chapter 2chapter 2 review of signal processing basics
TRANSCRIPT
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Chapter 2Chapter 2Review of Signal Processing Basics
Marc T. Thompson, Ph.D.Thompson Consulting, Inc.
9 Jacob Gates RoadHarvard MA 01451Harvard, MA 01451
Phone: (978) 456-7722 Fax: (240) 414-2655
E il @ h dEmail: [email protected]: http://www.thompsonrd.com
Slides to accompany Intuitive Analog Circuit Design by Marc T. Thompson© 2006-2008, M. Thompson
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Summary• Review of 1st-order systems• Relationship between bandwidth and risetime• Relationship between bandwidth and risetime• 2nd order systems• Resonance, damping and quality factor
E th d• Energy methods• Transfer functions, pole/zero plots and Bode plots• Calculating risetime for systems in cascadeg y• Comments on PSPICE
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Laplace Notation• Basic idea: Laplace transform converts differential
equation to algebraic equation
• Laplace method is used in sinusoidal steady state after all startup transients have died out
Circuit domain Laplace (s) domain Resistance, R R Inductance L LsCapacitance C 1Capac ta ce C
Cs1
Circuit Laplace transformed circuit
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System Function H(s)• Find “transfer function” H(s) (also called “transfer
function”) by solving Laplace transformed circuitfunction ) by solving Laplace transformed circuit
1)(1
1
1
)()()(
21
22
+++
=+
==CsRR
CsR
RRCs
R
svsvsH o
1)()( 2121
++++ CsRRCs
RRsvi
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Does this System Function Make Sense Intuitively?
+R1
+
-vo(s)+vi(s) -
1/(Cs)R2
At very low frequencies (s 0), the capacitor is an open-circuit:
At very high frequencies (s ∞), the capacitor is an short circuit:short-circuit:
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Does this System Function Make Sense Intuitively?
+R1
+
-vo(s)+vi(s) -
1/(Cs)R2
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First-Order Systemy
• Voltage-driven RC lowpass filter
Vtt−τ )1()(
eRVti
eVtvt
r
o
=
−=−τ
τ
)(
)1()(
RCRr
=τ
)(
τω 1
h =Time constant Bandwidth [Hz]
ττ 2.2=R
πω2
hhf =
35.0=τ
RisetimeBandwidth [rad/s]
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R f=τ
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Another First-Order SystemAnother First Order System• Current-driven RC
eIRtvt
t
o −=−
−τ )1()(
RCeIti
t
r
==
τ
τ)(
ττ 2.2=Rτ
ω 1h =
R f35.0
=τR
πω2
hhf = hf
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First-Order Systems --- Some Details
• Frequency response:
• Phase response:Phase response:
• -3 dB bandwidth:
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10 - 90% Risetime• Defined as the time it takes a step response to transition
from 10% of final value to 90% of final valueThi l t i f fi t d t ith h t• This plot is for a first-order system with no overshoot or ringing
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Relationship Between Risetime and Bandwidthp
• Exact for a first-order system:
• Approximate for higher-order systems
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First-Order System Step and Frequency Response
Step Response
1
Am
plitu
de
0.4
0.6
0.8
Time (sec.)0 1 2 3 4 5 6
0
0.2
BodeDiagrams
Mag
nitu
de (d
B)
Bode Diagrams
-20
-10
0
-3 dB point= 1 rad/sec
F ( d/ )
Pha
se (d
eg);
10-1
100
101
-80-60-40-20 Angle = -45 deg.
