illustrations we use quantitative mathematical models of physical systems to design and analyze...
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Illustrations
We use quantitative mathematical models of physical systems to design and analyze control systems. The dynamic behavior is generally described by ordinary differential equations. We will consider a wide range of systems, including mechanical, hydraulic, and electrical. Since most physical systems are nonlinear, we will discuss linearization approximations, which allow us to use Laplace transform methods.
We will then proceed to obtain the input–output relationship for components and subsystems in the form of transfer functions. The transfer function blocks can be organized into block diagrams or signal-flow graphs to graphically depict the interconnections. Block diagrams (and signal-flow graphs) are very convenient and natural tools for designing and analyzing complicated control systems
Chapter 2: Mathematical Models of Systems Objectives
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Illustrations
Introduction Six Step Approach to Dynamic System Problems
• Define the system and its components
• Formulate the mathematical model and list the necessary assumptions
• Write the differential equations describing the model
• Solve the equations for the desired output variables
• Examine the solutions and the assumptions
• If necessary, reanalyze or redesign the system
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Illustrations
Differential Equation of Physical Systems
Ta t( ) Ts t( ) 0
Ta t( ) Ts t( )
t( ) s t( ) a t( )
Ta t( ) = through - variable
angular rate difference = across-variable
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Illustrations
Differential Equation of Physical Systems
v21 Ltid
d E
1
2L i
2
v211
k tFd
d E
1
2
F2
k
211
k tTd
d E
1
2
T2
k
P21 ItQd
d E
1
2I Q
2
Electrical Inductance
Translational Spring
Rotational Spring
Fluid Inertia
Describing Equation Energy or Power
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Illustrations
Differential Equation of Physical SystemsElectrical Capacitance
Translational Mass
Rotational Mass
Fluid Capacitance
Thermal Capacitance
i Ctv21
d
d E
1
2M v21
2
F Mtv2
d
d E
1
2M v2
2
T Jt2
d
d E
1
2J 2
2
Q CftP21
d
d E
1
2Cf P21
2
q CttT2
d
d E Ct T2
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Illustrations
Differential Equation of Physical SystemsElectrical Resistance
Translational Damper
Rotational Damper
Fluid Resistance
Thermal Resistance
F b v21 P b v212
i1
Rv21 P
1
Rv21
2
T b 21 P b 212
Q1
Rf
P21 P1
Rf
P212
q1
Rt
T21 P1
Rt
T21
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Illustrations
Differential Equation of Physical Systems
M2
ty t( )d
d
2 b
ty t( )d
d k y t( ) r t( )
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Illustrations
Differential Equation of Physical Systems
v t( )
RC
tv t( )d
d
1
L 0
t
tv t( )
d r t( )
y t( ) K 1 e 1 t
sin 1 t 1
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Illustrations
Differential Equation of Physical Systems
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Illustrations
Differential Equation of Physical Systems
K2 1 2 .5 2 10 2 2
y t( ) K2 e2 t
sin 2 t 2
y1 t( ) K2 e2 t
y2 t( ) K2 e2 t
0 1 2 3 4 5 6 71
0
1
y t( )
y1 t( )
y2 t( )
t
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Illustrations
Linear Approximations
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Illustrations
Linear Approximations
Linear Systems - Necessary condition
Principle of Superposition
Property of Homogeneity
Taylor Serieshttp://www.maths.abdn.ac.uk/%7Eigc/tch/ma2001/notes/node46.html
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Illustrations
Linear Approximations – Example 2.1
M 200gm g 9.8m
s2 L 100cm 0 0rad
15 16
T0 M g L sin 0
T1 M g L sin
T2 M g L cos 0 0 T0
4 3 2 1 0 1 2 3 410
5
0
5
10
T1 ( )
T2 ( )
Students are encouraged to investigate linear approximation accuracy for different values of0
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Illustrations
The Laplace Transform
Historical Perspective - Heaviside’s Operators
Origin of Operational Calculus (1887)
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Illustrations
pt
d
d
1
p 0
t
u1
d
iv
Z p( )Z p( ) R L p
i1
R L pH t( )
1
L p 1R
L p
H t( )
1
R
R
L
1
p
R
L
21
p2
R
L
31
p3
.....
