SAMPLE CURRICULUM - Cyber.org · 2016. 1. 29. · , find 2(LM× ). Exercise 7 Given N = − 57 11 2 and O = 10 01, find NO× . Exercise 8 Alyssa and Jonathan have completed a math

SAMPLE CURRICULUM

AMES: Matrix Multiplication

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1

Teacher Notes: Contents• Matrix multiplication• Commutativity in matrix multiplication• Distributivity in matrix multiplication• Associativity in matrix multiplication• Identity Matrix

Lesson Information

This material is based upon work supported by the U.S. Department of Homeland Security under Grant Award Number, 2013-PD-127-000001, Modification #2.

Standards• CCSS-M: N-VM.C.8 Add, subtract, and multiply matrices of appropriate dimensions.• CCSS-M: N-VM.C.9 Understand that, unlike multiplication of numbers, matrix multiplication for square matrices is not a commutative operation, but still satisfies the associative and distributive properties.• CCSS-M: N-VM.C.10 Understand that the zero and identity matrices play a role in matrix addition and multiplication similar to the role of 0 and 1 in the real numbers. The determinant of a square matrix is nonzero if and only if the matrix has a multiplicative inverse.

Objectives• Introduce students to matrix multiplication

procedures, as well as helping students determine when matrices are able to be multiplied.

• Students discover whether matrix multiplication is commutative, distributive, or associative.

• Introduce students to the identity matrix for multiplication.

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www.NICERC.orgAMES: Matrix Multiplication


Teacher Notes: MATRIX MULTIPLICATION

As seen in the previous lesson, rows and columns are important when dealing with matrices. Multiplication of matrices is another example of that importance. Here’s the main idea with multiplying - think row times column, like this:

Row 1 Row1 Row1Row 2

Column 1Column1

Column 2Colum

× =

nn2Column2Row2 Row2Column1

Is the pattern apparent? Look closely at the numbers that are part of each position. If the name of the resulting matrix is A, then A11 is synonymous with the row1column1 spot. The numbers of the row and column match the subscript that indicates position.

Multiple Rows and Columns

Thinking more in depth, you may realize that rows and columns could have multiple numbers. So, how do you multiply a string of numbers against another string of numbers? Take the first number in Row 1, multiply it by the first number in Column 1, add it to the second number in the row times the second number in the column, and continue the process for that row and column. This sum will be in the position of the Row # and Column # in the resulting matrix (e.g. A11). Repeat the steps for each position. The visual below may assist in comprehending.

a b a b a bc d c d c d

eg

e ge g

fh

f hf h

+ ++ +

×

=

Let’s try one with numbers.

Exercise 1

LetY =− −

2 3 4 1 2 3 47 1 3 0 5 2 0

and Z =

−

−

03

22023

. What is Z Y× ?

Exercise 1 Solution:

Row 1 × Column 1 = 2(0) + 3(-3) + 4(2) + 1(2) +

2(0) + 3(2) + 4(-3) = -5

Row 2 × Column 1 = 7(0) - 1(-3) + 3(2) +0(2) +

5(0) - 2(2) + 0(-3) = 5

Row1 Column1Row2 Column1

××

=

−

55

*Special note: The final answer is a 2 × 1 matrix,

but it could also be called a vector! A vector

transformed into a matrix is a column matrix

consisting of the vector components.




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3

Ability to Multiply Matrices

Exercise 2

What is 1

179

1272

×

?

In Exercise 2, a 3×1 could not be multiplied by another 3×1. While in Exercise 1, a 1×7 matrix is multiplied against a 7×2 matrix. For the arbitrary example, a 2×2 times 2×2 is shown. Is there a common thread between these?

3 11 3× ×× Can NOT multiply

1 27 7× ×× Can multiply Produces a 1 2× matrix

2 22 2× ×× Can multiply Produces a 2 2× matrix

The column dimension of the first matrix and the row dimension of the second matrix must be the same in order to multiply. Also, the row dimension of the first matrix by the column dimension of the second matrix is the dimension of the matrix produced when multiplying the two matrices.