at -3dB point
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Frequency (rad/sec)
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“Group Delay”
11)(+
=s
sHτ 1
11)(
ωτω
+=
jjH
( )1
2
tan)(
1)(1)(
ωτω
ωτω
∠
+=
−jH
jH
( )
( )21)()(
tan)(
ωττ
ωωω
ωτω
+=
∠−=
−=∠
djHdjG
jH
( )Group delay is a measure of how much time delay the frequency components in a signal undergo. Mathematically, the group delay of a
t i th ti d i ti f th h ith t t Tsystem is the negative derivative of the phase with respect to omega. To find group delay for the first-order system, we make use of the identity:
( )dduu
dd
21
11tan+
=−
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( )dxudx 21+
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First-Order System --- Low and High Frequency Behavior
•• ClosedClosed--form solution for frequency response:form solution for frequency response:
11)(+
=s
sHτ
1)( jH
ω
ωτω
2 1)(1)(
1)(
=
+=
jH
jjH
( )ωτω
ωτ1
2
tan)(
1)(−−=∠
+
jH
• How does this response behave at frequencies well above and well below the pole frequency ?
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y
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First-Order System --- High Frequency Behavior
For high frequencies, where ωτ >> 1
tan-1(x) = π/2 – 1/x + 1/(3x3) - … for x > 1
ωτω
ωτ
1)(1≈
>>jH
ωτπω ωτ
12
)( 1 +−≈∠ >>jH
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First-Order System --- Low Frequency Behavior
For low frequencies, where ωτ << 1
tan-1(x) = x – x3/3 + x5/5 - … for x < 1 and 21
11 x
x−≈
+for x << 1
21
ωτω
ωτω
ωτ
ωτ
−≈∠
≈−≈
<<
<<
1
21
)(
1)(211)(
jH
jH
Well below the pole frequency, the magnitude is approximately 1.The phase is approximately linear phase, behaving like an approximate time delay “Group delay” is negative derivative of angle
ωτ <<1)( j
approximate time delay. Group delay is negative derivative of angle with respect to frequency, or:
τω ωτ ≈<<1)( jG
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ωτ <<1)( j
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Step Response of Single Pole
Step ResponseStepresponseofsinglepoleat -1r/s
• Single pole at -1 rad/sec.• Note that risetime = 2.2 sec
0.8
0.9
1
Step response of single pole at 1 r/s
plitu
de
0.5
0.6
0.7
Am
02
0.3
0.4
0 1 2 3 4 5 6
0
0.1
0.210-90% risetime= 2.2 sec
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Time (sec.)
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Step Response of Single Pole with High Frequency Pole Added
• Poles at -1 rad/sec. and -10 rad/sec.• Note the time delay of approximately 0 1 secNote the time delay of approximately 0.1 sec.
Step Response
0.9
1
Compare to system with real-axis poles at -1 and -10 r/s
e
0.6
0.7
0.8
Am
plitu
de
0.3
0.4
0.5
System with extra pole added
0 1 2 3 4 5 6
0
0.1
0.2
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Time (sec.)
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Bode Plot Single Pole with High Frequency Pole Added
• Poles at -1 rad/sec. and -10 rad/sec.Compare to system with real-axis poles at -1 and -10 r/s
-20
-10
0
Single pole Single pole + additional high freq. pole
); M
agni
tude
(dB)
-60
-50
-40
-30p
Pha
se (d
eg)
-100-80-60-40-20
Frequency (rad/sec)
10-1
100
101
102
-160-140-120
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q y( )
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Second-Order Mechanical Systemy
Spring force:
Newton’s law for moving mass:
Differential equation for mass motion:
Guess a solution of the form:
Differential equation for mass motion:
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Second-Order Mechanical Systemy
Put this proposed solution into the differential equation:
This solution works if:
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Second-Order Electrical Systemy
1
22
2
2
22 2121
11
1
1
)()()(
nn
n
i
o
ssssRCsLCsC
LsR
CssvsvsH
ωζωω
ζ ++=
++=
++=
++==
n LC1
=ω
2nnCs ωω
Natural frequency
o
n
ZR
CL
RRC21
21
2===
ωςDamping ratio
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Second-Order System Frequency Response
⎟⎟⎞
⎜⎜⎛−+
=2
121)(
ωςωω
jjH
=
⎟⎟⎠
⎜⎜⎝
+ 2
1)(
1
ω
ωω nn
jH
⎟⎟⎠
⎞⎜⎜⎝
⎛−+⎟⎟
⎠
⎞⎜⎜⎝
⎛2
2
22
12)(
ωω
ωςω
nn
j
⎟⎟⎞
⎜⎜⎛
−=⎟⎟⎠
⎞⎜⎜⎝
⎛
−=∠ −− 11 2tan
2
tan)( ςωωωςω
ω nnjH ⎟⎟⎠
⎜⎜⎝ −
=
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=∠ 22
2
2tan
1tan)(
ωωωω
ωn
n
jH
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Second-Order System Frequency Response
Frequency response for natural frequency = 1 and various damping ratios
dB)
20
-10
0
10
20
30
Resonance
ase
(deg
); M
agni
tude
(d
-40
-30
-20
0
Pha
-150
-100
-50
Frequency (rad/sec)
10-1
100
101
2 70701<= ςforM
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707.021 2 <−= ςςωω fornp707.0
12 2<
−= ς
ςςforM p
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Resonant Circuit --- Pole/Zero Plotseso a C cu o e/ e o o sωj
x njω+
ωj ωj
σ σxnω−
2 poles
σxx
x njω−
Zero damping
2 poles
C i i l d i OverdampedZero dampingζ=0
Critical dampingζ=1
Overdampedζ>>1
Signal Processing Basics
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Second-Order System Frequency Response at Natural Frequency
• Now, what happens if we excite this system exactly at the t l f ? Th inatural frequency, or ω = ωn. ? The response is:
121)(ςωω
== n
sH
2)( π
ωω −=∠ = nsH
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Quality Factor, or “Q”Quality factor is defined as:
diss
stored
PE
ω
where Estored is the peak stored energy in the system and Pdiss is the average power dissipationis the average power dissipation.
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Q of Series Resonant RLC
LC=
1ω
LIE
LC
pkstored =2
21
RIP pkdiss =2
21
2
ZCL
LIQ o
pk==⎟
⎟⎞
⎜⎜⎛
⎟⎞
⎜⎛=
2
21
1RRRILC
Qpk
==⎟⎟⎟
⎠⎜⎜⎜
⎝
⎟⎠
⎜⎝
=2
21
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Q of Parallel Resonant RLC=
1ω
pkstored CVE
LC
= 2
21
pkdiss R
VP =
2
21
2
pk
ZRR
V
CV
CQ ==
⎟⎟⎟⎞
⎜⎜⎜⎛
⎟⎠
⎞⎜⎝
⎛= 2
2
21
1
opk ZCL
RVLC
Q⎟⎟
⎠⎜⎜
⎝⎠⎝
2
21
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Relationship Between Damping Ratio and “Quality Factor” Q
• A second order system can also be characterized by it’s “Q lit F t ” Q“Quality Factor” or Q.
QsH ==1)( QsH
n==
= ςωω 2)(
11)(H
• Use Q in transfer function of series resonant circuit:
112)(
2
2
2
2
++=
++=
Qssss
sH
nnnn ωωωζ
ω
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Qnnnn
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Series Resonant Circuit at ResonanceSeries Resonant Circuit at ResonanceThe magnitude of this transfer function is:
222
1)(
⎟⎞
⎜⎛
⎟⎞
⎜⎛
=sHωω
21 ⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−
Qnn ωω
ωω
Exactly at resonance (ω = ωn), the magnitude of the transfer function is:
QsHn=
=ωω)(
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Second-Order System Step Response• Shown for varying values of damping ratio.
Step Response
16
1.8
2Step response for natural frequency = 1 and various damping ratios
Damping ratio = 0.1
tude
1
1.2
1.4
1.6
Am
plit
0.6
0.8
1
0 1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
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Time (sec.)
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Second-Order System Step Responsey p pStep Response
1.6
1.8
2Step response for natural frequency = 1 and various damping ratios
Am
plitu
de
0.8
1
1.2
1.4
0 1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
0.6
Time (sec.)
