PEElectrical Review Course Class 3 Transients

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PE-Electrical Review Course - Class 3 (Transients)
Class 3 Transients and Frequency
Response Objectives This review session is
designed to review material and provide practical
examples such that the student will be able to
1) Analyze first-order and second-order circuits
using differential equation methods. 2) Recognize
the differences between overdamped, underdamped,
and critically damped circuits. 3) Find Laplace
transforms of common functions and find
inverse-Laplace transforms of common expressions
using Partial Fractions Expansion. 4) Analyze
circuits using Laplace transforms
methods. 5) Determine the transfer function for a
given circuit. 6) Sketch the frequency response
(Bode log-magnitude (LM) and phase plots) for
transfer functions. 7) Recognize basic filter
types from their transfer functions or frequency
response. 8) Use impedance and frequency scaling
on filters to obtain desired cutoff frequencies.
Reading material 1) EE Ref. Manual, 6th Ed.,
Camara, Chapters 30-32 Transients and
Frequency Response Appendices 15.F and 15.G
Laplace Transforms and Laplace Transform
Properties 2) Handout Extra Problems for Week
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PE-Electrical Review Course - Class 3 (Transients)
Transients Transients are temporary responses
that occur in circuits containing inductors
and/or capacitors when switches are thrown or
other changes to the circuits occur. In general
the transient response of a circuit can be
determined using the following methods 1)
Differential Equations (DE) 2) Laplace Transforms
Transient Analysis of First-Order Circuits (using
differential equation methods) First a brief
review of key facts concerning inductors and
capacitors
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PE-Electrical Review Course - Class 3 (Transients)
Order of a Circuit Order of circuit order of
differential equation (DE) number of
independent Ls Cs For example,
1st-order circuit - has one C or one L
2nd-order circuit - has 2
Cs, 2 Ls, or one C and one L
etc.
General form of a first-order response In
general, the solution has two parts x(t) xh
xp homogeneous particular solutions (common
terms in math texts) x(t) xn xf natural
forced responses (common terms in engineering
texts)
Finding the forced response The forced response
is sometimes also called the steady-state
response. If a change is made to the circuit
(such as a switch being thrown), then the circuit
will go through some sort of transient response
and will eventually settle down (reach
steady-state). So, xf xss x(?) for circuits
with DC sources xf xss phasor value for AC
sources (sinusoidal steady state) So, for a
circuit with DC sources, find the steady-state
response by analyzing the circuit at t ? (note
that Cs will appear as open-circuits and Ls as
short-circuits).
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PE-Electrical Review Course - Class 3 (Transients)
Example 1
Find the steady-state value of v(t) in the
circuit shown below.
Example 2
Find the steady-state value of i(t) in the
circuit shown below.
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PE-Electrical Review Course - Class 3 (Transients)
Finding the complete response of a first-order
circuit There are two types of 1st-order
circuits a) RC circuit b) RL circuit Analysis
of a 1st-order circuit will yield a DE of the
form Solution of the DE will have the
form where t (Greek letter Tau) time
constant and the response generally reaches
steady-state at (since
e-5t/t e-5 ? 0)
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PE-Electrical Review Course - Class 3 (Transients)
General procedure for analyzing 1st-order
circuits (It is usually easiest to find
capacitor voltages or inductor currents and then
to use them to find other variables if
necessary.) 1) Find the steady-state solution.
Analyze the circuit at t ? for DC circuits
) 2) Find the initial condition, x(0). Analyze
the circuit at t 0-. Recall that 3) Find the
time-constant, t , using 4) Assume that the
solution has the form Solve for the
constant, A, using the initial condition, x(0).
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PE-Electrical Review Course - Class 3 (Transients)
Find the complete response for v(t) in the
circuit shown below. (Note the steady-state
portion was found in Example 1.)
Example 3
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PE-Electrical Review Course - Class 3 (Transients)
Find the complete response for i(t) in the
circuit shown below. (Note the steady-state
portion was found in Example 2.)
Example 4
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PE-Electrical Review Course - Class 3 (Transients)
Transient Analysis of Second-Order
Circuits (using differential equation
methods) As with 1st-order circuits, the
solution will have the form x(t) xn xss and
the steady-state response is found as seen
previously. The natural response, however, is
quite different.
The Natural Response to a 2nd-Order Circuit A
general 2nd-order DE has the form The
right-hand side of the DE is set to zero and the
characteristic equation is determined Using
the quadratic equation to find the roots yields
the characteristic roots (or natural
frequencies)
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PE-Electrical Review Course - Class 3 (Transients)
The characteristic roots can be real (and
distinct), repeated, or complex. This leads to
three types of natural responses 1) overdamped
(real, distinct roots s1 and s2) Form
2) critically damped (repeated roots s s1
s2) Form 3) underdamped (complex roots s1,
s2 ? ? j?) Form
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PE-Electrical Review Course - Class 3 (Transients)

