Download WEEK 1 SUMMARY - Oregon State University

Are oscillations ubiquitous
or are they merely a paradigm?
Superposition of
1
brain neuron activity
LRC Resonance Lab
• Average 74/100 (note BlackBoard Score 5 pts lower, pre-lab)
(scores 85-95: all elements excellent
scores <69  I was not convinced you found a physical to explain
your data)
• NOT about finding a mathematical model for LRC circuit
• Lab is about finding a physical model to explain your
observations of (i) current enhancement (ii) phase shift ;
Why do these occur? What’s happening the circuit?
What is the origin of resonance?
• Explain why your results are physically interesting and useful
WEEK 1 SUMMARY
Energy approach to equation of motion:
i.e. find trajectory x(t) if U(x) is known
•
dt =
dx
2
±
E - U(x)]
[
m
• Special case of U(x)=(½)kx2
found period T (indep of A),
found x(t) = A cos(wt+f)
• Found 3 other forms of A cos(wt+f)
• Learned to apply initial conditions to
determine A, f, and also the arbitrary
parameters in other 3 forms
• Harder case of U(q)= MgLcm(1-cos q)
found -> HO for small theta
found T numerically
measured T for pendulum
• Learned to argue qualitatively about time to
move certain distances, comparing T for diff U
• Did NOT learn equation of motion q(t)
Energy
Energy approach to equation of motion:
position
Complex numbers:
• rectangular and polar form and Argand diag.
• complex conjugate
• Euler relation
exp(if) = cos f + isin f
• solving one complex equation is actually
solving 2 simultaneous equations
WEEK 2 SUMMARY
Free, undamped oscillators
k
m
L
No friction
k
m
x
I
C
q
1
q=q
LC
mx = -kx
r; r = L
q
Common notation for all
T
m
mg
g
q»- q
L
y +w y = 0
2
0
Force approach to equation of motion of
FREE, UNDAMPED HARMONIC
OSCILLATOR:
i.e. find trajectory q(t) if F(q) is known
• Special case of F(q)=-sin(q) -> small angle approx:
F(q)=-q =>2nd order DE,
Found sinusoidal motion
q (t) = Ceiw 0t + C *e-iw 0t
q (t) = A cos (w 0t + f )
Applied initial conditions as before.
E = K +U = 12 Iq 2 + 12 mgLq 2
Free, damped oscillators
k
friction
m
x
mx = -kx - bx
r = Lcm
q
T
m
mg
1
LI + q + RI = 0
C
R
1
q+ q+
q=0
L
LC
Common notation for all
g
q » - q - b'q
L
y + 2by + w y = 0
2
0
Force approach to equation of motion of
FREE, DAMPED OSCILLATOR
• Add damping force to eqn of motion
• Found decaying sinusoid
x(t) = Ce
=e
- b t +iw1t
- bt
x(t) = Ae
éëCe
- bt
* - b t-iw 1t
+C e
+iw 1t
* -iw 1t
+C e
[cos(w1t + d )]
ùû
FREE, DAMPED OSCILLATOR
• Damping time t=1/b
• measures number of oscillations in Q = p
decay time
• apply initial conditions, energy decay
x(t) = Ae
- bt
t
w0
=
T 2b
[cos(w1t + d )]
WEEK 3 SUMMARY
DRIVEN, DAMPED OSCILLATOR
L
I
C
Vocoswt
R
q
V0 e - Lq - - Rq = 0
C
V0 iw t
2
q + 2b q + w 0 q = e
L
iw t
q ( t ) = Re éë q0 e e ùû
ifq iw t
-2 bw
q0 =
; tanfq = 2
1/2
2
2
w
w
é(w 2 - w 2 ) + 4 b 2w 2 ù
0
14
ë 0
û
V0 L
"Resonance"
CHARGE
Charge
Amplitude
|q0|
q0 =
w0
Charge
Phase fq
0
V0 L
é( w 2 - w
ë 0
)
2 2
+ 4 b 2w 2 ù
û
Driving Frequency w------>
é -2bw ù
fq = arctan ê 2
2 ú
ëw 0 - w û
-π/2
-π
w0
1/2
dq ( t )
I (t ) =
= iw q ( t )
dt
“Resonance”
CURRENT
Current
Amplitude
|I0|
I0 =
w0
Current
Phase fI
π/2
0
-π/2
wV0 L
é (w 2 - w
ë 0
)
2 2
1/2
+ 4 b 2w 2 ù
û
Driving Frequency w------>
p
é -2 bw ù
fI = + arctan ê 2
2 ú
2
ëw 0 - w û
ADMITTANCE
I
Y (w ) =
Vapp
NOT time dependent, but IS freq
dependent.
