Section 1: Sinusoidal Steady-State Analysis

SECTION 1: SINUSOIDAL STEADY-STATE ANALYSIS

ENGR 202 – Electrical Fundamentals II 2 Sinusoids

K. Webb ENGR 202 Sinusoidal Signals 3

 Sinusoidal signals are of particular interest in the field of electrical engineering

= cos + = cos( + )

𝑣𝑣 𝑡𝑡 𝑉𝑉𝑝𝑝 𝜔𝜔𝜔𝜔 𝜙𝜙 𝑉𝑉𝑝𝑝 2𝜋𝜋 ⋅ 𝑓𝑓 ⋅ 𝑡𝑡 𝜙𝜙  Sinusoidal signals defined by three parameters:  Amplitude:  Frequency: or 𝑉𝑉𝑝𝑝  Phase: 𝜔𝜔 𝑓𝑓 K. Webb ENGR 202 𝜙𝜙 Amplitude 4

 Amplitude of a sinusoid is its peak = sin + = sin + voltage, 𝑣𝑣 𝑡𝑡 𝑉𝑉𝑝𝑝 � 𝜔𝜔𝜔𝜔 𝜙𝜙 𝑉𝑉𝑝𝑝 � 2𝜋𝜋𝜋𝜋𝜋𝜋 𝜙𝜙

 Peak-to-peak voltage, 𝑉𝑉𝑝𝑝 , is twice the = 170

amplitude 𝑉𝑉 𝑝𝑝𝑝𝑝 𝑉𝑉𝑝𝑝 𝑉𝑉 𝑉𝑉 0  = 2 34

 = = 𝑉𝑉𝑝𝑝𝑝𝑝 𝑉𝑉𝑝𝑝 𝑝𝑝𝑝𝑝 𝑉𝑉 𝑉𝑉𝑝𝑝𝑝𝑝 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚 − 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚

K. Webb ENGR 202 Frequency 5

 Period ( )  Duration of one cycle 𝑇𝑇  Frequency ( )  Number of periods per second 𝑓𝑓 1 =

𝑓𝑓  Ordinary frequency𝑇𝑇 ,  Units: hertz (Hz), sec-1, cycles/sec 𝑓𝑓  Angular frequency,  Units: rad/sec 𝜔𝜔 = , = = 16 𝜔𝜔 𝑇𝑇 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝜔𝜔 2𝜋𝜋𝜋𝜋 𝑓𝑓 2𝜋𝜋

K. Webb ENGR 202 Phase 6

 Phase  Angular constant in signal expression, = sin + 𝜙𝜙  Requires a time reference 𝑣𝑣 𝑡𝑡 𝑉𝑉𝑝𝑝 𝜔𝜔𝜔𝜔 𝜙𝜙  Interested in relative, not absolute, phase  Here,  leads  lags 𝑣𝑣1 𝑡𝑡 𝑣𝑣2 𝑡𝑡  Units:𝑣𝑣2 𝑡𝑡 radians𝑣𝑣1 𝑡𝑡  Not technically correct, but OK to express in degrees, e.g.: = 170 sin + 34°

K. Webb ENGR 202 𝑣𝑣 𝑡𝑡 𝑉𝑉 2𝜋𝜋 ⋅ 60𝐻𝐻𝐻𝐻 ⋅ 𝑡𝑡 Sinusoidal Steady-State Analysis 7

 Often interested in the response of linear systems to sinusoidal inputs  Voltages and currents in electrical systems  Forces, torques, velocities, etc. in mechanical systems  For linear systems excited by a sinusoidal input  Output is sinusoidal  Same frequency  In general, different amplitude  In general, different phase

 We can simplify the analysis of linear systems by using phasor representation of sinusoids

K. Webb ENGR 202 Phasors 8

 Phasor  A complex number representing the amplitude and phase of a sinusoidal signal  Frequency is not included  Remains constant and is accounted for separately  System characteristics (frequency-dependent) evaluated at the frequency of interest as first step in the analysis  Phasors are complex numbers  Before applying phasors to the analysis of electrical circuits, we’ll first review the properties of complex numbers

K. Webb ENGR 202 9 Complex Numbers

K. Webb ENGR 202 Complex Numbers 10

 A complex number can be represented as = +

 : real part (a real 𝑧𝑧number)𝑥𝑥 𝑗𝑗𝑗𝑗  : imaginary part (a real number) 𝑥𝑥  = 1 is the imaginary unit 𝑦𝑦  Complex𝑗𝑗 − numbers can be represented three ways:  Cartesian form: = +  Polar form: = 𝑧𝑧 𝑥𝑥 𝑗𝑗𝑗𝑗  Exponential form: = 𝑧𝑧 𝑟𝑟푟𝑟𝑟 𝑗𝑗𝑗𝑗 K. Webb 𝑧𝑧 𝑟𝑟𝑒𝑒 ENGR 202 Complex Numbers as Vectors 11

