Circuits II: Phasors & Complex Numbersbisonacademy.com/ECE111/Videos/09 Circutis II.pdf · j50...
Transcript of Circuits II: Phasors & Complex Numbersbisonacademy.com/ECE111/Videos/09 Circutis II.pdf · j50...
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Circuits II: Phasors
& Complex NumbersECE 111 Introduction to ECE
Jake Glower - Week #9
Please visit Bison Academy for corresponding
lecture notes, homework sets, and solutions
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Numbers make a difference. They can create and end empires.
The number zero is an odd concept.
It isn't necessary: Romans had an extensiveeconomy without the number zero
Addition and multipliation is much easier withthe number zero
Roman Numbers Arabic Numbers
XXVII+ CIX
--------------?
27+ 106
------------?
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Negative NumbersWeird concept: negative one apple?
The Dutch invented the double-entrybookkeeping system
Keep track of expenses (debits)
Keep track of assets (credits)
Only invest in profitable ventures
With the double-entry bookkeeping systemthe Dutch were able to challenge England,France, and Spain for world dominance.
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"We will bury you"Nikita Khrushchev
Accounting can decide the fate of empires
In the 1960's, the Soviet Union thought theireconomy was growing 20% per year
vs. 4% for the United States
At that rate, by 2000 the Soviet Union would crushthe U.S. ecomomically
Exponential growth is a powerful thing...
Actually, the Soviet economy was shrinking 1-2%per year
Eventually, the size of the goverment (based upon20% growth) could not be supported
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Real Numbers & DC Signals1st half of Circuits I
At DC, real numbers canrepresent
Resistors
Voltages
Currents
When working with DC,real numbers are all youneed.
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AC Signals2nd half of Circuits I
All of Circuits II
You need 2 terms to represent a sine wave
Frequency is usually known
x(t) = a cos (ωt) + b sin (ωt)
or
x(t) = r cos (ωt + θ)
Real numbers don't work well
Only one degree of freedom
Two real numbers are needed to represent asine wave 0 1 2 3 4 5 6 7 8 9 10
-1.2
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
Time
cos(t)sin(t)
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Complex Numbers
Complex numbers have two terms
Real Part & Complex Part
Allows you to represent cosine() and sine() terms with a single number
Let
j = −1
A complex number can be written as
x = a + jb = c∠θ
a2 + b2 = c2
tan θ =ba
a + jb
b
a
c
real
imag
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Complex Number Math
Addition:
The real parts add and the complex parts add.
(a1 + jb1) + (a2 + jb2) = (a1 + a2) + j(b1 + b2)
Subtraction:
The real parts subtract and the complex parts subtract
(a1 + jb1) − (a2 + jb2) = (a1 − a2) + j(b1 − b2)
Multiplication:
Multiply out: results in a complex number
(a1 + jb1)(a2 + jb2) = a1a2 + ja1b2 + ja2b1 + j2b1b2
= (a1a2 − b1b2) + j(a1b2 + a2b1)
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Division: Division also results in a complex number but takes even morecomputations:
a1+jb1
a2+jb2
=
a1+jb1
a2+jb2
a2−jb2
a2−jb2
=
(a1a2+b1b2)+j(−a1b2+a2b1)
a22+b2
2
=
a1a2+b1b2
a22+b2
2
+ j
−a1b2+a2b1
a22+b2
2
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HP Calculators to the Rescue!Free42 is a free app on your cell phone
HP35s is a $60 from Amazon
Download it
Get familiar with it
Use it
HP calculators are worth 10 points on midtermsfrom my experience
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Handout (part 1)
Determine the result of the following operations with complex numbers
2 + j3 2 + j3 2 + j3 2 + j3
+ 9 + j4 - 9 + j4 * 9 + j4 ------------
----------- ---------- ----------- 9 + j4
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Phasor Representation of cosine()
Euler's identity states that the complex exponential is
ejωt = cos (ωt) + j sin (ωt)
Real part gives you cosinecos (ωt) = real(ejωt
)
Phasor representation of cosine is 1 + j0
1 ↔ cos (ωt)
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Phasor representation of a sine wave:
If you multiply by (a + jb)
(a + jb)ejωt = (a + jb)(cos (ωt) + j sin (ωt))
= (a cos (ωt) − b sin (ωt)) + j(. .. )
Take the real part
a + jb ↔ a cos (ωt) − b sin (ωt)
You can also represent voltages in polar form:
r∠θ ↔ r cos (ωt + θ)
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Example: Determine the phasor representation for x(t)
The frequency comes from the period
T = 6.28ms
f =1
T= 159.2Hz = 159.2
cycles
second
ω = 2πf = 1000radsec
The amplitude is 5.000 (polar form)
The delay is 57 degrees
θ = −1ms delay
6.28ms period 2π
θ = −1.0 radian = -57.30
meaning
X = 5∠ − 57.30
x(t) = 5 cos (1000t − 57.30)
0 1 2 3 4 5 6 7 8 9 10
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Time (ms)
Vp = 5.0V
Delay = 57 deg
= 1 radian
Period = 6.28ms
x(t)
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Phasor Representation for Impedance's
Resistors: V = IR
R → R
Inductors: V = LdI
dt
V = Ld
dt(ejωt
) = jωL ejωt = jωL ⋅ I
L → jωL
Capacitors: V =1
C ∫ I dt
V =1
C ∫ (e jωt); dt =
1
jωC e jωt =
1
jωC I
C →1
jωC
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Simplification of RLC circuitsSame as resisitor circuits
Now with complex numbers
Example: Determine the impedance Zab.
