eecs2200 chapter 7 lect - york university · 2015-10-20 · 1 eecs2200 electric circuits chapter 7...
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EECS2200 Electric Circuits
CChhaapptteerr 77 Sinusoidal Steady State Analysis
Objectives n Understanding phasor concept and be able to
perform phasor transform and inverse phasor transform.
n Be able to transform a circuit with sinusoidal source into the frequency domain using phasor transform
n Know how to use circuits analysis techniques to solve circuits in the frequency domain.
n Be able to use phasor in analyzing circuits with ideal transformers.
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EECS2200 Electric Circuits
Sinusoidal Review
v(t) = Vmcos(ωt + φ) V
Review of Sinusoids
Vm : Amplitude or magnitude ω : angular frequency [rad/s] t : time [s]
φ : phase angle [deg] f = ω /2π : frequency [Hz] T = 1/f : period [s]
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Phase shift of φφ/ωω v(t) =Vm cos(ωt)v(t) =Vm cos(ωt +φ)
Evaluating a Sinusoid at Time t Find i(t) = 0.2 cos(50πt + 45°) at=5 ms. i(0.005) = 0.2 cos(50πt + 45°) A 50π has the units: rad/s 50πt has the units: radians 45° has the units degrees Need to convert 50πt to degrees, or 45 to radians, and set your calculator appropriately! 2π/360° = 50π(0.005)/x° → x = 45° Thus, i(0.005) = 0.2 cos(45° + 45°) A
= 0.2 cos(90°) = 0 A
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RMS value n RMS (root mean square) value of a periodic function is
the square root of the mean value of the squared function
2
)(cos1 0
0
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mrms
Tt
tmrms
VV
dttVT
V
=
+= ∫+
φω
EECS2200 Electric Circuits
Sinusoidal Response
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The equation for i(t) is: L didt+ iR =Vm cos(ωt +ϕ )
The solution is:
i(t) = −VmR2 +ω 2L2
cos(ϕ −θ )e−(R/L )t +Vm
R2 +ω 2L2cos(ωt +ϕ −θ )
θ = tan−1(ωL / R)
Sinusoid Steady-State Analysis Suppose vs(t) = Vmcos(ωt + φ) V, find i(t).
For the response below, which term(s) go to 0 as t → ∞?
A. The first term B. The second term C. Both terms D. Neither term
)cos()(
)cos()(
)(22
)(22
θφωω
θφω
−++
−+−
+
−= − t
LRVe
LRVti mtLRm
Activity 1
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SSiinnuussooiidd SStteeaaddyy--SSttaattee AAnnaallyyssiiss
The first term is the transient term – it decays to 0 as t goes to infinity
The second term is the steady-state term – it persists for all time greater than 0. This term n Is sinusoidal n Has the same frequency as the input voltage n Has a different magnitude and phase angle compared
to the input voltage
)cos()(
)cos()(
)(22
)(22
θφωω
θφω
−++
−+−
+
−= − t
LRVe
LRVti mtLRm
SSiinnuussooiidd SStteeaaddyy--SSttaattee AAnnaallyyssiiss n The circuit analysis techniques we are
studying will allow us to calculate the sinusoidal steady-state response of the circuit – that is, the circuit’s response to a sinusoidal input once the transient response has effectively decayed to 0.
n This is also called the AC Steady-State (ACSS) response.
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EECS2200 Electric Circuits
Phasor
Phasor n A complex number in polar form, with a
magnitude and phase angle n Derived from a sinusoid using the phasor
transform Euler's identity: e± jθ = cosθ ± j sinθ
⇒ cosθ =Re e jθ{ }Now v(t) =Vm cos(ωt +ϕ )
=VmRe e j (ωt+ϕ ){ }=VmRe e jωte jϕ{ }=Re Vme
jϕe jωt{ }
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Phasor Transform n Extracts a sinusoid’s magnitude and phase
angle n Transforms a function of time into a function of
frequency
n Inverse phasor transform n Turns a phasor back into a sinusoid, if someone tells
you the frequency
V =P Vm cos(ωt +ϕ ){ }=Vmejϕ =Vm∠ϕ
V =Vm cosφ + jVm sinφ
{ } { } )cos()( φωφ +=∠== tVVtv mm-1-1 PP V
Activity 2 Suppose
i(t) = 36 cos(4πt + 45°) mA Then the phasor transform of i(t) is A. 36 mA B. 36∠45° mA C. 36∠4πt mA
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Phasor Transform in Circuit Analysis
n Adding or subtracting sinusoids in the time domain is hard (you need trigonometric identities!) But if you phasor-transform the sinusoids, it is easy to combine them.
