Download pdf - [email protected] Dept ECE, UCB 425, ECEE 232, x24661 ...ecee.colorado.edu/~ecen5696/LECTURES/FO19_1DFT-2x2.pdf · Computational Fourier Optics: A MATLAB Tutorial, 2011 J. Schmidt

ECEN 5696 Fourier Optics

Professor Kelvin WagnerDept ECE, UCB 425, ECEE 232, x24661

[email protected]

What you will learn by completing this specialization

• Fourier transforms in time and space. 1D, 2D, 3D and 4D

• From Maxwell’s equations to diffraction and imaging

•Numerical techniques in wave opticsAberations and Beam Propagation

•Holography and Optical Information Processing

• Spatio-Temporal Fourier Optics and multiple wavelengths

Kelvin Wagner, University of Colorado Fourier Optics Fall 2019 1

Fourier Optics Learning Objectives

• Review Fourier transforms and develop deep intuitive understanding

• Generalize the Fourier transform to 2-D images and fields

• Construct arbitrary solutions to Maxwell’s Eqn as a superposition of plane waves

• Understand how waves propagate through space and are focused by lenses

• Develop a clear intuition for the propagation of plane waves and Gaussian beams

• Compare, contrast, and analyze coherent and incoherent imaging systems

• Formulate a wave theory of aberations and visualize them

• Develop numerical techniques for optical beam propagation as one line of code

• Discover the use of optical correlations for pattern recognition

• Invent holography to record and transform optical fields

• Extend the ideas of holography to computer generated and digital holography

• Further generalize the Fourier approach to the case of broadband fields

• Utilize the Fourier decomposition to invent and evaluate novel optical systems


Suggested References for AdditionalReading

Texts and suggested references:J. Goodman , Introduction to Fourier Optics, 3rd EdJ. Shamir, Optical Systems & ProcessesJ. Gaskill, Linear Systems, Fourier Transforms, and OpticsT. Cathey, Optical Information Processing and HolographyB. Saleh, Fundamentals of Photonics Chapter 4D. Brady, Optical Imaging and Spectroscopy, 2009D. Voelz, Computational Fourier Optics: A MATLAB Tutorial, 2011J. Schmidt Numerical Simulation of Optical Wave Propagation, 2011N. George Fourier Optics, 2012 on-line short manuscriptR.K. Tyson, Principles and Applications of Fourier Optics, 2014Kedar Khare, Fourier Optics and Computational Imaging , 2016


Course Outline

Linear Systems and Fourier TransformsSampling Theory and the Fast Fourier Transform (FFT)

2-D Systems and Transforms, OperatorsWave Propagation, momentum spaceDiffraction Theory

Beam Propagation MethodFranhoffer and Fresnel DiffractionCoherent Optical ImagingIncoherent ImagingWave theory of aberrationsHolography

Computer Generated HolographyDigital Holography

Optical Information ProcessingSynthetic Aperture Radar (SAR) and TomographyVolume Holography: 3-D Fourier TransformsSpatial and Temporal Fourier Optics: 4-D Fourier transformsVector Effects, Subwavelength Structures


Review of 1-D Fourier TransformsLearning Objectives and Outcomes

•Remember integral definition of Fourier transform– define operator representation

•Recognize equivalence of 1-D spatial FT and temporal FT

•Review some FT pairs of compact and singular functions

•Visually identify 1-D Fourier transform pair plots

• Summarize the properties of 1-D Fourier transforms


4 types of Fourier TransformsR.A. Roberts, C.T. Mullis, Digital Signal processing, Addison Wesley 1987


The 1-D Temporal Fourier Transform:Definitions Valid for finite support or

periodic signalsForward temporal Fourier transform (Hz)

G(f) =

ˆ

g(t)e−i2πftdt = Fg(t)

Inverse transformg(t) =

ˆ

G(f)ei2πftdf = F−1G(f)

Alternate definition using angular radian frequency ω = 2πf

Forward temporal Fourier transform (rad/sec)

G(ω) =

ˆ

g(t)e−iωtdt = Fg(t) ω = 2πf

Inverse transform

g(t) =1

2π

ˆ

G(ω)eiωtdω = F−1G(ω) ≡ F−1G(ω) dω = 2πdf

Note that these FT functions are scaled versions of each other G(ω) = G(ω/2π)


Exactly Analogous 1-D Spatial FourierTransform: Definitions

Can similarly define FT in space using spatial frequency, u = fx [lines/mm], analogousto temporal frequency f [Hz], or use wavevector, kx [rad/mm], analogous to angularfrequency ω [rad/sec].

Forward 1-D spatial Fourier transform

G(u) =

ˆ

g(x)e−i2πuxdx = Fxg(x)

Inverse 1-D spatial Fourier transform

g(x) =

ˆ

G(u)ei2πuxdu = F−1x G(u)

or in terms of wavevector kx

G(kx) =

ˆ

g(x)e−ikxxdx = Fg(x) ≡ Fxg(x)

g(x) =1

2π

ˆ

G(kx)eikxxdkx = F−1G(kx) ≡ F−1x G(kx) ≡ F−1kx

G(kx)

Note that these FT functions are scaled versions of each other G(kx) = G(kx/2π)


1-dimensional Fourier transforms

Well defined continuous functionsrect(t) = Π(t) sinc(f)

rect( tT ) = Π(tT

) T sinc(Tf)

sinc(t) rect(f)

tri(t) = Λ(t) sinc2(f)

sinc2(t) tri(f)

e−πt2 e−πf

2

e±iπt2 e±iπ/4e∓iπf

2

e−|t| 21+(2πf)2

e−|t|H(t) 11+i2πf = 1−i2πf

1+(2πf)2

sechπt sechπf

jinc(t) = J1(πt)2t

√1− (2f)2Π(f)

1|t|1/2 1

|f |1/2

Singular functions in t or f

δ(t) 1(f)

δ(ta

)= |a|δ(t) |a|1(f)δ(t− t0) e−i2πt0f

1iπt sgn(−f)

u(t) = H(t) 12δ(f) + 1

i2πf

ei2πf0t δ(f − f0)cos(2πf0t) 1

2 [δ(f−f0)+δ(f+f0)]

sin(2πf0t) 12i [δ(f−f0)−δ(f+f0)]

comb(t) comb(f)

comb(tT

) |T |comb(Tf)

tk (−1i2π

)kδ(k)(f)

(1i2π

)kδ(k)(t) fk


Visual Fourier Transform DictionaryDiscontinuous Functions

rect(t)=P(t)

-1 0 1 2time

sinc(t)


Visual Fourier Transform DictionaryExponentially Decaying Functions

gausssian(t)

-2 0 2 4time

gaussian(f)

-1 0 1 2freq

exp(t)

-2 0 2 4time

e-|t|

-2 0 2 4time


Visual Fourier Transform DictionaryImpulsive Functions


Visual Fourier Transform DictionarySingular Functions


The Temporal Fourier Transform:Properties

Linearityag(t) + bh(t) aG(f) + bH(f)

Conjugationg∗(±t) G∗(∓f)

Scale

g(αt)1

|α|G(f

α

)

Shiftg(t− t0) e−i2πft0G(f)

Modulationei2πf0tg(t) G(f − f0)

Derivative and Integration

dn

dxng(t) (i2πf)nG(f)

ˆ t

−∞g(τ )dτ

1

i2πfG(f) +

G(0)

2δ(f)


The 1-D Fourier Transform: Properties 2

Parseval’s Theoremˆ ∞

−∞|g(t)|2dt =

ˆ ∞

−∞|G(f)|2df

Convolutionˆ

g(t′)h(t− t′)dt′ G(f)H(f)

g(t)h(t)

ˆ

G(t′)H(t− t′)dt′

Correlationˆ

g(t′)h∗(t′ − t)dt′ G(f)H∗(f)ˆ

g(t′)g∗(t′ − t)dt′ |G(f)|2

Fourier Integral

FtF−1t g(t) = F−1t Ftg(t) = g(t)

FtFtg(t) = g(−t)Kelvin Wagner, University of Colorado Fourier Optics Fall 2019 15

Operation x(t) = F−1f X(f) = F−1X (ω) Ff⇐⇒ X(f) =

ˆ ∞

−∞x(t)e−i2πftdt

F⇐⇒ X (ω) =ˆ ∞

−∞x(t)e−iωtdt

Linearity Ax(t) +By(t)Ff⇐⇒ AX(f) +BY (f)

F⇐⇒ AX (ω) +BY(ω)Conjugation x∗(t)

Ff⇐⇒ X∗(−f) F⇐⇒ X ∗(−ω)Scale x(at)

Ff⇐⇒ 1|a|X

(fa

) F⇐⇒ 1|a|X

(ωa

)

Mirror x(−t) Ff⇐⇒ X(−f) F⇐⇒ 2πX (−ω)Duality X(t) or X (t) Ff⇐⇒ x(−f) F⇐⇒ x(−ω)

