Lect. 2 Review material

Lect. 2 Review material
ECEN 2260 Circuits as Systems
Spring 2016
Lecture 2
Review Material
Sections 1 through 5 are to be studied thoroughly
during the first week of the semester.
Section 6 will be discussed during one of the HW
review sessions.
Harry Hilgers
Last revised on 12/26/2015
1.
2.
3.
4.
5.
6.
Complex Numbers Review
Review of Phasors
Dynamical Analogies
Additional reviews, examples and tips
A RC Time Constant Tutorial
P-I Control Tutorial
1
1.
Complex Numbers Review
Fundamentals.
z = complex number = x + jy
x = ‖z‖cos ()
2
M = Magnitude = √x 2 + y 2 = ‖z‖
y
 = tan−1 ( )
x
y = ‖z‖sin ()
x = Real(z)
‖z‖ = M = Magnitude(z)
y = Imaginary(z)
 = Angle(z) or Argument(z)
z = x + jy = Rectangular Form
‖z‖(cos  + jsin ) = Trigonometric Form
‖z‖ej = Exponential Form
‖z‖
z̅ = z ∗ = x − jy = ‖z‖ej(−) = ‖z‖
() = Polar Form
(−) = Complex Conjugate
‖ej ‖ = 1
j Im
ej = 
y
z = x + jy
Im
j
M
φ
ejφ
φ
x
Re
1
-1
Re
-φ
Unit Circle
M
_
z = z = x - jy
-j
*
-y
2
Addition and Subtraction:
z1 = x1 + jy1
z2 = x2 + jy2
z1 + z2 = (x1 + x2 ) + j(y1 +y2 )
z1 − z2 = (x1 − x2 ) + j(y1 −y2 )
Multiplication and Division:
z1 z2 = (x1 + jy1 )(x2 + jy2 ) = (x1 x2 − y1 y2 ) + j(x1 y2 + x2 y1 )
= √(x1 x2 − y1 y2 )2 + (x1 y2 + x2 y1 )2
() , where  = tan−1 (
x1 y2 +x2 y1
x1 x2 −y1 y2
)
The above method of multiplication invites algebra errors. In general it is best to
convert to polar or exponential form:
z1 = ‖z1 ‖
(1 )
z1 z2 = ‖z1 ‖‖z2 ‖
z2 = ‖z2 ‖
(2 )
(1 + 2 ) = ‖z1 ‖‖z2 ‖ej(1+1 )
Ditto for division:
‖z1 ‖ j( − )
z1 ‖z1 ‖
=
e 1 1
(1 − 2 ) =
‖z2 ‖
z2 ‖z2 ‖
3
Derive trig formulas using Euler’s Expression:
Example:
ej1 = cos1 + j sin1
ej2 = cos2 + j sin2
ej1 ej2 = ej(1+2) = cos(1 + 2 ) + j sin(1 + 2 ) (1)
ej1 ej2 = (cos1 + j sin1 )(cos2 + j sin2 ) =
ej1 ej2 = (cos1 cos2 − sin1 sin2 ) + j(sin1 cos2 + cos1 sin2 ) (2)
Equate Equations (1) and (2) Real parts:
cos(1 + 2 ) = cos1 cos2 − sin1 sin2
Equate Equations (1) and (2) Imaginary parts:
sin(1 + 2 ) = sin1 cos2 + cos1 sin2
4
Roots of Complex Numbers:
1
ᵒ
ᵒ
Find the roots of z = (−1)3 Recall: j = ej(90 +n360)
ᵒ
ᵒ
ᵒ
ᵒ
z = (−1)1/3 = (j2 )1/3 = j2/3 = (ej(90 +n360) )2/3 , z = ej(60 +n240) , n = 0,1,2,3 … ..
ᵒ
z0 = ej60
ᵒ
ᵒ
ᵒ
z1 = ej300 = e−j60
ᵒ
z2 = ej540 = ej180 (= −1)
Im
j
z0
Unit Circle
z2
60°
-1
1
Re
-60°
z1
-j
(3) roots
Application: The three phase voltages and currents in the power distribution grid.
5
2.