• Illustration The transient response to a
  2nd-order circuit must follow one of the forms
  indicated above (overdamped, critically damped,
  or underdamped). Consider the circuit shown
  below. The steady-state value of v(t) is 10V.
  How does it get there?
• Discuss the possible responses for v(t)
• Define the terms damping, rise time, ringing,
  and overshoot

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PE-Electrical Review Course - Class 3 (Transients)

• Illustration When is each of the 3 types of
  responses desired? Discuss the following cases.
• An elevator
• A cruise-control circuit
• The output of a logic gate
• The start up voltage waveform for a DC power
  supply

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PE-Electrical Review Course - Class 3 (Transients)
General procedure for analyzing 2nd-order
circuits (It is usually easiest to find
capacitor voltages or inductor currents and then
to use them to find other variables if
necessary.) 1) Find the steady-state solution,
xss . Analyze the circuit at t ? for DC
circuits ) 2) Find two initial condition, x(0)
and x(0). A) Find x(0) by analyzing the
circuit at t 0-. Recall that B) Find x(0)
by analyzing the circuit at t 0. Recall that
3) Write the differential equation 4) Find the
natural response, xn. 5) Find the total solution,
x(t) xn xss. Solve for the two constants in
the solution using the two initial conditions.
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PE-Electrical Review Course - Class 3 (Transients)
Example 5
Find the complete response for v(t) in the
circuit shown below.
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PE-Electrical Review Course - Class 3 (Transients)
Example 5
(continued)
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PE-Electrical Review Course - Class 3 (Transients)
Example 5
(continued)
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PE-Electrical Review Course - Class 3 (Transients)
Laplace Transform Analysis of Circuits (using
differential equation methods) Laplace
transforms could have been used to analyze the
circuits covered so far instead of using
differential equation methods. Laplace
transforms also have many other uses in
Electrical Engineering. The uses of Laplace
transforms include 1) Circuit analysis 2)
Transfer functions 3) Frequency response 4)
Applications in specific areas, such as Control
Theory and Communications First a brief review
of Laplace transforms (also see Appendices 15.F
and 15.G in the text)
F(s) L f(t) the Laplace transform of
f(t) f(t) L -1F(s) the inverse Laplace
transform of f(t)
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PE-Electrical Review Course - Class 3 (Transients)
Finding Laplace transforms Laplace Transforms can
be found in 2 ways 1) By definition 2)
Using tables of transform pairs (recommended for
the PE exam)
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PE-Electrical Review Course - Class 3 (Transients)

• Finding inverse Laplace transforms
• Use the table of Laplace Transforms for simple
  functions. For more complicated functions, use
  Partial Fractions Expansion (PFE).
• The purpose of Partial Fractions Expansion is to
  take an expression for F(s) and break it up into
  parts whose inverse transforms can be found using
  a table of Laplace transforms.