Time shift x(t− T ) Ff⇐⇒ e−i2πfTX(f)F⇐⇒ e−iωTX (ω)

Frequency shift e−i2πf0tx(t) = e−iω0tx(t)Ff⇐⇒ X(f − f0) F⇐⇒ X (ω − ωs)

Time differentiation dx(t)dt

Ff⇐⇒ i2πfX(f)F⇐⇒ iωX (ω)

d2x(t)dt2

Ff⇐⇒ (i2πf)2X(f)F⇐⇒ iω)2X (ω)

dnx(t)dtn

Ff⇐⇒ (i2πf)nX(f)F⇐⇒ (iω)nX (ω)

frequency differentiation (−i)ntnx(t) Ff⇐⇒ 1(2π)n

dnX(f)dfn

F⇐⇒ dnX (ω)dωn

Time Integrationˆ t

−∞x(τ)dτ = u ∗ x Ff⇐⇒ X(f)

i2πf+ 1

2X(0)δ(f)

F⇐⇒ X (ω)iω

+ πX (0)δ(ω)

frequency integration x(t)t

Ff⇐⇒ˆ ∞

f

X(f ′)df ′ F⇐⇒ˆ ∞

ω

X (ω′)dω′

Convolution x(t) ∗ y(t) =ˆ ∞

−∞x(τ)y(t− τ)dτ Ff⇐⇒ X(f)Y (f)

F⇐⇒ X (ω)Y(ω)

Correlation x(t) ⋆ y(t) =

ˆ ∞

−∞x(τ)y∗(τ − t)dτ Ff⇐⇒ X(f)Y ∗(f)

F⇐⇒ X (ω)Y∗(ω)

Frequency Convolution x(t)y(t)Ff⇐⇒ X(f) ∗ Y (f)

F⇐⇒ 12πX (ω) ∗ Y(ω)

Time periodicity x(t)=combT (t)∗p(t)Ff⇐⇒ 1

Tcomb1/T (f)P (f)

F⇐⇒ 1Tcombω0(ω)P(ω)

Real & Even x(t)=Eℜx(t)

Ff⇐⇒ Real&Even F⇐⇒ Real&Even

Real & Odd x(t)=Oℜx(t)

Ff⇐⇒ Imagnary&Odd F⇐⇒ Imagnary&Odd

Decomposition of Real Ex(t) Ff⇐⇒ ℜX(f) F⇐⇒ ℜX (ω)Signals into Even & Odd Ox(t) Ff⇐⇒ iℑX(f) F⇐⇒ iℑX (ω)

Name x(t) = F−1f X(f) = F−1X (ω) Ff⇐⇒ X(f) =

ˆ ∞

−∞x(t)e−i2πftdt

F⇐⇒ X (ω) =ˆ ∞


rect Π(t) = rect(t)Ff⇐⇒ sinc(f)

F⇐⇒ sinc(

ω2π

)

sinc sinc(t)= sinπtπt

or snc(t)= sin tt

Ff⇐⇒ Π(f) = rect(f) or πrect(πf)F⇐⇒ rect

(ω2π

)or πrect

(ω2

)

Causal exponential e−αtu(t)Ff⇐⇒ 1

α+i2πf

F⇐⇒ 1α+iω

tent e−α|t| Ff⇐⇒ 2αα2+(2πf)2

F⇐⇒ 2αα+ω2

Triangle = tri(t) Λ(t) = Π(t) ∗ Π(t) =

1−|t| |t|<10 |t|>1

Ff⇐⇒ sinc2fF⇐⇒ sinc2

(ω2π

)

Gaussian e−πt2 e−12(t/σ)2 Ff⇐⇒ e−πf2

σ√2πe−2π2f2σ2 F⇐⇒ e−ω2/4π σ

√2πe−ω2σ2/2

Delta δ(t/T ) = |T |δ(f) Ff⇐⇒ |T |1(f) F⇐⇒ |T |1(ω)sign sgn(t) = t

|t| = −1 t<0

1 t>0

Ff⇐⇒ 1iπf

F⇐⇒ 2iω

step u(t) =

0 t<01 t>0

Ff⇐⇒ 1i2πf

+ 12δ(f)

F⇐⇒ 1iω

+ πδ(ω)

Hilbert kernel iπt

Ff⇐⇒ sgn(f)F⇐⇒ sgn(ω)

Constant KFf⇐⇒ Kδ(f)

F⇐⇒ K2πδ(ω)

Cosine cos(2πf0t) = cos(ω0t)Ff⇐⇒ 1

2[δ(f + f0) + δ(f − f0)] F⇐⇒ π[δ(ω + ω0) + δ(ω − ω0)]

Sine sin(2πf0t) = sin(ω0t)Ff⇐⇒ 1

2i[δ(f − f0)− δ(f + f0)]

F⇐⇒ iπ[δ(ω + ω0)− δ(ω − ω0)]

Complex exponential ei2πf0t = eiω0tFf⇐⇒ δ(f − f0) F⇐⇒ 2πδ(ω − ω0)

Causal Cosine cos(2πf0t)u(t) = cos(ω0t)u(t)Ff⇐⇒ 1

2[δ(f+f0)+δ(f−f0)] + if

2π(f20−f2)

F⇐⇒ π[δ(ω + ω0) + δ(ω − ω0)] +iω

ω20−ω2

Causal Decaying Sine e−at sin(ω0t)u(t)Ff⇐⇒ 2πf0

(2πf0)2+(a+i2πf)2)

F⇐⇒ ω0

ω20+(a+iω)2

Chirp eiat2= eiπbt

2 Ff⇐⇒√

ibe−iπf2/b F⇐⇒

√iπae−iω2/4a

CombT combT (t) =∑

n

δ(t− nT ) Ff⇐⇒ 1T

∑

m

δ(f − m

T

)= 1

Tcomb 1

T(f)

F⇐⇒ 2πT

∑

m

δ(ω − m2π

T

)= 2π

Tcomb 2π

T(f)

ramp tu(t)Ff⇐⇒ i

4πδ′(f)− 1

4π2f2

F⇐⇒ iπδ′(ω)− 1ω2

power tnFf⇐⇒

(i2π

)nδ(n)(f)

F⇐⇒ in(2π)δ(n)(ω)

Bessel J0(t)Ff⇐⇒ 2√

1−(2πf)2

F⇐⇒ 2√1−ω2 |ω| < 1 , 0 |ω| > 1

Jinc jinc(t) = J1(t)/2tFf⇐⇒

√1− (2πf)2

F⇐⇒√1− ω2 |ω| < 1 , 0 |ω| > 1

Periodic wave x(t) =∑

n

p(t− nT ) Ff⇐⇒ 1T

∑

n

P(nT

)δ(f − n/T ) F⇐⇒ 2π

T

∑

n

P

(2πn

T

)δ(ω − n2π/T )

x(t) = x(t + T ) =∑

n

Xnei2πnt/T Ff⇐⇒ =

∑

n

Xnδ(f − n/T ) F⇐⇒ = 2π∑

n

Xnδ(ω − n2π/T )

Fourier Transform Pairs:Sin,Rect,SincLearning Objectives and Outcomes

• Practice manipulation of simple Fourier integralsComplex exponentialsSine and CosineDelta functions and the sifting property

•Convergence factorsEvaluate diverging integrals as limits

•Rect and sinc and bandlimited superpositions


FT of complex exponentials

The most basic and fundamental FT pairs are the complex exponentials

Fei2πf0t

=

ˆ ∞

−∞ei2πf0te−i2πftdt =

ˆ ∞

−∞e−i2π(f−f0)tdt = δ(f − f0)

and

Fe−i2πf0t

=

ˆ ∞

−∞e−i2πf0te−i2πftdt =

ˆ ∞

−∞e−i2π(f+f0)tdt = δ(f + f0)

which produce singular FT spectra since when f 6= f0 the integrand is oscillatory andintegrates to zero, but when matched, f = f0 the integral is unbounded.