0 1 2 3 4 5 6 7 8 9 10
21 1
tansin1)(ς
ωςω
⎟⎟⎞
⎜⎜⎛
⎟⎟⎞
⎜⎜⎛ −
+−= −− t
tetvn LCn
1=ω
2
2
1
tansin1
1)(
ςωω
ςω
ς ⎟⎠
⎜⎝
⎟⎟
⎠⎜⎜
⎝+
−−= do ttv
LR
2=ς
Damped natural frequency
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1 ςωω −= nd C2Damped natural frequency
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Logarithmic Decrement
• Method of estimating damping ratio based on measurementmeasurement
)sin( 1111 φωςω +
=− tCex d
tn
)sin( 1222 φωςω +− tCex d
tn
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Logarithmic Decrement (cont )Logarithmic Decrement (cont.)• Simplify:
11ntex −ςω
• TD = oscillation period
)(2
11 Dn Ttex +−= ςω
TD oscillation period = (2π)/ωd
• Simplify againx
DnTexx ςω=
2
1
⎞⎛⎞⎛⎞⎛• Take log of both sides: ⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛
−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−==≡⎟⎟
⎠
⎞⎜⎜⎝
⎛22
2
1
12
1ln
ςπς
ςωςςωδ D
dDn TT
xx
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Logarithmic Decrement (cont )Logarithmic Decrement (cont.)• The logarithmic decrement δ is the natural log of the ratio
of any two successive oscillation amplitudes.
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Logarithmic Decrement (cont )Logarithmic Decrement (cont.)• Comment: If we
measure how the amplitude decreased cycle-by-cycle, we can find dampingcan find damping ratio
• Example: In a free vibration test thevibration test, the ratio of amplitudes is 2.5 to 1 on
⎞⎛successive oscillations... 145.0
12916.0
15.2ln
2=⇒
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−==⎟
⎠⎞
⎜⎝⎛= ς
ςπςδ
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Logarithmic Decrement (cont )Logarithmic Decrement (cont.)• If damping ratio is very low,
πςδ 22l 1 ⎟⎞
⎜⎛
⎟⎞
⎜⎛ x
• In the case of low damping, it may not be easy to make
πςς
πςδ 212ln
22
1 ≈⎟⎟
⎠⎜⎜
⎝ −=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
xx
In the case of low damping, it may not be easy to make measurement of successive oscillations. We can make measurements at time t1 and at N cycles later. Note that:
⎞⎛⎞⎛⎞⎛⎞⎛⎟⎟⎠
⎞⎜⎜⎝
⎛•••⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
++ 14
3
3
2
2
1
1
1
N
N
N xx
xx
xx
xx
xx
• Take log of both sides
δNxxxxx N =⎟⎟⎠
⎞⎜⎜⎝
⎛+•••⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛ 3211 lnlnlnlnln
10/30/2008 Signal Processing Basics 38
xxxxx NN⎟⎠
⎜⎝
⎟⎠
⎜⎝
⎟⎠
⎜⎝
⎟⎠
⎜⎝
⎟⎠
⎜⎝ ++ 14321
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Logarithmic Decrement (cont )Logarithmic Decrement (cont.)• Take log of both sides
xxxxx ⎟⎞
⎜⎛
⎟⎞
⎜⎛
⎟⎞
⎜⎛
⎟⎞
⎜⎛
⎟⎞
⎜⎛
δNxx
xx
xx
xx
xx
N
N
N
=⎟⎟⎠
⎞⎜⎜⎝
⎛+•••⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
++ 14
3
3
2
2
1
1
1 lnlnlnlnln
• This means we can find the logarithmic decrement as:
⎟⎞
⎜⎛
⎟⎞
⎜⎛ 1l1 xδ ⎟⎟
⎠⎜⎜⎝
⎟⎠⎞
⎜⎝⎛=
+1
1lnNxN
δ
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A Potpourri of Resonant CircuitsA Potpourri of Resonant Circuits• Both damped and undamped• All circuits simulated with PSIM
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Example 1: Undamped Resonant CircuitExample 1: Undamped Resonant Circuit• L = 1/(2π); C = 1/(2π). Initial conditions: capacitor voltage
= 0 and inductor current = 1 Amp
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Example 1: What Do We Know About This Ci it?Circuit?
