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6.003: Signals and Systems
CT Fourier Transform
November 12, 2009
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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.
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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
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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
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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
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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
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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
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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
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Fourier Transforms
Stretching time compresses frequency.
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?
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Check Yourself
Stretching time compresses frequency.
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?
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Check Yourself
Stretching time compresses frequency.
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?
x2(2) = x1(1)
x2(t) = x1(at)
Therefore a = 1/2, or more generally, a < 1.
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Check Yourself
Stretching time compresses frequency.
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?a < 1
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Fourier Transforms
Find a general scaling rule.
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).
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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
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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
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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 !
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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
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Fourier Transforms in Physics: Diffraction
A diffraction grating breaks a laser beam input into multiple beams.
Demonstration.
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Fourier Transforms in Physics: Diffraction
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.
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Fourier Transforms in Physics: Diffraction
CD demonstration.
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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
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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
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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
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Fourier Transforms in Physics: Diffraction
DVD demonstration.
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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×
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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
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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×
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Fourier Transforms in Physics: Diffraction
Macroscopic information in the far field provides microscopic (invis-
ible) information about the grating.
θ
λ
D
sin θ = λD
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Fourier Transforms in Physics: Crystallography
What if the target is more complicated than a grating?
target
image?
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Fourier Transforms in Physics: Crystallography
Part of image at angle θ has contributions for all parts of the target.
target
image?
θ
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Fourier Transforms in Physics: Crystallography
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
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Fourier Transforms in Physics: Crystallography
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) !
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Fourier Transforms in Physics: Diffraction
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
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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
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Two Dimensions
Demonstration: 2D grating.
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Microscope
Images from even the best microscopes are blurred.
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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.
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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.
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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.
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Microscope
Blurring can be represented by convolving the image with the optical
“point-spread-function” (3D impulse response).
target image
∗ =
Blurring is inversely related to the diameter of the lens.
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Microscope
Blurring can be represented by convolving the image with the optical
“point-spread-function” (3D impulse response).
target image
∗ =
Blurring is inversely related to the diameter of the lens.
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Microscope
Blurring can be represented by convolving the image with the optical
“point-spread-function” (3D impulse response).
target image
∗ =
Blurring is inversely related to the diameter of the lens.
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Microscope
Measuring the “impulse response” of a microscope.
Image diameter ≈ 6 times target diameter: target → impulse.
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Microscope
Images at different focal planes can be assembled to form a three-
dimensional impulse response (point-spread function).
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Microscope
Blurring along the optical axis is better visualized by resampling the
three-dimensional impulse response.
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Microscope
Blurring is much greater along the optical axis than it is across the
optical axis.
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An Historic Fourier Transform
Taken by Rosalind Franklin, this image sparked Watson and Crick’s
insight into the double helix.
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An Historic Fourier Transform
This is an x-ray crystallographic image of DNA, and it shows the
Fourier transform of the structure of DNA.
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An Historic Fourier Transform
High-frequency bands indicate repreating structure of base pairs.
b1/b
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An Historic Fourier Transform
Low-frequency bands indicate a lower frequency repeating structure.
h 1/h
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An Historic Fourier Transform
Tilt of low-frequency bands indicates tilt of low-frequency repeating
structure: the double helix!
θ
θ
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Simulation
Easy to calculate relation between structure and Fourier transform.
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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.