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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 freeman@mit.edu 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
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?
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?
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.
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
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).
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.
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.
Fourier Transforms in Physics: Diffraction
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
Fourier Transforms in Physics: Diffraction
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×
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
Fourier Transforms in Physics: Crystallography
What if the target is more complicated than a grating?
target
image?
Fourier Transforms in Physics: Crystallography
Part of image at angle θ has contributions for all parts of the target.
target
image?
θ
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
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) !
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
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
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
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
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.
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.
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.
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.
An Historic Fourier Transform
This is an x-ray crystallographic image of DNA, and it shows the
Fourier transform of the structure of DNA.
An Historic Fourier Transform
High-frequency bands indicate repreating structure of base pairs.
b1/b
An Historic Fourier Transform
Low-frequency bands indicate a lower frequency repeating structure.
h 1/h
An Historic Fourier Transform
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.
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