“Resonance”
Admittance
Amplitude
|Y0|
Y =
w0
Addmittanceπ/2
Phase fI
0
-π/2
w L
é (w 2 - w
ë 0
)
2 2
1/2
+ 4 b 2w 2 ù
û
Driving Frequency w------>
p
é -2 bw ù
fI = + arctan ê 2
2 ú
2
ëw 0 - w û
DRIVEN, DAMPED OSCILLATOR
• can also rewrite diff eq in terms of I and solve
directly (same result of course)
L
I
Vocoswt
C
R
q
V0 e - Lq - - Rq = 0
C
V0 iw t
2
q + 2b q + w 0 q = e
L
V0 iw t
2
Þ q + 2 b q + w 0 q = iw e
L
I
V0 iw t
I + 2b I +
= iw e
LC
L
iw t
FOURIER SERIES – periodic functions are sums of sines and
cosines of integer multiples of a fundamental frequency. These
“basis functions are orthonormal
f ( t ) = å an cosnw t + bn sin nw t
n
T
2
sin ( pw t ) sin ( qw t ) dt = d pq
ò
T 0
T
2
cos ( pw t ) cos ( qw t ) dt = d pq
ò
T
ODD functions f(t) = -f(-t). Their Fourier representation
must also be in terms of odd functions, namely sines.
Suppose we have an odd periodic function f(t) like our
sawtooth wave and you have to find its Fourier series
å bn sin(nwt )
n=1,2...
Then the unknown coefficients can be evaluated this way
Integrate over the period of the fundamental
Here’s the coefficient of
the sin(wnt) term!
Plot it on your spectrum!
2
bn =
T
normalize properly
T
ò f (t)sin ( nw t ) dt
0
the function
the harmonic
Time
domain
(
f ( t ) = å 2 sin nw f t
np
n
Frequency
domain
)
2A
f (t ) = A t 0 < t < Tf
Tf
B
frequency
Fundamantal freq = 2π/T
coefficient of the sin(wnt)
term!
Integrate over the period of the fundamental
2
bn =
T
normalize properly
T
ò f (t)sin ( nw t ) dt
0
the function
the harmonic
DRIVING AN OCILLATOR WITH A PERIODIC FORCING
FUNCTION THAT IS NOT A PURE SINE
Oscillator response
Important to know where the
fundamental freq of the
forcing function lies in relation
to the oscillator max response
freq!
B
Forcing function
frequency
DRIVING AN OCILLATOR WITH A PERIODIC FORCING
FUNCTION THAT IS NOT A PURE SINE
Vapp = V0 e
iw f t
+ 2V0 e
i 2w f t
iw t
(given)
i 2w t
q + 2 b q + w 02 q = V0 e f + 2V0 e f (Kirchoff)
Þ q = qw + q2w (linear diff eq - superposition)
Þ I = I w f + I 2w f
I =q
I = Yw f Vapp,w f + Y2w f Vapp,2w
I=
+
wf L
1/2 e
2
é w 2 - w 2 + 4 b 2w 2 ù
f
f ú
êë 0
û
(
)
( 2w )
f
(
é w 2 - ( 2w )2
f
êë 0
)
L
æp
é -2 bw f ùö
iç +arctan ê 2 2 ú÷
çè 2
êë w 0 -w f úû÷ø
1/2 e
2
2
+ 4 b 2 ( 2w ) f ù
úû
V0 e
iw f t
æ
é -2 b 2 w ùö
p
f ú÷
ç
i +arctan ê 2
ê w 0 -( 2 w )2 ú÷
ç2
f ûø
è
ë
2V0 e (
)
i 2w f t
FT - you
know this
time
Black box
Z(w)
frequency
Observe
what (LRC)
black box
does to an
impulse
function
Black box
Z(w)
?