Multiplication Properties

In Matrix Addition and Subtraction, the commutative, distributive, and associative properties remained as they would in the addition and subtraction of numbers. Is the same true for multiplication?

Exercise 3

Let A =−

7 43 5

and B =−

2 61 8

. Multiply A B× , then multiply B A× .


To begin, look at row 1 times column 1.

Row 1 is 1 and column 1

is 1272

.

Usually, when we multiply matrices, the first number

in the row goes with the first number in the column,

the second number in the row goes with the second

number in the column, and so on and so forth.

However, in this problem, row 1 only has a single number to correspond

with 3 numbers in column 1; therefore, these two

matrices cannot be multiplied.

Teacher Notes:

Exercise 3 Solutions:

A B× =−

−

18 101 58

B A× =−

−

32 2217 44

The two resulting matrices are different, so we learn

that matrix multiplication is not commutative ( A B×

does not always equalB A× ).

For full solution, see page 5.



4



Teacher Notes: Exercise 4

Let C =

10 1 69 2 4

, D =

5012

, and F = − 12 1 . Compute ( )C D F× × and C D F× ×( ) .

Exercise 5

If X =−

12 103 2

and Y =−

1 50 1

, does 5 5 5( )X Y X Y+ = + ?

Let’s look at an interesting case for matrix multiplication in the next exercise, and evaluate what we find.

Exercise 6

If M =−

5 1822 3

and N =

1 00 1

, then what is M N× ?

Check out the exercise above. Matrix M is exactly the same as the final answer. This is because N is a special kind of matrix called an identity matrix – a square matrix consisting of only ones in the diagonal and zeros in every other position. This type of matrix has several special properties (which will be covered more in a following lesson), but as seen in the exercise, multiplying any matrix by the identity results in the original matrix. Let’s look at another example.

Exercise 7

Let C = 2 12 4 and D =

1 0 00 1 00 0 1

. Find C D× .


( )C D F× × =−−

636 53564 47

C D F× ×( ) =−−

636 53564 47

The final answer for each computation was the same;

matrix multiplication is associative.



5( )X Y+ =−

65 2515 5

5 5X Y+ =−

65 2515 5

The third property, distribution, holds when

dealing with matrices.



M N× =−

5 1822 3



C D× = 2 12 4



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Exercise 3 Full Solution:

A B× =+ − +

− + − − +

7 2 4 1 7 6 4 83 2 5 1 3 6 5 8( ) ( ) ( ) ( )( ) ( ) ( ) ( )

=−

−

18 101 58

B A× =+ − − + −+ − +

2 7 6 3 2 4 6 51 7 8 3 1 4 8 5( ) ( ) ( ) ( )( ) ( ) ( ) ( )

=−

−

32 2217 44

The two resulting matrices are different, so we learn that matrix multiplication is not commutative (A×B does not always equal B×A).


1. C D× =+ ++ +

10 5 1 0 69 5 2 0 4

1212

( ) ( ) ( )( ) ( ) ( )

=

5347

( )C D F× × =

× −

5347

12 1 =−−

53 12 53 147 12 47 1

( ) ( )( ) ( )

=−−

636 53564 47

2. D F× =−−−

5 12 5 10 12 0 1

12 112

12

( ) ( )( ) ( )( ) ( )

=−

−

60 50 06 1

2

C D F× ×( ) =

×

−

−

10 1 69 2 4

60 50 06 1

2

=+ + − + + −+ + − +

10 60 1 0 6 6 10 5 1 0 69 60 2 0 4 6 9 5 2

12( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) (00 4 12) ( )+ −

=

−−

636 53564 47

The final answer for each computation was the same; matrix multiplication is associative.


5( )X Y+ =+ − ++ + −

=

−

5

12 1 10 53 0 2 1

513 53 1

=−

65 2515 5

5 5X Y+ =−

+ −

=

−

+ −

5

12 103 2

51 50 1

60 5015 10

5 250 5

=−

65 2515 5

The third property, distribution, holds when dealing with matrices.