Review of Phasors
Euler: ej(ωt+φ) = cos(ωt + φ) + j sin(ωt + φ)
The representation of a sinusoid 𝐯(𝐭) = 𝐕𝐦 𝐜𝐨𝐬( 𝛚𝐭 + 𝛗) = 𝐑𝐞(𝐕𝐦 𝐞𝐣(𝛚𝐭+𝛗) ) =
⃗ = 𝐕𝐦 𝐞𝐣 𝛗 (𝐨𝐫 𝐕
̂ 𝐨𝐫 𝐕
⃗ 𝐨𝐫 𝐕
̅) is called a
𝐑𝐞(𝐕𝐦 𝐞𝐣𝛗 𝐞𝐣𝛚𝐭 ) by a complex number 𝐕
“phasor”.
⃗ ejωt )
A phasor transforms back to the time domain by: v(t) = Re(V
Phasor analysis is the ideal tool for solving linear circuits in sinusoidal steady
state.
Phasor analysis leads to “Impedances”:
R
jωL
1
jωC
Impedance 𝐙 is a complex quantity (NOT a phasor) of the form Z = R + jX
R = the real part = Resistance
X = the imaginary part = Reactance
1
X may be inductive (XL = ωL) or capacitive (XC = − ) ( Note: no j )
ωC
All of the Circuits Introduction analysis techniques (series and parallel
combinations, voltage and current division, Thevenin and Norton equivalent
circuits, superposition, etc.) also apply to phasor analysis.
6
The voltages generated by our Power Plants are sinusoidal 60 Hz (50 Hz in many
other countries) and the impedances of the distribution system can be closely
approximated with phasor analysis. (The word “approximated” is used, because
our distribution system is not always “steady state” and often includes higher
harmonics).
Phasor operation is identical to classical vector operation.
A power distribution grid from the power-plant synchronous generators to the
final user (a factory, your home, etc.) consists of impedances of transmission
lines, transformers and switching devices. The phasor voltages and currents of
this power gridsystem are often represented by a vector diagram that shows the
phase differences between the transmission system voltages and currents. (This is
a topic covered in ECEN 3170 - Energy Conversion 1).
Review in your text the derivations of:
Resistor
+
Capacitor
L
R
R
Inductor
R
̅𝐑 = 𝐈̅𝐑 𝐑
𝐕
̅R = IR̅ R = R
V
(IR̅  0°)
̅𝐑 and 𝐈̅𝐑 are in-phase
𝐕
+
L
+
L
-
̅𝐋 = (𝐣𝛚𝐋)𝐈̅𝐋
𝐕
̅L = ωL IL̅ 
V
C
C
-
̅L = ωL (IL̅ 
V
C
̅𝐂 =
𝐕
0° )(j)
90°
𝟏
𝐈̅
𝐣𝛚𝐂 𝐂
1
(IC̅ 
ωC
1
̅C =
V
I̅ 
ωC C
̅C =
V
0° )(j−1 )
(−90°)
̅L leads IL̅ by 90°
V
̅C lags IC̅ by 90°
V
̅𝐋 by 𝟗𝟎°
𝐈̅𝐋 lags 𝐕
̅𝐂 by 𝟗𝟎°
𝐈̅𝐂 leads 𝐕
7
A non-loop/node equations approach to solving a circuit:
Objective:
1. Look at the circuit by breaking it into functional blocks.
2. Avoid useless manipulations that invite algebra errors.
3. Check units.
4. Substitute numerical values at the very end. You cannot check units after
the numbers have been substituted.
Apply these four objectives on the following phasor circuit example:
1. Find the expressions in terms of the circuit components for the input
impedance and all steady state voltages and currents (magnitude and
phase). Then compute the numerical values.
