6.003: Signals and Systemsweb.mit.edu/6.003/F09/www/handouts/lec18.pdf · Homework 10 will not collected or graded. Solutions will be posted. Closed book: 3 page of notes (81 2 ×11

6.003: Signals and Systems

CT Fourier Transform

November 12, 2009

Mid-term Examination #3

Wednesday, November 18, 7:30-9:30pm, Walker Memorial

(this exam is after drop date).

No recitations on the day of the exam.

Coverage: cumulative with more emphasis on recent material

lectures 1–18

homeworks 1–10

Homework 10 will not collected or graded. Solutions will be posted.

Closed book: 3 page of notes (812 × 11 inches; front and back).

Designed as 1-hour exam; two hours to complete.

Review sessions during open office hours.

Conflict? Contact [email protected] by Friday, November 13, 2009.

CT Fourier Transform

Representing signals by their frequency content.

X(jω)=∫ ∞−∞x(t)e−jωtdt (“analysis” equation)

x(t)= 12π

∫ ∞−∞X(jω)ejωtdω (“synthesis” equation)

• generalizes Fourier series to represent aperiodic signals.

• equals Laplace transform X(s)|s=ω if ROC includes jω axis.

→ inherits properties of Laplace transform.

• complex-valued function of real domain ω.

• simple ”inverse” relation

→ more general than table-lookup method for inverse Laplace.

→ “duality.”

• filtering.

• applications in physics

Visualizing the Fourier Transform

Fourier transform is a function of a single variable: frequency ω.

Time representation:

−1 1

x1(t)

1

t

Frequency representation:

2

π

X1(jω) = 2 sinωω

ω

Check Yourself

Signal x2(t) and its Fourier transform X2(jω) are shown below.

−2 2

x2(t)

1

t

b

ω0

X2(jω)

ω

Which is true?

1. b = 2 and ω0 = π/22. b = 2 and ω0 = 2π3. b = 4 and ω0 = π/24. b = 4 and ω0 = 2π5. none of the above

Check Yourself

Find the Fourier transform.

X2(jω) =∫ 2

−2e−jωtdt = e

−jωt

−jω

∣∣∣∣2−2

= 2 sin 2ωω

= 4 sin 2ω2ω

4

π/2ω

Check Yourself

Signal x2(t) and its Fourier transform X2(jω) are shown below.

−2 2

x2(t)

1

t

b

ω0

X2(jω)

ω

Which is true? 3

1. b = 2 and ω0 = π/22. b = 2 and ω0 = 2π3. b = 4 and ω0 = π/24. b = 4 and ω0 = 2π5. none of the above

Fourier Transforms

Stretching time compresses frequency.

−1 1

x1(t)

1

t

2

π

X1(jω) = 2 sinωω

ω

−2 2

x2(t)

1

t

4

π/2

X2(jω) = 4 sin 2ω2ω

ω

Fourier Transforms


Find a general scaling rule.

Let x2(t) = x1(at).

If time is stretched in going from x1 to x2, is a > 1 or a < 1?

Check Yourself



Let x2(t) = x1(at).


Check Yourself



Let x2(t) = x1(at).


x2(2) = x1(1)

x2(t) = x1(at)

Therefore a = 1/2, or more generally, a < 1.

Check Yourself



Let x2(t) = x1(at).

If time is stretched in going from x1 to x2, is a > 1 or a < 1?a < 1

Fourier Transforms


Let x2(t) = x1(at).

X2(jω) =∫ ∞−∞x2(t)e−jωtdt =

∫ ∞−∞x1(at)e−jωtdt

Let τ = at (a > 0).

X2(jω) =∫ ∞−∞x1(τ)e−jωτ/a 1

adτ = 1

aX1

(jω

a

)

If a < 0 the sign of dτ would change along with the limits of integra-

tion. In general,

x1(at) ↔ 1|a|X1

(jω

a

).

If time is stretched (a < 1) then frequency is compressed and ampli-

tude increases (preserving area).

Moments

The value of X(jω) at ω = 0 is the integral of x(t) over time t.

X(jω)|ω=0 =∫ ∞−∞x(t)e−jωtdt =

∫ ∞−∞x(t)ej0tdt =

∫ ∞−∞x(t) dt

−1 1

x1(t)

1

t

area = 2 2

π

X1(jω) = 2 sinωω

ω

Moments

The value of x(0) is the integral of X(jω) divided by 2π.

x(0) = 12π

∫ ∞−∞X(jω) e jωtdω = 1

2π

∫ ∞−∞X(jω) dω

−1 1

x1(t)

1

t++

−− ++ −−

2

π

X1(jω) = 2 sinωω

ω

area

2π= 1

Moments

The value of x(0) is the integral of X(jω) divided by 2π.

x(0) = 12π

∫ ∞−∞X(jω) e jωtdω = 1

2π

∫ ∞−∞X(jω) dω

−1 1

x1(t)

1

t++

−− ++ −−

2

π

X1(jω) = 2 sinωω

ω

area

2π= 1

2

πω

equal areas !

Stretching to the Limit

Moment relations + scaling → new way to think about an impulse.

Find Fourier transform of x2(t) = lima→0x1(at).