 A complex number can be represented as a vector in the complex plane  Complex plane  Real axis – horizontal  Imaginary axis – vertical  A vector from the origin to  Real part,  Imaginary part, 𝑧𝑧 𝑥𝑥 = + 𝑦𝑦  Vector has𝑧𝑧 a magnitude𝑥𝑥 𝑗𝑗𝑗𝑗 ,  And an angle, 𝑟𝑟 = 𝜃𝜃 𝑧𝑧 𝑟𝑟푟𝑟𝑟 K. Webb ENGR 202 Cartesian Form Polar Form 12  Cartesian form Polar↔ form = + = → 𝑧𝑧 = 𝑥𝑥 =𝑗𝑗𝑗𝑗 𝑟𝑟푟+𝑟𝑟 = arg =2 2 𝑟𝑟 𝑧𝑧 𝑥𝑥 𝑦𝑦 𝜃𝜃 = tan 𝑧𝑧 ∠𝑧𝑧 −1 𝑦𝑦 𝜃𝜃  Polar form Cartesian𝑥𝑥 form = cos → = sin 𝑥𝑥 𝑟𝑟 𝜃𝜃 𝑦𝑦 𝑟𝑟 𝜃𝜃 K. Webb ENGR 202 Complex Numbers – Addition/Subtraction

 Addition and subtraction of complex numbers  Best done in Cartesian form  Real parts add/subtract  Imaginary parts add/subtract  For example: = + = + 𝑧𝑧1 𝑥𝑥1 𝑗𝑗𝑦𝑦1 + = + + + 𝑧𝑧2 𝑥𝑥2 𝑗𝑗𝑦𝑦2 = + 𝑧𝑧1 𝑧𝑧2 𝑥𝑥1 𝑥𝑥2 𝑗𝑗 𝑦𝑦1 𝑦𝑦2 1 2 1 2 1 2 K. Webb 𝑧𝑧 − 𝑧𝑧 𝑥𝑥 − 𝑥𝑥 𝑗𝑗 𝑦𝑦 − 𝑦𝑦 ENGR 202 Complex Numbers – Multiplication/Division

 Multiplication and division of complex numbers  Best done in polar form  Magnitudes multiply/divide  Angles add/subtract  For example: = = 𝑧𝑧1 𝑟𝑟1∠𝜃𝜃1 = + 𝑧𝑧2 𝑟𝑟2∠𝜃𝜃2 𝑧𝑧1 ⋅=𝑧𝑧2 𝑟𝑟1𝑟𝑟2∠ 𝜃𝜃1 𝜃𝜃2 𝑧𝑧1 𝑟𝑟1 ∠ 𝜃𝜃1 − 𝜃𝜃2 K. Webb 𝑧𝑧2 𝑟𝑟2 ENGR 202 Complex Conjugate 15

 Conjugate of a complex number  Number that results from negating the imaginary part = + = 𝑧𝑧 𝑥𝑥 𝑗𝑗𝑗𝑗 ∗  Or, equivalently,𝑧𝑧 𝑥𝑥 − 𝑗𝑗𝑗𝑗 from negating the angle = = 𝑧𝑧 𝑟𝑟푟𝑟𝑟 ∗ K. Webb 𝑧𝑧 𝑟𝑟푟 − 𝜃𝜃 ENGR 202 Complex Fractions 16

 Multiplying a number by its complex conjugate yields the squared magnitude of that number  A real number = + = + ∗ 2 2 𝑧𝑧 ⋅ 𝑧𝑧 = 𝑥𝑥 𝑗𝑗𝑗𝑗 𝑥𝑥 − =𝑗𝑗𝑗𝑗 𝑥𝑥 𝑦𝑦= ∗ 2 2  Rationalizing𝑧𝑧 ⋅ the𝑧𝑧 denominator𝑟𝑟푟𝑟𝑟 ⋅ 𝑟𝑟푟 − 𝜃𝜃 of𝑟𝑟 a ∠complex𝜃𝜃 − 𝜃𝜃 fraction:𝑟𝑟  Multiply numerator and denominator by the complex conjugate of the denominator + = + 𝑥𝑥1 𝑗𝑗𝑦𝑦1 𝑥𝑥2 − 𝑗𝑗𝑦𝑦2 𝑧𝑧 2 + 2 ⋅ 2 2 = 𝑥𝑥 𝑗𝑗𝑦𝑦 𝑥𝑥 +− 𝑗𝑗𝑦𝑦 + + 𝑥𝑥1𝑥𝑥2 𝑦𝑦1𝑦𝑦2 𝑥𝑥2𝑦𝑦1 − 𝑥𝑥1𝑦𝑦2 𝑧𝑧 2 2 𝑗𝑗 2 2 K. Webb 𝑥𝑥2 𝑦𝑦2 𝑥𝑥2 𝑦𝑦2 ENGR 202 Complex Fractions 17