M N× =− −+ +

5 1 18 0 5 0 18 122 1 3 0 22 0 3 1

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

=−

5 1822 3


C D× = + + + + + + 2 1 12 0 4 0 2 0 12 1 4 0 2 0 12 0 4 1( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = 2 12 4




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Teacher Notes: Matrix Multiplication Practice Exercises

Exercise 1

Given Matrices A =

1 37 2

and B =

6 43 1

, find A B× and B A× .

Exercise 2

Given Matrices C =−

−

1 87 25 9

and D =−

41

, find C D× .

Exercise 3

Given E = 7 3 4 1 and F =

−

−

2158

, find E F× .

Exercise 4

Given G =−

1 32 4

and H =−

−

7 45 21 9

, find G H× .


A B× =

15 748 30

B A× =

34 2610 11


C D× = −−

123011


E F× = 1


Not possible.



2



Exercise 5

Given I =−

−

3 5 25 8 4

, J =−

2 91 46 3

, and K =−

31

, find I J K× ×( ) and ( )I J K× × .

Exercise 6

Given L =−

−

5 12 4

and M =−

−

5 12 4

, find 2( )L M× .

Exercise 7

Given N =−

5 711 2

and O =

1 00 1

, find N O× .

Exercise 8

Alyssa and Jonathan have completed a math class consisting of two tests and a final. Alyssa made a 90 on the first test, a 95 on the second test, and a 100 on the final. Jonathan made a 75 on the first test, a 100 on the second test, and an 85 on the final. Each test is worth 25 percent of the grade for the class, and the final is worth 50 percent of the grade. Using matrix multiplication, determine what grades Alyssa and Jonathan made in the class. (The matrices have been set up for you below.)

90 95 10075 100 85

252550

×

.

.

.

Teacher Notes:


J K× = −

151

15

I J K× ×( ) =−

20143

I J× =−

11 5326 65

( )I J K× × =−

20143


L M× =−

−

27 918 18

2( )L M× =−

−

54 1836 36


N O× =−

5 711 2


AlyssaJonathan

=

96 2586 25

.

.

Therefore, Alyssa finished the class with a 96.25, and

Jonathan finished the class with an 86.25.


AMES: Applications of Complex Numbers


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1

Notes:

Contents• DC Voltage Simulation• AC Voltage Simulation• Complex Numbers and Alternating Current• Basic Circuit Diagrams• Kirchhoff ’s Voltage Law (KVL)• Ohm’s Law

Lesson Information

This material is based upon work supported by the U.S. Department of Homeland Security under Grant Award Number, 2013-PD-127-000001, Modification #2.

Materials Per Student• 1 computer with access to https://goo.gl/gmEPq5In an ideal environment,

each student would have their own computer.

However, students could easily share or the instructor can show the

simulation to the students via a projector.


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www.NICERC.orgAMES: Applications of Complex Numbers


Teacher Notes:

Many people may think that the applications of complex numbers, like their name suggests, are imaginary, but the applications of complex numbers are real and can significantly simplify many problems. The practical applications of complex numbers exist in many fields, including physics, biology, economics, chemistry, engineering, statistics, and mathematics. However, the majority of these applications require the use of calculus. One application that does not use calculus involves circuits.

To begin learning about complex numbers in circuits, we will use the PhET Interactive Simulation called “Circuit Construction Kit (AC+DC), Virtual Lab.” To access this simulation, visit https://goo.gl/gmEPq5 and click “Run Now!” Once the simulation is running, it will look similar to the image below.

The first thing we will do is pause the simulation using the double bar pause icon at the bottom of the window. In the image above, the simulation is paused.

INTRODUCTION

Another option is to use the physical items to

actually build the circuit. This would require light

bulbs, a voltmeter (or multimeter), an AC/

DC power supply, and wires to connect the

previous component. For safety reasons, this is not recommended if the

instructors/facilitators do not have prior experience

with AC/DC power supplies.


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Before we build a circuit with DC voltage, answer the following questions.

1. What does DC stand for?

2. What are common sources of DC voltage?

Now let’s build a DC circuit with two wires, a light bulb, and a battery. This can be done by dragging and dropping the components from the right side of the simulation to the blue area. The completed circuit will resemble the image below.