The inverse transform is of a singular function, but we can just use sifting property

F−1δ(f − f0) =ˆ ∞

−∞δ(f − f0)ei2πftdt = ei2πf0t

Using linearity of the FT to combine these results we get

Fcos(2πf0t) = F12[e

i2πf0t + e−i2πf0t]= 1

2[δ(f − f0) + δ(f + f0)]

and

Fsin(2πf0t) = F

12i[ei2πf0t − e−i2πf0t]

= 1

2i[δ(f − f0)− δ(f + f0)]


Fourier Transform of Impulse Pair: Cosineand Sine Duality gives FT of symetric

temporal impulse pairP. Kraniauskas, Transforms in Signals and Systems, Addison-Wesley, 1992


Fourier Transform of complex phase factortimes cosine eiπ/4 cos(ωot) and eiπ/8 cos(ωot+π/4)

eiπ/4 cos(ωot)F⇐⇒ eiπ/4

2 [δ(ω − ωo) + δ(ω + ωo)]

eiπ/8 cos(ωo + π/4)tF⇐⇒ ei(π/8+π/4)

2 δ(ω − ωo) + ei(π/8−π/4)2 δ(ω + ωo)


FT of complex exponential

Or we can use equivalent ω-transform notation

Feiω0t

=

ˆ ∞

−∞eiω0te−iωtdt =

ˆ ∞

−∞ei(ω0−ω)tdt = 2πδ(ω0 − ω)

and scaled inverse transform

F−1δ(ω − ω0) =1

2π

ˆ ∞

−∞δ(ω − ω0)e

iωtdt =1

2πeiω0t

FT of the delta impulseDuality or direct integration can be used to evaluate

Fδ(t) =ˆ ∞

−∞δ(t)e−iωtdt = 1

and for a shifted impulse

Fδ(t− T ) =ˆ ∞

−∞δ(t− T )e−iωtdt = e−iωT

which also comes out directly from time shifting property


Phase shifts of complex exponentials

Positive and Negative Linear Phase Factorseiωot

F⇐⇒ δ(ω − ωo) e−iωot F⇐⇒ δ(ω + ωo)

Phase Shifting Complex Exponential by multiplying by eiφ

iei2ωotF⇐⇒ iδ(ω − 2ωo) −iei2ωot F⇐⇒ −iδ(ω−2ωo)


FT of causal decreasing exponential

Fe−atu(t)

=

ˆ ∞

−∞u(t)e−ate−i2πftdt=

ˆ ∞

0

e(−a−i2πf)tdt=e(−a−i2πf)t

−a− i2πf

∣∣∣∣∞

0

= 0−1−a−i2πf =

1

a + i2πf

As long as e−a∞ → 0 which requires a > 0 and a convergent causal exponential

This could be performed with the ω transfom definition to obtain the equivalent result

Fe−atu(t)

=

ˆ ∞

−∞e−ate−iωtdt =

ˆ ∞

0

e(−a−iω)tdt =e(−a−iω)t

−a− iω

∣∣∣∣∞

0

=0− 1

−a− iω =1

a + iω

Which should be obvious from just the substitution ω = 2πf

Note that a causal sinusoid does not FT to a delta, and there is spectral spreading. Sincethe integral is not convergent, use the modulation theorem and known FT of u(t)

Fe−i2πf0tu(t)

=

ˆ ∞

0

e−i2π(f+f0)tdt =e−i2π(f+f0)t

−i2π(f + f0)

∣∣∣∣∞

0

=?

= δ(f + f0) ∗[12δ(f) +

1

i2πf

]= 1

2δ(f + f0) +1

i2π(f + f0)


Example Fourier Transform of rect

rect

(t

2T

)= Π

(t

2T

)=

0 t > T1 t < T

X(iω)=

ˆ ∞

−∞Π

(t

2T

)e−iωtdt =

ˆ T

−Te−iωtdt =

1

−iωe−iωt∣∣∣∣T

−T=

1

−iω(e−iωT − e+iωT

)

=2

ω

1

2i

(eiωT − e−iωT

)=

2

ωsinωT =

2

2πf

T

Tsin 2πft = 2T

sinπ2Tf

π2Tf= 2T sinc2Tf

where sinc(f) = sin πfπf

or using alternative sncω = sinωω

we get X(ω) = 2T snc (ωT )

Π

(t

1

) 1 · sinc(1 · f)

Π

(t

T

) T sinc(Tf)

1T

1T

1T

-

.636

.128

-.212t f

1.0

T


Sinc and its width


Pictorial Fourier Transform of a symmetricrect

P. Kraniauskas, Transforms in Signals and Systems, Addison-Wesley, 1992


Pictorial Inverse Fourier Transformproducing a temporal rect in the limit of

infinite bandwidthP. Kraniauskas, Transforms in Signals and Systems, Addison-Wesley, 1992


Duality Pictorial Inverse Fourier Transformof a rect in frequency



Time lapse movie of phasor evolution ofeach Fourier component of rect in

frequencyP. Kraniauskas, Transforms in Signals and Systems, Addison-Wesley, 1992


FT of sinc

Fsinc(t)=ˆ ∞

−∞sinc(t)e−i2πftdt=

ˆ ∞

−∞

sin πt

πtcos(2πft) dt− i

0 =´

oddˆ ∞

−∞

sinπt

πtsin(2πft) dt

=

ˆ ∞

−∞

sin(πt + 2πft)

2πt+sin(πt− 2πft)

2πtdt

=

ˆ ∞

−∞

sin[π(1 + 2f)t]

(1 + 2f)t

(1 + 2f)

2+sin[π(1− 2f)t]

(1− 2f)t

(1− 2f)

2dt

=1 + 2f

2

ˆ ∞

−∞

sin[π(1 + 2f)t]

(1 + 2f)tdt +

1− 2f

2

ˆ ∞

−∞

sin[π(1− 2f)t]

(1− 2f)tdt

=1 + 2f

2|1 + 2f | +1− 2f

2|1− 2f | = Π(t) = rect (t)

Where we used ˆ ∞

−∞sinc(at) dt =

1

|a|1+2f

2|1+2f | =12sgn(f + 1

2)1−2f

2|1−2f | =12sgn(−(f − 1

2))

Or directly by duality since

f|f|

1+2f

|1+2f|1-2f

|1-2f|

-2

23

1 2-1-2

12

-1

3

1 2-1-2

12

-1

3

-1-2 1 2

1

-1

-1-2 1 2

1

-1-1-2 1 2

1

-1-1-2 1 2

1

-1-1-2 1 2

1

-1

-1-2 1 2

1

-1

-1-2 1 2

1

-1

Π(t) sinc(f) ⇒ sinc(t) Π(−f) = Π(f)


Fourier Transform Pairs:Step,GaussianLearning Objectives and Outcomes

• Practice manipulation of simple 1-D Fourier integralsSign and Step functionsGaussian function needed for lasersComplex Gaussian as generalization of real Gaussian


FT of sign function

The sign function is defined as

sgn(t) = 2u(t)− 1 = u(t)− u(−t) =−1 t < 01 t > 0

and Fsgn(t) not absolutely convergent since´∞−∞ |sgn(t)|dt→∞

Must define FT in terms of series of functions that in the limit become sgn(t)

sgn(t)= lima→∞

sa(t) where sa(t)=

−et/a t < 0e−t/a t > 0

For finite a we have a well defined FT

Fsa(t)=ˆ ∞

0

e−t/ae−i2πftdt−ˆ 0

−∞et/ae−i2πftdt=

ˆ ∞

0

e(−1/a−i2πf)tdt−ˆ 0

−∞e(1/a−i2πf)tdt

=e(−1/a−i2πf)t

−1/a− i2πf

∣∣∣∣∞

0

− e(1/a−i2πf)t

1/a− i2πf

∣∣∣∣0

−∞=

0− 1

−1/a− i2πf −1− 0

1/a− i2πf =−i4πf

(1/a)2 + (2πf)2

Giving the FT of sgn in the limit a→∞

Fsgn(t) = Flima→∞

sa(t)=

−i4πf(0)2 + (2πf)2

=1

iπf


FT of sign function

Note that an easier route is to use the derivative and integration properties

d

dt(sgn(t) +K) = 2δ(t)

Fd

dt(sgn(t) +K)

= iωFsgn(t) +K = 2

Thus

iωFsgn(t) + iω2πKδ(ω) = 2 ⇒ Fsgn(t) = 2

iω− 2πKδ(ω)

The constant K must be zero since sgn(t) + sgn(−t) = 0

2

iω− 2πKδ(ω) +

2

−iω − 2πKδ(ω) = 0 ⇒ K = 0

Fsgn(t) = 2

iω

And from this we can get the FT of u(t) = 12 +

12sgn(t)

Fu(t) = πδ(ω) +1

iω


FT of Heavyside step function

´∞−∞ |u(t)|dt→∞ not convergent so Laplace transform 1/s not defined along iω axis.

Must instead define in terms of series of functions that in the limit become u(t)

u(t) = lima→∞

ua(t) where ua(t) =

0 t < 0

e−t/a t > 0

Or we can use that u(t) = 12 +

12sgn(t) with known FT and linearity of FT

Fu(t) = F12+ 1

2sgn(t)

= 1

2δ(f) +

1

i2πf

Now using the scaling of a delta to write δ(f) = δ(ω2π

)= 2πδ(ω)

Fu(t) = πδ(ω) +1

iω

Duality can be invoked to evaluate1

iπt sgn(−f)

12δ(t) +

1

i2πt u(−f)


FT of 1/iπt

F

1

iπt

=

ˆ ∞

−∞

1

iπte−i2πftdt =

0 =´

evenoddˆ ∞

−∞

1

iπtcos(2πft)dt− i

ˆ ∞

−∞

1

iπtsin(2πft)dt

=−i2fi

ˆ ∞

−∞

sin(2πft)

2πftdt = −2f

ˆ ∞

−∞sinc(2ft)dt =

−2f|2f | = sgn(−f)

where the infinite integral (eg area) of a scaled sinc of width 1/2f is just given by thewidth 1/2f , but when f is negative this area of the even sinc is −1/2f , thus

ˆ ∞

−∞sinc(2ft)dt =

1

2|f |

or duality can be invoked from the known FT Fsgn(t) = 1/iπt to evaluate

1

iπt sgn(−f)

But in this case the FT of 1/iπt is actually the convergent one, so duality should beinvoked instead to find FT of sgn(t).


