Transcript Solid State III, Lecture 23

Lecture 1 - Background from 1A
Revision of key concepts with application
to driven oscillators:
 Aims:
 Review of complex numbers:
Addition;
Multiplication.
 Revision of oscillator dynamics:
Free oscillator - damping regimes;
Driven oscillator - resonance.
 Concept of impedance.
 Superposed vibrations.
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Complex representation
 Complex nos. and the Argand diagram:
 Use complex number A, where the real part
represents the physical quantity.
A  Aoei  a1  ia2
Amplitude
Phase
Amplitude follows from:
Phase follows from:
Ao2  a12  a22
tan   a2 a1
 Harmonic oscillation:
z  Ae it  Aoei t  
Displacement
z  Ae it
x   z  
Velocity
z  i  Ae it
x   z     Ao sin t   
Accelerati on
z  i 2 Ae it
x  z   2 Ao cost   
2
Ao cost   
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Manipulation of Complex Nos. I
 Addition

A  Aoeio  A1ei1  A2ei 2
Ao2  A12  A22  2 A1 A2 cos 2  1 
A1 sin 1   A2 sin  2 
tan o  
A1 cos1   A2 cos 2 
  A  Ao coso   A1 cos1   A2 cos 2 
 The real part of the sum is the sum of the real
parts.
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Manipulation of Complex Nos. II
 Multiplication
 WARNING:
 z1z2    z1  z2 
One cannot simply multiply the two complex
numbers.
 Example (i): To calculate (velocity)2 .
Take velocity v = Voeit with Vo real.
2
2
2
Instantaneous value: v   Vo cos t 
Mean value:
1
2
1
2
1
2
v 2  Vo2  vv  v 2
 Example (ii). Power, (Force . Velocity).
Take f = Foei(t+) with Fo real.
Instantaneous value:
 f v   FoVo cost    cost 
Mean value:
1
 FoVo cos2t     cos 
2
 
1
1
FoVo cos    fv
2
2
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The damped oscillator
 Equation of motion

mx  sx  bx
Restoring force
1.1
Dissipation (damping)
Rearranging gives
x  2x   o2 x  0
2  b m
 o2  s m
1.2
Natural resonant frequency
 Two independent solutions of the form x=Aept.
Substitution gives the two values of p, (i.e. p1,
p2), from roots of quadratic:
p 2  2p   o2  0
p1 / 2     2   o2
 General solution to [1.2]

x  e t A1e qt  A2e  qt

q 2   2   o2
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Damping régimes
 Heavy damping
q 2  0 or    o
  q t
 A2e  q t
 x  A1e
Sum of decaying exponentials.
2
 Critical damping
q  0 or    o
t
x

e
 A1  A2t 

Swiftest return to equilibrium.
 Light damping
q  i1, where 12  o2 ,
t
A1ei1t  A2ei1t
 xe


Damped vibration.
x0   0,
x 0   1,  o  1
Heavy :
 6
Critical :   1
Light :
6
  0.02
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Driven Oscillator
 Oscillatory applied force (frequency ):
 Force:

 f    Feit

 Equation of motion:
F

x  2x  o2 x   eit 
m

Use complex variable, z, to describe
displacement: i.e. x   z 
F
z  2z   o2 z  eit
m
1.3
 Steady state solution MUST be an oscillation at
it
frequency . So z  Ae
 A gives the magnitude and phase of the
“displacement response”. Substitute z into [1.3]
F
to get
2
2
A    i 2   o 

A

F m


 o2   2  i 2
 The “velocity response” is
z  iAe it 
m

.
F m
eit

  2   o2  
 2  i

 





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Impedance
 Mechanical impedance
 Note, the velocity response is proportional to
the driving force, i.e.
Force =constant(complex) x velocity
Mechanical impedance
 Z = force applied / velocity response

  2   o2  

Z  m 2  i
 





 In general it is complex and, evidently,
frequency dependent.
 Electrical impedance
 Z=applied voltage/current response
 Example, series electrical circuit:
Z  R  iL  1 iC
We can write the mechanical impedance in a
similar form:
Z  2m  im  imo2   b  im  s i
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