Example 6A
Find f(t) for the following function
Example 6B
Find f(t) for the following function
Example 6C
Find f(t) for the following function
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PE-Electrical Review Course - Class 3 (Transients)
Example 6D
Find f(t) for the function shown below.
(Note that no function like F(s) is found in the
table of Laplace transforms).
F(s) should be decomposed using Partial Fraction
Expansion as follows
(Note that all three parts of F(s) now
correspond to functions in the table of Laplace
transforms).
Finding the coefficients in the PFE
expression Three methods will be presented for
finding A, B, and C in the example above. 1)
Finding a common denominator 2) Residue
method 3) Using a software or a calculator
(TI-85, 86, 89, 92 or HP-48) Not allowed
on the PE exam!
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PE-Electrical Review Course - Class 3 (Transients)
Method 1 Finding a common denominator
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PE-Electrical Review Course - Class 3 (Transients)
Method 2 Residue method
Method 3 Calculator method For the calculator
method and for more examples, refer to the
handout Partial Fraction Expansion provided by
the instructor.
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PE-Electrical Review Course - Class 3 (Transients)
Circuit Analysis using inverse Laplace
transforms In order to analyze circuits with
Laplace transforms, it is necessary to know how
to represent each type of component in the
s-domain.
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PE-Electrical Review Course - Class 3 (Transients)
Procedure Circuit Analysis using the
Laplace-transformed Circuit 1) Form the
Laplace-transformed circuit (using the
appropriate circuit models) 2) Analyze the
circuit as you might analyze a DC circuit (using
any circuit analysis method) 3) The final result
will be a function of s (such as V(s) or I(s)).
Use an inverse Laplace transform to find v(t)
or i(t).
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PE-Electrical Review Course - Class 3 (Transients)
Example 7
Find i(t) in the circuit shown below using
Laplace transforms.
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PE-Electrical Review Course - Class 3 (Transients)
Transfer Functions A transfer function, H(s),
can be used to describe a system or circuit in
the s-domain in terms of its input and output as
illustrated below.

• H(s) is defined more specifically as
• Notes
• Transfer functions are always defined with zero
  initial conditions
• Y(s) and X(s) typically represent voltages or
  currents
• The input and the output must be designated by
  the user.

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PE-Electrical Review Course - Class 3 (Transients)
(transfer function)(Laplace transform of the
input)
so y(t) L 1 H(s)?X(s)

• Although y(t) can be found for any input x(t),
  there are two special cases of interest
• Impulse response the output to a circuit when
  the input x(t) ?(t)

If x(t) ?(t), then X(s) 1, so Y(s) H(s)?1
H(s), so y(t) L 1 H(s) h(t). So h(t)
impulse response L 1 H(s)

• Unit step response the output to a circuit
  when the input x(t) u(t)

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PE-Electrical Review Course - Class 3 (Transients)
A) Find the transfer function H(s)
Vout(s)/Vin(s) B) Determine the impulse
response C) Determine the unit step response
Example 8
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PE-Electrical Review Course - Class 3 (Transients)
Frequency Response Recall that a transfer
function H(s) is defined as
In general, s ? jw. For frequency
applications we use s jw (so ? 0). So we
define
Since H(jw) can be thought of as a complex number
that is a function of frequency, it can be placed
into polar form as follows
When we use the term "frequency response", we
often are referring to information that is
conveyed using the following graphs
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PE-Electrical Review Course - Class 3 (Transients)
Bode Plots A Bode plot is a quick estimate of the
of the frequency response. Since the estimate
uses straight lines, it is often called a
straight-line approximation.

• There are two types of Bode plots
• The Bode straight-line approximation to LM
  versus w (with w on a log scale)
• The Bode straight-line approximation to the
  phase plot, ?(w) versus w (with w on a log scale)