H t( )
1
pn
H t( )tn
n
i1
R
R
Lt
R
L
2t2
2
R
L
3t3
3 ..
i1
R1 e
R
L
t
Expanded in a power series
v = H(t)
Historical Perspective - Heaviside’s OperatorsOrigin of Operational Calculus (1887)
(*) Oliver Heaviside: Sage in Solitude, Paul J. Nahin, IEEE Press 1987.
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Illustrations
The Laplace Transform
Definition
L f t( )( )0
tf t( ) es t
d = F(s)
Here the complex frequency is s j w
The Laplace Transform exists when
0
tf t( ) es t
d this means that the integral converges
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Illustrations
The Laplace Transform
Determine the Laplace transform for the functions
a) f1 t( ) 1 for t 0
F1 s( )0
tes t
d = 1
s e
s t( )1
s
b) f2 t( ) ea t( )
F2 s( )0
tea t( )
es t( )
d = 1
s 1 e
s a( ) t[ ] F2 s( )1
s a
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Illustrations
note that the initial condition is included in the transformationsF(s) - f(0+)=Ltf t( )d
d
s0
tf t( ) es t( )
d-f(0+) +=
0
tf t( ) s es t( )
df t( ) es t( )=
0
vu
d
we obtain
v f t( )anddu s es t( ) dt
and, from which
dv df t( )u es t( )where
u v uv
d=vu
dby the use of
Ltf t( )d
d
0
ttf t( ) e
s t( )d
d
d
Evaluate the laplace transform of the derivative of a function
The Laplace Transform
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Illustrations
The Laplace TransformPractical Example - Consider the circuit.
The KVL equation is
4 i t( ) 2ti t( )d
d 0 assume i(0+) = 5 A
Applying the Laplace Transform, we have
0
t4 i t( ) 2ti t( )d
d
es t( )
d 0 40
ti t( ) es t( )
d 2
0
tti t( ) e
s t( )d
d
d 0
4 I s( ) 2 s I s( ) i 0( )( ) 0 4 I s( ) 2 s I s( ) 10 0
transforming back to the time domain, with our present knowledge of Laplace transform, we may say thatI s( )
5
s 2
0 1 20
2
4
6
i t( )
t
t 0 0.01 2( )
i t( ) 5 e2 t( )
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Illustrations
The Partial-Fraction Expansion (or Heaviside expansion theorem)
Suppose that
The partial fraction expansion indicates that F(s) consists of
a sum of terms, each of which is a factor of the denominator.
The values of K1 and K2 are determined by combining the
individual fractions by means of the lowest common
denominator and comparing the resultant numerator
coefficients with those of the coefficients of the numerator
before separation in different terms.
F s( )s z1
s p1( ) s p2( )
or
F s( )K1
s p1
K2
s p2
Evaluation of Ki in the manner just described requires the simultaneous solution of n equations.
