Transcript Document

Q factor of an
underdamped
oscillator
Damping time or "1/e" time is t = 1/b > 1/w0
(>> 1/w0 if b is very small)
How many T0 periods elapse in the damping time?
This number (times π) is the Quality factor or Q of the
system.
t
w0
Q=p =
T0 2 b
large if b is small compared to w0
Max Amplitude
Current
Amplitude
|I0|
1
of max amp
2
Dw
w0
Q=
Dw
Driving Frequency------>
Find frequencies where POWER
drops to half maximum (current drops
to 0.707 of max). These define w.
Find resonant frequency, w0
Homework problem to show the two definitions are the same.
QUETSION: FM radio stations have broadcast frequencies of
approximately 100 MHz. Most radios use a series LRC circuit similar
to the one you used in the lab as part of the receiver electronics.
Estimate the spacing of the broadcast frequencies of FM stations if
typical receivers have a Q of 500 or better. Explain your reasoning,
and include a graph.
station 1
w
station 2
w
w0
f0
Qº
=
Dw Df
QUETSION: FM radio stations have broadcast frequencies of
approximately 100 MHz. Most radios use a series LRC circuit similar
to the one you used in the lab as part of the receiver electronics.
Estimate the spacing of the broadcast frequencies of FM stations if
typical receivers have a Q of 500 or better. Explain your reasoning,
f0
and include a graph.
Q=
station 1
w
Df
= 500
station 2
f0 100 MHz
Df = =
= 0.2 MHz
Q
500
Therefore, stations 99.3
and 99.5 FM are allowed,
but 99.3 and 99.4 FM are
not!
They have cross-talk!
You should be able to:
• Calculate & plot the magnitude and phase of 1/Z
• Convert between the mag/phase and Re/Im forms
• Draw phasor diagrams of Vext, I, 1/Z (or Z)
• Express 1/Z (or Z) in terms of R, L, C or w0, b
You should be able to discuss:
• The amplitude of the response and resonance
• The phase of the response
• The nature of the behavior at all frequencies
• The transfer of the series LCR circuit analysis to
analogous oscillatory systems
Blackboard Example
I = YV
V ( t ) = V0 cos (w 0t )
Y (w )
V ( t ) = - V0 cos ( 2w 0t )
w
f I (w )
V ( t ) = V0 cos ( 3w 0t )
w0
2w0
3w0
t
q ( t ) = Re éë q0 e ùû
iw t
iw t
Vext = Re éëV0 e ùû
i (fq + p /2 ) iw t ù
é
I(t) = q ( t ) = Re w q0 e
e
ë
û
I0 =
wV0 L
(
é w2 -w
0
ë
)
2 2
1/2
+ 4b w ù
û
2
-2bw
f I = + arctan 2
2
2
w0 - w
2
p
8
I0 =
wV0 L
(
é w -w
ë
2
0
)
2 2
1/2
+ 4b w ù
û
2
2
What is the best variable to plot for the LRC lab?
ADMITTANCE
Amplitude
of I/Vapp
Frequency, w ->
p
-2bw
f I = + arctan 2
2
2
w0 - w
1
-wL
tan f = w C
R
Phase of I
(rel to Vapp)
Frequency, w ->
Response of a damped oscillator to a periodic
driving force of arbitrary shape:
We have learned that a damped oscillator produces a sinusoidal
response to a sinusoidal driving force. In the language of your
lab example: The LRC circuit responds to a sinusoidal driving
voltage with a sinusoidal current (at the same frequency). That
current is related to the voltage by the “admittance” Y by
I = V Y = V(1/Z)
Y is determined by the circuit parameters L, R, C, and is also
dependent on the frequency of the driving force. We have
learned that Y can be written as a complex number whose
amplitude tells how large the current is for a given V and whose
phase tells how much I(t) is shifted wrt V(t). (see previous
notes and your class group exercise for derivation)
The following page shows this graphically for a given circuit
and for 3 different driving voltages.
Response of a damped oscillator to a periodic
driving force of arbitrary shape:
If a system is a LINEAR system, it means that if several
sinusoidal driving forces are added and applied at once, then
the response is just the sum of the individual responses. It
seems obvious that the circuit is linear, but there are many
systems that are non-linear.
The following page shows this graphically for the same circuit
the driving voltage that is the sum of the three previous ones,
and the resulting current which is also the sum of the three
previous currents. No longer pure sinusoids!