2. Deduce the value of the input impedance at DC (ω → 0) and at very high
frequency (ω → ∞) by just looking at the circuit; i.e. without calculating
any limits.
v1 (t) = 100 cos(2000t) , L = 250mH, C = 0.5μF, R = 3k𝛀
Note that ω = 2000 rad/s
iL(t)
+ v (t)
1
-
L
iC(t)
C
iR(t)
R
+
v2(t)
-
8
Solution:
Draw phasor representation of the circuit:
L
+
-
jXL = jωL
C
1
jXC = j(-1/ωC)
R
+
R
2
Note:
XL = Inductive Reactance = ωL (no j)
XC = Capacitive Reactance = -1/ωC (no j)
9
1. Functional “blocks” :
a) Input Impedance:
 Impedance of L in series with impedances of C parallel with R.
b) Voltage divider: Impedance of C parallel with R over the input impedance.
c) Current divider between C and R.
a) Input Impedance: Zin = jXL + R‖jXC , IL =
⃗2 = V
⃗1
b) Voltage divider: V
c) Current dividers: IC =
R‖jXC
jXL +R⫽‖jXC
R
IL
, IR =
R+jX
C
IL
⃗V
⃗1
Zin
jXC
R+jXC
= ⃗V1
1
jXL + R⫽‖jXC
(= IL − IC )
2. Avoid useless manipulations:
The above results look understandable, tidy and uncluttered; so no further
“simplification” manipulations are necessary.
3. Check Units:
Verify left side vs. right side units. Convince yourself.
4. Substitution of numerical values.
We now need to do some numerical calculations. Even though there are many
computer based computation tools available to us, we still need to be able to
make computations (estimations) by hand.
Why?
We must be able to verify (estimate) the computer output results. So if your
computer program computes an output current of 10 amps at an output
voltage of 100 V for an LF356 op-amp, then you should recognize that this is
not possible. Maybe it should be 10 mA at 10 V.
Just recall that “garbage-in = garbage-out”.
10
Compute the reactance values:
R = 3000 Ω
XL = ωL = (2000)(250)10−3 = 500 Ω
1
1
XC = − = − (2000)(0.5)10−6 = −1000 Ω
ωC
Compute the different impedance values:
R‖jXC =
jRXC
3000(−j1000) −3000j 3 + j −9000j + 3000
=
=
=
R + jXC
3000 − j1000
3−j 3+j
10
= 300 − 900j Ω
Input Impedance: Zin = jXL + R‖jXC = j500 + 300 − 900j = 300 − 400j Ω
R‖jXC
300 − 900j 3 − 9j 3 + 4j 9 + 12j − 27j + 36
=
=
=
jXL + R‖jXC
300 − 400j 3 − 4j 3 + 4j
9 + 16
=
45 − 15j 9 − 3j
=
Ω
25
5
(−j1000)
jXC
−j 3 + j −3j + 1
=
=
=
= 0.1 − 0.3j Ω
R + jXC 3000 − j1000 3 − j 3 + j
10
11
Compute the current and voltage phasors:
⃗V1
1
100 0ᵒ
100 0ᵒ
IL =
= ⃗V1
=
=(
)
Zin
jXL + R‖jXC
300 − 400j
ᵒ
500 −53.1
= 0.2 53.1ᵒ A
IC = IL
( 0.6
= 0.19
R
= ( 0.2
R+jXC
3+j
53.1ᵒ )
10
ᵒ
53.1ᵒ )
= (0.06
3000
3000−j1000
ᵒ (3
53.1 )
= ( 0.2
+ j) = (0.06
53.1ᵒ )
3 3+j
3−j 3+J
ᵒ
=
53.1 ) (√10
18.4ᵒ )
71.5 A
IR = IL − IC = 0.2
53.1ᵒ − 0.19
71.5ᵒ This method of computing IR looks cumbersome
as we will need to convert from polar to rectangular and back to polar forms. So let us use the
result of the current division:
IR = IL
⃗2 = V
⃗1
V
jXC
= (0.2 53.1ᵒ ) (0.1 − 0.3j)
R + jXC
= (0.2 53.1ᵒ ) (0.316
− 71.5ᵒ ) = 0.063
−18.4ᵒ A
R‖jXC
9 − 3j
=
(100 0ᵒ ) = 60(3 − j)
jXL + R‖jXC
5
= (60)(√10) −18.4ᵒ = 189 −18.4ᵒ V
1. Check for yourself the results by computing the output voltage with
⃗V2 = IR R and ⃗V2 = IC (jXC )
2. Draw a phasor diagram showing all phasor currents and voltages. Such a
diagram may be on a test/quiz.