 Fractions or ratios are, of course, simply division  Very common form, so worth emphasizing  Magnitude of a ratio of complex numbers

= = 𝑧𝑧1 𝑧𝑧1 𝑧𝑧 → 𝑧𝑧  Angle of a ratio of complex𝑧𝑧2 numbers 𝑧𝑧2

= = 𝑧𝑧1 𝑧𝑧 → ∠𝑧𝑧 ∠𝑧𝑧1 − ∠𝑧𝑧2 𝑧𝑧2  Calculators and complex numbers  Manipulation of complex numbers by hand is tedious and error-prone  Your calculators can perform complex arithmetic  They will operate in both Cartesian and polar form, and will convert between the two  Learn to use them – correctly

K. Webb ENGR 202 18 Phasors

K. Webb ENGR 202 Euler’s Identity 19

 Fundamental to phasor notation is Euler’s identity: = cos + sin 𝑗𝑗𝑗𝑗𝑗𝑗 where is the imaginary𝑒𝑒 unit,𝜔𝜔 and𝜔𝜔 𝑗𝑗is angular𝜔𝜔𝜔𝜔 frequency  It follows that 𝑗𝑗 𝜔𝜔 cos = 𝑗𝑗𝑗𝑗𝑗𝑗 sin 𝜔𝜔𝜔𝜔 = 𝑅𝑅𝑅𝑅 𝑒𝑒 and 𝑗𝑗𝑗𝑗𝑗𝑗 𝜔𝜔𝜔𝜔 𝐼𝐼𝐼𝐼 +𝑒𝑒 cos = 𝑗𝑗𝑗𝑗𝑗𝑗 2 −𝑗𝑗𝑗𝑗𝑗𝑗 𝑒𝑒 𝑒𝑒 𝜔𝜔𝜔𝜔 sin = 𝑗𝑗𝑗𝑗𝑗𝑗 −𝑗𝑗𝑗𝑗𝑗𝑗 𝑒𝑒 − 𝑒𝑒 𝜔𝜔𝜔𝜔 K. Webb 2𝑗𝑗 ENGR 202 Phasors 20

 Consider a sinusoidal voltage = cos +

 Using Euler’s identity,𝑣𝑣 𝑡𝑡 we𝑉𝑉𝑝𝑝 can represent𝜔𝜔𝜔𝜔 𝜙𝜙 this as = = where 𝑗𝑗 𝜔𝜔𝜔𝜔+𝜙𝜙 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗𝑗𝑗𝑗 𝑣𝑣 𝑡𝑡 𝑅𝑅𝑅𝑅 𝑉𝑉𝑝𝑝𝑒𝑒 𝑅𝑅𝑅𝑅 𝑉𝑉𝑝𝑝𝑒𝑒 𝑒𝑒  represents magnitude  represents phase 𝑝𝑝  𝑉𝑉𝑗𝑗𝑗𝑗 represents a sinusoid of frequency 𝑒𝑒𝑗𝑗𝑗𝑗𝑗𝑗  Grouping𝑒𝑒 the first two terms together,𝜔𝜔 we have = 𝑗𝑗𝑗𝑗𝑗𝑗 where is the phasor𝑣𝑣 representation𝑡𝑡 𝑅𝑅𝑅𝑅 𝐕𝐕𝑒𝑒 of K. Webb 𝐕𝐕 𝑣𝑣 𝑡𝑡 ENGR 202 Phasors 21 = 𝑗𝑗𝑗𝑗𝑗𝑗  The phasor representation𝑣𝑣 𝑡𝑡 of 𝑅𝑅𝑅𝑅 𝐕𝐕𝑒𝑒

=𝑣𝑣 𝑡𝑡 𝑗𝑗𝑗𝑗  A representation of magnitude𝐕𝐕 𝑉𝑉and𝑝𝑝𝑒𝑒 phase only  Time-harmonic portion ( ) has been dropped 𝑗𝑗𝑗𝑗𝑗𝑗 Time-domain 𝑒𝑒 Phasor-domain representation: representation:

= sin + = = 𝑗𝑗𝑗𝑗 𝑣𝑣 𝑡𝑡 𝑉𝑉𝑝𝑝 𝜔𝜔𝜔𝜔 𝜙𝜙 𝐕𝐕 𝑉𝑉𝑝𝑝𝑒𝑒 𝑉𝑉𝑝𝑝∠𝜙𝜙  Phasors greatly simplify sinusoidal steady-state analysis  Messy trigonometric functions are eliminated  Differentiation and integration transformed to algebraic operations

