \cdot r_t \cdot f} ).$$

Furthermore, the rectangular momentum  ${r(t)}$  and the triangular momentum  ${d(t)}$  are also shown in the graph, both of which can be interpreted as limiting cases of the trapezoidal momentum  ${x(t)}$ .




Hints:


Questions

1

What is the equivalent impulse duration and the rolloff factor of  ${x(t)}$?

$\Delta t \ = \ $

 $\text{ms}$
$r_t\hspace{0.3cm} = \ $

2

Which statements are true regarding the spectral function  ${X(f)}$ ?

The spectral value at frequency  $f = 0$  is equal to  $20 \,\text{mV/Hz}$.
For the phase function the values  $0$  or  $\pi$  $(180^{\circ})$  are possible.
${X(f)}$  only has zeros at all multiples of  $100 \,\text{Hz}$  auf.

3

Which statements are true regarding the spectral function  ${R(f)}$  ?

The spectral value at frequency  $f = 0$  is equal to  ${X(f = 0)}$.
The values $0$ or  $\pi$  $(180^{\circ})$  are possible for the phase function.
${R(f)}$  only has zeros at all multiples of   $100 \,\text{Hz}$ auf..

4

Welche Aussagen sind hinsichtlich der Spektralfunktion  ${D(f)}$  zutreffend?

The spectral value at frequency  $f = 0$  is equal to  ${X(f = 0)}$.
The values $0$ or  $\pi$  $(180^{\circ})$  are possible for the phase function.
${D(f)}$  only has zeros at all multiples of   $100 \,\text{Hz}$ auf.


Solution

(1)  The equivalent pulse duration is  $\Delta t = t_1 + t_2 \;\underline{= 10 \,\text{ms}}$  and the rolloff factor  $r_t = 2/10 \;\underline{= 0.2}$.


(2)  Proposed solutions 2 and 3 are correct:

  • The spectral value at  $f = 0$  is  $A \cdot \Delta t = 10 \,\text{mV/Hz}$.
  • Since  ${X(f)}$  is real and can assume both positive and negative values, only the two phase values  $0$  und  $\pi$  are possible.
  • Zeros exist due to the first si-function at all multiples of  $1/\Delta t = 100\, \text{Hz}$.
  • The second si function leads to zero crossings at intervals of  $1/(r_t \cdot \Delta t) = 500 \,\text{Hz}$. These coincide exactly with the zeros of the first si-function.


(3)  All proposed solutions are correct:

  • With the equivalent pulse duration  $\Delta t = 10 \,\text{ms}$  and the rolloff factor  $r_t = 0$  one obtains:   $R( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ( {{\rm{\pi }} \cdot \Delta t \cdot f} ).$
  • It follows that  $R( f = 0) = A \cdot \Delta t = X( f = 0).$


(4)  Proposed solutions 2 and 3 are correct:

  • For the triangular pulse, the rolloff factor is  $r_t = 1$.
  • The equivalent pulse duration is  $\Delta t = 10 \,\text{ms}$. It follows that   $D( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ^2 ( {{\rm{\pi }} \cdot \Delta t \cdot f} )$  and  $D( f = 0) = A \cdot \Delta t = X( f = 0)$.
  • Since  ${D(f)}$  cannot become negative, the phase  $[{\rm arc} \; {D(f)}]$  is always zero. The phase value  $\pi$  $(180°)$  is therefore not possible with the triangular form.