Aufgaben:Exercise 1.3Z: Calculating with Complex Numbers II - Revision history

Guenter at 12:28, 24 May 2021

2021-05-24T12:28:39Z

Guenter at 16:18, 17 May 2021

2021-05-17T16:18:08Z

Guenter at 11:36, 12 April 2021

2021-04-12T11:36:56Z

Guenter at 15:37, 9 April 2021

2021-04-09T15:37:12Z

Guenter at 15:28, 9 April 2021

2021-04-09T15:28:32Z

Guenter at 15:19, 9 April 2021

2021-04-09T15:19:14Z

Noah at 09:11, 26 March 2021

2021-03-26T09:11:59Z

Javier: Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

2021-03-23T12:38:57Z

Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

Noah at 22:50, 26 December 2020

2020-12-26T22:50:58Z

Noah at 22:43, 26 December 2020

2020-12-26T22:43:02Z

← Older revision		Revision as of 12:28, 24 May 2021
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−	{{quiz-Header\|Buchseite=~~Signaldarstellung~~/~~Prinzip der Nachrichtenübertragung~~}}	+	{{quiz-Header\|Buchseite=Signal_Representation/Calculating_With_Complex_Numbers}}

	[[File:P_ID802__Sig_Z_1_3.png\|right\|frame\|Considered numbers <br>in the complex plane]]		[[File:P_ID802__Sig_Z_1_3.png\|right\|frame\|Considered numbers <br>in the complex plane]]

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Signaldarstellung/Prinzip der Nachrichtenübertragung}}
-[[File:P_ID802__Sig_Z_1_3.png|right|frame|Considered numbers in the complex plane]]
+[[File:P_ID802__Sig_Z_1_3.png|right|frame|Considered numbers <br>in the complex plane]]
 The following three complex quantities are shown in the complex plane to the right:

← Older revision		Revision as of 11:36, 12 April 2021
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−	[[Category:Signal Representation: Exercises\|^1.3 Calculating ~~With~~ Complex Numbers	+	[[Category:Signal Representation: Exercises\|^1.3 Calculating with Complex Numbers
	^]]		^]]

@@ Line 69: / Line 69: @@
 '''(1)'''&nbsp;  The magnitude can be calculated according to the&nbsp; [https://en.wikipedia.org/wiki/Pythagoras Pythagorean ]&nbsp;theorem:
 :$$|z_1| = \sqrt{x_1^2 + y_1^2}= \sqrt{4^2 + 3^2}\hspace{0.15cm}\underline{ = 5}.$$
-*For the phase angle, the following applies according to the page&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_Amplidute_and_Phase|Representation by Amplidute and 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]]:
 :$$\phi_1 = \arctan \frac{y_1}{x_1}= \arctan \frac{3}{4}\hspace{0.15cm}\underline{ = 36.9^{\circ}}.$$

@@ Line 41: / Line 41: @@
-{Calculate&nbsp; $z_5 = x_5 + {\rm j} \cdot y_5$&nbsp;.
+{Calculate&nbsp; $z_5 = z_1 + 2 \cdot z_2 - {z_3}/{2} = x_5 + {\rm j} \cdot y_5$&nbsp;.
 |type="{}"}
 $x_5\ = \ $ { 0. }
@@ Line 69: / Line 69: @@
 '''(1)'''&nbsp;  The magnitude can be calculated according to the&nbsp; [https://en.wikipedia.org/wiki/Pythagoras Pythagorean ]&nbsp;theorem:
 :$$|z_1| = \sqrt{x_1^2 + y_1^2}= \sqrt{4^2 + 3^2}\hspace{0.15cm}\underline{ = 5}.$$
-*For the phase angle, the following applies according to the page&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#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_Amplidute_and_Phase|Representation by Amplidute and Phase]]:
 :$$\phi_1 = \arctan \frac{y_1}{x_1}= \arctan \frac{3}{4}\hspace{0.15cm}\underline{ = 36.9^{\circ}}.$$

@@ Line 7: / Line 7: @@
 : $$ z_2 = -2 ,$$
 : $$z_3 = 6\cdot{\rm j} .$$
-Within the framework of this task, the following quantities are to be calculated::
+Within the framework of this task, the following quantities are to be calculated:
 : $$z_4 = z_1 \cdot z_1^{\star},$$
 : $$z_5 = z_1 + 2 \cdot z_2 - {z_3}/{2},$$
@@ Line 19: / Line 19: @@
 ''Hints:''
-*This exercise belongs to the chapter&nbsp;[[Signal_Representation/Calculating_With_Complex_Numbers|Calculating With Complex Numbers]].
+*This exercise belongs to the chapter&nbsp;[[Signal_Representation/Calculating_With_Complex_Numbers|Calculating with Complex Numbers]].
-*The topic is also covered in the teaching video&nbsp;[[Rechnen_mit_komplexen_Zahlen_(Lernvideo)|Rechnen mit komplexen Zahlen]]&nbsp;.
+*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; $-\hspace{-0.05cm}180^{\circ} < \phi ≤ +180^{\circ}$.