K. Webb ENGR 202 Voltage & Current in the Phasor Domain

 We will use phasors to simplify analysis of electrical circuits  Need an understanding of electrical component behavior in the phasor domain  Relationships between voltage phasors and current phasors for Rs, Ls, and Cs

 Resistor  Voltage across a resistor given by =

𝑣𝑣 𝑡𝑡 = 𝑖𝑖 𝑡𝑡cos𝑅𝑅 +  Converting to phasor𝑖𝑖 𝑡𝑡 form𝐼𝐼𝑝𝑝 𝜔𝜔𝜔𝜔 𝜙𝜙 = 𝑗𝑗𝑗𝑗 𝐕𝐕 𝐼𝐼𝑝𝑝𝑒𝑒 𝑅𝑅 = = R 𝐕𝐕  Ohm’s law in phasor𝐕𝐕 form𝐈𝐈𝑅𝑅 𝐈𝐈

K. Webb ENGR 202 V-I Relationships in the Phasor Domain 23

 Capacitor  Current through the capacitor given by

= 𝑑𝑑𝑑𝑑 𝑖𝑖 𝑡𝑡 𝐶𝐶 = 𝑑𝑑𝑑𝑑 cos + 𝑑𝑑 𝑖𝑖 𝑡𝑡 = 𝐶𝐶 𝑉𝑉𝑝𝑝sin 𝜔𝜔𝜔𝜔+ 𝜙𝜙 𝑑𝑑𝑑𝑑  Applying a trig identity:𝑖𝑖 𝑡𝑡 −𝜔𝜔𝜔𝜔𝑉𝑉𝑝𝑝 𝜔𝜔𝜔𝜔 𝜙𝜙 sin = cos + 90° gives − =𝐴𝐴 cos𝐴𝐴 + + 90°

 Converting to phasor𝑖𝑖 𝑡𝑡 form𝜔𝜔𝜔𝜔𝑉𝑉𝑝𝑝 𝜔𝜔𝜔𝜔 𝜙𝜙 = = 𝑗𝑗 𝜙𝜙+90° 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗� 𝐈𝐈 𝜔𝜔𝜔𝜔𝑉𝑉𝑝𝑝𝑒𝑒 𝜔𝜔𝜔𝜔𝑉𝑉𝑝𝑝𝑒𝑒 𝑒𝑒 K. Webb ENGR 202 V-I Relationships - Capacitor 24

 Current phasor = = 𝑗𝑗 𝜙𝜙+90° 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗�  Voltage phasor𝐈𝐈 𝜔𝜔𝜔𝜔 is𝑉𝑉𝑝𝑝𝑒𝑒 𝜔𝜔𝜔𝜔𝑉𝑉𝑝𝑝𝑒𝑒 𝑒𝑒 = so 𝑗𝑗𝑗𝑗 𝐕𝐕 𝑉𝑉𝑝𝑝𝑒𝑒 = 𝑗𝑗𝑗�  Recognizing𝐈𝐈 that𝜔𝜔𝜔𝜔𝐕𝐕 𝑒𝑒 = , we have 𝑗𝑗𝑗� 𝑒𝑒 𝑗𝑗 1 = =

K. Webb 𝐈𝐈 𝑗𝑗𝑗𝑗𝑗𝑗𝐕𝐕 𝐕𝐕 𝐈𝐈 ENGR 202 𝑗𝑗𝑗𝑗𝑗𝑗 V-I Relationships - Inductor 25

 Inductor  Voltage across an inductor given by

= 𝑑𝑑𝑖𝑖 𝑣𝑣 𝑡𝑡 𝐿𝐿 = 𝑑𝑑𝑑𝑑 cos + 𝑑𝑑 𝑣𝑣 𝑡𝑡 = 𝐿𝐿 𝐼𝐼𝑝𝑝sin 𝜔𝜔𝜔𝜔+ 𝜙𝜙 = cos + + 90° 𝑑𝑑𝑑𝑑  Converting to𝑣𝑣 𝑡𝑡phasor−𝜔𝜔 form𝐿𝐿𝐼𝐼𝑝𝑝 𝜔𝜔𝜔𝜔 𝜙𝜙 𝜔𝜔𝐿𝐿𝐼𝐼𝑝𝑝 𝜔𝜔𝜔𝜔 𝜙𝜙 = = 𝑗𝑗 𝜙𝜙+90° 𝑗𝑗𝑗𝑗 𝑗𝑗𝑗�  Again, recognizing𝐕𝐕 𝜔𝜔𝐿𝐿 that𝐼𝐼𝑝𝑝𝑒𝑒 = , gives𝜔𝜔𝐿𝐿𝐼𝐼𝑝𝑝𝑒𝑒 𝑒𝑒 𝑗𝑗𝑗� 𝑒𝑒 𝑗𝑗 1 = =

