Difference between revisions of "Aufgaben:Exercise 1.3Z: Calculating with Complex Numbers II"
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| − | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Signal_Representation/Calculating_With_Complex_Numbers}} |
| − | |||
| − | [[File:P_ID802__Sig_Z_1_3.png|right| | + | [[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: | |
| − | : z1=4+3j, | + | : $$z_1 = 4 + 3\cdot {\rm j},$$ |
: z2=−2, | : z2=−2, | ||
| − | : z3=6j. | + | : $$z_3 = 6\cdot{\rm j} .$$ |
| − | + | Within the framework of this task, the following quantities are to be calculated: | |
: z4=z1⋅z⋆1, | : z4=z1⋅z⋆1, | ||
| − | : $$z_5 = z_1 + 2 \cdot z_2 - | + | : $$z_5 = z_1 + 2 \cdot z_2 - {z_3}/{2},$$ |
: z6=z1⋅z2, | : z6=z1⋅z2, | ||
| − | : $$z_7 = | + | : $$z_7 = {z_3}/{z_1}.$$ |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| + | ''Hints:'' | ||
| + | *This exercise belongs to the chapter [[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> [[Rechnen_mit_komplexen_Zahlen_(Lernvideo)|Rechnen mit komplexen Zahlen]] ⇒ "Arithmetic operations involving complex numbers". | ||
| + | *Enter the phase values in the range of −180∘<ϕ≤+180∘. | ||
| + | |||
| − | === | + | |
| + | |||
| + | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
| − | { | + | {Enter the magnitude and phase of z1 . |
|type="{}"} | |type="{}"} | ||
| − | |z1| | + | $|z_1|\ = \ ${ 5 3% } |
| − | ϕ1 | + | $\phi_1\ = \ { 36.9 3% }\hspace{0.2cm}\text{deg}$ |
| − | { | + | {What is z4=z1⋅z⋆1=x4+j⋅y4? |
|type="{}"} | |type="{}"} | ||
| − | x4 | + | $x_4\ = \ $ { 25 3% } |
| − | y4 | + | $y_4\ = \ $ { 0. } |
| − | { | + | {Calculate $z_5 = z_1 + 2 \cdot z_2 - {z_3}/{2} = x_5 + {\rm j} \cdot y_5$ . |
|type="{}"} | |type="{}"} | ||
| − | x5 | + | $x_5\ = \ $ { 0. } |
| − | y5 | + | $y_5\ = \ $ { 0. } |
| − | { | + | {Specify the magnitude and phase of z6=z1⋅z2 $($range $\pm 180^{\circ})$. |
|type="{}"} | |type="{}"} | ||
| − | |z6| | + | $|z_6|\ = \ $ { 10 3% } |
| − | $\phi_6 | + | $\phi_6\ = \ $ { -145--140 } $\hspace{0.2cm}\text{deg}$ |
| − | { | + | {What is the phase value of the purely imaginary number z3? |
|type="{}"} | |type="{}"} | ||
| − | ϕ3 | + | $\phi_3 \ = \ { 90 3% }\hspace{0.2cm}\text{deg}$ |
| − | { | + | {Calculate the magnitude and phase of z7=z3/z1 $($range $\pm 180^{\circ})$. |
|type="{}"} | |type="{}"} | ||
| − | |z7| | + | $|z_7| \ = \ $ { 1.2 3% } |
| − | ϕ7 | + | $\phi_7 \ = \ { 53.1 3% }\hspace{0.2cm}\text{deg}$ |
| − | |||
| − | |||
</quiz> | </quiz> | ||
| − | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''1 | + | '''(1)''' The magnitude can be calculated according to the [https://en.wikipedia.org/wiki/Pythagoras Pythagorean ] theorem: |
:|z1|=√x21+y21=√42+32=5_. | :|z1|=√x21+y21=√42+32=5_. | ||
| − | + | *For the phase angle, the following applies according to the page [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_Magnitude_and_Phase|Representation by Magnitude and Phase]]: | |
:ϕ1=arctany1x1=arctan34=36.9∘_. | :ϕ1=arctany1x1=arctan34=36.9∘_. | ||
| − | '''2 | + | |
| + | |||
| + | '''(2)''' Multiplying z1 by its conjugate complex z⋆1 yields the purely real quantity z4, 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,$$ | y_1^2}= |z_1|^2 = 25,$$ | ||