FT Gaussian and Complex Gaussian

g(t) = e−πt2 |

´∞−∞ e

−πt2dt = 1 n(t) = e−(t/τ)2

G(f) =

ˆ

e−πt2e−i2πftdt =

ˆ

e−π(t2+i2ft)dt

note (t + if)2 = t2 + 2ift− f 2

= e−πf2ˆ

e−π(t+if)2dt = e−πf

2ˆ

e−πt′2dt′

︸︷︷︸= e−πf

2

t′ = t + if dt′ = dt 1= area of normalized unit GaussianScaled Gaussian

g(t) = e−π(t/τ)2

τe−π(τf)2

Complex Gaussian (Chirp with Gaussian amplitude modulation)1/τ 2 = a + ib when a = 0 this is a chirp

e−π(a+ib)t2

1√a + ib

e−πf2/(a+ib)

e−iπbt2

1√ibe+iπf

2/b


Fourier Transform PropertiesLearning Objectives and Outcomes

•Recognize and use Linearity in evaluating FT

•Understand Shift and Scale theorems

•Gain insight from Duality

•Understand the Convolution Theorem

•Apply FT Symmetries


Linearity

The Fourier integral is clearly linear, so that

Fax(t) + by(t) = aFx(t) + bFy(t)and

F−1aX(f) + bY (f) = aF−1X(f) + bF−1Y (f)

Linearity is an implicit assumption in the decomposition of functions into even andodd parts. Similarly implicit in decomposition of signals or transforms into real andimaginary parts

Example

Fe−|t|

= F

e−tu(t) + etu(−t)

=

1

1 + j2πf+

1

1− j2πf =1−

j2πf + (1 +j2πf)

1 + (2πf)2

=2

1 + (2πf)2

Note this follows from the Laplace transform evaluated at s = 2πf


Shift Theorem

x(t− T ) e−i2πfTX(f)

ˆ ∞

−∞e−i2πfTX(f)ei2πftdf =

ˆ ∞

−∞X(f)ei2πf(

t′︷︸︸︷t− T )df = x(t′) = x(t− T )

Π(t) sinc(f)

Π(t− T ) sinc(f)e−i2πfT

Π(t− 1/2) sinc(f)e−iπf


Fourier Transform of Shifted FunctionsP. Kraniauskas, Transforms in Signals and Systems, Addison-Wesley, 1992


Conjugation

x∗(t) X∗(−ω)Proof

X∗(ω) =

(ˆ ∞


)∗=

ˆ ∞

−∞x∗(t)e+iωtdt

let ω′ = −ωX∗(−ω′) =

ˆ ∞

−∞x∗(t)e−iω

′tdt

So when x(t) is real x(t) = x∗(t)

X∗(−ω) =ˆ ∞

−∞x(t)e−iωtdt = X(ω)

So we can write X(ω) in terms of real and imaginary parts

X(ω) = ℜX(ω) + iℑX(ω) = Xr(ω) + iXi(ω)

Xr(ω) + iXi(ω) = Xr(−ω)− iXi(−ω)So we can conclude for real x(t) we have Hermition X(ω)

Xr(ω) = Xr(−ω) eveniXi(ω) = −iXi(−ω) odd


Cosine and Sine parts of Fourier TransformP. Kraniauskas, Transforms in Signals and Systems, Addison-Wesley, 1992

inverse Fourier transform of unitheight rect pulse

Note the even character of cosωt

Sum along ridge does not cancel

sine carrier surface

Odd symetry of sinωt

Symmetric sums along ω cancelfor all t giving 0 real part of f(t)for even ℑF (jω)


Scaling

x(at)1

|a|X(f

a

)

ˆ ∞

−∞x(at)e−i2πftdt =

ˆ ∞

−∞x(τ )e−i2πfτ/adτ/a τ = at dτ = adt a > 0

=

ˆ −∞

+∞x(τ )e−i2πfτ/adτ/a =

1

|a|

ˆ ∞

−∞x(τ )e−i2πfτ/adτ a < 0

=1

|a|

ˆ ∞

−∞x(τ )e

−i2π(fa

)τdτ =

1

|a|X(f

a

)

special case for mirroring in time

x(−t) X(−f)

Normalized time width T form of the scaling thm

x

(t

T

) |T |X(fT )


Fourier Transform of Scaled FunctionsP. Kraniauskas, Transforms in Signals and Systems, Addison-Wesley, 1992


Shifted and scaled functions

Find FT of shifted and scaled rectangle by breaking it down into steps

x(t) = 25Π

(t− 4

10

)

apply scale theorem

Π

(t

10

)F⇐⇒10sinc(10f )

apply shift theorem to that result

Π

(t− 4

10

)F⇐⇒10sinc(10f ) e−i2π4f

apply linearity

25Π

(t− 4

10

)F⇐⇒250sinc(10f ) e−i2π4f


Shifted and scaled functions

Find FT of composite function containing shifted, scaled,phase shifted, and weighted functions.

sinc

[t− 7

2

]ei2π11t + 4Π[3t + 15]e−i2π13t

Looks complicated. Don’t panic, just break it down into in-dividual parts.

First note the interesting duality

sinc(t) + Π(t)F⇐⇒Π(f ) + sinc(f )

This works for even functions

For general functions there is a more complete duality

x(t) +X(t) + x(−t) +X(−t) F⇐⇒X(f ) + x(−f ) +X(−f ) + x(f )Kelvin Wagner, University of Colorado Fourier Optics Fall 2019 47

linearity combined with shifted and scaledfunctions

Find FT of composite function

sinc

[t− 7

2

]ei2π11t + 4Π[3t + 15]e−i2π13t

Build up the complicated functions from its parts

sinc

(t

2

)F⇐⇒2Π(2f)

Π(3t)F⇐⇒ 1

3sinc

(f

3

)

sinc

(t− 7

2

)F⇐⇒2Π(2f)e−i2π7f

Π[3(t + 5)]F⇐⇒ 1

3sinc

(f

3

)ei2π5f

sinc

(t− 7

2

)ei2π11t

F⇐⇒2Π[2(f − 11)]e−i2π7(f−11)

Π[3(t + 5)]e−i2π13tF⇐⇒ 1

3sinc

(f + 13

3

)ei2π5(f+13)

sinc[t−72

]ei2π11t+4Π[3t+15]e−i2π13t

F⇐⇒2Π[2(f−11)

]e−i2π7(f−11)+ 4

3sinc(f+133

)ei2π5(f+13)


Duality and the Transform of a Transform

Suppose h(t) H(f) but H() is just a function (eg sinc) and that we want to knowthe Fourier transform of that function of timelet g(t) = H(t)

FH(t) = Fg(t) = G(f) =

ˆ ∞

−∞g(t)e−i2πftdt =

ˆ ∞

−∞H(t)e−i2πftdt

=

ˆ ∞

−∞H(t′)ei2π(−f)t

′dt′ = h(−f)

This means thatFFh(t) = h(−t)

Duality suggests that property of time domain has dual when applied to frequencydomain

dX(f)

df=d

df

ˆ ∞

−∞x(t)e−i2πftdt =

ˆ ∞

−∞x(t)(−i2πt)e−i2πftdt

−i2πtx(t) dX(f)

dfDifferentiation in frequency

e+i2πf0tx(t) X(f − f0) Modulation theorem


DualityP. Kraniauskas, Transforms in Signals and Systems, Addison-Wesley, 1992


Duality: Shifting in time and Shifting inFrequency



Convolution Theorem

y(t) =

ˆ ∞

−∞x(τ )h(t− τ )dτ =

ˆ ∞

−∞h(τ ′)x(t− τ ′)dτ ′

Fy(t)=Y (f)=X(f)H(f)=H(f)X(f)

Transform domain shift by τ

Y (iω) =

ˆ

[ˆ ∞

−∞x(τ )h(t− τ )dτ

]e−iωtdt =

ˆ

x(τ )

[ˆ

e−iωt︷︸︸︷h(t− τ ) dt

]dτ

=

ˆ

x(τ )[H(ω)e−iωτ

]dτ = H(ω)

ˆ

x(τ )e−iωτdτ = H(ω)X(ω)

y(t) = h(t) ∗ x(t) F⇐⇒ Y(ω) = H(ω)X(ω)

Y (f) = H(f)X(f)