Activity 3
i1(t) = 20cos(ωt −30°)A, i2 (t) = 40cos(ωt + 60°) AFind i(t) = i1(t)+ i2 (t)
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Solution based on trig. identities
i1(t) = 20cos(ωt −30°)A, i2 (t) = 40cos(ωt + 60°) Ai1(t) = 20cosωt cos30°+ 20sinωt sin30°i2 (t) = 40cosωt cos60°− 40sinωt sin60°i1 + i2 = (20cos30°+ 40cos60°)cosωt + (20sin30°− 40sin60°)sinωt= 37.32cosωt − 24.64sinωt
= 44.72 37.3244.72
cosωt − 24.6444.72
sinωt⎛
⎝⎜
⎞
⎠⎟
= 44.72(cos33.43°cosωt − sin33.43°sinωt)= 44.72cos(ωt +33.43°)
Solution based on Phasor Transform
i1(t) = 20∠−30° = 20cos(−30°)+ j20sin(−30°)= 20 * 0.866− j20 * 0.5=17.32− j10i2 (t) = 40∠60° = 40cos(60°)+ j40sin(60°)= 40 * 0.5+ j40 * 0.866 = 20+ j34.64I = 37.32+ j24.64 = 44.72∠33.43° A∴i(t) =P −1 I{ }= 44.72cos(ωt +33.43°) A
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EECS2200 Electric Circuits
Impedance of R,L,C
RLC in Time and Phasor Domains
Let’s consider the equations that relate voltage and current in a resistor, inductor, and capacitor in the time domain, and in the phasor domain.
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Resistor
Time domain Phasor (frequency) domain
P →
)cos()()()cos()(
θω
θω
+==∴
+=
tRItRitvtIti
m
m
θ
θ
∠==∴
∠=
m
m
RIRI
IVI
Note that the voltage and current phasors for a resistor have the same phase angle – therefore we say that the voltage and current in a resistor are “in phase”.
Inductor
Time domain Phasor (frequency) domain
P →
)90cos()sin(
)()()cos()(
°−+−=
+−=
=∴
+=
θωω
θωω
θω
tLItLI
dttdiLtvtIti
m
m
m
I
VI
LjeLIjjeLI
eeLIeLI
I
jm
jm
jjm
jm
m
ωω
ω
ω
ω
θ
θ
θ
θ
θ
==
−−=
−=
−=∴
∠=
°−
°−
)(
90
)90(
In an inductor, the voltage leads the current by 90°
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Phase Relationship of I & V at L
v=cos
i=sin
Voltage leads
Capacitor
Time domain Phasor (frequency) domain
P →
)90cos()sin(
)()()cos()(
°−+−=
+−=
=∴
+=
θωω
θωω
θω
tCVtCV
dttdvCtitVtv
m
m
m
V
IV
CjeCVjjeCV
eeCVeCV
V
jm
jm
jjm
jm
m
ωω
ω
ω
ω
θ
θ
θ
θ
θ
==
−−=
−=
−=∴
∠=
°−
°−
)(
90
)90(
In a capacitor, the voltage lags the current by 90°
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Phase Relationship of I & V at C
Voltage lags
v=cos
i=sin
Impedance
n In the time domain Resistor: v(t) = Ri(t) inductor: v(t) = Ldi(t)/dt Capacitor: i(t) = Cdv(t)/dt
n In the phasor domain V = ZI Z is impedance, defined as the ratio of V to I Z has the units Ohms [Ω] Resistor: ZR = R Inductor: ZL = jωL Capacitor: ZC = 1/jωC = -j/ωC
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Summary
n In the time domain n Ohm’s law (only for resistors): v = Ri n KVL (around a loop): v1 + v2 + … + vn = 0 n KCL (at a node): i1 + i2 + … + in = 0 n These three laws lead to all other time-domain
circuit analysis techniques
n In the phasor domain n Ohm’s law (for R, L, C): V = ZI n KVL (around a loop): V1 + V2 + … + Vn = 0 n KCL (at a node): I1 + I2 + … + In = 0 n These three laws mean we can use all time-domain
circuit analysis techniques in the phasor domain!
EECS2200 Electric Circuits
Sinusoid Steady-State Analysis
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SSiinnuussooiidd SStteeaaddyy--SSttaattee AAnnaallyyssiiss
Circuit in the time domain
Circuit in the phasor domain
Solution in the time domain
Solution in the phasor domain
P
P -1
Easy – algebra!
Hard – calculus!
Steps in ACSS Analysis: 1. Redraw the circuit (the phasor transform does
not change the components or their connections).
2. Phasor transform all known v(t) and i(t). 3. Represent unknown voltages and currents with
V and I. 4. Replace component values with impedance (Z)
values. 5. Use any circuit analysis method(s) to write
equations and solve them with a calculator. 6. Inverse-transform the result, which is a phasor,
back to the time domain.