3. The above showed a computational methodology. So numbers to three
decimals were used. Often this accuracy is not needed. What will be
needed is for you to be able to make an approximation of a RELATIVE value
of a variable/component with respect to other variables/components. So
go back to the above circuit and see how close you can estimate the final
12
results. Sketching a phasor-diagram (vector diagram) is a very useful tool
for this.
Behavior of circuit near DC:
L
+
-
C
1
ω → 0:
jωL → “short circuit”
1
→ “open circuit”
jXL = jωL
R
+
jωC
R
jXC = j(-1/ωC)
2
Thus:
‖Z‖ → R and
-
Z → 0°ᵒ
Note:
XL = Inductive Reactance = ωL (no j)
XC = Capacitive Reactance = -1/ωC (no j)
Behavior of circuit at high frequency:
L
+
-
jXL = jωL
C
1
jXC = j(-1/ωC)
R
+
R
2
Note:
XL = Inductive Reactance = ωL (no j)
XC = Capacitive Reactance = -1/ωC (no j)
As ω → ∞: jωL → ∞
1
→ “short circuit”
jωC
1
Zin = jωL + R‖(
)
jωC
Since jωL → ∞ , the inductor
dominates. Thus:
‖Z‖ → ωL → ∞ and
Z → 90°ᵒ since
90°ᵒ
jωL →
13
For the above circuit problem we were asked to compute all phasor voltages and
currents. We obtained the answers with a minimum of “clutter” by using the
inherent functionality of the circuit.
Suppose we were asked to show how the magnitude and phase of the input
impedance depends on ω. We would then have no choice but doing some “ugly”
complex algebra manipulations. However note how we can minimize the
“ugliness” by using polar multiplication/division as indicated earlier.
R
1
R
jωC
Zin (jω) = jωL + R‖ (
= jωL +
) = jωL +
1
jωC
1 + jωRC
R+
jωC
jωL + (jωL)(jωRC) + R
(R − ω2 RLC) + j(ωL)
num
=
=
=
1 + jωRC
1 + j(ωRC)
denom
Magnitude:
‖num‖ √(R − ω2 RLC)2 + (ωL)2
num
‖Zin (jω)‖ = ‖
‖=
=
‖den‖
denom
√1 + (ωRC)2
Phase:
Zin (jω) =
num
denom
Zin (jω) = tan−1 (
=
num −
denom ⟹
ωL
ωRC
−1
)
−
tan
(
)
(R − ω2 RLC)
1
These results show how the impedance magnitude and phase are somewhat
complicated functions of ω. We will be addressing this topic very extensively
during this semester.
14
3.
Dynamical Analogies
nd
Ref.: Dynamical Analogies by Harry F. Olson, 2 edition, 1943/1958, free down loadable. Highly Recommended.
A mass M moving at
velocity u due to force F
An inductor L with a
current i due to voltage v
+
v
-
F
u (velocity)
M
i
Newton’s Law
𝐝𝐮
𝐅= 𝐌
𝐝𝐭
A compliance C (= spring Ks)
compressed by distance x
due to force F
F x
L
𝐯= 𝐋
A capacitor C charged by
charge q due to voltage v
v
+
Ks =
𝐝𝐢
𝐝𝐭
i
1
-
C
Compliance
1
dx ⟹ dx = CdF ⟹
C
dx
dF
= C
⟹
dt
dt
𝐝𝐅
𝐮= 𝐂
𝐝𝐭
dF = K s dx =
A force F due to a friction B as a result
of an object moving at velocity u
u (velocity)
𝐢= 𝐂
A voltage v across a resistance R due to
a current i
+
v
i
F
𝐝𝐯
𝐝𝐭
-
R
B
𝐅 = 𝐁𝐮
𝐯 = 𝐑𝐢
15
4.
Additional reviews, examples and tips
A. Approximations
Often a circuit can be simplified by looking at approximations.
For example: A 10 ohm resistor in parallel with a 1000 ohm resistor still “equals”
10 ohms; especially when the resistors have 10% tolerances.
Initial approximations may reduce a complicated looking circuit into some very
elementary functional blocks. This in turn may turn a “spaghetti” looking “mess”
into a very logical and readable diagram.
B. Units
What is wrong with the following equation?
I2 =
jωLI1 − V1
V1 +R+jωL+
1
jωC
A
This equation is complete nonsense. Volts are added to ohms!!!!!!!!!!!!!