Modulation Theorem : Duality - convolution in frequency domain

r(t) = s(t)p(t)F⇐⇒ 1

2π

ˆ

S(µ)P(ω − µ)dµ =1

2πS(ω) ∗ P (ω)

ˆ

S(f ′)P (f − f ′)df ′ = S(f) ∗ P (f)


Convolution Theorem Examples

Tri(t) = Λ(t) = Π(t) ∗ Π(t) = rect (t) ∗ rect (t)

FΛ(t) = FΠ(t)FΠ(t) = sinc2(f)

Simple proofs of convolution properitesCommutative

x(t) ∗ y(t) = y(t) ∗ x(t) since X(f)Y (f) = Y (f)X(f)

Associative

x(t)∗[y(t)∗z(t)

]=[x(t)∗y(t)

]∗z(t) since X(f)

[Y (f)Z(f)

]=[X(f)Y (f)

]Z(f)

Distributive

x(t) ∗[y(t) + z(t)

]=x(t) ∗ y(t) + x(t) ∗ z(t) since X

[Y + Z

]=XY +XZ


Convolution Theorem Examples

Find the FT of the convolution of x(t) = 10 sin 2πf0t with y(t) = 2δ(t + 4)

Do convolution first then FT

10 sin(2πf0t) ∗ 2δ(t + 4) = 20 sin[2πf0(t + 4)]F⇐⇒20

1

2i(δ(f − f0)− δ(f + f0))e

i2π4f

Do FT first to avoid having to do convolution

10 sin(2πf0t) ∗ 2δ(t + 4)F⇐⇒10

1

2i(δ(f − f0)− δ(f + f0)) · 2ei2π4f

where we used10 sin(2πf0t)

F⇐⇒101

2i(δ(f − f0)− δ(f + f0))

2δ(t + 4)F⇐⇒ei2π4f


Convolution Theorem Example: Timeshifting as convolution with shifted impulse



Correlation Theorem

The correlation function measures the similarity between functions

x(t) ⋆ y(t) =

ˆ ∞

−∞x(τ )y∗(τ − t)dτ =

ˆ ∞

−∞x(τ ′ + t)y∗(τ ′)dτ ′

introducing the impulse response h(t) = y∗(−t) we can write this as a convolution

x(t)∗h(t)=ˆ ∞

−∞x(τ )h(t−τ )dτ =

ˆ ∞

−∞x(τ )y∗(−[t−τ ])dτ =

ˆ ∞

−∞x(τ )y∗(τ−t)dτ =x(t)⋆y(t)

Now we can use the convolution theorem and the relation y∗(−t) ⇐⇒ Y ∗(ω) to writethe correlation theorem

x(t) ⋆ y(t) = x(t) ∗ h(t) ⇐⇒ X(ω)H(ω) = X(ω)Y ∗(ω) X(f)Y ∗(f)

when x(t) = y(t− T ) and for a flat bandwidth B code (|Y (f)|2 = Π[fB

]) we get

y(t−T )⋆y(t) = F−1Y (f)e−i2πfT · Y (f)∗

= F−1

|Y (f)|2e−i2πfT

= sinc[B(t−T )]

Autocorrelation Theorem

x(t) ⋆ x(t) =

ˆ ∞

−∞x∗(τ )x(τ + t)dτ ⇐⇒ |X(f)|2


Modulation Theorem

Feiω0tf(t)

= F (ω − ω0)

Now using standard identities cosω0t =12[e

iω0t + e−iω0t] we get

Fcos(ω0t)f(t) =1

2

[F (ω − ω0) + F (ω + ω0)

]

And using sinω0t =12i[eiω0t − e−iω0t] we get

Fsin(ω0t)f(t) =1

2i

[F (ω − ω0)− F (ω + ω0)

]

1T

1T

1T

-

.636

.128

-.212t f

1.0

T

-f0 f0t f


Differentiation and Integration

Differentiationdx(t)

dt=d

dt

ˆ ∞

−∞X(f)ei2πftdt =

ˆ ∞

−∞

d

dt

X(f)ei2πft

dt =

ˆ ∞

−∞X(f)

d

dt

ei2πft

dt

=

ˆ ∞

−∞[X(f)(i2πf)]ei2πftdt

dx(t)

dt i2πfX(f)

dx(t)

dt iωX(ω)

Integration

y(t) =

ˆ t

−∞x(τ )dτ = x(t) ∗ u(t) = u(t) ∗ x(t)

=

ˆ ∞

−∞u(τ )x(t− τ )dτ =

ˆ ∞

0

x(t− τ )dτ =

ˆ ∞

−∞x(τ )u(t− τ )dτ =

ˆ t

−∞x(τ )dτ

Now using the convolution theorem

Y (f) = X(f)U (f) = X(f)

[1

−i2πf + 12δ(f)

]=

X(f)

−i2πf +X(0)

2δ(f)

ˆ t

−∞x(τ )dτ

X(f)

−i2πf +X(0)

2δ(f)

X(ω)

−iω +X(0)πδ(ω)


FT of Triangle Function and its DerivativeP. Kraniauskas, Transforms in Signals and Systems, Addison-Wesley, 1992


Pictorial FT of IntegrationP. Kraniauskas, Transforms in Signals and Systems, Addison-Wesley, 1992


Differentiation in Frequency

Duality in its f and ω form gives

X(t)F⇐⇒x(−f) X(t)

F⇐⇒2πx(−ω)

Differentiation in timedx(t)

dt

F⇐⇒ i2πfX(f)dx(t)

dt

F⇐⇒ iωX(ω)

Dual is differentiation in frequency

−i2πtx(t) F⇐⇒ dX(f)

df− itx(t) F⇐⇒ dX(ω)

dω

Now we can use this to evaluate FT of causal ramp, remember

u(t)F⇐⇒πδ(ω) +

1

iω

Thus

tu(t)F⇐⇒ i

dU (ω)

dω= i

d

dω

[πδ(ω) +

1

iω

]= iπδ′(ω)− ω−2


Parseval’s Theorem(Rayleigh’s or Plancherel’s Thm)

Power P (t) = v2(t)R

= i2(t)R so Energy delivered to R = 1Ω load isW =

´∞−∞P (t)dt=

´∞−∞f

2(t)dt=´∞−∞|x(t)|2dt where f(t) is either real v(t) or i(t)

Total Energy can be calculated by integrating energy per unit time, |x(t)|2, across timeOr by integrating energy per unit frequency, |X(f)|2, across all frequencies

ˆ ∞

−∞|x(t)|2dt =

ˆ ∞

−∞|X(f)|2df

Proof

E =

ˆ ∞

−∞x(t)x∗(t)dt =

ˆ ∞

−∞x(t)x∗(t)e−i2πf

′tdt

∣∣∣∣f ′=0

= X(f) ∗X∗(−f)∣∣∣f ′=0

=

ˆ ∞

−∞X(f)X∗(−(f ′ − f))df

∣∣∣∣f ′=0

=

ˆ ∞

−∞X(f)X∗(f − f ′)df

∣∣∣∣f ′=0

=

ˆ ∞

−∞X(f)X∗(f)df

For periodic x(t) with period T0 = 1/f0

1

T0

ˆ t0+T0

t0

|x(t)|2dt =∞∑

n=−∞|Xn|2 where X(f) =

∞∑

n=−∞Xnδ(f − nf0)

The Xn are the Fourier series coefficients


Power Theoremˆ ∞

−∞x(t)y∗(t)dt =

ˆ ∞

−∞X(f)Y ∗(f)df

Compact Proofˆ ∞

−∞x(t)y∗(t)dt = Fx(t)y∗(t)

∣∣∣0= X(f) ∗ Y ∗(−f)

∣∣∣0=

ˆ ∞

−∞X(f)Y ∗(f)df

For real x and y, FT are Hermitian (evenXr(f) = Xr(−f) and oddXi(f) = −Xi(−f))ˆ ∞

−∞x(t)y(t)dt =

ˆ ∞

−∞Xr(f)Y

∗r (f) +Xi(f)Y

∗i (f)df

This tells us something about the geometry of the Fourier Transform function spaceThe inner product between signals in function space,

´∞−∞ x(t)y

∗(t)dt, is preserved bythe FT, thus FT is a rotation operator in∞-dimensional function space since rotationspreserve angles between vectors.Since FFx(t) = x(−t) we have

FFFFx(t) = x(t)

And the FT rotation is a 90 rotation in function space


The Fourier Transform of Even and OddFunctions

Even(real) Even(real)12[g(t) + g(−t)] 1

2[G(f) +G(−f)]

Odd(real) Odd(imaginary)12[g(t)− g(−t)] 1

2[G(f)−G(−f)])

real Hermitian

g(t) G(f) such that G(−f) = G∗(f)


Fourier Transform of Odd Function isImaginary Spectrum



Real and Imaginary parts of Even functionFourier Transform



Real and Imaginary parts of Odd functionFourier Transform



Even and Odd Function Properties andFourier Transform Relations

• g(t) = e(t) + o(t)

