Difference between revisions of "Aufgaben:Exercise 1.3Z: Calculating with Complex Numbers II"

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{{quiz-Header|Buchseite=Signaldarstellung/Prinzip der Nachrichtenübertragung}}
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{{quiz-Header|Buchseite=Signal_Representation/Calculating_With_Complex_Numbers}}
  
[[File:P_ID802__Sig_Z_1_3.png|right|frame|Betrachtete Zahlen in der komplexen Ebene]]
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[[File:P_ID802__Sig_Z_1_3.png|right|frame|Considered numbers <br>in the complex plane]]
Ausgegangen wird von drei komplexen Zahlen, die rechts in der komplexen Ebene dargestellt sind:
+
The following three complex quantities are shown in the complex plane to the right:
  
: $$z_1 = 4 + 3{\rm j},$$
+
: $$z_1 = 4 + 3\cdot {\rm j},$$
 
: $$ z_2 = -2 ,$$
 
: $$ z_2 = -2 ,$$
: $$z_3 = 6{\rm j} .$$
+
: $$z_3 = 6\cdot{\rm j} .$$
Im Rahmen dieser Aufgabe sollen berechnet werden:
+
Within the framework of this task, the following quantities are to be calculated:
 
: $$z_4 = z_1 \cdot z_1^{\star},$$
 
: $$z_4 = z_1 \cdot z_1^{\star},$$
 
: $$z_5 = z_1 + 2 \cdot z_2 - {z_3}/{2},$$
 
: $$z_5 = z_1 + 2 \cdot z_2 - {z_3}/{2},$$
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''Hinweise:''
 
*Die Aufgabe gehört zum Kapitel [[Signaldarstellung/Zum_Rechnen_mit_komplexen_Zahlen|Zum Rechnen mit komplexen Zahlen]].
 
*Die Thematik wird auch im Lernvideo [[Rechnen_mit_komplexen_Zahlen_(Lernvideo)|Rechnen mit komplexen Zahlen ]] behandelt.
 
*Geben Sie Phasenwerte stets im Bereich $-\hspace{-0.05cm}180^{\circ} < \phi ≤ +180^{\circ}$ ein.
 
*Sollte die Eingabe des Zahlenwertes &bdquo;0&rdquo; erforderlich sein, so geben Sie bitte &bdquo;0.&rdquo; ein.
 
  
  
  
===Fragebogen===
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''Hints:''
 +
*This exercise belongs to the chapter&nbsp;[[Signal_Representation/Calculating_With_Complex_Numbers|Calculating with Complex Numbers]].
 +
*The topic of this task is also covered in the (German language) learning video <br> &nbsp; &nbsp;  &nbsp;[[Rechnen_mit_komplexen_Zahlen_(Lernvideo)|Rechnen mit komplexen Zahlen]] &nbsp; &rArr; &nbsp; "Arithmetic operations involving complex numbers".
 +
*Enter the phase values in the range of&nbsp; $-\hspace{-0.05cm}180^{\circ} < \phi ≤ +180^{\circ}$.
 +
 +
 
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Geben Sie $z_1$ nach Betrag und Phase an.
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{Enter the magnitude and phase of&nbsp; $z_1$&nbsp;.
 
|type="{}"}
 
|type="{}"}
 
$|z_1|\ = \ ${ 5 3% }
 
$|z_1|\ = \ ${ 5 3% }
$\phi_1\ = \ $ { 36.9 3% } $\hspace{0.1cm}\text{Grad}$
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$\phi_1\ = \ $ { 36.9 3% } $\hspace{0.2cm}\text{deg}$
  
  
{Wie lautet $z_4 = z_1 \cdot z_1^{\star} = x_4 + \text{j} \cdot y_4$?
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{What is&nbsp; $z_4 = z_1 \cdot z_1^{\star} = x_4 + \text{j} \cdot y_4$?
 