Are these
connected
by FT??
They’d
better be you find
out!
This was
harmonic
response
expt - you
know what
black box
(LRC) does
to a single
freq
Series LRC Circuit
Which admittance plot |Y(w)| corresponds to which free decay
plot I(t)?
(All plots scaled to 1 at maximum value)
(A) i <-> 1, ii <-> 2
(B) i <-> 2, ii(i)<-> 1
(ii)
(1)
(2)
Series LRC Circuit
If the admittance |Y(w)| and the phase fI response of a series LCR circuit
are as given on the left below, then which oscilloscope trace on the right
corresponds to a circuit driven below resonance? Red is drive voltage, blue
is current, represented by VR.
(A)
(B)
(C)
(D)
 Suppose that, if you apply a (red) sinusoidal
voltage across a series LRC circuit, you measure the
(blue) voltage response across the resistor.
 Now, if you now apply a (red) square-wave
voltage with the same period to the same circuit,
and you measure the (blue) voltage response
across the resistor, will you get this ….?
 Or this?
 Or this?
Or something else?
 Suppose that, if you apply a (red) sinusoidal
voltage across a series LRC circuit, you measure the
(blue) voltage response across the resistor.
 Now, if you now apply a (red) square-wave
voltage at the same frequency the same circuit,
and you measure the (blue) voltage response
across the resistor, will you get this ….
 Or this?
 Or this?
Or something else?
The last note of the piece Mozart chose for his piano
recital was a sustained middle C (256 Hz). While he
listened for the last note to fade away to a ghost of its
initial amplitude, he had time calculate the quality
factor Q of that oscillation. If the energy died away to
1% of its initial value after 15 seconds, approximately
what was Q?
•
•
•
•
•
133
834
2600
5200
None of the above
a. Which the following are not Fourier Transform pairs?
(i.e. Fourier Transforms of eachother)
i. cos(wot) and (w-wo)+wwo))
ii. cos(wot) and i(w-wo)+wwo))
iii. square wave impulse and sinc function
iv. sin(wot)[q(t+t/2)+q(t-t/2)] and sinc function
v. exp(-iwt) and exp(-iwt)
eiq = cosq + i sinq
f (t) = Acos (w t + f )
f (t) = Bp cos w t + Bq sin w t
f (t) = Ceiw t + C *e-iw t
f (t) = Re éë Deiw t ùû
cos ( a - b ) x cos ( a + b ) x
; a ¹ ±b
2 ( a - b)
2 ( a + b)
sin ( a - b ) x sin ( a + b ) x
ò cos ( ax ) cos (bx ) dx = 2 ( a - b) + 2 ( a + b ) ; a ¹ ±b
ò sin ( ax ) cos (bx ) dx = -
y + 2by + w 02y = 0
t=
1
b
w
w
Q= 0 = 0
2 b Dw
f (t) =
sin ( a - b ) x
ò sin ( ax ) sin (bx ) dx = 2 ( a - b )
¥
a0 ¥
+ å éë an cos ( nw t ) + bn sin ( nw t ) ùû = å cn einw t
2 n=1
n=-¥
2
an =
T
bn =
cn =
2
T
1
T
Y (w ) =
T
ò f (t)cos(nw t)dt
0
T
ò f (t)sin(nw t)dt
0
T
ò f (t)e
-inw t
dt
0
I out
= Y (w ) eifI (w )
Vin
Y (w ) =
w /L
é(w 2 - w 2 )2 + 4 b 2w 2 ù
ë 0
û
1 1
1 1
2
2
cos q = + cos2q sin q = - cos 2q
2 2
2 2
1/2
f I (w ) =
p
-2bw
+ arctan 2
2
w0 - w 2
-
sin ( a + b ) x
; a ¹ ±b
2 (a + b)