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Announcements 10/20/10

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Announcements 10/20/10. Prayer Term project proposals due on Saturday night! Email to me: proposal in body of email, 650 word max. See website for guidelines, grading, ideas, and examples of past projects. If in a partnership, just one email from the two of you - PowerPoint PPT Presentation
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Announcements 10/20/10 Prayer Term project proposals due on Saturday night! Email to me: proposal in body of email, 650 word max. See website for guidelines, grading, ideas, and examples of past projects. a. If in a partnership, just one email from the two of you Exam 2 starts a week from tomorrow. a. Exam 2 optional review session: vote on times by tomorrow evening. Survey link sent out this morning. Anyone not get the Fourier transform handout last lecture?
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Page 1: Announcements 10/20/10

Announcements 10/20/10 Prayer Term project proposals due on Saturday

night! Email to me: proposal in body of email, 650 word max. See website for guidelines, grading, ideas, and examples of past projects.

a. If in a partnership, just one email from the two of you

Exam 2 starts a week from tomorrow.a. Exam 2 optional review session: vote on

times by tomorrow evening. Survey link sent out this morning.

Anyone not get the Fourier transform handout last lecture?

Page 2: Announcements 10/20/10

Quick Writing Ralph doesn’t understand what a transform

is. As discussed last lecture and in today’s reading, how would you describe the “transform” of a function to him?

Page 3: Announcements 10/20/10

Reading Quiz In the Fourier transform of a periodic

function, which frequency components will be present?

a. Just the fundamental frequency, f0 = 1/period

b. f0 and potentially all integer multiples of f0

c. A finite number of discrete frequencies centered on f0

d. An infinite number of frequencies near f0, spaced infinitely close together

Page 4: Announcements 10/20/10

Fourier Theorem Any function periodic on a distance L can be written

as a sum of sines and cosines like this:

Notation issues: a. a0, an, bn = how “much”

at that frequencya. Time vs distanceb. a0 vs a0/2c. 2/L = k (or k0)… compare 2/T = (or 0 )d. Durfee:

– an and bn reversed– Uses 0 instead of L

The trick: finding the “Fourier coefficients”, an and bn

01 1

2 2( ) cos sinn n

n n

nx nxf x a a b

L L

01

compare to: ( ) nn

n

f x a a x

Page 5: Announcements 10/20/10

How to find the coefficients

What does mean?

What does mean?

0

0

1( )

L

a f x dxL

0

2 2( )cos

L

nnx

a f x dxL L

0

2 2( )sin

L

nnx

b f x dxL L

01 1

2 2( ) cos sinn n

n n

nx nxf x a a b

L L

0

0

1( )

L

a f x dxL

1

0

2 2( )cos

Lx

a f x dxL L

Let’s wait a minute for derivation.

Page 6: Announcements 10/20/10

Example: square wave

f(x) = 1, from 0 to L/2 f(x) = -1, from L/2 to L

(then repeats) a0 = ? an = ? b1 = ? b2 = ? bn = ?

0

0

1( )

L

a f x dxL

0

2 2( )cos

L

nnx

a f x dxL L

0

2 2( )sin

L

nnx

b f x dxL L

01 1

2 2( ) cos sinn n

n n

nx nxf x a a b

L L

004/Could work out each bn individually, but why?

4/(n), only odd terms

Page 7: Announcements 10/20/10

Square wave, cont.

Plots with Mathematica:http://www.physics.byu.edu/faculty/colton/courses/phy123-fall10/lectures/lecture 22 - square wave

Fourier.nb

1(odd only)

4 2( ) sin

n

nxf x

n L

4 2 4 6 4 10( ) sin sin sin ...

3 5

x x xf x

L L L

Page 8: Announcements 10/20/10

Deriving the coefficient equations

To derive equation for a0, just integrate LHS and RHS from 0 to L. To derive equation for an, multiply LHS and RHS by cos(2mx/L),

then integrate from 0 to L.(To derive equation for bn, multiply LHS and RHS by sin(2mx/L), then integrate from 0 to L.)

Recognize that when n and m are different, cos(2mx/L)cos(2nx/L) integrates to 0. (Same for sines.)

Graphical “proof” with MathematicaOtherwise integrates to (1/2)L (and m=n). (Same for sines.)

Recognize that sin(2mx/L)cos(2nx/L) always integrates to 0.

0

0

1( )

L

a f x dxL

0

2 2( )cos

L

nnx

a f x dxL L

0

2 2( )sin

L

nnx

b f x dxL L

01 1

2 2( ) cos sinn n

n n

nx nxf x a a b

L L

Page 9: Announcements 10/20/10

0N 1N 2N

3N 10N 500N

1 1 2sin

2

nx

n L

Sawtooth Wave, like HW 22-1

(The next few slides from Dr. Durfee)

Page 10: Announcements 10/20/10

The Spectrum of a Saw-tooth WaveThe Spectrum of a Saw-tooth Wave

0 10 20 30 40 50 60-0.4

-0.2

0

0.2

0.4

0.6

Am

plitu

de

[m]

k [rad/m]

Page 11: Announcements 10/20/10

The Spectrum of a Saw-tooth WaveThe Spectrum of a Saw-tooth Wave

0 10 20 30 40 50 600

0.1

0.2

0.3

0.4

0.5

0.6

Am

plitu

de [

m]

k [rad/m]0 10 20 30 40 50 60

-pi/2

-pi/4

0

Pha

se [

rad]

Page 12: Announcements 10/20/10

Electronic “Low-pass filter” “Low pass filter” = circuit which

preferentially lets lower frequencies through.

?Circuit

What comes out?

How to solve: (1) Decompose wave into Fourier series(2) Apply filter to each freq. individually(3) Add up results in infinite series again

Page 13: Announcements 10/20/10

Low-Pass Filter – before filterLow-Pass Filter – before filter

0 10 20 30 40 50 600

0.1

0.2

0.3

0.4

0.5

0.6

Am

plitu

de [

m]

k [rad/m]0 10 20 30 40 50 60

0

0.1

0.2

0.3

0.4

0.5

0.6

Am

plitu

de [

m]

k [rad/m]0 10 20 30 40 50 60

0

0.1

0.2

0.3

0.4

0.5

0.6

Am

plitu

de [

m]

k [rad/m]0 10 20 30 40 50 60

-pi

-3 pi/4

-pi/2

-pi/4

0

Pha

se [

rad]

Page 14: Announcements 10/20/10

Low-Pass Filter – after filterLow-Pass Filter – after filter

0 10 20 30 40 50 600

0.1

0.2

0.3

0.4

0.5

0.6

Am

plitu

de [

m]

k [rad/m]0 10 20 30 40 50 60

0

0.1

0.2

0.3

0.4

0.5

0.6

Am

plitu

de [

m]

k [rad/m]0 10 20 30 40 50 60

0

0.1

0.2

0.3

0.4

0.5

0.6

Am

plitu

de [

m]

k [rad/m]0 10 20 30 40 50 60

-pi

-3 pi/4

-pi/2

-pi/4

0

Pha

se [

rad]

Page 15: Announcements 10/20/10

Low Pass FilterLow Pass Filter

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

y an

d y fil

tere

d

[m]

x [m]

Page 16: Announcements 10/20/10

Actual Data from OscilloscopeActual Data from Oscilloscope


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