|type="{}"}
 
|type="{}"}
 
$x_4\ = \ $ { 25 3% }
 
$x_4\ = \ $ { 25 3% }
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{Berechnen Sie $z_5 = x_5 + {\rm j} \cdot y_5$ entsprechend der Angabenseite.
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{Calculate&nbsp; $z_5 = z_1 + 2 \cdot z_2 - {z_3}/{2} = x_5 + {\rm j} \cdot y_5$&nbsp;.
 
|type="{}"}
 
|type="{}"}
 
$x_5\ = \ $ { 0. }
 
$x_5\ = \ $ { 0. }
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{Geben Sie $z_6 = z_1 \cdot z_2$ nach Betrag und Phase (im Bereich $\pm 180^{\circ}$) an.
+
{Specify the magnitude and phase of&nbsp; $z_6 = z_1 \cdot z_2$&nbsp; &nbsp; $($range&nbsp; $\pm 180^{\circ})$.
 
|type="{}"}
 
|type="{}"}
 
$|z_6|\ = \ $ { 10 3% }
 
$|z_6|\ = \ $ { 10 3% }
$\phi_6\ = \ $  { -145--140 } $\hspace{0.1cm}\text{Grad}$
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$\phi_6\ = \ $  { -145--140 } $\hspace{0.2cm}\text{deg}$
  
  
{Welchen Phasenwert besitzt die rein imaginäre Zahl $z_3$?
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{What is the phase value of the purely imaginary number&nbsp; $z_3$?
 
|type="{}"}
 
|type="{}"}
$\phi_3 \ = \ $ { 90 3% } $\hspace{0.1cm}\text{Grad}$
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$\phi_3 \ = \ $ { 90 3% } $\hspace{0.2cm}\text{deg}$
  
  
{Berechnen Sie $z_7 = z_3/z_1$ nach Betrag und Phase (im Bereich $\pm 180^{\circ}$).
+
{Calculate the magnitude and phase of&nbsp; $z_7 = z_3/z_1$&nbsp; &nbsp; $($range&nbsp; $\pm 180^{\circ})$.
 
|type="{}"}
 
|type="{}"}
 
$|z_7| \ = \ $ { 1.2 3% }
 
$|z_7| \ = \ $ { 1.2 3% }
$\phi_7 \ = \ $ { 53.1 3% } $\hspace{0.1cm}\text{Grad}$
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$\phi_7 \ = \ $ { 53.1 3% } $\hspace{0.2cm}\text{deg}$
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp;  Der Betrag kann nach dem Satz von [https://de.wikipedia.org/wiki/Pythagoras Pythagoras] berechnet werden:
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'''(1)'''&nbsp;  The magnitude can be calculated according to the&nbsp; [https://en.wikipedia.org/wiki/Pythagoras Pythagorean ]&nbsp;theorem:
 
:$$|z_1| = \sqrt{x_1^2 + y_1^2}= \sqrt{4^2 + 3^2}\hspace{0.15cm}\underline{ = 5}.$$
 
:$$|z_1| = \sqrt{x_1^2 + y_1^2}= \sqrt{4^2 + 3^2}\hspace{0.15cm}\underline{ = 5}.$$
Für den Phasenwinkel gilt entsprechend der Seite [[Signaldarstellung/Zum_Rechnen_mit_komplexen_Zahlen#Darstellung_nach_Betrag_und_Phase|Darstellung nach Betrag und Phase]] :
+
*For the phase angle, the following applies according to the page&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_Magnitude_and_Phase|Representation by Magnitude and Phase]]:
 
:$$\phi_1 = \arctan \frac{y_1}{x_1}= \arctan \frac{3}{4}\hspace{0.15cm}\underline{ = 36.9^{\circ}}.$$
 
:$$\phi_1 = \arctan \frac{y_1}{x_1}= \arctan \frac{3}{4}\hspace{0.15cm}\underline{ = 36.9^{\circ}}.$$
  