Once the circuit is built, click the play button. The electrons (blue spheres) will begin to flow around the circuit and the light bulb will light up. Now we can measure the DC voltage over the entire circuit by using the voltmeter in the “Tools” section of the simulator. Click the check box next to “Voltmeter” and it will appear near the circuit.

To measure the voltage across this circuit, place the red lead of the voltmeter on one of the wires and place the black lead on the other wire. Answer the following questions .

1. How much voltage was measured?

2. In terms of voltage, what kind of battery is being used in this DC circuit?

3. What is the magnitude of the measured voltage?

Pause the simulation, remove the voltmeter, and add another battery to the circuit to reflect the circuit below.

DC VOLTAGE

Solutions:

1. Direct current.

2. Batteries, Solar Cells

DC stands for direct current but when DC is

used as an adjective, it is understood to just mean

direct (e.g. DC voltage means direct voltage).

Solutions:

1. -9V or 9V

2. A 9 Volt battery.

3. 9 Volts



4



Resume the simulation with the play button and measure the voltage again by placing one of the voltmeter’s leads on one wire and the other lead on the other wire. Answer the following questions.

1. How much voltage was measured?

2. What was the magnitude of the measured voltage?

3. How much larger/smaller was the magnitude of the voltage?

4. Since the voltage has been measured with two batteries, make a hypothesis about what would happen if a 3rd battery was added or a 4th battery.

5. Test the hypothesis. Is it true?

6. What conclusions can be made about how to add voltage/power sources in a DC circuit?

Before building an AC circuit, answer the following questions.

1. What does AC stand for (in the context of electricity)?

2. What are common uses for AC?

If the simulation is not paused, be sure to do so before continuing. Remove the voltmeter and all DC power sources from the circuit. Once the circuit no longer has a power supply, connect an AC voltage source. The circuit should look similar to the circuit pictured below.

Now press the play button and then use the voltmeter to measure the voltage of the power source. This can be done by placing the voltmeter leads as shown in the image below.

Solutions:

1. -18V or 18V

2. 18 Volts

3. 9 Volts

For 4, 5, and 6, answers may vary. Ultimately, the students should observe

that DC voltage can be added (or subtracted)

as one would expect. For example, three 9 Volt power sources can be

placed adjacent to each other and the voltage

behaves the same as if the circuit had one 27 Volt

power source.

AC VOLTAGE

Solutions:

1. Alternating Current

2. Powering homes and businesses. Wall outlets

supply AC Voltage.




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Answer the following questions.

1. How are the electrons moving?

2. How does the electron motion differ from direct current?

3. What is the measurement for voltage?

4. What is the magnitude for voltage?

In this case the voltage seems to be alternating between -10 and 10. Thus, we claim that this is a 10 Volt AC power supply and that the magnitude of the voltage is 10. Pause the simulation and complete the following.

1. Pause the simulation, and remove the voltmeter.

2. Place another AC power source adjacent to the first one as pictured.

3. Before pressing play, make a hypothesis about the magnitude of the total voltage with two 10V AC power supplies.

4. Press play and use the voltmeter to measure the voltage.

5. What is the measurement for the voltage?

6. What is the magnitude for the voltage?

7. Repeat steps 1 through 6 to place another AC voltage source. Make a conclusion about how AC voltage sources are added together.

Solutions:

1. They should be alternating direction.

2. The electrons in direct current flow in one

direction. The electrons in alternating current flow in

an alternating pattern.

3. It alternates between -10 and 10 volts.

4. The magnitude is 10 volts.

Solutions:

5. The students may get different measurements

based on what the measurement was when

they paused the simulation in step 1. An explanation is on the following pages.

6. The magnitude will vary from student to student as

well.

7. The students should conclude that AC voltage does not add in the same

way that DC voltage adds.



6



In the previous activity, we observed that alternating current behaves quite differently from direct current. The first observation was with the motion of the electrons. Instead of flowing around the circuit, they stayed relatively close to where they began, moved in one direction, and then moved back in the other direction.

Another observation was that the voltage had a similar periodicity to it. It would measure 10V, 0V, -10V, 0V; 10V, 0V, -10V, 0V; and repeat this pattern indefinitely. However, in the DC voltage simulation, it would remain a constant 9V.