Drawing Bode plots To draw a Bode plot for any
H(s), we need to 1) Recognize the different
types of terms that can occur in H(s) (or
H(jw)) 2) Learn how to draw the log-magnitude
and phase plots for each type of term.
H(s) has the general form where zi are the
zeros of H(s) and pi are the poles of H(s)
Example 9
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PE-Electrical Review Course - Class 3 (Transients)
5 types of terms in H(jw) 1) K (a constant) 2)
jw (a pole or zero) 3) (a pole
or a zero) 4) Any of the terms raised to a
positive integer power. 5) Each term is now
examined in detail. 1. Constants terms in
H(jw) If H(jw) K K/0? Then LM
20log(K) and ?(w) 0? , so the LM and phase
responses are
So a constant term in H(jw) adds a constant DC
level to the LM and has no effect on the phase.
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PE-Electrical Review Course - Class 3 (Transients)
2. jw terms in H(jw) A) jw is a zero
If H(jw) jw w/90? Then LM 20log(w) and
?(w) 90? , so the LM and phase responses are
So a jw zero term in H(jw) adds an upward slope
of 20dB/dec or 6dB/oct to the LM plot. And a jw
zero term in H(jw) adds a constant 90? to the
phase plot.
B) jw is a pole
So a jw zero term in H(jw) adds an downward slope
of -20dB/dec or -6dB/oct to the LM plot. And a jw
zero term in H(jw) adds a constant -90? to the
phase plot.
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PE-Electrical Review Course - Class 3 (Transients)
3. 1 jw/w1 terms in H(jw) A) 1 jw/w1 is a
zero The straight-line approximations are
So a 1 jw/w1 zero term in H(jw) causes an
upward break at w w1 in the LM plot. There
is a 0dB effect before the break and a slope of
20dB/dec or 6dB/oct after the break.. And a 1
jw/w1 zero term in H(jw) adds 90? to the phase
plot over a 2 decade range beginning a decade
before w1 and ending a decade after w1 .
B) 1 jw/w1 is a pole The straight-line
approximations are
So a 1 jw/w1 pole term in H(jw) causes an
downward break at w w1 in the LM plot. There
is a 0dB effect before the break and a slope of
-20dB/dec or -6dB/oct after the break.. And a 1
jw/w1 pole term in H(jw) adds -90? to the phase
plot over a 2 decade range beginning a decade
before w1 and ending a decade after w1 .
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PE-Electrical Review Course - Class 3 (Transients)

• 4. Any of the terms raised to a positive integer
  power
• A zero raised to the Nth power will have the
  following effect on the LM plot
• Has a 0dB contribution before its break frequency
• Will increase at a rate of 20NdB/dec after the
  break
• There will be an error of 3NdB at the break
  between the Bode straight-line
• approximation and the exact LM
• A zero raised to the Nth power will have the
  following effect on the phase plot
• Has a 0 degree contribution until 1 decade before
  its break frequency
• Will increase at a rate of 45Ndeg/dec for two
  decades (from 0.1w1 to 10w1).
• The total final phase contribution will be 90N
  degrees.

5. Complex terms in H(jw) A second order term in
H(s) with complex roots has the general form s2
2?s wo2 . In the straight-line approximation
it acts like a double term (s wo)2. They are
different however, in the actual response.
Whereas the the double term (s wo)2 will round
off the corner on the LM plot by 2(3dB) 6dB,
the complex term will have a complex peak with
a value of LM 20log(2z) for a zero or LM
20log(1/(2z)) for a pole, where z is the
damping ratio defined below (and Q is the
Quality factor.
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PE-Electrical Review Course - Class 3 (Transients)
Example 10
Sketch the Bode LM and phase plots for
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PE-Electrical Review Course - Class 3 (Transients)
Example 11A
Sketch the Bode LM and phase plots for
Example 11B
Sketch the Bode LM and phase plots for
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PE-Electrical Review Course - Class 3 (Transients)
Example 12
Sketch the Bode LM plot for
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PE-Electrical Review Course - Class 3 (Transients)
Example 13
Sketch the Bode LM plot for
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PE-Electrical Review Course - Class 3 (Transients)
Filters A filter is a circuit designed to have a
particular frequency response, perhaps to alter
the frequency characteristics of some signal.
There are 4 basic filter types
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PE-Electrical Review Course - Class 3 (Transients)

• Impedance and Frequency Scaling
• Filter tables
• List component values for various types of
  filters
• Often only list cutoff frequencies of 1 Hz or
  1 kHz
• Often use inconvenient component values (such
  as 1 ohm or 1 Farad)
• Frequency scaling can be used to scale the
  cutoff frequencies to the desired values
• Impedance scaling can be used to find useful
  component values

Impedance Scaling to scale the impedance by a
factor KZ
Frequency Scaling to scale the frequency by a
factor KF
Note that impedance scaling has no effect on the
frequency response.
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PE-Electrical Review Course - Class 3 (Transients)
Example 14
A) Show that the circuit below is a LPF with a
cutoff frequency of 1 rad/s.
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PE-Electrical Review Course - Class 3 (Transients)
B) Use impedance and frequency scaling for a new
cutoff freq. of 500 rad/s and C 1 uF.