'''(2)'''&nbsp; Die Multiplikation von $z_1$ mit deren Konjugiert-Komplexen $z_1^{\star}$ ergibt die rein reelle Größe $z_4$, wie die folgenden Gleichungen zeigen:
+
 
 +
'''(2)'''&nbsp; Multiplying&nbsp; $z_1$&nbsp; by its conjugate complex&nbsp; $z_1^{\star}$&nbsp; yields the purely real quantity&nbsp; $z_4$, as the following equations show:
 
:$$z_4 = (x_1 + {\rm j} \cdot y_1)(x_1 - {\rm j} \cdot y_1)= {x_1^2 +
 
:$$z_4 = (x_1 + {\rm j} \cdot y_1)(x_1 - {\rm j} \cdot y_1)= {x_1^2 +
 
y_1^2}= |z_1|^2 = 25,$$
 
y_1^2}= |z_1|^2 = 25,$$
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\Rightarrow\hspace{0.3cm} x_4 \hspace{0.1cm}\underline{=  25}, \hspace{0.25cm}y_4 \hspace{0.15cm}\underline{=  0}.$$
 
\Rightarrow\hspace{0.3cm} x_4 \hspace{0.1cm}\underline{=  25}, \hspace{0.25cm}y_4 \hspace{0.15cm}\underline{=  0}.$$
  
'''(3)'''&nbsp; Aufgeteilt nach Real- und Imaginärteil kann geschrieben werden:
+
 
 +
'''(3)'''&nbsp; By dividing into real and imaginary part one can write:
 
:$$x_5 = x_1 + 2 \cdot x_2 - {x_3}/{2} = 4 + 2 \cdot(-2) -0 \hspace{0.15cm}\underline{= 0},$$
 
:$$x_5 = x_1 + 2 \cdot x_2 - {x_3}/{2} = 4 + 2 \cdot(-2) -0 \hspace{0.15cm}\underline{= 0},$$
 
:$$y_5 = y_1 + 2 \cdot y_2 - {y_3}/{2} = 3 + 2 \cdot 0 - \frac{6}{2} \hspace{0.1cm}\underline{=0}.$$
 
:$$y_5 = y_1 + 2 \cdot y_2 - {y_3}/{2} = 3 + 2 \cdot 0 - \frac{6}{2} \hspace{0.1cm}\underline{=0}.$$
  
'''(4)'''&nbsp; Schreibt man $z_2$ nach Betrag und Phase &nbsp; &rArr; &nbsp; $|z_2| = 2, \phi_2 = 180^{\circ}$, so erhält man für das Produkt:
+
 
 +
'''(4)'''&nbsp; If one writes&nbsp; $z_2$&nbsp; as magnitude and phase&nbsp; &rArr; &nbsp; $|z_2| = 2, \ \phi_2 = 180^{\circ}$, one obtains for the product:
 
:$$|z_6| = |z_1| \cdot |z_2|= 5 \cdot 2 \hspace{0.15cm}\underline{= 10},$$
 
:$$|z_6| = |z_1| \cdot |z_2|= 5 \cdot 2 \hspace{0.15cm}\underline{= 10},$$
 
:$$\phi_6 = \phi_1 + \phi_2 = 36.9^{\circ} + 180^{\circ} =
 
:$$\phi_6 = \phi_1 + \phi_2 = 36.9^{\circ} + 180^{\circ} =
 
216.9^{\circ}\hspace{0.15cm}\underline{= -143.1^{\circ}}.$$
 
216.9^{\circ}\hspace{0.15cm}\underline{= -143.1^{\circ}}.$$
  
'''(5)'''&nbsp; Die Phase ist $\phi_3 = 90^{\circ}$ (siehe Grafik auf der Angabenseite), wie man formal nachweisen kann:
+
 
 +
'''(5)'''&nbsp; The phase is&nbsp; $\phi_3 = 90^{\circ}$&nbsp; (see graph above). This can be formally proven:
 
:$$\phi_3 = \arctan \left( \frac{6}{0}\right) = \arctan (\infty)
 