Lastly, the way operations (such as addition) were performed on DC voltage and AC voltage differed. For DC if we began with 9V and added 9V, it resulted in 18V. If we added 9 more volts, we would get 27V. However, for AC voltage we started with a 10V source, added another 10V source and seemed to get a random resulting power source. The reason this happens is the periodicity of alternating current. Mathematically, one of the best ways to describe this phenomena is with complex numbers. Let’s look at what is happening graphically on the complex plane. When working with circuits, it is customary to use the letter j in place of i because I is typically used to refer to current. Thus the imaginary axis below is labeled with j.

As we have previously seen, complex numbers can be represented with their absolute value, or magnitude, by plotting the number on the complex plane and connecting it to the origin. In the simulation with one AC voltage source, the magnitude of the voltage was 10V. Thus we can represent the voltage as the complex number 10 0+ j V. This can be graphically represented as shown above.

Moving forward in time with respect to the AC simulation, after displaying 10V there is a moment where the voltmeter showed 0 V. This can be represented as the complex number 0 10+ j V. This is graphically represented as shown below.

AC AND COMPLEX NUMBERS

j

10

j

10j

10




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7

Exercise 1 Assume the simulation moves forward in time, and the voltmeter displays -10V. Express this value as a complex number, find it’s magnitude, and represent this moment on a complex plane.

Exercise 2 Assume the simulation moves forward in time, and the voltmeter now displays 0V. Express this value as a complex number, find it’s magnitude, and represent this moment on a complex plane.

Exercise 3Assume the simulation moves forward in time, and the voltmeter now displays 10V. Express this value as a complex number, find it’s magnitude, and represent this moment on a complex plane.

Based on the two images on the previous page and exercises 1-3, we can see that in the mathematical representation of AC voltage, the magnitude of the complex number never actually changes. This can be represented with as circle, as shown below.

j

10j

10-10

-10j

As previously shown, when the voltmeter reads 0, the voltage can be representing using only the imaginary component of a complex number. But why would we care to represent the voltage with a complex number when the voltmeter reads 0 volts? The exercises that follow will guide us to making a conclusion about the importance of imaginary and complex numbers.

Before completing the exercises, we will need to be able to understand circuit diagrams and Kirchhoff ’s Voltage Law.

Circuit Diagrams

A circuit diagram is a set of images and lines that represent a circuit like the ones created in the simulation. The two diagrams below show some an AC circuit diagram and a DC circuit diagram, each with a power source and a load.

+

- +-

The table that follows further describes the symbols used in the examples that follow.

j

R

10j

10-10


− +10 0 j V

Magnitude is 10

j

R

10j

10-10

-10j


0 10− j V

Magnitude is 10. This representation of 0V points

downward because it falls at a different part of the

cycle 10, 0, -10, 0. Having one 0 point upward and

the other point downward gives a mathematical and graphical representation

for two different 0s.

j

R

10j

10-10

-10j


10 0+ j V

Magnitude is 10.



8



Symbol Description+- AC voltage source

+-Battery Cell

Since battery cells produce are DC volt-age, we will use this to mean a DC voltage

source.Resistor (Or a load)

This symbol is usually used to represent a resistor. The purpose of a resistor is to reduce the flow of current. Since we will not be concerned with what exactly is in

the circuit, we will use this more generally to mean a load. A load can be a light bulb, a resistor, or any other circuit component

that slows the current.

Kirchhoff’s Voltage Law

Kirchoff ’s Voltage Law can be stated as:

The sum of voltages around a closed loop is zero.

This means that the voltage drop across every component in a loop within a circuit along with the voltage sources adds to be 0. The image below provides a simple visual of 1 loop.

PowerSource Load+ =+ =

= −

010 0

10V x

x V

+-

10V

Load1 loop

In the loop above, the voltage drop across the load is not labeled. However, because of Kirchhoff ’s Voltage Law we know that it will be 10V since there is a 10V power source. Although this direct application of KVL is useful for determining the change in voltage for individual circuit components, we will use a property that KVL implies. That is, KVL shows that voltage can be added directly under the rules allowed in the set of complex numbers as long as the voltage sources are in the same loop. For example, the complex voltages in the circuit below can be added together to give us the resultant voltage source. The resultant voltage source is the sum of two or more sources.