← Older revision		Revision as of 12:38, 23 March 2021
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−	[[Category:~~Exercises for~~ Signal Representation\|^1.3 Calculating With Complex Numbers	+	[[Category:Signal Representation: Exercises\|^1.3 Calculating With Complex Numbers
	^]]		^]]

@@ Line 85: / Line 85: @@
-'''(4)'''&nbsp; Schreibt man&nbsp; $z_2$&nbsp; 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:
 :$$|z_6| = |z_1| \cdot |z_2|= 5 \cdot 2 \hspace{0.15cm}\underline{= 10},$$
 :$$\phi_6 = \phi_1 + \phi_2 = 36.9^{\circ} + 180^{\circ} =
@@ Line 91: / Line 91: @@
-'''(5)'''&nbsp; Die Phase ist&nbsp; $\phi_3 = 90^{\circ}$&nbsp; (siehe Grafik auf der Angabenseite), wie man formal nachweisen kann:
+'''(5)'''&nbsp; The phase is&nbsp; $\phi_3 = 90^{\circ}$&nbsp; (see graph above). This can be formally proven:
 :$$\phi_3 = \arctan \left( \frac{6}{0}\right) = \arctan (\infty)
 \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \phi_3 \hspace{0.15cm}\underline{= 90^{
@@ Line 97: / Line 97: @@
-'''(6)'''&nbsp; Zunächst die umständlichere Lösung:
+'''(6)'''&nbsp; First, the more inconvenient solution:
 :$$z_7 = \frac{z_3}{z_1}= \frac{6{\rm j}}{4 + 3{\rm j}} = \frac{6{\rm j}\cdot(4 - 3{\rm j})}{(4 + 3{\rm j})\cdot (4 - 3{\rm j})} =
    \frac{18 +24{\rm j}}{25} = 1.2 \cdot{\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 53.1^{ \circ}}.$$
-*Ein anderer Lösungsweg lautet:
+*An easier way of solving the problem is:
 :$$|z_7| = \frac{|z_3|}{|z_1|} = \frac{6}{5}\hspace{0.15cm}\underline{=1.2}, \hspace{0.3cm}\phi_7 = \phi_3 - \phi_1 = 90^{\circ} - 36.9^{\circ}
 \hspace{0.15cm}\underline{=53.1^{\circ}}.$$

@@ Line 32: / Line 32: @@
 |type="{}"}
 $|z_1|\ = \ ${ 5 3% }
-$\phi_1\ = \ $ { 36.9 3% } $\hspace{0.2cm}\text{Grad}$
+$\phi_1\ = \ $ { 36.9 3% } $\hspace{0.2cm}\text{deg}$
@@ Line 47: / Line 47: @@
-{Geben Sie&nbsp; $z_6 = z_1 \cdot z_2$&nbsp; nach Betrag und Phase an &nbsp; $($im Bereich&nbsp; $\pm 180^{\circ})$.
+{Specify the magnitude and phase of&nbsp; $z_6 = z_1 \cdot z_2$&nbsp; &nbsp; $($range&nbsp; $\pm 180^{\circ})$.
 |type="{}"}
 $|z_6|\ = \ $ { 10 3% }
-$\phi_6\ = \ $  { -145--140 } $\hspace{0.2cm}\text{Grad}$
+$\phi_6\ = \ $  { -145--140 } $\hspace{0.2cm}\text{deg}$
-{Welchen Phasenwert besitzt die rein imaginäre Zahl&nbsp; $z_3$?
+{What is the phase value of the purely imaginary number&nbsp; $z_3$?
 |type="{}"}
-$\phi_3 \ = \ $ { 90 3% } $\hspace{0.2cm}\text{Grad}$
+$\phi_3 \ = \ $ { 90 3% } $\hspace{0.2cm}\text{deg}$
-{Berechnen Sie&nbsp; $z_7 = z_3/z_1$&nbsp; nach Betrag und Phase&nbsp; $($im Bereich&nbsp; $\pm 180^{\circ})$.
+{Calculate the magnitude and phase of&nbsp; $z_7 = z_3/z_1$&nbsp; &nbsp; $($range&nbsp; $\pm 180^{\circ})$.
 |type="{}"}
 $|z_7| \ = \ $ { 1.2 3% }
-$\phi_7 \ = \ $ { 53.1 3% } $\hspace{0.2cm}\text{Grad}$
+$\phi_7 \ = \ $ { 53.1 3% } $\hspace{0.2cm}\text{deg}$
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Der Betrag kann nach dem Satz von&nbsp; [https://de.wikipedia.org/wiki/Pythagoras Pythagoras]&nbsp; berechnet werden:
+'''(1)'''&nbsp;  The magnitude can be calculated according to the&nbsp; [https://en.wikipedia.org/wiki/Pythagoras Pythagorean ]&nbsp;theorem:
 :$$|z_1| = \sqrt{x_1^2 + y_1^2}= \sqrt{4^2 + 3^2}\hspace{0.15cm}\underline{ = 5}.$$
-*Für den Phasenwinkel gilt entsprechend der Seite&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#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#Darstellung_nach_Betrag_und_Phase|Darstellung nach Betrag und Phase]] :
 :$$\phi_1 = \arctan \frac{y_1}{x_1}= \arctan \frac{3}{4}\hspace{0.15cm}\underline{ = 36.9^{\circ}}.$$
-'''(2)'''&nbsp; Die Multiplikation von&nbsp; $z_1$&nbsp; mit deren Konjugiert-Komplexen&nbsp; $z_1^{\star}$&nbsp; ergibt die rein reelle Größe&nbsp; $z_4$, 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:
 :$$z_4 = (x_1 + {\rm j} \cdot y_1)(x_1 - {\rm j} \cdot y_1)= {x_1^2 +
 y_1^2}= |z_1|^2 = 25,$$
@@ Line 80: / Line 80: @@
-'''(3)'''&nbsp; Aufgeteilt nach Real- und Imaginärteil kann geschrieben werden:
+'''(3)'''&nbsp; By dividing into real and imaginary part one can write:
 :$$x_5 = x_1 + 2 \cdot x_2 - {x_3}/{2} = 4 + 2 \cdot(-2) -0 \hspace{0.15cm}\underline{= 0},$$
 :$$y_5 = y_1 + 2 \cdot y_2 - {y_3}/{2} = 3 + 2 \cdot 0 - \frac{6}{2} \hspace{0.1cm}\underline{=0}.$$