# Lecture 5 Physics 2CL

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```					             Lecture 5: Physics 2CL
20 August 2007

• Experiment 4: Filters
– Band-Pass
– High-Pass

• Rejection of Data/Theory
– Chauvenet Criterion
− χ2 Test

• Quiz 4
Summer II 2007
Extra Credit: Nikola Tesla!

5 × 1015 % inflation!
Experiment 3: Filters
• Band-Pass – passes frequencies
within a specific range and
attenuates those outside of that
range

• High-Pass – passes frequencies
as high as a specific threshold
frequency and attenuates any
signal of lesser frequency

• Band-Pass = Low-Pass + High-
Pass
Experiment 3: Resonance
Definition:
Tendency of system to oscillate with a maximum amplitude
at a specific frequency

Other Examples:
Acoustical, Mechanical (e.g. mass on a spring,
pendulum), Tidal, Orbital

Electrical Resonance:
Occurs in an electric circuit when impedance between the
input and the output is minimum

Impedance - Resistance in AC (with phase)
Experiment 3: Electrical Resonance
Time-dependent signal   V (t ) = Vo sin(ωt )

1
Resistor V = IR            IR = Vo sin(ωt ) ⇒ I = Vo sin(ωt )
R
dI                    VoC
Inductor V = − L dI     − L = Vo sin(ωt ) ⇒ I =      cos(ωt )
dt       dt                     L
Voltage leads current by π/2
Q
Capacitor V = = ∫
Idt      Idt
C     C     ∫ C = Vo sin(ωt ) ⇒ I = VoC cos(ωt )
Current leads voltage by π/2
Experiment 3: Impedance Re-visited

V = Vo eiωt              e   iωt
= cos(ωt ) + i sin(ωt )

VR
VR = I R R ⇒ I R =
R
dVC             iωt  IC
= VC 0iωe =       ⇒ I C = iCωVC
dt                 C
1 iωt                 i
∫ VL dt = VL0 iω e = LI L ⇒ I L = − Lω VL
Experiment 3: Electrical Resonance
• Impedance
– Series:
Z Σ = Z1 + Z 2          ZR = R
– Parallel: 1                       Z L = i ωL
1   1
=   +
ZΣ   Z1 Z 2             ZC = 1         =− i
iCω          Cω
1 − iCω     − iCω         −i
+++          •     =               =
iCω − iCω − (i ) (Cω )
2      2
Cω
• Maximize parallel impedance/Minimize series
impedance

• In RLC, electric energy oscillates from magnetic
field (L) to electrical field (C)
Experiment 3: Resonance (Frequency Domain)
VR = RI m sin(ω t − ϕ )
Loop rule:
1
VC = −    I m cos(ω t − ϕ ) V = VR + VC + VL
ωC
all have different
VL = ω LI m cos(ω t − ϕ )        phases…

V 2 = I m R 2 + (I m X L − I m X C )2
2

X L ≡ ωL
V                                                                  Imω L
Im   =                                        1
Z                              XC   ≡
Z ≡ R2 + (X L − X C )
2                   ωC                   φ               V
φ
2                  φ
Impedance: Z = R 2 +  ωL − 1 
                                 Im
     ωC                          ωC                       Im R

Courtesy of Professor Basov
Experiment 3: Resonance (Frequency Domain)

Resistance of Inductor

V out   Z (ω )
G=       =
Vin      R in
Plot Gain vs. Frequency
Experiment 3: Resonance
vertical      horizontal       trigger

Ext input Ch 1
Ch2
ext
scale           scale

coupling
V/div        Sec/div             DC
AC
LF rej
HF rej

Ch1        Ch2
frequency

Bandwidth: difference
between frequencies at
which amplitude is 1 2
ωo
∆ω =
Q
Experiment 3: Resonance (Band-Pass)
• Plot Phase vs. Frequency
− 1  Im 
• Fit to expression ϕ = tan  
 Re 
 