5 30− j V+

-

-

A Load +− +3 4 j V

The resultant voltage is 5 30 3 4 2 26−( )+ − +( ) = −j j j V.

Exercise 4 solution:

KVL allows us to add the voltages directly since they

are in the same loop.

Let z j1 0 10= + and z j2 0 10= − .

Then z1 10= ,

z2 10= ,

z z j1 2 0 0+ = + ,

z z1 2 0+ = .




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9

Max Voltage Output

In the following exercises, we look at moments at which all of the voltage is considered to be in the imaginary component, thus the voltmeter would read 0.

Exercise 4

Two AC power sources are placed in a circuit as pictured with their voltages. Find the magnitude of each voltage source and the resulting voltage source.

0 10+ j V+

-

-

A Load + 0 10− j V

Exercise 5

Two AC power sources are placed in a circuit as pictured with their voltages. Find the magnitude of each voltage source and the resulting voltage source.

0 10+ j V+

-

-

A Load + 0 10+ j V

Exercise 6

Answer the following questions to compare the voltage sources in exercises 4 and 5.

1. What value would the voltmeter read for each voltage source in exercise 4?

2. What value would the voltmeter read for each voltage source in exercise 5?

3. Recall that after exercises 1, 2, and 3 we concluded that the magnitude of the voltage source never changes. As time passes, what is the maximum value the voltmeter would read for the entire circuit in exercise 4?

4. What is the maximum value the voltmeter would read for the entire circuit in exercise 5?

5. In general, what is the maximum output equivalent to?

6. Make a conclusion about the importance of the complex and imaginary numbers concerning the magnitude and periodic nature of AC voltage (e.g. in exercises 1, 2, 3 the pattern 10V, 0V, -10V, 0V | 10V, 0V, -10V, 0V| 10V, 0V, -10V, 0V etc.).


KVL allows us to add the voltages directly since they

are in the same loop.

Let z j1 0 10= + and z j2 0 10= + .

Then z1 10= ,

z2 10= ,

z z j1 2 0 20+ = + ,

z z1 2 20+ = .

Exercise 6 solutions:

1. 0V

2. 0V

3. 0V

4. 20V

5. The magnitude of the complex number that

represents the voltage.

6. During moments where the voltmeter reads 0, the

imaginary component maintains the value of

maximum output for the power source. The complex number allows us to know

where the AC voltage source is in its repeating

pattern, or period, (e.g. the first time the voltage source outputs 0 in the pattern 10, 0, -10, 0; verses the second

time it outputs 0).



10



Exercise 7

Find the maximum voltage output in the circuit pictured below.

3 4+ j V+

-A Load

Exercise 8


7 9− j V+

-

-

A Load +− +3 6 j V

Exercise 9


8 3+ j V+

-

-

A Load +2 1+ j V

Exercise 10


− +1 8 j V+

-

-

-

A Load+

+

6 15+ j V

− −10 11j V


3 4 3 4 52 2+ = ( ) + ( ) =j

4 3 4 3 52 2− = ( ) + −( ) =j V


7 9 3 6 4 3−( ) + − +( ) = −j j j

4 3 5− =j V


8 3 2 1 10 4+( ) + +( ) = +j j j

10 4 10 77+ ≈j . V


− + =5 12 13j V




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11

Thus far in this lesson, we have mentioned the periodic nature of AC voltage by recognizing a pattern (previously it was 10, 0, -10, 0 | 10, 0, -10, 0) with respect to voltage output. Let’s look at this pattern in a bit more detail to understand how AC voltage can be graphically represented with respect to time.

Let’s begin by assuming we have an alternating voltage source that at time t=0 seconds outputs 0 volts, at 2 seconds outputs 5 volts, at 4 seconds outputs 0 volts, at 6 seconds outputs -5 volts, and then begins to repeat this pattern. This is represented with the table and graph below.