Notice that the shape of the current and the voltage are not the
same anymore! It’s not true that there’s one simple scaling
factor and one phase shift!
V ( t ) = V0 cos (w 0t ) - V0 cos ( 2w 0t ) + V0 cos ( 3w 0t )
Y (w )
f I (w )
Vapp
t
V ( t ) = V0 cos (w 0t ) - V0 cos ( 2w 0t ) + V0 cos ( 3w 0t )
I ≠ Y Vapp
Y (w )
f I (w )
t
- V0 Y
2w 0
(
I ( t ) = V0 Y
)
cos 2w 0t + f2w 0 + V0 Y
3w 0
w0
(
cos w 0t + fw 0
(
cos 3w 0t + f3w 0
)
)
Response of a damped oscillator to a periodic
driving force of arbitrary shape:
So now you can see how useful the admittance function is, and
why it is important to know how its magnitude and phase shift
vary with frequency: if there is a driving force that can be
expressed as the sum of sinusoids, we simply use the
admittance function to find the response at each of the driving
frequencies, and add the responses to get the net response.
It turns out that any periodic driving force can be decomposed
into such a sum of sinusoidal components, and obviously it
must be our job to learn how to break down such a periodic
function into its component sinusoids. This technique is called
Fourier analysis.
What if V(t) is this?
t
V (t ) = 1 0 < t < 2p
t
p
And periodic repetitions
Use Fourier Analysis!
Fourier Analysis
• is easy
• is a sensible thing to do
• has a bad reputation (unjustly)
• does not involve impossible integrals
Before we tackle the problem we just posed, backtrack a
bit to review some terminology involving Fourier series
f t =1cost -cos2t +1cos3t
3
æ
ç
ç
è
ö
÷
÷
ø
æ
ç
ç
è
ö
÷
÷
ø
æ
ç
çç
è
ö
÷
÷÷
ø
æ
ç
çç
è
ö
÷
÷÷
ø
f(t)
time
f(t)
time
A
Fourier
Coefficients
Height = amp
position = freq
“A” = code for
cosine
frequency
g(t ) =1sin(t ) - sin(2t ) + 1 sin(3t )
3
g(t)
time
g(t)
B
Fourier
Coefficients
time
Height = amp
position = freq
“B” = code for
sine
frequency
h(t ) = f (t) + g(t)
h(t)
time
Fourier
Spectrum
A
frequency
B
frequency
Look at fundamental:
Compute:
1cos (w t ) + 1sin (w t )
p
æ -1ö
2
2
1 + 1 = 2;arctan ç ÷ = è 1ø
4
Aha! Fundamental also
i (w t - p / 4 )
é
ù
Re ë 2e
û
Fourier
Spectrum
A
frequency
B
frequency
-ip /4 iw t
i 2w t
i 3w t
é
h(t ) = f (t) + g(t) = Re ë 2e e + ? e + ? e ùû
h(t)
time
Fourier
Spectrum
A
frequency
B
frequency
What have we learned so far?
• Odd periodic functions with period T=2π/w can be
represented by a series of sin(nwt) functions
• Even periodic functions with period T=2π/w can be
represented by a series of cos(nwt) functions
• If these functions represent physical motion, then we
think of the motion as the superposition of the motions
of SHOs with increasing frequencies (each a multiple of
the fundamental one)
• In these special odd/even cases the coefficient of the
sin(nwt) (or cos) term represents the amplitude of that
particular SHO
• In these special odd/even cases, we can plot the size of
a coefficient of the nwt term at the nwt frequency to give
an alternative representation of the function
What have we learned so far?
• Periodic functions with period T=2π/w that are neither
even nor odd can be represented by a series of sin(nwt)
and cos(nwt) functions, all with different coefficients
• If these functions represent physical motion, then we
think of the motion as the superposition of the motions
of SHOs with increasing frequencies (each a multiple of
the fundamental one)
• In these general cases the coefficients of the sin(nwt)
cos(nwt) terms must be combined to represents the
amplitude and phase of that particular SHO
• In these general cases, we must make two plots to give
an alternative representation of the function. We can
chose to plot sin coeff and cos coeff, or magnitude and
phase
What have not learned yet?
• How to find the coefficients if the function is not
explicitly written in terms of sines and cosines (we will)
• How to deal with exponential functions
• Lots of practice needed, of course ….