• L = 1/(2π); C = 1/(2π)• Zo = 1 Ohmo• ωo = 2π rad/sec; fo = 1 Hz
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Example 1: Undamped Resonant Circuit RResponse
• L = 1/(2π); C = 1/(2π)
• Note ratio of voltage/current = 1 ohm
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Example 1: Series Resonant Step ResponseExample 1: Series Resonant Step Response
• L = 1/(2π); C = 1/(2π). Assume zero initial conditions( ); ( )
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Example 1: Series Resonant Step ResponseExample 1: Series Resonant Step Response• Zo = 1 Ohm• Natural frequency: ω = 2π rad/sec; f = 1 HzNatural frequency: ωo 2π rad/sec; fo 1 Hz
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Example 1: Series Resonant Step ResponseExample 1: Series Resonant Step Response• Zo = 1 Ohm• ωo = 2π rad/sec; fo = 1 Hz
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Example 1: Series Resonant Step ResponseExample 1: Series Resonant Step Response• What is inductor current? Remember Zo = 1 Ohm
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Example 1: Series Resonant Step RResponse
• What is inductor current? Remember Zo = 1 Ohm
Peak-peak = 20V
Peak-peak = 20A
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Example 2: Now Add 0.1 Ω ResistorExample 2: Now Add 0.1 Ω Resistor
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Example 2: Now Add 0.1 Ω ResistorExample 2: Now Add 0.1 Ω Resistor• R << Zo, so damping is small
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Example 2: Now Add 0.1 Ω ResistorExample 2: Now Add 0.1 Ω Resistor
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Example 3: Change the 0.1 Ω to 1 ΩExample 3: Change the 0.1 Ω to 1 Ω• R = Zo in this case
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Example 3: Change the 0.1 Ω to 1 ΩExample 3: Change the 0.1 Ω to 1 Ω
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Example 4: Change the 1 Ω to 10 ΩExample 4: Change the 1 Ω to 10 Ω• R >> Zo in this case
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Example 4: Series Resonant Step ResponseExample 4: Series Resonant Step Response• What’s the initial inductor current, approximately (and
why)?why)?
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Example 4: Series Resonant Step ResponseExample 4: Series Resonant Step Response• What’s the initial inductor current, approximately?
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Example 4: Response for R >> ZoExample 4: Response for R Zo
• At low frequencies, this behaves like an RC lowpass filter• At high frequencies when capacitor is almost a short thisAt high frequencies, when capacitor is almost a short, this
behaves as a LPF with L/R time constant• Why is this?
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Second Order System --- Pole Location Variation with Damping
Critically Very underdamped
ωj
j
dampedOverdamped
ωjωj
σ
x njω+
σxnω−
2 polesσxx
x njω−
2 poles
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Overdamped Case1)( 2=sH
Overdampedωj• At low frequencies, for RCs >> LCs2,
R L th
1)( 2 ++ RCsLCs
sH
σxx
or R >> Ls, then …
11)(+
≈RCs
sHσxx
• At high frequencies, with RCs>>1, then
1+RCs
then
⎟⎞
⎜⎛ +
≈+
≈1
11)( 2
sLRCsRCsLCssH
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⎟⎠
⎜⎝
+1sR
RCs
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Example 5: Series Resonant Circuit with Di dDiode
• Q: With a positive voltage step, what does the diode do in this circuit?this circuit?
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Example 5: Series Resonant Circuit with Di dDiode
• A: Nothing…it’s always reverse biased
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Example 6: Now, Flip the DiodeExample 6: Now, Flip the Diode• Q: What does the diode do in this circuit?