:$$\phi_3 = \arctan \left( \frac{6}{0}\right) = \arctan (\infty)
 
\hspace{0.2cm}\Rightarrow \hspace{0.2cm} \phi_3 \hspace{0.15cm}\underline{= 90^{
 
\hspace{0.2cm}\Rightarrow \hspace{0.2cm} \phi_3 \hspace{0.15cm}\underline{= 90^{
 
  \circ}}.$$
 
  \circ}}.$$
  
'''(6)'''&nbsp; Zunächst die umständlichere Lösung:
+
 
 +
'''(6)'''&nbsp; First, the more inconvenient solution:
 
:$$z_7 = \frac{z_3}{z_1}= \frac{6{\rm j}}{4 + 3{\rm j}} = \frac{6{\rm j}\cdot(4 - 3{\rm j})}{(4 + 3{\rm j})\cdot (4 - 3{\rm j})} =
 
:$$z_7 = \frac{z_3}{z_1}= \frac{6{\rm j}}{4 + 3{\rm j}} = \frac{6{\rm j}\cdot(4 - 3{\rm j})}{(4 + 3{\rm j})\cdot (4 - 3{\rm j})} =
   \frac{18 +24{\rm j}}{25} = 1.2 \cdot{\rm e}^{{\rm j} 53.1^{ \circ}}.$$
+
   \frac{18 +24{\rm j}}{25} = 1.2 \cdot{\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 53.1^{ \circ}}.$$
Ein anderer Lösungsweg lautet:
+
*An easier way of solving the problem is:
 
:$$|z_7| = \frac{|z_3|}{|z_1|} = \frac{6}{5}\hspace{0.15cm}\underline{=1.2}, \hspace{0.3cm}\phi_7 = \phi_3 - \phi_1 = 90^{\circ} - 36.9^{\circ}
 
:$$|z_7| = \frac{|z_3|}{|z_1|} = \frac{6}{5}\hspace{0.15cm}\underline{=1.2}, \hspace{0.3cm}\phi_7 = \phi_3 - \phi_1 = 90^{\circ} - 36.9^{\circ}
 
\hspace{0.15cm}\underline{=53.1^{\circ}}.$$
 
\hspace{0.15cm}\underline{=53.1^{\circ}}.$$
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[[Category:Aufgaben zu Signaldarstellung|^1. Grundbegriffe der Nachrichtentechnik^]]
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[[Category:Signal Representation: Exercises|^1.3 Calculating with Complex Numbers
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^]]

Latest revision as of 13:28, 24 May 2021

Considered numbers
in the complex plane

The following three complex quantities are shown in the complex plane to the right:

$$z_1 = 4 + 3\cdot {\rm j},$$
$$ z_2 = -2 ,$$
$$z_3 = 6\cdot{\rm j} .$$

Within the framework of this task, the following quantities are to be calculated:

$$z_4 = z_1 \cdot z_1^{\star},$$
$$z_5 = z_1 + 2 \cdot z_2 - {z_3}/{2},$$
$$z_6 = z_1 \cdot z_2,$$
$$z_7 = {z_3}/{z_1}.$$




Hints:

  • This exercise belongs to the chapter Calculating with Complex Numbers.
  • The topic of this task is also covered in the (German language) learning video
         Rechnen mit komplexen Zahlen   ⇒   "Arithmetic operations involving complex numbers".
  • Enter the phase values in the range of  $-\hspace{-0.05cm}180^{\circ} < \phi ≤ +180^{\circ}$.



Questions

1

Enter the magnitude and phase of  $z_1$ .

$|z_1|\ = \ $

$\phi_1\ = \ $

$\hspace{0.2cm}\text{deg}$

2

What is  $z_4 = z_1 \cdot z_1^{\star} = x_4 + \text{j} \cdot y_4$?

$x_4\ = \ $

$y_4\ = \ $

3

Calculate  $z_5 = z_1 + 2 \cdot z_2 - {z_3}/{2} = x_5 + {\rm j} \cdot y_5$ .