You can catch a majority of the errors with units-expressions. At one time in the
past we (the USA tax payers) lost a space craft because of an error in units.
Note: Suppose during an exam you discover that the units don’t mach and no
time to fix it. If you simply note that you are aware of this error, then I will be
much more lenient with the grading.
16
C. Circuit Manipulation
A resistance R1 in series with a parallel combination of R2 and R3:
R2
R1
100 Ω
1 kΩ
470 Ω
R3
Req
The equivalent resistance is R eq = R1 + R 2 ‖R 3
R 2 ‖R 3 = parallel combination of R 2 and R 3 =
1
1
1
+
R2 R3
= inverse addition
We could “simplify” as follows:
R eq = R1 + R 2 ‖R 3 = R1 +
R2R3
R1 R 2 + R1 R 3 + R 2 R 3
=
R2 + R3
R2 + R3
However, even though the manipulation is correct, it is useless and is an
opportunity to make algebra errors.
The original form 𝐑 𝐞𝐪 = 𝐑 𝟏 + 𝐑 𝟐 ‖𝐑 𝟑 is much more understandable.
The ‖ operator ⫽can be viewed as inverse addition and is an operator similar to
“+”, “-“, etc.
Computation of R eq is actually easier using the original expression:
R eq = R1 + R 2 ‖R 3 = 470Ω + 100Ω ‖ 1kΩ = 470Ω + 91Ω = 561Ω
17
D. Some more phasor circuits using functionality to solve them
For each of the following circuits, write an expression for the output voltage
⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
phasor V
out in terms of the input voltage phasor Vin and component values. Do
this by inspection without resorting to loop/node equations.
1. An Op-Amp amplifier
R2
Inverting Amplifier Expression:
R1
+
-
+
+
in
out
⃗ out = V
⃗ in (−
V
R2
)
R1
-
2. Another Op-Amp amplifier
C2
R2
R1
+
-
in
C1
+
+
out
-
Inverting Amplifier Expression with series
and parallel impedances:
1
R 2 ‖(
)
jωC
2
⃗ out = V
⃗ in (−
V
)
1
R1 + (
)
jωC1
Can multiply out into polar numerator and
denominator forms as before to find amplifier
magnitude and phase as function of ω.
18
3. A voltage divider
R1
+
+
-
C
in
R2
out
-
Use voltage divider expression:
⃗ out = V
⃗ in
V
R2‖
1
jωC
R1 + R 2 ‖
1
jωC
19
4. A more complicated voltage divider
R2
R1
+
+
-
in
C1
C2
out
-
Use voltage divider expression twice:
⃗ out = V
⃗ in
V
1
1
1
‖(R 2 +
)
jωC1
jωC2
jωC2
[
].[
]
1
1
1
R1 +
‖(R 2 +
)
R2 +
jωC1
jωC2
jωC2
The first bracketed term divides the input voltage into the voltage across C1
including the loading of R2 and C2. The second bracketed term divides the voltage
across C1 into the output voltage.
Another approach to solving this circuit would be using the Thevenin equivalent
⃗ in . Then combine the Thevenin equivalent circuit with
for the input circuit R1-C1-V
R2-C2 , followed by voltage division to find ⃗Vout . You should do this for yourself;
this could be on a quiz.
20
Suppose now that the magnitude of the R2-C2 impedance is much greater than the
magnitude of the C1 impedance; i.e. a “large” impedance in parallel with a
“small” impedance.
Recall the earlier statement:
A 10 ohm resistor in parallel with a 1000 ohm resistor “is” still 10 ohms; especially
if the resistors have 10% tolerances.
So in our case here we may be able to approximate the final result by ignoring the
loading impedance of R2-C2.
The use of the parallel symbols ⫽ in the first bracketed term makes it pretty clear
1
that we can simply omit the second term (R 2 +
) in both the numerator and
jωC2
denominator. So in that case
⃗ out ≅ V
⃗ in
V
1
jωC1
1
jωC2
[
].[
]
1
1
R1 +
R2 +
jωC1
jωC2
This method of approximation would have been practically impossible if we
would have “simplified” the expressions by multiplying out the parallel terms.