•´ T

−T e(t)dt = 2´ T

0 e(t)dt

•∑N−N e[n] = e[0]+2

∑N1 e[n]

•´ T

−T o(t)dt = 0

•∑N−N o[n] = 0

• o + o = o

• e + e = e

• e× e = e

• o× o = e

• o× e = o

• ddto = e

• ddte = o

• e ∗ e = e

• o ∗ o = e

• e ∗ o = o

Even Even

Odd Odd

Real and even Real and even

Real and odd imaginary and odd

Imaginary and even Imaginary and even

Imaginary and odd Real and odd

Complex and even Complex and even

Complex and odd Complex and odd

Real even plus Imaginary odd Real

Real odd plus Imaginary even Imaginary

Real & asymmetrical Complex & Hermitian

Imaginary & asymmetrical Complex & anti-Hermitian


Fourier Transform Properties of Even andOdd Function

R.N. Bracewell, The Fourier Transform and its Applications, Mcgraw-Hill, 2000


Causal function Fourier TransformP. Kraniauskas, Transforms in Signals and Systems, Addison-Wesley, 1992


Real and Imaginary parts of Causalfunction Fourier Transform



Delta and other singularity functions

Delta function is the limiting form of any compact function as itis scaled narower and taller

δ(t) = limA→∞

AΠ [At] = limw→0

1

wΠ

[t

w

]= lim

T→0

1

Te−π(t/T )

2

Or the delta function can be defined by its integral representationˆ

e−i2π(t−t0)fdf = δ(t− t0) δ(t) =1

2π

ˆ ∞

−∞eiωtdω

Or a delta can be defined as the derivative of a step function

δ(t) =d

dtu(t) = 1

2

d

dtsgn(t)

Or a delta function can be defined by its properties

δ(t− t0) = 0 t 6= t0 δ(t)→∞ at t→ 0

sifting :

1

1

u(t)

δ(t)=u’(t) 1

1

Nature abhors a naked singularity, δ() functions should only appear under an integralˆ

f(t)δ(t− t0)dt = f(t0)


Delta function properties, continued

scaling: How do you stretch or compress something that is infinitely narrow?

δ(−t) = δ(t) An even function

δ

(t

b

)= |b|δ(t)

δ

(t− t0b

)= |b|δ(t− t0)

δ(at− t0) =1

|a|δ(t− t0/a)

delta of functions with simple roots ti (gives above scaling relation for g(t) = at)

δ(g(t)) =∑

i

|g′(ti)|−1δ(t− ti)

Properties in products

f(t)δ(t− t0) = f(t0)δ(t− t0)ˆ

δ(τ − t0)δ(t− τ )dτ = δ(t− t0)


Scaling: How do you stretch (b>1,a<1) orcompress (b<1,a>1) something that is

infinitely narrow?δ(tb

)= |b|δ(t) δ(at) = 1

|a|δ(t)

Proof: first consider b > 0 T ′=bT = |b|Tδ(t) = lim

T→0

1T e−π( tT )

2

⇒ δ(tb

)= lim

T→0

1

Te−π(t/bT

)2= lim

T ′→0

b

T ′e−π(

tT ′)

2

= bδ(t)

Now consider b < 0, since delta is even this must give the same result: δ(bt) = δ(|b|t).Or introduce T ′ = −bT = |b|T inside the Gaussian exponential without changing it.

δ(tb

)= |b|δ(t)

Alternative proof using defining integral property: First consider a > 0 and intro-duce t′ = at = |a|t into scaled delta sifting integral so t = t′/a and dt = dt′/a

ˆ ∞

−∞g(t)δ(at)dt =

ˆ ∞

−∞g(t′a

)δ(t′)dt

′a=

1

ag(0)

Now for a < 0 we should use t′ = −at = |a|t to avoid flipping g(), but this changessign of limits, however sign of 1/a changes them back givingˆ ∞

−∞g(t)δ(at)dt =

1

|a|g(0) =1

|a|

ˆ ∞

−∞g(t)δ(t)dt ⇒ δ(at) =

1

|a|δ(t)

which is the sifting behavior of an unscaled delta, with amplitude normalization by 1|a|.


Derivatives and integrals of delta functions

δ(k)(t) =dkδ(t)

dtk=dk+1u(t)

dtk+1

where the unit step, u(t) = δ(−1)(t) = H(t) =

0 t < 012 t = 01 t > 0

The derivatives are hard to visualize so instead defined in terms of integral propertiesConsider a function f(t) with derivatives defined up to at least kth order defined (con-tinuous at t0)

f (k)(t) =dkf(t)

dtk

Then the defining properties of derivatives of δ(t)

δ(k)(t− t0) = 0 t 6= t0

ˆ ∞

−∞δ(k)(t)dt = 0

ˆ t2

t1

f(t)δ(k)(t− t0)dt = (−1)kf (k)(t0) t1 < t0 < t2

unlike the delta function, the integrals are identically zero for k 6= 0

δ(t) = limb→0

1

bΠ

[t

b

]δ(1)(t) = lim

b→0

1

b[δ(t + b/2)− δ(t− b/2)]


FT of delta functionssifting property of delta allows easy calculation of FT

Fδ(t) =ˆ ∞

−∞δ(t)e−i2πftdt = e−i2πf0 = 1(f)

FT of doublet, acts like a differentiator of the integrand at the point it fires

Fδ(1)(t)

=

ˆ ∞

−∞δ(1)(t)e−i2πftdt =

d

dte−i2πft|t=0 = (−i2πf)e−i2πft|t=0 = (−i2πf)

Higher order doublets

Fδ(n)(t)

=

ˆ ∞

−∞δ(n)(t)e−i2πftdt =

dn

dtne−i2πft|t=0 = (−i2πf)ne−i2πft|t=0 = (−i2πf)n

And the function whose derivative is a delta function behaves in the opposite ways

F12sgn(t)

=

1

i2πf

whileFu(t) = F

12 +

12sgn(t)

= 1

2δ(f) +1

i2πf


Comb function(Shah III(t))

comb(t) =

∞∑

n=−∞δ(t− n)

scaled and shifted

comb

(t− t0T

)= |T |

∞∑

n=−∞δ(t− t0 − nT )

comb(tT

)=

∞∑

n=−∞δ(tT− n

)= |T |

∞∑

n=−∞δ(t− nT )

combT (t) =1

|T |comb(t

T

)=

∞∑

n=−∞δ(t− nT )

Finite number of comb teeth

Π

(t

T

)comb∆(t)

Sampling

f(t)

[1

|T |comb(t− t0T

)]=

∞∑

n=−∞f(t0+nT )δ(t−t0−nT )

1

0 1 2 3-1

T

to-T to to+T

big T

small T

density remains constant

T

f(t)

T0 2T

1

0 2∆-∆ ∆-2∆ Τ/2−Τ/2


Fourier Transform of the comb

comb (t) =

∞∑

n=−∞δ(t− n · 1) comb (f) =

∞∑

k=−∞δ(f − k · 1)

comb is limit of weighted envelope under which there are periodically placed narrowbumps

c(t) = τ−1e−πτ2t2

∞∑

n=−∞e−π(t−n·1)

2/τ2 limτ→0

c(t) = comb (t)

envelope timewidth 1/τ

τ

t f

envelope frequencywidth 1/τ

FT of the product of envelope of width 1τ

with∞ array of Gaussian bumps of width τconvolution of Gaussian of width τ in frequency with∞ sum of FT of shifted Gaussian

e−π(t/τ−1)2 τ−1e−π(τ

−1f)2

e−π(t−n1τ )

2

e−i2πn1fτe−π(fτ)2

∞∑

n=−∞e−π(

t−n1τ )

2

τe−π(fτ)2∞∑

n=−∞e−i2πn1f


comb FT

this can instead be expanded as a Fourier series

τ−1∞∑

n=−∞e−π(

t−n1τ )

2

=∞∑

m=−∞e−πτ

2m2cos 2πmt

c(t) = e−πτ2t2

∞∑

m=−∞e−πτ

2m2cos 2πmt =

∞∑

m=−∞e−πτ

2t2e−πτ2m2e−i2πmt

⇒ C(f) =∞∑

m=−∞e−πτ

2m2τ−1e

−π(f−m/1

τ

)2

Thus

comb (f) = limτ→0

C(f) =∞∑

m=−∞δ(f −m1)

So, in the limit the temporal comb function, comb (t), Fourier transforms to the fre-quency comb, comb (f). True even though this function not only has singularities likea δ(t), it has an infinite number of such singularities!