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Example 6: Series Resonant Circuit with Di dDiode
• A: It shorts out the capacitor; the circuit behaves like a series LR circuit with L/R time constant 159 secseries LR circuit with L/R time constant 159 sec.
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Example 7: Second-Order System Step Response with Extra PoleResponse, with Extra Pole
• Second order system with ωn = 1 r/s, damping ratio = 0.5• Extra pole added at -10 r/sp
Step ResponseCompare step response of wn=1, damp=0.5 2nd order system with pole added at -10 r/s
e
0.8
1
Effects of extra pole
Am
plitu
de
0.4
0.6Effects of extra pole
0 2 4 6 8 10 12
0
0.2
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Time (sec.)
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Example 8: Second-Order System Step Response, with Extra Pole
• Second orderStep Response
Compare step response of wn=1, damp=0.5 2nd order system with pole added at -2 r/s
Second order system with ωn = 1 r/s, damping
ti 0 5 08
1
ratio = 0.5• Extra pole added at -2 r/s A
mpl
itude
0.6
0.8
0.2
0.4
Time (sec.)
0 2 4 6 8 10 12
0
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Widely-Spaced Pole ApproximationWidely-Spaced Pole Approximation
For a transfer function of the form:
BA2
1 BAss ++2
If the poles are on the real axis and are widely spaced, we can approximate them by:
As fast −≈
ABsslow
fast
/−≈
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Widely-Spaced Pole Approximation• Assume that we have 3 real-axis polesp
11)(sH ( ) ( ) 1)1)(1)(1()(
3212
3231213
321321 +++++++=
+++=
sssssssH
τττττττττττττττ
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Widely-Spaced Pole Approximation• We can re-write this transfer function:
23
1)( =sH
3213
12
23
3 1)(
τττ=+++
asasasa
3231212
τττττττττ
++=++=
aa
3211 τττ ++=a
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Widely-Spaced Pole Approximation• Pole locations can be approximated if the poles are widely spaced:widely spaced:
111−≈−≈−≈plow
213231212
13211
1ττττττττττττ
−≈−≈++
−≈−≈
++ap
aplow
13211
33213213
1τττττττττττ
++
≈≈≈≈
aa
phigh
221
1
323121
321
2
1
τττττττττ−≈−≈
++≈≈
apmedium
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Widely-Spaced Pole Approximation• In the general case with k poles, if they are widely spaced and on the real axis:spaced and on the real axis:
1ka 1−≈ −
k
kk a
p
1=oa
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Widely-Spaced Pole Approximation ---Sanity Check
• Examine system with poles at -1, -10, -100, -1000 and 10000 r/sand -10000 r/s
1111.110122.110122.110111.1101)( 213346510 ++×+×+×+
= −−−− ssssssH
srp
10122.1
/110,1110
10111.1
3
10
6
5
×
−≈×
−≈
−
−
−
srp
srp
/10010122.110122.1
/010,110111.110122.1
3
1
3
64
−≈××
−≈
−≈××
−≈
−
−
−
• The approximation does
srp
srp
/901
/9.910122.1
111.112
−≈−≈
−≈×
−≈ −
approximation does a decent job
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srp /9.0111.12 ≈≈
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Undamped Resonant Circuitp
Lv
dtdi cL = C
idtdv Lc −
=
2
LCv
dtdi
Cdtvd cLc −=−=
12
2
Guess that the voltage v(t) is sinusoidal with v(t) = Vosinωt. Putting this into the equation for capacitor voltage results in:
)i (1)i (2 tt )sin()sin(2 tLC
t ωωω −=−
This means that the resonant frequency is the standard (as expected) q y ( p )resonance:
LCr12 =ω
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Energy Methodsgy
Storage Mode Relationship Comments Capacitor/electric field 21 CVE
pstorage
2
2CVEelec =
Inductor/magnetic field storage dVBLIE
omag ∫==
μ221 2
2
Kinetic energy 2
21 MvEk =
Rotary energy 2
21 ωIEr =
I ≡ mass moment of inertia (kg m2)2r (kg- m )
Spring 2
21 kxEspring = k ≡ spring constant (N/m)
Potential energy hMgE Δ=Δ Δh ≡ height changeote t a e e gy hMgEp ΔΔ Δh height changeThermal energy TCE THT Δ=Δ CTH ≡ thermal capacitance
(J/K)
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Energy Methodsgy