-j100 + 300 = 300 - j100
(200) (300 − j100) =
1
200+
1
300−j100
−1
= 123.07 − j15.38
(200) + (123.07 − j15.38) = 323.07 − j15.38
(j300) (323.08 − j15.38) = 156.85 + j161.83
(100) + (156.85 + j161.83) = 256.85 + j161.83
Answer:
Zab = 256.85 + j161.83
100 200 300
-j100200j300
a
b
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Example: Determine Zab
Solution
(10) + (−j60) = 10 − j60
(10 − j60) (j50) = 125.00 + j175.00
(125.00 + j175.00) + (40) = 165.00 + j175.00
(165.00 + j175.00) (30 − j20) = 31.42 − j14.98
answer:
Zab = 31.42 − j14.98
-j20 30
40
j50
10-j60
a b
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Handout
Problem #2
a
b
20 j30 40
50 j60-j70
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Circuit Analysis with Phasors
Same as DC
Now with complex numbers
Example: Determine y(t)
V0 = 10 sin (628t)
Step 1: Replace each term with phasor value
V0 → 0 − j10
R → 400
C →1
jωC= −j159Ω
Step 2: Solve ( Voltage division )
Y =
−j159
−j159+400 (0 − j10) = −3.436 − j1.368
y(t) = −3.436 cos(628t) + 1.368 sin(628t)
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ValidationUse CircuitLab
Input the circuit
Run a Time Domain simulation
The output (orange line) is easier tosee in polar form
Y = −3.436 − j1.368
= 3.698∠ − 1580
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Example 2: 3-Stage RC Circuit
Step 1: Convert to phasor form
V0 → 0 − j10
ω = 100
C1 →1
jωC= −j50Ω
C2 →1
jωC= −j40Ω
C3 →1
jωC= −j33.33Ω
5 10 15
100 150 200
200uF 250uF 300uF+
-10 sin(100t)
V0 V1 V2 V3
-j50 -j40 -j33.33
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Step 2: Write N equations for N unknowns
V0 = 0 − j10
V1−V0
5 +
V1
100 +
V1
−j50 +
V1−V2
10 = 0
V2−V1
10 +
V2
150 +
V2
−j40 +
V2−V3
15 = 0
V3−V2
15 +
V3
200 +
V3
−j33.33 = 0
5 10 15
100 150 200
200uF 250uF 300uF+
-10 sin(100t)
V0 V1 V2 V3
-j50 -j40 -j33.33
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Step 3: Solve. First, group terms
V0 = −j10
−1
5V0 +
1
5+
1
100+
1
−j50+
1
10V1 +
−1
10V2 = 0
−1
10V1 +
1
10+
1
150+
1
−j40+
1
15V2 +
−1
15V3 = 0
−1
15V2 +
1
15+
1
200+
1
−j33.33V3 = 0
Place in matrix form
1 0 0 0
−1
5
1
5+
1
100+
1
−j50+
1
10
−1
10 0
0
−1
10
1
10+
1
150+
1
−j40+
1
15
−1
15
0 0
−1
15
1
15+
1
200+
1
−j33.33
V0
V1
V2
V3
=
−j10
0
0
0
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Put into MATLAB and solve
a1 = [1,0,0,0];
a2 = [-1/5,1/5+1/100+1/(-j*50)+1/10,-1/10,0];
a3 = [0,-1/10,1/10+1/150+1/(-j*40)+1/15,-1/15];
a4 = [0,0,-1/15,1/15+1/200+1/(-j*33.33)];
A = [a1;a2;a3;a4]
1.0000 0 0 0
-0.2000 0.3100 + 0.0200i -0.1000 0 0 -0.1000 0.1733 + 0.0250i -0.0667
0 0 -0.0667 0.0717 + 0.0300i
B = [-j*10;0;0;0];
V = inv(A)*B
V0 0 -10.0000i
V1 -1.6314 - 8.0724i
V2 -3.4430 - 5.3506i
V3 -4.4982 - 3.0942i
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V0 0 -10.0000i
V1 -1.6314 - 8.0724i
V2 -3.4430 - 5.3506i
V3 -4.4982 - 3.0942i
meaning
V0 = 10 sin(100t)
V1 = −1.6314 cos (100t) + 8.0742 sin (100t)
V2 = −3.4430 cos (100t) + 5.5306 sin (100t)
V3 = −4.4982 cos (100t) + 3.0942 sin (100t)
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CircuitLab Simulation
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|Vin| |V1| |V2| |V3|
Calculated 10.0000V 8.2356V 6.3627V 5.4596V
CircuitLab 10.00V 8.231V 6.348V 5.435V
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Handout
Problem #3:
Convert to phasors
Write the voltage node equations
20 40
50
0.1H
0.2H
0.01F
8 cos(100t)
V0 V1 V2 V3
+
-