$x_5\ = \ $

$y_5\ = \ $

4

Specify the magnitude and phase of  $z_6 = z_1 \cdot z_2$    $($range  $\pm 180^{\circ})$.

$|z_6|\ = \ $

$\phi_6\ = \ $

$\hspace{0.2cm}\text{deg}$

5

What is the phase value of the purely imaginary number  $z_3$?

$\phi_3 \ = \ $

$\hspace{0.2cm}\text{deg}$

6

Calculate the magnitude and phase of  $z_7 = z_3/z_1$    $($range  $\pm 180^{\circ})$.

$|z_7| \ = \ $

$\phi_7 \ = \ $

$\hspace{0.2cm}\text{deg}$


Solution

(1)  The magnitude can be calculated according to the  Pythagorean  theorem:

$$|z_1| = \sqrt{x_1^2 + y_1^2}= \sqrt{4^2 + 3^2}\hspace{0.15cm}\underline{ = 5}.$$
$$\phi_1 = \arctan \frac{y_1}{x_1}= \arctan \frac{3}{4}\hspace{0.15cm}\underline{ = 36.9^{\circ}}.$$


(2)  Multiplying  $z_1$  by its conjugate complex  $z_1^{\star}$  yields the purely real quantity  $z_4$, as the following equations show:

$$z_4 = (x_1 + {\rm j} \cdot y_1)(x_1 - {\rm j} \cdot y_1)= {x_1^2 + y_1^2}= |z_1|^2 = 25,$$
$$z_4 = |z_1| \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}\phi_1} \cdot |z_1| \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \phi_1}= |z_1|^2 = 25\hspace{0.3cm} \Rightarrow\hspace{0.3cm} x_4 \hspace{0.1cm}\underline{= 25}, \hspace{0.25cm}y_4 \hspace{0.15cm}\underline{= 0}.$$


(3)  By dividing into real and imaginary part one can write:

$$x_5 = x_1 + 2 \cdot x_2 - {x_3}/{2} = 4 + 2 \cdot(-2) -0 \hspace{0.15cm}\underline{= 0},$$
$$y_5 = y_1 + 2 \cdot y_2 - {y_3}/{2} = 3 + 2 \cdot 0 - \frac{6}{2} \hspace{0.1cm}\underline{=0}.$$


(4)  If one writes  $z_2$  as magnitude and phase  ⇒   $|z_2| = 2, \ \phi_2 = 180^{\circ}$, one obtains for the product:

$$|z_6| = |z_1| \cdot |z_2|= 5 \cdot 2 \hspace{0.15cm}\underline{= 10},$$
$$\phi_6 = \phi_1 + \phi_2 = 36.9^{\circ} + 180^{\circ} = 216.9^{\circ}\hspace{0.15cm}\underline{= -143.1^{\circ}}.$$


(5)  The phase is  $\phi_3 = 90^{\circ}$  (see graph above). This can be formally proven:

$$\phi_3 = \arctan \left( \frac{6}{0}\right) = \arctan (\infty) \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \phi_3 \hspace{0.15cm}\underline{= 90^{ \circ}}.$$


(6)  First, the more inconvenient solution:

$$z_7 = \frac{z_3}{z_1}= \frac{6{\rm j}}{4 + 3{\rm j}} = \frac{6{\rm j}\cdot(4 - 3{\rm j})}{(4 + 3{\rm j})\cdot (4 - 3{\rm j})} = \frac{18 +24{\rm j}}{25} = 1.2 \cdot{\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 53.1^{ \circ}}.$$
  • An easier way of solving the problem is:
$$|z_7| = \frac{|z_3|}{|z_1|} = \frac{6}{5}\hspace{0.15cm}\underline{=1.2}, \hspace{0.3cm}\phi_7 = \phi_3 - \phi_1 = 90^{\circ} - 36.9^{\circ} \hspace{0.15cm}\underline{=53.1^{\circ}}.$$