| − | :$$z_4 = |z_1| \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}\phi_1} \cdot |z_1| \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} \phi_1}= |z_1|^2 = 25 | + | :$$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}.$$ | |
| − | '''3 | + | |
| − | :$$x_5 = x_1 + 2 \cdot x_2 - | + | |
| − | :$$y_5 = y_1 + 2 \cdot y_2 - | + | '''(3)''' By dividing into real and imaginary part one can write: |
| − | '''4 | + | :$$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}.$$ | ||
| + | |||
| + | |||
| + | '''(4)''' If one writes z2 as magnitude and phase ⇒ $|z_2| = 2, \ \phi_2 = 180^{\circ}$, one obtains for the product: | ||
:|z6|=|z1|⋅|z2|=5⋅2=10_, | :|z6|=|z1|⋅|z2|=5⋅2=10_, | ||
:$$\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}}.$$ | ||
| − | '''5 | + | |
| − | :$$\ | + | |
| − | \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \ | + | '''(5)''' The phase is ϕ3=90∘ (see graph above). This can be formally proven: |
| + | :$$\phi_3 = \arctan \left( \frac{6}{0}\right) = \arctan (\infty) | ||
| + | \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \phi_3 \hspace{0.15cm}\underline{= 90^{ | ||
\circ}}.$$ | \circ}}.$$ | ||
| − | '''6 | + | |
| + | |||
| + | '''(6)''' First, the more inconvenient solution: | ||
:$$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})} = | :$$z_7 = \frac{z_3}{z_1}= \frac{6{\rm j}}{4 + 3{\rm j}} = \frac{6{\rm j}\cdot(4 - 3{\rm j})}{(4 + 3{\rm j})\cdot (4 - 3{\rm j})} = | ||
| − | \frac{18 +24{\rm j}}{25} = 1.2 \cdot{\rm e}^{{\rm j} 53.1^{ \circ}}.$$ | + | \frac{18 +24{\rm j}}{25} = 1.2 \cdot{\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 53.1^{ \circ}}.$$ |
| − | + | *An easier way of solving the problem is: | |
:$$|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: | + | [[Category:Signal Representation: Exercises|^1.3 Calculating with Complex Numbers |
| + | ^]] | ||
Latest revision as of 14:28, 24 May 2021
Considered numbers
in the complex plane
in the complex plane
The following three complex quantities are shown in the complex plane to the right:
- z1=4+3⋅j,
- z2=−2,
- z3=6⋅j.
Within the framework of this task, the following quantities are to be calculated:
- z4=z1⋅z⋆1,
- z5=z1+2⋅z2−z3/2,
- z6=z1⋅z2,
- z7=z3/z1.
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 −180∘<ϕ≤+180∘.
Questions
Solution
(1) The magnitude can be calculated according to the Pythagorean theorem:
- |z1|=√x21+y21=√42+32=5_.
- For the phase angle, the following applies according to the page Representation by Magnitude and Phase:
- ϕ1=arctany1x1=arctan34=36.9∘_.
(2) Multiplying z1 by its conjugate complex z⋆1 yields the purely real quantity z4, as the following equations show:
- z4=(x1+j⋅y1)(x1−j⋅y1)=x21+y21=|z1|2=25,
- z4=|z1|⋅ej⋅ϕ1⋅|z1|⋅e−j⋅ϕ1=|z1|2=25⇒x4=25_,y4=0_.
(3) By dividing into real and imaginary part one can write:
- x5=x1+2⋅x2−x3/2=4+2⋅(−2)−0=0_,
- y5=y1+2⋅y2−y3/2=3+2⋅0−62=0_.
(4) If one writes z2 as magnitude and phase ⇒ |z2|=2, ϕ2=180∘, one obtains for the product:
- |z6|=|z1|⋅|z2|=5⋅2=10_,
- ϕ6=ϕ1+ϕ2=36.9∘+180∘=216.9∘=−143.1∘_.
(5) The phase is ϕ3=90∘ (see graph above). This can be formally proven:
- ϕ3=arctan(60)=arctan(∞)⇒ϕ3=90∘_.
(6) First, the more inconvenient solution:
- z7=z3z1=6j4+3j=6j⋅(4−3j)(4+3j)⋅(4−3j)=18+24j25=1.2⋅ej⋅53.1∘.
- An easier way of solving the problem is:
- |z7|=|z3||z1|=65=1.2_,ϕ7=ϕ3−ϕ1=90∘−36.9∘=53.1∘_.