An alternative method is to multiply both sides of the equation by (s + pi) then setting s= - pi, the
right-hand side is zero except for Ki so that
Kis pi( ) s z1( )
s p1( ) s p2( )s = - pi
The Laplace Transform
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Illustrations
The Laplace Transform
s -> 0t -> infinite
Lim s F s( )( )Lim f t( )( )7. Final-value Theorem
s -> infinitet -> 0
Lim s F s( )( )Lim f t( )( ) f 0( )6. Initial-value Theorem
0
sF s( )
df t( )
t5. Frequency Integration
F s a( )f t( ) ea t( )4. Frequency shifting
sF s( )d
dt f t( )3. Frequency differentiation
f at( )2. Time scaling
1
aF
s
a
f t T( ) u t T( )1. Time delaye
s T( )F s( )
Property Time Domain Frequency Domain
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Illustrations
Useful Transform Pairs
The Laplace Transform
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Illustrations
The Laplace Transform
y s( )
sb
M
yo
s2 b
M
sk
M
s 2 n s
22 n n
2
s1 n n 2
1n
k
M
b
2 k M s2 n n
21
Roots
RealReal repeatedImaginary (conjugates)Complex (conjugates)
s1 n j n 1 2
s2 n j n 1 2
Consider the mass-spring-damper system
Y s( )Ms b( ) yo
Ms2
bs k
equation 2.21
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Illustrations
The Laplace Transform
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Illustrations
The Transfer Function of Linear Systems
V1 s( ) R1
Cs
I s( ) Z1 s( ) R
Z2 s( )1
CsV2 s( )1
Cs
I s( )
V2 s( )
V1 s( )
1
Cs
R1
Cs
Z2 s( )
Z1 s( ) Z2 s( )
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Illustrations
The Transfer Function of Linear Systems
Example 2.2
2ty t( )d
d
24
ty t( )d
d 3 y t( ) 2 r t( )
Initial Conditions: Y 0( ) 1ty 0( )d
d0 r t( ) 1
The Laplace transform yields:
s2 Y s( ) s y 0( ) 4 s Y s( ) y 0( )( ) 3Y s( ) 2 R s( )
Since R(s)=1/s and y(0)=1, we obtain:
Y s( )s 4( )
s2 4s 3 2
s s2 4s 3
The partial fraction expansion yields:
Y s( )
3
2
s 1( )
12
s 3( )
1s 1( )
1
3
s 3( )
2
3
s
Therefore the transient response is:
y t( )3
2e t
1
2e 3 t
1 e t 1
3e 3 t
2
3
The steady-state response is:
ty t( )lim
2
3
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Illustrations
The Transfer Function of Linear Systems
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Illustrations
The Transfer Function of Linear Systems
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Illustrations
The Transfer Function of Linear Systems
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Illustrations
The Transfer Function of Linear Systems
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Illustrations
Kf if
Tm K1 Kf if t( ) ia t( )
field controled motor - Lapalce Transform
Tm s( ) K1 Kf Ia If s( )
Vf s( ) Rf Lf s If s( )
Tm s( ) TL s( ) Td s( )
TL s( ) J s2 s( ) b s s( )
rearranging equations
TL s( ) Tm s( ) Td s( )
Tm s( ) Km If s( )
If s( )Vf s( )
Rf Lf s
The Transfer Function of Linear Systems
Td s( ) 0
s( )
Vf s( )
Km
s J s b( ) Lf s Rf
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Illustrations
The Transfer Function of Linear Systems
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Illustrations
The Transfer Function of Linear Systems
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Illustrations
The Transfer Function of Linear Systems
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Illustrations
The Transfer Function of Linear Systems
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Illustrations
The Transfer Function of Linear Systems
V 2 s( )
V 1 s( )
1RCs
V 2 s( )
V 1 s( )RCs
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Illustrations
The Transfer Function of Linear Systems
V 2 s( )
V 1 s( )
R 2 R 1 C s 1 R 1
V 2 s( )
V 1 s( )
R 1 C 1 s 1 R 2 C 2 s 1 R 1 C 2 s
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Illustrations
The Transfer Function of Linear Systems
s( )
V f s( )
K m
s J s b( ) L f s R f
s( )
V a s( )
K m
s R a L a s J s b( ) K b K m
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Illustrations
The Transfer Function of Linear Systems
Vo s( )
Vc s( )
K
Rc Rq
s c 1 s q 1
c
Lc
Rc
q
Lq
Rq
For the unloaded case:
id 0 c q
0.05s c 0.5s
V12 Vq V34 Vd
s( )
Vc s( )
Km
s s 1
J
b m( )
m = slope of linearized torque-speed curve (normally negative)
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Illustrations
The Transfer Function of Linear SystemsY s( )
X s( )
K
s Ms B( )
KA kx
kp
B bA2
kp
kxx
gd
dkp
Pgd
dg g x P( ) flow
A = area of piston
Gear Ratio = n = N1/N2
N2 L N1 m
L n m
L n m
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Illustrations
The Transfer Function of Linear Systems
V2 s( )
V1 s( )
R2
R
R2
R1 R2
R2
R
max
V2 s( ) ks 1 s( ) 2 s( ) V2 s( ) ks error s( )
ks
Vbattery
max
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Illustrations
The Transfer Function of Linear Systems