Time, t (seconds) Output (Volts)0 02 54 06 -58 0

10 512 014 -516 0

Out

put (

V)

t (sec)2 4 6 8 10 12 14 16

5

-5

Exercise 11

Answer the following questions about the table and graph above.

1. What type of function can model this?

2. What is the period of this function (how long does it take to begin repeating)?

3. What is the amplitude of this function (distance from the horizontal to the crest of a wave, or half the total height)?

4. Describe the relationship between the amplitude, maximum voltage output, and magnitude of the voltage.

Now let’s consider if we have two voltage sources that output 0 volts at two different times, one 4 seconds later than the other. Let’s assume both sources have an amplitude of 5 volts. They can be plotted as follows.

Out

put (

V)

t (sec)2 4 6 8 10 12 14 16

5

-5

17 18


1. Sinusoidal, Sine Function, Cosine Function

2. 8 seconds

3. 5 Volts

4. They are all the same.



12



Exercise 12

Let source 1 be represented by the red (solid) line and let source 2 be represented by the blue (dashed) line. Answer the following questions concerning the previous graph.

1. How much later than source 1 does source 2 output 0 volts?

2. How much later than source 1 does source 2 output 5 volts?

3. What is the output voltage of source 1 at six seconds?

4. What is the output voltage of source 2 at six seconds?

5. What is the sum of their output at 6 seconds?

6. Choose a different time between 4 and 16, what is the sum of the voltages?

7. If source 1 was in a loop by itself, what would the maximum voltage output be?

8. If source 2 was in a loop by itself, what would the maximum voltage output be?

9. If these two sources were in the same loop, what would the maximum voltage output be?

The delay of one source with respect to another can be referred to as phase shift. In exercise 12, the phase shift of source 2 with respect to source 1 causes the total output to always be 0.

Let’s look at how phase shift is represented with respect to complex numbers.

Exercise 13

Consider the same two voltage sources at any given time. Which of the following graphs represents them.

j

5j

5-5

-5j

j

5j

5-5

-5j

j

5j

5-5

-5jj

5j

5-5

-5j

In the complex plane representation, we can see that the two voltage sources always have an angle of 180 degrees, or π. Thus, we can establish the phase shift with respect to a source as the angle between that source and another voltage source.


1. 4 seconds

2. 4 seconds

3. -5 Volts

4. 5 Volts

5. 0 Volts

6. 0 Volts

7. 5 Volts

8. 5 Volts

9. 0 Volts


All of the graphs. Each graph represents the two

sources at a different time.




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13

Now we have a way to refer to a power source using its maximum voltage output and phase shift.

Exercise 14

On the complex plane, plot the given voltage source combinations.

1. Source 1 has a maximum voltage output of 3 volts. Source 2 has a maximum output of 3 volts and is phase shifted 45 degrees with respect to source 1.

2. Source 1 has a voltage of 2 2 2 2+ j . Source 2 has a max voltage output of 4 Volts and is phase shifted 90 degrees with respect to source 1.

3. Source 1 has a voltage of − +5 12 j . Source 2 has a max voltage output of 10 volts and is phase shifted 135 degrees with respect to source 3. Source 3 has a max voltage of 11 volts and is phase shifted 60 degrees with respect to source 1.

To simplify this concept a bit more, we can set the positive real axis as the basis for the phase shift and refer to all of the voltages with an amplitude and a phase shift. For example, the voltage sources below are each labeled with their amplitude and phase shift with respect to the positive real axis. Note that there are multiple ways one voltage source can be represented.

j

2

2 45� � V or 2 315� �− Vj

5

5 210� � V or 5 150� �− V

When given voltages in the notation introduced above, assume that each phase shift is with respect to the positive real axis.

Exercise 15

Answer the following:

1. How does this new notation (with amplitude and phase shift) relate to the polar form of complex numbers?

2. From the two images above, represent the sources 2 45� � and 5 210� � as an ordered pair of polar coordinates.

3. Find the corresponding complex coordinates for 2 45� � and 5 210� � .

j

3j

3-3

-3j

j

4j

4-4

-4j

13

j

1110


Part 1

Part 2

Part 3


1. r is the amplitude and θis the phase shift.

