ECE410 Spring 2012

Lecture #32 AC Circuits I

Homework Due 3/16/2012

• Chapter 7 – Problems 33, 35, 48, 50, 55, 70, 73, 74, 78, 89

• PSpice – Model the circuit from problem 89 from your homework assignment. This circuit is shown below as well as the input waveform:

Homework (cont)• Plot voltage v0 vs. time.

• Ignore the switch in the circuit and model the input waveform using VPWL, a piecewise linear voltage source. Using this source you can specify the voltage at multiple times (t1,t2,t3,…) and the corresponding voltages at those times (v1, v2, v3, …) and the points will be connected by straight lines. In order to model the vertical transitions use two times that are close together.. (let t1=0, v1=0; t2=1ns, v2=-200mV) to simulate the initial voltage change from 0V to -200mV at time zero.

• My suggestion:

• Run the simulation for 750ms• Use a LM741 or similar op-amp in your simulation

Time Voltage0.000000000 00.000000001 -0.20.250000000 -0.20.250000001 20.500000000 20.500000001 0

Midterm Exam #3

• Friday April 27th

• Will Cover Chapters 6, 7 and 9– Capacitance– Inductance– RC and RL circuits– Sinusoidal Steady State Analysis

• Use 1-3x5 Notecard (both sides)• Can bring a calculator and writing utensil• Be ready to go at the beginning of the class period

AC Waveforms

• AC waveforms are sinusoidal and can be described by either a sine or cosine.. We will use the cosine convention.

tVtv m cos)(

Vm = Amplitudeω = angular frequency (equal to 2πf or 2π/T)φ = phase angle

phase

Effect of Phase Angle

• The phase angle shifts the cosine wave to the right or left in time. tVtv m cos)(

Unshifted cosine

Cosine with non-zero phase angle

Cosine is equal to 1 when the phase is equal to zero. Thus:

t

t 0

The wave with phase angle φ with be shifted in time by:

Sine-Cosine Relations

• The following are useful relationships for AC circuits:

• Shifting a sine by 90 degrees results in ±cosine• Shifting a cosine by 90 degrees results in ±sine

)2

cos()2

cos()sin(

)2

sin()2

sin()cos(

ttt

ttt

RMS Values

• AC sinusoids have an average of zero, so we use RMS (root of the mean squared function) to get a measure for effective average value.

Tt

t

mrms dttVT

V0

0

2cos1

2m

rms

VV

For sinusoidal signals

Tt

t

rms dttVT

V0

0

2)(1

RMS calculation for a non-sinusoid

• Find RMS value for a Square Wave:

0 0.5 1 1.5 2 2.5

-1.5

-1

-0.5

0

0.5

1

1.5

Let Amplitude = Vmax

Tt

t

rms dttVT

V0

0

2)(1

T

T

T

rms dtVT

dtVT

V2/

2max

2/

0

2max

11

22

2max2

max

2max V

VV

Vrms

2

10

2

1 2max

2max

2max

TVTV

T

TV

TVrms

maxVVrms

Complex Number Review

• Complex numbers involve numbers with imaginary terms (involving j=sqrt(-1))

• They can be expressed in polar form or rectangular form– Rectangular x = A+jB– Polar

• Complex numbers can be plotted in the complex plane

jeCx

Complex Plane

Real

Imaginary A complex number is a vector in the complex plane

It can be expressed in terms of it’s real and imaginary components… this is the rectangular form:

X = A + jB

Or it can be expressed in terms of it’s angle to the real axis and it’s length… this is the polar form:

φ

A

B

C

jeCx

22 BAC

B

A1tanwhere

Conversion Between Polar and Rectangular Notation

Changing from Polar to Rectangular

• Given length C and angle φChanging from Rectangular to Polar

• Given Real component A and imaginary component B

22 BAC

A

B1tan

jBAx

A

Bj

eBAjBA1tan

22

jCex

cosCA

sinCB

sincos jCCCe j

Real and Imaginary Parts

• The Real Part of a complex number is it’s vector component in real direction

• The Imaginary Part of a complex number is it’s component in the imaginary direction

Real

Imaginary

φ

A

B

C AjBA

BjBA

Sinusoidal Response

• What happens if we drive a circuit with capacitance or inductance with a sinusoidal voltage source?

tVv ms cos

tVRidt

diL m cos

tLR

Ve

LR

Vi m

tL

R

m coscos222222

Full Solution:

Characteristics of Response

• The transient solution only lasts for a short time• The steady state solution is a sinusoid• The steady state sinusoid has the same frequency

as the sinusoid used in the driving voltage.• The Amplitude and phase angle of the steady state

response differ from the driving voltage source

tLR

Ve

LR

Vi m

tL

R

m coscos222222

Transient Response (dies out with time) Steady State Response

Implications of the Steady State Response

• The steady state response of a sinusoidal driving voltage is a sinusoid with the same frequency

• We therefor only need to keep track of the amplitudes and phase angles of the voltages and currents in our circuit.

• ω is constant in all expressions and can be ignored.

Phasors and Phasor Transform

• The phasor allows us to simplify dealing with sinusoids by looking at them in the complex domain (also called frequency domain)

sincos je j Euler’s Identity

jtjm

tjmm eeVeVtVv )()cos(

The Phasor Representation or Phasor Transform of a Sinusoidal waveform drops the frequency term and the Real Designation:

V jmm eVtV )cos(P

Phasor Continued

• A phasor can be expressed in either polar or rectangular form as we showed earlier when discussing complex numbers

• People often get tired of writing the exponential form and have developed the following shorthand:

sincos mmj

m jVVeV V

mj

m VeV

Inverse Phasor Transform

• The inverse phasor transfrom converts a phasor back into a sinusoidal voltage waveform.

• Note that there is nothing in the phasor itself that allows you to know what ω is. It must be independently known for your circuit

)cos(1 tVeeVeV mjtj

mj

m-P

Usefulness of the Phasor

• The phasor transform is useful because it applies directly to the sum of sinusoidal voltages.

• If we have a sum of sinusoidal voltages:

• We can also represent it as a sum of phasors:nvvvvv ...321

nVVVVV ...321

Adding Sinusoids with and without phasor

• Try adding• Try with trig identities to express this as a

single sinusoid• Try doing the same with phasors– I will work through both on the board– Hint… phasors are a LOT easier

60cos4030cos20 tt

sincos jCCCe j

A

Bj

eBAjBA1tan

22

Remember our expressions from our complex number review:

Hints for using phasors

• Phasors are easiest to add in rectangular form– When adding (and subtracting) phasors… convert them all to

rectangular format first, it will save you time• Phasor are easiest to multiply and divide in polar form• When converting from a phasor to a sinusoid through

the inverse phasor transform, always make sure the phasor is in polar form

• Remember, as annoying as it might be to convert back and forth from polar to rectangular form, it is much easier than trying to directly deal with the sinusoidal expression

Sample Problems

• Find the phasor transform of the following:

• Find the time-domain expression for the following:

)3020000sin(100)4520000cos(300

)13.53cos(10)87.36cos(5

)201000sin(10

)40377cos(170

tt

tt

t

t

)15308020(

)30504520(

546.18

jV

I

V