K. Webb 𝐕𝐕 𝑗𝑗𝑗𝑗𝐿𝐿𝐈𝐈 𝐈𝐈 𝐕𝐕 ENGR 202 𝑗𝑗𝑗𝑗𝑗𝑗 26 Impedance

K. Webb ENGR 202 Impedance 27

 For resistors, Ohm’s law gives the ratio of the voltage phasor to the current phasor as

= 𝐕𝐕  is, of course, resistance 𝑅𝑅 𝐈𝐈  A special case of impedance 𝑅𝑅  Impedance,

𝑍𝑍 = 𝐕𝐕  The ratio of the voltage phasor𝑍𝑍 to the current phasor for a component or network 𝐈𝐈  Units: ohms ( )  In general, complex-valued Ω K. Webb ENGR 202 Impedance 28

 Resistor impedance:

= = 𝐕𝐕 𝑍𝑍 𝑅𝑅  Capacitor impedance: 𝐈𝐈 1 = = 𝐕𝐕 𝑍𝑍  Inductor impedance: 𝐈𝐈 𝑗𝑗𝑗𝑗𝑗𝑗

= = 𝐕𝐕 𝑍𝑍 𝑗𝑗𝑗𝑗𝑗𝑗  In general, Ohm’s law can be applied𝐈𝐈 to any component or network in the phasor domain

= = 𝐕𝐕 𝐕𝐕 𝐈𝐈𝑍𝑍 𝐈𝐈 𝑍𝑍 K. Webb ENGR 202 29 Capacitor Impedance

K. Webb ENGR 202 Capacitor Impedance 30 1 1 = = −𝑗𝑗𝑗� 𝑍𝑍 𝑒𝑒 = 𝑗𝑗𝑗𝑗𝑗𝑗= 𝜔𝜔𝜔𝜔 𝐈𝐈 −𝑗𝑗𝑗� 𝐕𝐕= 𝐈𝐈𝑍𝑍 𝑒𝑒 𝜔𝜔𝜔𝜔 𝑗𝑗𝑗�  In the time𝐈𝐈 𝜔𝜔domain,𝜔𝜔𝐕𝐕𝑒𝑒 this translates to = cos +

𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑝𝑝 cos𝜔𝜔𝜔𝜔 𝜙𝜙+ + 90° 𝑝𝑝  Current𝑖𝑖 through𝑡𝑡 𝑉𝑉 𝜔𝜔 a𝜔𝜔 capacitor𝜔𝜔𝜔𝜔 𝜙𝜙 leads the voltage across a capacitor by 90°

K. Webb ENGR 202 Capacitor Impedance – Phasor Diagram 31

 Phasor diagram for a capacitor  Voltage and current phasors drawn as vectors in the complex plane  Current always leads voltage by 90°

K. Webb ENGR 202 Capacitor Impedance – Time Domain 32

 Current leads voltage by 90°

= 90°

Δ𝜙𝜙

K. Webb ENGR 202 Capacitor Impedance – Frequency Domain

Capacitor approaches an open circuit at DC

Capacitor approaches a short circuit at very high frequencies

K. Webb ENGR 202 34 Inductor Impedance

K. Webb ENGR 202 Inductor Impedance 35 = = = = 𝑗𝑗𝑗� 𝑍𝑍 𝑗𝑗𝑗𝑗𝐿𝐿 𝜔𝜔𝜔𝜔 𝑒𝑒 𝑗𝑗𝑗� 𝐕𝐕= 𝐈𝐈𝑍𝑍 𝐈𝐈𝜔𝜔𝜔𝜔 𝑒𝑒 𝐕𝐕 −𝑗𝑗𝑗� 𝐈𝐈 𝑒𝑒  In the time 𝜔𝜔domain,𝜔𝜔 this translates to = cos +

𝑝𝑝 𝑣𝑣 𝑡𝑡 = 𝑉𝑉 cos 𝜔𝜔𝜔𝜔 +𝜙𝜙 90° 𝑉𝑉𝑝𝑝 𝑖𝑖 𝑡𝑡 𝜔𝜔𝜔𝜔 𝜙𝜙 −  Current through𝜔𝜔𝜔𝜔 an inductor lags the voltage across an inductor by 90°

K. Webb ENGR 202 Inductor Impedance – Phasor Diagram 36

 Phasor diagram for an inductor  Voltage and current phasors drawn as vectors in the complex plane  Current always lags voltage by 90°