−1           x 
2

ϕ = tan      ax  1 −    − bx   
         c         
                     
Experiment 3: High Pass

• Plot Gain vs. Frequency

• Fit the data to an
appropriate form of

Z fb
G=
Z = x + iy                  Z in
Z =   x2 + y2
Rejection of Data: Chauvenet’s Criterion
suspicious
x − x suspect = tσ               x − x suspect = 1 .6 = 2σ

P ( X − 2σ > x ∩ X + 2σ < x ) = 1 − P ( X − 2σ < x < X + 2σ )
P ( X − 2σ > x ∩ X + 2σ < x ) = 1 − 0 .95

n = N × P ( X − 2σ > x ∩ X + 2σ < x )
1
n<           suspicious value can be rejected
2
n = 0.3               1.8 is improbable and rejectable
+++ The probability that a measurement                 If the method/technique is suspicious,
will be as deviant as the suspect value must           data are disposable without comparison
be scaled by the total number of measurements          to other data.
in order to determine the expected fraction of such
deviants in the relevant sample size.
χ2                                        ∑ (y               − f (x j ))
N
Test                                            j
2

j =1
χ =
2

σ   2
y

30

20
y(x)

10                       How good is the agreement
between theory and data?
0
0      5   10       15    20        25
x
Courtesy of Professor Basov
χ2 Test (continued - 2)

∑ (y           − f (x j ))
N
2
j                           Nσ      2

≅                =N
j =1                                           y
χ =
2

σ    2
y                   σ  2
y

~2 = χ ≅ 1
2

30
χ
d # of degrees of
freedom
20
d=N-c
y(x)

10                                                   # of data
points    # of parameters
0
computed from data
0    5     10            15       20       25
x                                            (# of constraints)
Courtesy of Professor Basov
χ2 TEST for FIT
1         − x2
d −1
χ2 distribution: H d , χ 2
~
~      =       x e 2
const

~     ~
Prob( χ 2 ≥ χ 02 )

Hd,χ2
GX,σ

tσ

12 20 28 36 44 52 60         0.0   1.0        2.0      3.0        4.0
~
χ   2
0
Prob outside tσ            from data
Courtesy of Professor Basov
(y − f (x ) )                   Probability of χ2
2
n
χ =∑
2                i           i

i =1            σ2
y
Quantitative measure of agreement between
χ           2
χ =
%2                                  observed data and their expected distribution
nd.o.f.                        Table D

Probability of obtaining
a value of χ2 greater or
equal to χ02 , assuming
the measurements are
governed by the
expected distribution

~   ~                          Probability that for d degrees of freedom reduced χ2 is as
Prob d ( χ ≥ χ )2           2
o           large as the reduced observed χ2

disagreement is “significant” if ProbN(χ2 = χ02) < 5 %                         reject the expected
disagreement is “highly significant” if ProbN(χ2 = χ02) < 1 %                  distribution
Courtesy of Professor Avi Yagil
Example: Application of χ2 Test
Die is tossed 600 x

Expectation: each face has
equivalent likelihood of
showing up
v         1      2    3    4             5     6 _
Verification of expectation             n        91    137 111 87                80 94
by computing the χ2                     exp     100    100 100 100              100 100
∆2       81    1369 121 169             400 36
σ        10     10 10 10                 10 10
This term is the squared difference     χ2i      0.81   13.7 1.21 1.69           4.0 0.36
between observation and expectation.
Total χ2                  21.76
In computation of χ2 the ∆2 term is divided by           ndof                       5
expectation. σ is square root of expectation            reduced χ2               4.35
(Ey = σy2)                                        Adapted from presentation by Professor Avi Yagil
Application of χ2 Test: Usage of Table D
Just calculated:
Total χ2         21.77
ndof             5
Reduced χ2       4.35

Die is loaded at 99.9% Confidence Level

Courtesy of Professor Avi Yagil

```
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