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
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
------------?
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.
"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
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.
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)
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
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)
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
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
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
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)
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)
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)
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
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
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
Handout
Problem #2
a
b
20 j30 40
50 j60-j70
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)
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
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
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
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
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
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)
CircuitLab Simulation
|Vin| |V1| |V2| |V3|
Calculated 10.0000V 8.2356V 6.3627V 5.4596V
CircuitLab 10.00V 8.231V 6.348V 5.435V
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
+
-
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