K. Webb ENGR 202 Inductor Impedance – Time Domain 37

 Current lags voltage by 90°

= 90°

Δ𝜙𝜙

K. Webb ENGR 202 Inductor Impedance – Frequency Domain

Inductor approaches an open circuit at very high frequencies

Inductor approaches a short circuit at DC

K. Webb ENGR 202 Summary 39 Capacitor Inductor  Impedance:  Impedance:

1 = = 𝑍𝑍𝐿𝐿 𝑗𝑗𝑗𝑗𝑗𝑗 𝑍𝑍𝑐𝑐  V-I phase relationship:  V-I phase relationship:𝑗𝑗𝑗𝑗𝑗𝑗 Current lags voltage by 90° Current leads voltage by 90° = cos = cos 𝑝𝑝 𝑣𝑣 𝑡𝑡 = 𝑉𝑉 cos 𝜔𝜔𝜔𝜔 90° = 𝑝𝑝 cos + 90° 𝑣𝑣 𝑡𝑡 𝑉𝑉 𝜔𝜔𝜔𝜔 𝑉𝑉𝑝𝑝 𝑝𝑝 𝑖𝑖 𝑡𝑡 𝜔𝜔𝜔𝜔 − K. Webb𝑖𝑖 𝑡𝑡 𝑉𝑉 𝜔𝜔𝜔𝜔 𝜔𝜔𝜔𝜔 𝜔𝜔𝐿𝐿 ENGR 202 ELI the ICE Man 40

 Mnemonic for phase relation between current (I) and voltage (E) in inductors (L) and capacitors (C)

E L I the I C E man

Voltage Current leads current leads voltage in an inductor in a capacitor

K. Webb ENGR 202 41 Example Problems

K. Webb ENGR 202 Convert each of the following time-domain signals to phasor form. = cos + 12°

𝑣𝑣 𝑡𝑡 = 6𝑉𝑉 ⋅ sin2𝜋𝜋 ⋅1008𝑘𝑘𝑘𝑘𝑘𝑘 ⋅ 𝑡𝑡 38° 𝑖𝑖 𝑡𝑡 200𝑚𝑚𝑚𝑚 ⋅ ⋅ 𝑡𝑡 −

K. Webb ENGR 202 Convert the following circuit to the phasor domain.

K. Webb ENGR 202 K. Webb ENGR 202 The following current is applied to the capacitor. = cos Find the voltage across the capacitor, . 𝑖𝑖 𝑡𝑡 100𝑚𝑚𝑚𝑚 ⋅ 2𝜋𝜋 ⋅ 50𝑘𝑘𝑘𝑘𝑘𝑘 ⋅ 𝑡𝑡 𝑣𝑣 𝑡𝑡

K. Webb ENGR 202 K. Webb ENGR 202 The following voltage is applied to the inductor. = cos Find the current through the inductor,𝑣𝑣 𝑡𝑡 4𝑉𝑉 ⋅. 2𝜋𝜋 ⋅ 800𝐻𝐻𝐻𝐻 ⋅ 𝑡𝑡

𝑖𝑖 𝑡𝑡

K. Webb ENGR 202 A test voltage is applied to the input of an electrical network. = sin 5 The input current is measured. 𝑣𝑣 𝑡𝑡 1𝑉𝑉 ⋅ 2𝜋𝜋 ⋅ 𝑘𝑘𝑘𝑘𝑧𝑧 ⋅ 𝑡𝑡 = sin 46° What is the circuit’s input impedance, ? 𝑖𝑖 𝑡𝑡 268𝑚𝑚𝑚𝑚 ⋅ 2𝜋𝜋 ⋅ 5𝑘𝑘𝑘𝑘𝑘𝑘 ⋅ 𝑡𝑡 − 𝑍𝑍𝑖𝑖𝑖𝑖

K. Webb ENGR 202 49 Impedance of Arbitrary Networks

K. Webb ENGR 202 Impedance 50

 So far, we’ve looked at impedance of individual components  Resistors =  Purely real 𝑍𝑍 𝑅𝑅  Capacitors 1 =

 Purely imaginary, purely𝑍𝑍 reactive 𝑗𝑗𝑗𝑗𝑗𝑗  Inductors =

 Purely imaginary, purely reactive 𝑍𝑍 𝑗𝑗𝑗𝑗𝑗𝑗

K. Webb ENGR 202 Impedance 51

 Also want to be able to characterize the impedance of electrical networks  Multiple components  Some resistive, some reactive  In general, impedance is a complex value = + where  is resistance 𝑍𝑍 𝑅𝑅 𝑗𝑗𝑗𝑗  is reactance 𝑅𝑅  So, in𝑋𝑋 ENGR 201 we dealt with impedance all along  Resistance is an impedance whose reactance (imaginary part) is zero  A purely real impedance

K. Webb ENGR 202 Reactance 52

 For capacitor and inductors, impedance is purely reactive  Resistive part is zero = and =

where is𝑍𝑍 𝑐𝑐capacitive𝑗𝑗𝑋𝑋𝑐𝑐 reactance𝑍𝑍𝐿𝐿 𝑗𝑗𝑋𝑋𝐿𝐿 𝑋𝑋𝑐𝑐 = 1 and is inductive𝑋𝑋reactance𝑐𝑐 − 𝜔𝜔𝜔𝜔