By using energy methods we can find the ratio of maximum capacitor voltage to maximum inductor current. Assuming that the capacitor is initially charged t V lt d b i th t it t d E ½CV2 dto Vo volts, and remembering that capacitor stored energy Ec = ½CV2 and inductor stored energy is EL = ½LI2, we can write the following:
22 11 22
21
21
oo LICV =
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Energy Methods
What does this mean about the magnitude of the inductor current ? Well, we can solve for the ratio of Vo/Io resulting in:
o ZLV≡= o
o
ZCI
≡
The term “Zo” is defined as the characteristic impedance of a resonant circuit. Let’s
th t h i d t it i it ith C 1 i F d d L 1assume that we have an inductor-capacitor circuit with C = 1 microFarad and L = 1 microHenry. This means that the resonant frequency is 106 radians/second (or 166.7 kHz) and that the characteristic impedance is 1 Ohm.
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Example 9: Simulation
• Initial conditions at t = 0: capacitor voltage = 1; inductor current = 0.
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Transfer Functions, Pole-Zero and Bode Plots
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Transfer Functions, Pole-Zero and Bode Plots
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Transfer Functions, Pole-Zero and Bode Plots
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Risetime for Systems in Cascade
For multiple systems in cascade, the risetimes do not simply add; for instance, for N systems wired in series, each with its own risetime τR1, τR2, … τRN, the overall y , R1, R2, RN,risetime of the cascade τR is:
223
22
21 RNRRRR τττττ ⋅⋅⋅+++≈
Note that this equation works if each system is buffered/isolated from the next sytem; i.e. the systems don’t load each other down.
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Example 10: Risetime for Systems in Cascade
For vo1: sec220)10)(1000)(2.2(2.2 7
1 μτ === −RCR For vo2: Consider TR2, which is risetime of 2nd RC circuit by itself:
7
( )( ) sec311sec2202
sec220)10)(1000)(2.2(2.22
22
1,
72
μμτττ
μτ
≈≈+≈
=== −
RRsystemR
R RC
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Example 10: Risetime for Systems in Cascade
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Example 13: Risetime for 3 Systems in CascadeExample 13: Risetime for 3 Systems in Cascade
FFor vo3: ( )( ) sec381sec22032
32
22
1, μμττττ ==++≈ RRRsystemR
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Example 13: Risetime for 3 Systems in CascadeExample 13: Risetime for 3 Systems in Cascade
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Comments on PSPICE• PSPICE is a useful tool but beware of some pitfallsPSPICE is a useful tool, but beware of some pitfalls
– GIGO problem– Models
S ti th f t ’ d l i t• Sometimes the manufacturers’ models are incorrect – Tolerances
• Set tolerances sufficiently small. However, this will increase simulation time
– Convergence issues• Transient response is not guaranteed to convergeTransient response is not guaranteed to converge
– Maximum time step• Choose a time step small enough to capture the
detail that you need If you make the time step toodetail that you need. If you make the time step too small, simulation time can be very large
– Parasitics
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Comments on PSPICE (cont.)• Continued ...
– Parasitic elements can cause your PSPICE– Parasitic elements can cause your PSPICE simulation not to match protoboard results. Parasitic capacitance and inductance; resistive h t th tshunt paths, etc.
– You should always do a sanity check to verify that your PSPICE result is reasonabley
– PSPICE simulation is not a replacement for good design and hand calculations.
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References
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