FT of periodic signals

Arbitrary periodic waveform with period T can be produced by a superposition ofharmonic tones weighted appropriately

X(f) =∑

k

akδ (f − k/T )a1

a5a3-1 T

1T 3

T

5T

FT FSak

⇒ x(t) =

ˆ ∞

−∞X(f)ei2πftdf =

∑

k

ak

ˆ ∞

−∞δ (f − k/T ) ei2πftdf =

∑

k

akei2π kT t

Can synthesize a time domain periodic waveform by convolving a comb with element*

*

=

=

Now use convolution thm

p(t) ∗∑

δ(t− nT ) F⇐⇒P (f) · comb (fT )

a1

a5a3-1 T

1T 3

T

5T

FT FSak


Array Theorem

*

comb

window

element

# zeroes =N-1 # subpeaks =N-2To represent diffraction gratings and other finite width periodic structuresConsider a finite width periodic structure

g(x) = w(x)·[f(x) ∗ 1

Xcomb

( xX

)] G(u) = W (u)∗[F (u)·comb (uX)]

2Dg(x, y) = w(x, y) ·

[f(x, y) ∗ ∗ 1

XYcomb

( xX,y

Y

)]

G(u, v) =W (u, v) ∗ [F (u, v) · comb (uX, vY )]


Fourier Transform WidthsLearning Objectives and Outcomes

• Learn about relation between moments and widths offunctions

• Study variance as a measure of width of common func-tions

•Recognize uncertainty principle as a Fourier theorem


Moments of the Fourier transform

Moments m0 =

ˆ

g(t)dt = G(0)

m1 =

ˆ

tg(t)dt =−12πi

G′(0)

mn =

ˆ

tnf(t)dt =

(−12πi

)ndnG(f)

dfn

∣∣∣∣f=0

Centroid center of gravity〈g〉 =

´∞−∞ tg(t)dt´∞−∞ g(t)dt

=−G′(0)2πiG(0)

=m1

m0

Moment of Inertia 2nd moment proved with derivative thm´∞−∞(−i2πt)2g(t)e−i2πftdt = G”(f)

m2 =

ˆ ∞

−∞t2g(t)dt = − 1

4π2G”(0) curvature at origin in f

Mean Square

〈t2〉 =´∞−∞ t

2g(t)dt´∞−∞ g(t)dt

=m2

m0= − 1

4π2G”(0)

G(0)


Variance and Equivalent Widths offunctions and transforms

Instead of using area of function (which can be 0)Define Variance

σ2g = 〈(t−〈t〉)2〉= 〈t2〉−〈t〉2=´∞−∞(t− 〈t〉)2g(t)dt´∞−∞ g(t)dt

= − G”(0)

4π2G(0)+

[G′(0)]2

4π2[G(0)]2=m2

m0−m

11

m20

Example

Gaus(t/τ )F⇐⇒τGaus(fτ ) m0 = τ, m1 = 0, t = m1

m0= 0, m2 =

G”(0)

−4π2 , σ2 =

τ 2

2π

To have finite 2nd moment, must die out more rapidly than x−2

sinc2(t) Λ(f) cusp has∞ 2nd derivative ⇒ m2 →∞

These measures breakdown for oscillatory functions when the area is 0.To cope with this case consider the centroid and variance of energy density

σ|g|2 = (∆x)2 =

´∞−∞ x

2|g(x)|2dx´∞−∞ |g(x)|2dx

−[´∞−∞ x|g(x)|2dx´∞−∞ |g(x)|2dx

]

This behaves more like our intuition and will be utilized in the uncertainty relationcalculation


Table of various widths

〈t〉 = m1m0

〈t2〉 = m2m0

G”(0) σ|g|2 = (∆t)2

Π(t) 0 1/12 −π2/3 1/12

Λ(t) 0 1/6 −2π2/3 1/4sinc(t) 0 osc 0 ∞sinc2(t) 0 ∞ ∞ .030...

e−xu(x) 1 2 −8π2 1/4

e−πx2a 0 1/2πa2 −2π/a3 1/4πa2

11+x2

0 ∞ ∞ ...

sech (πt) 0 14 −π2 1/12


Uncertainty Principle: Equivalent width ofthe Power in time and Fourier domains

Key limitation on QM, imaging, antennas, radar, etc

∆t∆f ≥ 1

4π∆t∆ω ≥ 1

2

where the variance of the function g(t) = |s(t)|2 in t and transform, G(f) = |S(f)|2, inf are the mean square width about the centroids of power 〈t〉 and 〈f〉

σ2g = (∆t)2 = 〈(t− 〈t〉)2〉 =´∞−∞(t− 〈t〉)2g(t)dt´∞−∞ g(t)dt

= − G”(0)

4π2G(0)+

[G′(0)]2

4π2[G(0)]2

σ2G = (∆f)2 = 〈(f − 〈f〉)2〉 =´∞−∞(f − 〈f〉)2G(f)df´∞−∞G(f)df

Gaussian is minimum uncertainty wavepacket. σ2g is variance of power in t, σ2G in f

s(t) = e−π(at)2= e−

12(t/τ)

2g(t) = |s(t)|2 = e−2πa

2t2= e−(t/τ)2

σ2g=(∆t)2=1

4πa2= 1

2τ2

S(f) =e−π(fa

)2

aG(f)= |S(f)|2= a−2e−2πf

2/a2 σ2G=(∆f)2=a2

4π= 1

2

1

(2π)2τ 2

(∆t)2(∆f)2 =1

4πa2a2

4π=

1

(4π)2⇒ ∆t ·∆f =

1

4πKelvin Wagner, University of Colorado Fourier Optics Fall 2019 86

Examples of Uncertainty Principle

CW signal observed for a finite time T has frequency uncertainty ≈ 1/T

1T

1T

1T

-

.636

.128

-.212t f

1.0

T

-f0 f0t f

Antenna size A gives beamwidth ≈ λ/A

Telescope with diameter D gives angular resolution ≈ λ/D

Atom with lifetime T yields spectral linewidth ≈ 1/T


Proof of the Uncertainty relationfirst some identities

From the derivative thm dn

dxng(t) (i2πf)nG(f) and conjugation thm g∗(±t) G∗(∓f) we get that

g′(t) =d

dtg(t) (i2πf)G(f) g′∗(t) =

d

dtg∗(t) (−i2πf)G(−f)

Which can be combined with Parsevals Thm´∞−∞ |g(t)|2dt =

´∞−∞ |G(f)|2df to give

ˆ ∞

−∞|g′(t)|2dt = 4π2

ˆ ∞

−∞f 2|G(f)|2df

Schwarz inequality states that |~x|2|~y|2 ≥ |~x · ~y|2 or 4(~x · ~x ~y · ~y) ≥ |~x · ~y+ ~y · ~x|2 whichgives for functions

4

ˆ

g(t)·g∗(t)dtˆ

h(t)·h∗(t)dt ≥∣∣∣∣ˆ

g(t)·h∗(t) + g∗(t)·h(t)dt∣∣∣∣

Integration by parts´

u dv = uv −´

v du for functions which have a well definedwidth and decay to 0 as t→ ±∞ gives (u = t dv = g′dt→ du = dt v = g)

ˆ ∞

−∞tg′(t) dt = tg(t)

∣∣∞−∞ +

ˆ ∞

−∞g(t)dt =

ˆ ∞

−∞g(t)dt


Uncertainty Relation Proofcalculate variance of power in time and

frequencyConsider the power width in time and Fourier domains p(t)= |s(t)|2 and P (f)= |S(f)|2Transform both functions to be centered at the origin without changing the width

g(t) = p(t)−〈p〉 s(t) = s(t)−〈p〉 G(f) = P (f)−〈P 〉 S(f) = S(f)−〈P 〉Define variance of the powers (∆t)2= σ2p = σ2g=〈t2〉 and (∆f)2= σ2P = σ2G =〈f 2〉 so

(∆t)2(∆f)2 =

´

(t− 〈t〉)2p(t)dt´

p(t)dt

´

(f − 〈f〉)2P (f)df´

P (f)df=

´

t2g(t)dt´

g(t)dt

´

f 2G(f)df´

G(f)df

=

´

t2s(t)s∗(t)dt´

s(t)s∗(t)dt

´

f 2S(f)S∗(f)df´

|S(f)|2df =

´

ts(t) · ts∗(t)dt´

s(t)s∗(t)dt

14π2

´

s′(t)s′∗(t)dt´

|s(t)|2dt

≥14

∣∣´ ts∗(t) · s′(t) + ts(t) · s′∗(t)dt∣∣

4π2(´

s(t)s∗(t)dt)2 =

∣∣´ t ddt(s(t)s∗(t)) dt

∣∣2

16π2(´

s(t)s∗(t)dt)2

=

∣∣´ s(t)s∗(t)dt∣∣2

16π2(´

s(t)s∗(t)dt)2 =

1

16π2

Therefore(∆t)(∆f) = σgσG ≥

1

4π


Fourier Transform ApodizationLearning Objectives and Outcomes

•Recognize relation between discontinuities and assymp-totic Fourier domain rolloff

