EECS2200 Chapter 7 lect - York University · 2015-10-20 · 1 EECS2200 Electric Circuits Chapter 7...

1

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.

2


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]

3

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

4

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

22

mrms

Tt

tmrms

VV

dttVT

V

=

+= ∫+

φω


Sinusoidal Response

5

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

6

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.

7


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{ }

8

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

9

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)

10

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

11


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.

12

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


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


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°

14

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

15

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!


Sinusoid Steady-State Analysis

16

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

18

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

19

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


{ } 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


{ } 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


Source Transformation

23

Source Transformation

Example n  Find V0

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Example n  Source transform V to I

Example n  Source transform I to V

25

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.

26

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.

EECS2200 Chapter 7 lect - York University · 2015-10-20 · 1 EECS2200 Electric Circuits Chapter 7...

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