𝑋𝑋𝐿𝐿 =  Note that reactance𝑋𝑋𝐿𝐿is a 𝜔𝜔real𝜔𝜔 quantity  It is the imaginary part of impedance

 Units of reactance: ohms ( )

K. Webb Ω ENGR 202 Admittance 53

 Admittance, , is the inverse of impedance 1 𝑌𝑌 = = + where 𝑌𝑌 𝐺𝐺 𝑗𝑗𝑗𝑗 is conductance – the real part𝑍𝑍 is susceptance – the imaginary part 𝐺𝐺 𝐵𝐵 1 = = + + + + 𝑅𝑅 −𝑋𝑋  Conductance 𝑌𝑌 2 2 𝑗𝑗 2 2 𝑅𝑅 𝑗𝑗𝑗𝑗 𝑅𝑅 𝑋𝑋 𝑅𝑅 𝑋𝑋 = + 𝑅𝑅 𝐺𝐺 2 2  Note that 1/ unless = 0 𝑅𝑅 𝑋𝑋  Susceptance 𝐺𝐺 ≠ 𝑅𝑅 𝑋𝑋 = + −𝑋𝑋 𝐵𝐵 2 2  Units of , , and : Siemens (S) 𝑅𝑅 𝑋𝑋

K. Webb 𝑌𝑌 𝐺𝐺 𝐵𝐵 ENGR 202 Impedance of Arbitrary Networks 54

 In general, the impedance of arbitrary networks may be both resistive and reactive

= +

𝑍𝑍 = 𝑅𝑅1 𝑗𝑗𝑋𝑋1 where 𝑍𝑍 𝑍𝑍 ∠𝜃𝜃 = + 2 2 and 𝑍𝑍 𝑅𝑅1 𝑋𝑋1 = tan −1 𝑋𝑋1 𝜃𝜃 1 K. Webb 𝑅𝑅 ENGR 202 Impedances in Series 55

 Impedances in series add

= +

𝑍𝑍𝑒𝑒𝑒𝑒 = 𝑍𝑍1 +𝑍𝑍2 + +

𝑒𝑒𝑒𝑒 1 2 1 2 K. Webb 𝑍𝑍 𝑅𝑅 𝑅𝑅 𝑗𝑗 𝑋𝑋 𝑋𝑋 ENGR 202 Impedances in Parallel 56

 Admittances in parallel add

= +

𝑌𝑌𝑒𝑒𝑒𝑒 𝑌𝑌11 𝑌𝑌2 1 1 = = + −1 𝑍𝑍𝑒𝑒𝑒𝑒 𝑌𝑌𝑒𝑒𝑒𝑒 𝑍𝑍1 𝑍𝑍2 = 1+ 2 𝑒𝑒𝑒𝑒 𝑍𝑍 𝑍𝑍 𝑍𝑍 1 2 K. Webb 𝑍𝑍 𝑍𝑍 ENGR 202 57 Sinusoidal Steady-State Analysis

K. Webb ENGR 202 Sinusoidal Steady-State Analysis – Ex. 1 58

 Determine the current, = 1 cos 1 𝑖𝑖 𝑡𝑡 𝑠𝑠  First,𝑣𝑣 𝑡𝑡 convert𝑉𝑉 the2𝜋𝜋 circuit⋅ 𝑀𝑀 𝑀𝑀𝑀𝑀to the⋅ 𝑡𝑡 phasor domain  Express the source voltage as a phasor = 1

 Evaluate𝐕𝐕𝐬𝐬 impedances𝑉𝑉𝑉� at 1 MHz = 10 1 𝑅𝑅 = Ω = 1 10 𝑗𝑗 𝑍𝑍𝑐𝑐 − = 𝑗𝑗𝑗𝑗𝑗𝑗 .9 2𝜋𝜋 ⋅ 𝑀𝑀𝑀𝑀𝑀𝑀 ⋅ 𝑛𝑛𝑛𝑛

K. Webb 𝑍𝑍𝑐𝑐 −𝑗𝑗𝑗 Ω ENGR 202 Sinusoidal Steady-State Analysis – Ex. 1 59

 The load impedance is = + = 10 .9

𝑍𝑍 = 𝑅𝑅18. 𝑗𝑗𝑋𝑋𝑐𝑐 57.8° − 𝑗𝑗𝑗 Ω  Apply Ohm’s𝑍𝑍 law8∠ − to calculateΩ the current phasor 1 = = 18. 57.8° 𝐕𝐕 𝑉𝑉𝑉� 𝐈𝐈 = 53𝑍𝑍 . 8.∠8°− Ω