−1 1

x1(t)

1

t

2

π

X1(jω) = 2 sinωω

ω

−2 2

1

t

4

πω

1

t

2π

ω

Fourier Transforms in Physics: Diffraction

A diffraction grating breaks a laser beam input into multiple beams.

Demonstration.


The grating has a periodic structure (period = D).

θ

λ

D

sin θ = λD

The “far field” image is formed by interference of scattered light.

Viewed from angle θ, the scatterers are separated by D sin θ.

If this distance is an integer number of wavelengths λ→ constructive

interference.


CD demonstration.

Check Yourself

CD demonstration.

laser pointer

λ = 500 nm

CDscreen

3 feet

1 feet

What is the spacing of the tracks on the CD?

1. 160 nm 2. 1600 nm 3. 16µm 4. 160µm

Check Yourself

What is the spacing of the tracks on the CD?

grating tan θ θ sin θ D = 500nm

sin θmanufacturing spec.

CD 13 0.32 0.31 1613 nm 1600 nm

Check Yourself

Demonstration.

laser pointer

λ = 500 nm

CDscreen

3 feet

1 feet

What is the spacing of the tracks on the CD? 2.

1. 160 nm 2. 1600 nm 3. 16µm 4. 160µm


DVD demonstration.

Check Yourself

DVD demonstration.

laser pointer

λ = 500 nm

DVDscreen

1 feet

1 feet

What is track spacing on DVD divided by that for CD?

1. 4× 2. 2× 3. 12× 4. 1

4×

Check Yourself

What is spacing of tracks on DVD divided by that for CD?

grating tan θ θ sin θ D = 500nm

sin θmanufacturing spec.

CD 13 0.32 0.31 1613 nm 1600 nm

DVD 1 0.78 0.71 704 nm 740 nm

Check Yourself

DVD demonstration.

laser pointer

λ = 500 nm

DVDscreen

1 feet

1 feet

What is track spacing on DVD divided by that for CD? 3

1. 4× 2. 2× 3. 12× 4. 1

4×


Macroscopic information in the far field provides microscopic (invis-

ible) information about the grating.

θ

λ

D

sin θ = λD

Fourier Transforms in Physics: Crystallography

What if the target is more complicated than a grating?

target

image?


Part of image at angle θ has contributions for all parts of the target.

target

image?

θ


The phase of light scattered from different parts of the target un-

dergo different amounts of phase delay.

θx sin θ

x

Phase at a point x is delayed (i.e., negative) relative to that at 0:

φ = −2πx sin θλ


Total light F (θ) at angle θ is the integral of amount scattered from

each part of the target (f(x)) appropriately shifted in phase.

F (θ) =∫f(x)e−j2π

x sin θλ dx

Assume small angles so sin θ ≈ θ.

Let ω = 2π θλ .

Then the pattern of light at the detector is

F (ω) =∫f(x)e−jωxdx

which is the Fourier transform of f(x) !


There is a Fourier transform relation between this structure and the

far-field intensity pattern.

· · ·· · ·

grating ≈ impulse train with pitch D

t0 D

· · ·· · ·

far-field intensity ≈ impulse train with reciprocal pitch ∝ λD

ω0 2πD

Impulse Train

The Fourier transform of an impulse train is an impulse train.

· · ·· · ·

x(t) =∞∑k=−∞

δ(t− kT )

t0 T

1

· · ·· · ·

ak = 1 ∀ k

k

1

· · ·· · ·

X(jω) =∞∑k=−∞

2πTδ(ω − k2π

T)

ω0 2πT

2πT

Two Dimensions

Demonstration: 2D grating.

Microscope

Images from even the best microscopes are blurred.

Microscope

A perfect lens transforms a spherical wave of light from the target

into a spherical wave that converges to the image.

target image

Blurring is inversely related to the diameter of the lens.

Microscope



target image


Microscope



target image


Microscope

Blurring can be represented by convolving the image with the optical

“point-spread-function” (3D impulse response).

target image

∗ =


Microscope



target image

∗ =


Microscope



target image

∗ =


Microscope

Measuring the “impulse response” of a microscope.

Image diameter ≈ 6 times target diameter: target → impulse.

Microscope

Images at different focal planes can be assembled to form a three-

dimensional impulse response (point-spread function).

Microscope

Blurring along the optical axis is better visualized by resampling the

three-dimensional impulse response.

Microscope

Blurring is much greater along the optical axis than it is across the

optical axis.

An Historic Fourier Transform

Taken by Rosalind Franklin, this image sparked Watson and Crick’s

insight into the double helix.


This is an x-ray crystallographic image of DNA, and it shows the

Fourier transform of the structure of DNA.


High-frequency bands indicate repreating structure of base pairs.

b1/b


Low-frequency bands indicate a lower frequency repeating structure.

h 1/h


Tilt of low-frequency bands indicates tilt of low-frequency repeating

structure: the double helix!

θ

θ

Simulation

Easy to calculate relation between structure and Fourier transform.

Fourier Transform Summary

Represent signals by their frequency content.

Key to “filtering,” and to signal-processing in general.

Important in many physical phenomenon: x-ray crystallography.

Documents

6.003: Signals and Systemsweb.mit.edu/6.003/F09/www/handouts/lec18.pdf · Homework 10 will not collected or graded. Solutions will be posted. Closed book: 3 page of notes (81 2 ×11