V2 s( ) Kt s( ) Kt s s( )
Kt constant
V2 s( )
V1 s( )
ka
s 1
Ro = output resistanceCo = output capacitance
Ro Co 1s
and is often negligible for controller amplifier
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Illustrations
The Transfer Function of Linear Systems
T s( )
q s( )
1
Ct s Q S1
R
T To Te = temperature difference due to thermal process
Ct = thermal capacitance= fluid flow rate = constant= specific heat of water= thermal resistance of insulation= rate of heat flow of heating element
QSRt
q s( )
xo t( ) y t( ) xin t( )
Xo s( )
Xin s( )
s2
s2 b
M
sk
M
For low frequency oscillations, where n
Xo j Xin j
2
k
M
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Illustrations
The Transfer Function of Linear Systems
x r
converts radial motion to linear motion
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Illustrations
Block Diagram Models
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Illustrations
Block Diagram Models
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Illustrations
Block Diagram Models
Original Diagram Equivalent Diagram
Original Diagram Equivalent Diagram
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Illustrations
Block Diagram Models
Original Diagram Equivalent Diagram
Original Diagram Equivalent Diagram
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Illustrations
Block Diagram Models
Original Diagram Equivalent Diagram
Original Diagram Equivalent Diagram
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Illustrations
Block Diagram Models
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Illustrations
Block Diagram Models
Example 2.7
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Illustrations
Block Diagram Models Example 2.7
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Illustrations
Signal-Flow Graph Models
For complex systems, the block diagram method can become difficult to complete. By using the signal-flow graph model, the reduction procedure (used in the block diagram method) is not necessary to determine the relationship between system variables.
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Illustrations
Signal-Flow Graph Models
Y1 s( ) G11 s( ) R1 s( ) G12 s( ) R2 s( )
Y2 s( ) G21 s( ) R1 s( ) G22 s( ) R2 s( )
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Illustrations
Signal-Flow Graph Models
a11 x1 a12 x2 r1 x1
a21 x1 a22 x2 r2 x2
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Illustrations
Signal-Flow Graph Models
Example 2.8
Y s( )
R s( )
G 1 G 2 G 3 G 4 1 L 3 L 4 G 5 G 6 G 7 G 8 1 L 1 L 2
1 L 1 L 2 L 3 L 4 L 1 L 3 L 1 L 4 L 2 L 3 L 2 L 4
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Illustrations
Signal-Flow Graph Models
Example 2.10
Y s( )
R s( )
G 1 G 2 G 3 G 4
1 G 2 G 3 H 2 G 3 G 4 H 1 G 1 G 2 G 3 G 4 H 3
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Illustrations
Signal-Flow Graph Models
Y s( )
R s( )
P1 P2 2 P3
P1 G1 G2 G3 G4 G5 G6 P2 G1 G2 G7 G6 P3 G1 G2 G3 G4 G8
1 L1 L2 L3 L4 L5 L6 L7 L8 L5 L7 L5 L4 L3 L4
1 3 1 2 1 L5 1 G4 H4
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Illustrations
Design Examples
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Illustrations
Speed control of an electric traction motor.
Design Examples
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Illustrations
Design Examples
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Illustrations
Design Examples
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Illustrations
Design Examples
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Illustrations
Design Examples
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
error
The Simulation of Systems Using MATLAB
Sys1 = sysh2 / sysg4
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
error
The Simulation of Systems Using MATLAB
Num4=[0.1];
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
The Simulation of Systems Using MATLAB
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Illustrations
Sequential Design Example: Disk Drive Read System
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Illustrations
Sequential Design Example: Disk Drive Read System
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Illustrations
=
Sequential Design Example: Disk Drive Read System
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Illustrations
P2.11
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Illustrations
1
L c s R c
Vc
Ic
K1
1
L q s R q
Vq
K2
K3-Vb
+Vd
Km
Id
1
L d L a s R d R a
Tm
1
J s b
1
s
P2.11
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Illustrations
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Illustrations
http://www.jhu.edu/%7Esignals/sensitivity/index.htm
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Illustrations
http://www.jhu.edu/%7Esignals/