 Finally, 𝐈𝐈convert2 ∠to57 the time𝑚𝑚𝑚𝑚 domain = 53.2 cos( + 57.8°)

𝑖𝑖 𝑡𝑡 𝑚𝑚𝑚𝑚 2𝜋𝜋 ⋅ 1𝑀𝑀𝑀𝑀𝑀𝑀 ⋅ 𝑡𝑡 K. Webb ENGR 202 Sinusoidal Steady-State Analysis – Ex. 2 60

 Consider the following circuit, modeling a source driving a load through a transmission line

= 170 sin

𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉 2𝜋𝜋 ⋅ 60𝐻𝐻𝐻𝐻 ⋅ 𝑡𝑡  Determine:  The impedance, , at 60 Hz  Voltage across the load, 𝑖𝑖𝑖𝑖  Current delivered𝑍𝑍 to the load, 𝑣𝑣𝐿𝐿 𝑡𝑡 K. Webb 𝑖𝑖𝐿𝐿 𝑡𝑡 ENGR 202 Sinusoidal Steady-State Analysis – Ex. 2 61

 First, convert to the phasor domain and evaluate impedances at 60 Hz

 The line impedance is = + = 0.5 + .88

 The load impedance𝑍𝑍𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑅𝑅1 is𝑗𝑗𝑗𝑗𝐿𝐿1 𝑗𝑗� Ω 1 = + || = 3 + .65 ||

𝑍𝑍𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑅𝑅2 𝑗𝑗𝑗𝑗𝐿𝐿2 𝑗𝑗� Ω − 𝑗𝑗𝑗� Ω 1 𝑗𝑗𝑗𝑗𝑗𝑗 1 = + = 3.13 + .74 3 + .65 −1 𝑍𝑍𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑗𝑗� Ω K. Webb 𝑗𝑗� Ω −𝑗𝑗𝑗� Ω ENGR 202 Sinusoidal Steady-State Analysis – Ex. 2 62

 The impedance seen by the source is = + = 0.5 + .88 + 3.13 + .74 𝑍𝑍𝑖𝑖𝑖𝑖 𝑍𝑍𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑍𝑍𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 3.63 + .62 𝑍𝑍𝑖𝑖𝑖𝑖 𝑗𝑗� Ω 𝑗𝑗� Ω  In polar form𝑍𝑍𝑖𝑖𝑖𝑖: 𝑗𝑗� Ω = 8. .5°

 The impedance𝑍𝑍𝑖𝑖𝑖𝑖 driven44∠64 by theΩ source looks resistive and inductive  Resistive: non-zero resistance, ±90°  Inductive: positive reactance, positive angle ∠𝑍𝑍𝑖𝑖𝑖𝑖 ≠ K. Webb ENGR 202 Sinusoidal Steady-State Analysis – Ex. 2 63

 Apply voltage division to determine the voltage across the load = 𝑙𝑙𝑙𝑙+𝑙𝑙𝑙𝑙 𝐿𝐿 𝑆𝑆 𝑍𝑍 𝐕𝐕 𝐕𝐕 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 3.𝑙𝑙𝑙𝑙13𝑙𝑙𝑙𝑙+ .74 = 170𝑍𝑍 𝑍𝑍 3.63 + .62 𝐿𝐿 𝑗𝑗� Ω 𝐕𝐕 ∠0° 𝑉𝑉 6. .4° = 170 𝑗𝑗� Ω= 3.1° 8. .5° 54∠61 Ω 𝐕𝐕𝐿𝐿 ∠0° 𝑉𝑉 132∠ − 𝑉𝑉  Converting to time-domain44∠64formΩ = 132 sin 3.1°

𝐿𝐿 K. Webb 𝑣𝑣 𝑡𝑡 𝑉𝑉 2𝜋𝜋 ⋅ 60𝐻𝐻𝐻𝐻 ⋅ 𝑡𝑡 − ENGR 202 Sinusoidal Steady-State Analysis – Ex. 2 64

 Finally, calculate the current delivered to the load

= 𝐿𝐿 𝐿𝐿 𝐕𝐕 𝐈𝐈 132𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 3.1° = 𝑍𝑍 6. .4° ∠ − 𝑉𝑉 𝐈𝐈𝐿𝐿 = 20. 64.5° 54∠61 Ω  In time𝐈𝐈𝐿𝐿-domain1∠ −form: 𝐴𝐴 = 20.1 sin 64.5°

𝑖𝑖𝐿𝐿 𝑡𝑡 𝐴𝐴 2𝜋𝜋 ⋅ 60𝐻𝐻𝐻𝐻 ⋅ 𝑡𝑡 − K. Webb ENGR 202 65 Example Problems

K. Webb ENGR 202 Determine the input impedance and an equivalent circuit model for the following network at 50 kHz.

K. Webb ENGR 202 K. Webb ENGR 202