Again, the moral should be clear: Show functional blocks inside your equations
whenever and wherever possible.
21
Let us now suppose that the above circuit is a simple low voltage circuit on your
lab bread-board and let us further suppose that now the magnitude of the R2-C2
impedance is at the same order of magnitude as the C1 impedance. So we no
longer can ignore the loading by R2-C2. How could we change the circuit so that
we could still ignore the loading?
Recall the voltage-follower from Circuits 1:
R1
C1
I+ = 0
I- = 0
+
-
in
+
FB
R2
-
+
C1
C2
FB =
out
C1
(Ideal Op-Amp)
-
Before you started Circuits 1, this schematic might have looked like a bowl of
spaghetti. Now you should be able to see the three functional blocks: The R1-C1
impedance, the voltage follower and the R2-C2 impedance.
You should write out for yourself the equations that lead to the final input to
output expression:
⃗Vout ≅ ⃗Vin
1
jωC1
1
jωC2
[
].[
]
1
1
R1 +
R2 +
jωC1
jωC2
22
E. A first order transient problem
R1
ic(t)
+
- vin(t)
C
+
vout(t)
-
Objective:
For vin (t) equals a step at t=0 of magnitude VP volts, sketch vout (t) and ic (t) and
explain physically what happens, without formally solving the differential
equation.
Solution:
In equilibrium the cap behaves as an open circuit. Hence:
For t < 0 , vout (t) = vin (t) = 0 and ic (t) = 0
For t → ∞ ,
vout (t) → vin (t) = VP and ic (t) = 0
Recall from Circuits 1 that the variables in these types of circuits progress as
exponentials with a certain time-constant.
So in this circuit it stands to reason that the output voltage increases with an
exponential from 0 volt at t=0 to VP volts at t → ∞ . The only time constant
possible is  = RC sec.
23
Now let us look at the current. Recall that the charge q on the cap plates is
directly proportional to the voltage across the cap, where C is the proportionality
constant: qC (t) = Cvout (t) . To change the cap’s charge, a current must flow
dq
through the cap iC (t) = .
dt
v (t)−v
(t)
out
But in this circuit iC (t) is given by iC (t) = in
. In words: the voltage
R
across the resistor determines the capacitor current.
For t < 0, vout (t) = vin (t) = 0, ic (t) = 0 ⟹ The cap’s voltage does not change,
because the current is zero; i.e. vout (t) = 0 .
For t = 0+, vin (0+ ) = VP , vout (0+ ) = 0. ⟹ ic (0+ ) =
charge qC (t) increases ⟹ vout (t) increases.
VP −0
R
⟹ The capacitor
As vout (t) approaches vin (t) = VP , the voltage across R decreases
⟹ ic (t) decreases ⟹ the capacitor voltage vout (t) increases more slowly .
For yourself you should draw a sequence of plots for each of the above steps.
The end result will look like:
ic(t)
V(t)
Vin(t) - Vout(t)
Vin(t)
VP
VP
R
Vout(t)
Slope =
ic(t)
Vin(t) - Vout(t)
RC
0
0
0
The actual equations are:
t
t
vout (t) = 0 , for ⩽ t  0 , vout (t) = VP (1 − e−τ ) , for t  0
V
t
iC (t) = 0 , for t ⩽ 0 , iC (t) = ( P )e−τ , for t  0 , with τ = RC .
R
24
We obtained the previous results by simple logical reasoning using fundamental
electrical properties, without ever writing any node/loop equations. You should
strive to do this for each problem you are trying to solve. In other words: look at
the problem for a while and “absorb” its structure into your mind. Then, when
you do write equations, you have a better chance to see if the results make
physical sense. Last but not least: Check your units.
F. Another first order transient circuit problem
i1(t)
R1
+
ic(t)
+
- vin(t)
C
R2
vout(t)
-
For this circuit “F”, do for yourself the reasoning of circuit “E”. I could put this
exact circuit “F” on a quiz. So work together with a few classmates and really get
to know the above described scenario of functional reasoning.
G. Last but not least: Review complex analysis and phasors in your
text book. Work examples and exercises.
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5. A RC Time Constant Tutorial
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27
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6. P-I Control Tutorial
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