2. 2 45, ( ) and 5 210, ( )3.

2 45 2 45 2 2cos sin ( ) + ( ) = +i i

2 45 2 45 2 2cos sin ( ) + ( ) = +i i

5 210 5 210 5 32

52

cos sin ( ) + ( ) = −i i

5 210 5 210 5 32

52

cos sin ( ) + ( ) = −i i



14



Exercise 16

Assume that each pair of voltage sources in the following are connected as shown in the image below. Find the amplitude and phase shift of the sum of both sources. Plot both sources and their sum on the complex plane.

+

-

-

A Load +Source B

Source A

1. Source A: 8 30� � ; Source B: 8 90� �

2. Source A: 4 60� � ; Source B: 6 180� �

3. Source A: 12 315� � ; Source B: 2 135� �−

As the previous exercise shows, it can be rather tedious to add two sources given their polar form. The benefit of using this form comes with multiplication.

Ohm’s Law

Ohm’s Law relates voltage, current, and resistance through the equation V IR= where V is voltage, I is current, and R is resistance (impedance for AC cirucits). When we are working with AC power sources the voltage can be a complex number. Since voltage is a product of current and impedance (AC resistance), then those can be complex numbers as well.

Exercise 17

Given the following, solve for the unknown component in Ohm’s formula. Provide the solution in the same form as the givens (i.e. if the givens are polar, then the solution should be polar).

1. V = 22 6+ j and I = 4 2+ j . Find R.

2. V =12 315� � and I = 2 135� �− . Find R.

3. I =8 90� � and R = 4 60� � . Find V.

4. I = 5 8− j and R = − +9 2 j . Find V.


Part 1

Rjj

=+( )+( )

22 64 2

Rjj

jj

=+( )+( )

−( )−( )

22 64 2

4 24 2

R j= −5

Part 2

12 3152 135��

�

�−

R =122

315 135� � �− −( )( )R = 6 450 6 90� �� =

Part 3

V = 8 90 4 60� �� ( )( )

V = 8 4 90 60⋅ +( )� �

V = 32 150� �( )Part 4

V j j= −( ) − +( )5 8 9 2

V j= − +29 82

R =

Exercise 16 solution on the following page.


AMES: Applications of Complex Numbers www.NICERC.org

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Teacher Page:

j

8 8 3 13.9

8 30 8 30 4 3 4

8 90 8 90 0 8

4 3 4

cos sin

cos sin

( ) + ( ) = +

( )+ ( ) = +

+( )

j j

j j

j ++ +( ) = +

+ = ( ) + ( ) =

0 8 4 3 12

4 3 12 4 3 12 8 32 2

j j

j

For Phase Shift:Since both the real component and the imaginary component of the sum are positive, the resulting angle would be in Quadrant I.

phase shift =

=

−tan 1 124 3

60

Exercise 16 Part 1 Solution:

j

642 7 5.29

4 60 4 60 2 2 3

6 180 6 180 6 0

2 2 3

cos sin

cos sin

( ) + ( ) = +

( ) + ( ) = − +

+

j j

j j

jj j j

j

( ) + − +( ) = − +

− + =

6 0 4 2 3

4 2 3 2 7


For Phase Shift:For the sum, since the real component is negative and the imaginary component is positive, the resulting angle is in Quadrant II. Since arctangent is defined from -90 degrees to positive 90 degrees, we must add 180 degrees to get the solution in quadrant II.

phase shift =−

+ = − + =−tan 1 2 3

4180 40 180 140

j

122 2 37 12.17

12 315 12 315 6 2 6 2

2 135 2 135 2

cos sin

cos sin

( ) + ( ) = −

−( )+ −( ) = −j j

j −−

−( )+ − −( ) = −

− =

2

6 2 6 2 2 2 5 2 7 2

5 2 7 2 2 37

j

j j

j


For Phase Shift:For the sum, since the real component is positive and the imaginary component is negative, the resulting angle is in Quadrant IV.

phase shift = −

= −

−tan .1 7 25 2

54 46


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