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Example is(t) = 8cos(200,000t) A. Find v(t), i1(t), and v2(t) in the steady-state.
1. Redraw the circuit. 2. Phasor transform all
known voltages and currents.
3. Represent unknown voltages and currents with phasor symbols
4. Calculate the impedances of all resistors, inductors, and capacitors.
Ω−=−
=−
=
Ω===
5)1)(000,200(
8)40)(000,200(
jjCjZ
jjLjZ
C
L
µω
µω
Example
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5. Use DC circuit analysis techniques to find the phasor V. In this problem, we’ll simplify the circuit by
replacing all impedances by one equivalent impedance, and use Ohm’s law for phasors.
Zeq =10 || (6+ j8) ||− j51Zeq
=110
+1
6+ j8+1− j5
=10100
+6− j8100
+j20100
=4+ j325
Zeq =254+ j3
= (4− j3)Ω
V = ZI = (4− j3)(8∠0°) = (32− j24)V
Example
6. Inverse phasor-transform the result to get back to the time domain.
{ } V V V
)87.36000,200cos(4087.3640)(87.3640)2432(
1 °−=°−∠=
°−∠=−=− ttvj
PV
Example
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5. Use DC circuit analysis techniques to find the phasor I1. Use current division.
A )4.22.3(
)08(10
)34(
11
j
jZZ
seq
−=
°∠−
== II
Example
6. Inverse phasor-transform the result to get back to the time domain.
{ } A A A
)87.36000,200cos(487.364)(87.364)4.22.3(
11
1
°−=°−∠=
°−∠=−=− ttij
P
I
Example
20
5. Use DC circuit analysis techniques to find the phasor V2. Use voltage division.
V )032(
)2432(868
2
j
jjj
ZZZ
LR
L
+=
−+
=+
= VV
Example
6. Inverse phasor-transform the result to get back to the time domain.
{ } V V V
ttvj
000,200cos32032)(032)032(
12
2
=°∠=
°∠=+=−P
V
Example
21
Suppose we can vary the frequency of the current source in the previous example. What frequency will cause is(t) and v(t) to be in-phase in the steady-state? In-phase means: n Their phase angles are the same. n The equivalent impedance seen by the current source
has a phase angle of 0°. n The equivalent impedance seen by the current source is
purely resistive.
Activity 4
rad/s 000,50)40(3614036
0)36(10100)36(10
)36(10101
)36(10)36(10106036
)6)(6(10)6)(6)((10
)6)(6(10)6(10
)6)(6(10)6)(6(
61
1011
222
2222
22
22
2222
=∴−
=−
=∴
=++−∴=+
++−=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
+++−++
=
−+−+
+−+
−+
−+−+
=
++
+=
ωµ
ω
ωωωω
ωωω
ωωωωω
ωωωωω
ωωω
ωωωω
ωω
LCL
LCLL
LCLZ
LLCjLjL
LjLjLjLjCj
LjLjLj
LjLjLjLj
CjLjZ
eq
eq
IImm
Solution
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Ω=−+=∴
Ω−=−
=−
=
Ω===
=
420||)26(||10
20)1)(000,50(
2)40)(000,50(000,50
jjZ
jjCjZ
jjLjZ
eq
C
L
µω
µω
ω rad/s For
P
Solution
EECS2200 Electric Circuits
Source Transformation
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Source Transformation
Example n Find V0
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Example n Source transform V to I
Example n Source transform I to V
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Example
n Calculate V0
I0 =36− j1212− j16
=12(3− j)4(3− j4)
=3(3− j)(3+ j4)
9+16=3(13+ j9)25
=39+ j2725
=1.56+ j1.08
V0 = (10− j19)I0 = (10− j19)(1.56+ j1.08)= 36.12− j18.84
Example n Find Thevenin equivalent circuit.
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Example n A simplified version of circuit.
Example n Find VTh. First find I and Vx 100 =10I − j40I +120I +10VxVx =100−10I
⇒(130− j40)I +10Vx =100
10I +Vx =100
⎧⎨⎪
⎩⎪
I = −90030− j40
=18∠−126.87°
Vx =100−180∠−126.87° = 208+ j144VVTh =10Vx +120I= 2080+ j1440+120*18∠−126.87°= 784− j288 = 835.22∠− 20.17°
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Example n A circuit for calculating the Thévenin
equivalent impedance.
Example n Calculate ZTh.
Ia =VT
10− j40,Vx =10Ia
Ib =VT −10Vx120
=−VT (9+ j4)120(1− j4)
IT = Ia + Ib
=VT
10− j401− 9+ j4
12⎛
⎝⎜
⎞
⎠⎟
=VT (3− j4)12(10− j40)
ZTh =VTIT= 91.2− j38.4Ω
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Example n The Thévenin equivalent.