•Understand main lobe widening with apodization

•Recognize common apodization windows and Gaussianapodization


Low pass Filtersconsiderh(t) = e−atu(t)

F⇐⇒ 1

a + jω= H(ω)

magnitude and phase are given by

|H(ω)| = 1√a2 + ω2

∠H(ω) = − tan−1ω

aThe values of magnitude and phase variesfrom

ω |H(ω)| ∠H(ω)0 1 0

a 1/√2 −π/4

∞ 0 −π/2

now consider

h2(t) =0.5

be−b|t|

F⇐⇒ 1

b2 + ω2= H2(ω)

In this case the filter response is phase flat

ω |H2(ω)| ∠H2(ω)0 1 0b 1/2 0∞ 0 0

h2(t) real and even⇐⇒ FT real and even

But as a physial ciruit filter this is not re-alistic since it is non causal, but OK inspace h2(x)

F⇐⇒H2(kx)


Apodization : Lowering the feet (toes)

Fourier domain sidelobes can be lowered by multiplying by smooth weighting functionTypically use raised cosine or Hamming or Hanning window in DSPIn optics we multiply by a truncated gaussian

Discontinuities lead to 20dB/decade assymptotic rolloffDiscontinuities of the derivative lead to 40dB/decade assymptotic rolloffDiscontinuities of the 2nd derivative lead to 60dB/decade assymptotic rolloff


Unit Width Apodization Functions

• Smoother functions with same support have wider mainlobeBoth 3dB half width (.453,.656,.796) and first zero (1,2,3)


Examples of Apodization

Discontinuitysign(t)

1

iωasymptotics 20db/decade

Π(t) sinc(ω) asymptotics 20db/decade

Discontinuity of slopesign(t) ∗ Π(t)

1

(iω)

sinω/2

ω/2asymptotics 40db/decade

Λ(t) sinc2(ω) asymptotics 40db/decade

Discontinuity of second derivativeΛ(t) ∗ Π(t) sinc3(ω) asymptotics 60db/decade

name Mainlobe width Rolloff rate peak sidelobeΠ(t/T ) rect 2/T -6/oct -20/dec -13.3dBΛ(t/T

2) Bartlett 4/T -12/oct -40/dec -26.5dB

12[1 + cos 2πt

T ] Hanning 4/T -18/oct -60/dec -31.5dB.54 + .46 cos 2πt

T Hamming 4/T -6/oct -20/dec -42.7dB.42 + .5 cos 2πt

T + .08 cos 4πtT Blackman 6/T -18/oct -60/dec -58.1dB


Truncated Gaussian Impulse Response

w(x) = Π

(x

D− 1

2

)e−4T

2(x/D−.5)2

σ = 12T = ω0

D , truncation ratio T = D2ω0

, 2ω0 = 1/e2 intensity width.


Effective of Varying the Gaussiantruncation ratio

• Peak intensity drops

•Mainlobe widens

• 1st sidelobe drops

• Assymptotics stays at 20dB/decadeKelvin Wagner, University of Colorado Fourier Optics Fall 2019 96

Routes to computing Fourier Transform

• If possible, use properties and table of known transforms.– Decompose as sum, product, or convolution of known function or deriva-

tive/integral of known function

• If x(t) has finite support its FT exists, so can use integral definition– or can use Laplace transform

• IF Laplace transform X(s) has ROC that contains iω axis, its FT is X(s)∣∣s=iω

• If x(t) is periodic infinite energy but finite power signal, its FT is obtained from FSusing delta functions or equivalently using comb T convolved with a single periodp(t) whose FT can be obtained.

• If x(t) is none of the above, if it has discontinuities (eg u(t) or sgn(t)), or it hasdisconinuities and is not finite energe (eg u(t) cos(ω0t)), then either– add a convergence factor– or take limits of support– or take limits in the Laplace domain.

It is best in such cases to use properties and known transforms if possible.

• Last resort: Complex contour integration and residue calculus


Fourier Transform Properties Summary

• Linearity ax(t) + by(t) aX(f) + bY (f)

• Shift x(t− t0) e−i2πft0X(f)

• Scale x(at) 1|a|X

(fa

)

• Convolutionˆ

g(t′)h(t− t′)dt′ G(f)H(f)

• Correlationˆ

g(t′)h∗(t′ − t)dt′ G(f)H∗(f)

• Modulation g(t)h(t)

ˆ

G(f ′)H(f − f ′)df ′

• Differentiation dn

dtng(t) (i2πf)nG(f)

• Integrationˆ t

−∞g(τ )dτ

1

i2πfG(f) +

G(0)

2δ(f)

• Duality FFg(t) = g(−t)

• Parsevals’ thmˆ ∞

−∞|x(t)|2dt =

ˆ ∞

−∞|X(f)|2df


FT summary

Symmetries (Duality gives additional symmetries)

Real in t F⇐⇒ Hermitian in f

Even Real in t F⇐⇒ Even Real in f

Odd Real in t F⇐⇒ Odd Imaginary in f

Wide(narrow) in t F⇐⇒ Narrow(wide) in f

Periodic in t F⇐⇒ Sampled in f

Some transforms

rect (t)F⇐⇒ sinc(f)

comb (t)F⇐⇒comb (f)

1(t)F⇐⇒δ(f)

e−αtu(t)F⇐⇒ 1

α+i2πf

e−αt sin(ω0t)u(t)F⇐⇒ ω0

ω20+(α+iω)2= 2πf0

(2πf0)2+(α+i2πf)2

ω transforms are obtained from f transforms by substitution ω = 2πf and vice versa,and scaling any delta functions by 2π


Fourier Transform Summary

• CTFS is a special case of CTFT

• Frequency spectrum plotted as |X(f)| and ∠X(f) or ℜX(f) and ℑX(f)• Signals and Systems are more usefully described by their f domain properties

– Thinking in both domains yields the clearest insights and checks

• FT of impulse response of LTI system, h(t), yields Transfer function H(f)

Output spectrum Y (f) =X(f)H(f) so in time y(t) =F−1X(f)H(f)• Generalized CTFT allows periodic signals to be FT as frequency impulses

The FT of a periodic signal consist only of impulses regularly spaced in f at kT

• The more a signal is localized in one domain, the less it is localized in the other

• ∗ and × of signals and their transforms are dual operations in t and f

• Signal energy is conserved by the FT, as is signal similarity (dot product)– FT is a 90 rotation in function space and FT are thus unique

• Most FT can be done using tables of transforms and properties of transforms

• Existence of FT unless infinite number of discontinuities or∞ in finite interval


Linear Space Invariant SystemLearning Objectives and Outcomes

•Recognize shift invariance in space as cousin of LTI

•Understand operation of convolution in 1-D as convolu-tion movie

•Generalize convolution movie to the case of 2-D convolu-tions

•Appreciate imaging and diffraction as examples of 2-Dconvolution


Linear Time Invariant (LTI) Systems

Linear Circuits and linear fiber optics share fundamental simplifying propertiesLinearity, Causality, Memory

Such time domain linear systems are described by mathematical operators that interre-late the input, x(t), and output, y(t), waveforms as a convolution integral

y(t) = L x(t) = h(t)∗x(t) = x(t)∗h(t) =ˆ ∞

−∞h(t−τ )x(τ )dτ =

ˆ ∞

−∞h(τ )x(t−τ )dτ

Where h(t) is the impulse response and is causal if h(t) = 0 ∀ t < 0

Convolution MovieFor finite support functions the convolution operation can be thought of graphically:

1. plot x(τ ) versus dummy variable of integration τ2. Mirror image the impulse response as h(−τ ) (reflect about origin)3. Slide mirrored impulse far to the left (t≪ 0) to plot h(t− τ )4. Find the area of the product x(τ )h(t− τ ) giving the value of the convolution y(t)

For 0, 1 functions this is the width of the overlap region5. Now slide h(t− τ ) to the right and for every value of t where it overlaps with x(τ )

calculate the area of the product y(t) =´∞−∞ x(τ )h(t− τ )dτ

6. Plot y(t) for all values where the two functions overlap.


1-D Convolution Movie Sketch of Π( t2

)∗ Π(t)

-.5 .5

-1.5

-1

-.5

0

.5

1

1.5

-1 1

1

1

Π(t)

Π(t/2)

t

t

τ

τ

τ

τ

τ

τ

τ

τ

τ

-1.5 1.5-.5 .5

1

t

Π(t)*Π(t/2)1.5

Note the Π(t/2) is a rect of width 2, NOT a rect of width 1/2, since t has toget twice as large for the value of t/2 to reach ±1/2. Or by unit analysis, t/2is unitless (inside the argument of the rect function), so the width must be 2[s].Π(2t) = Π(t/.5) is a rect of width 1/2.

As the rect of width 1 slides into the rect of width 2, at a shift of t = −1.5 thearea increases linearly until the rect is fully contained in the wider rect starting att = −.5 at which point the full area of 1 remains constant at 1 until t = .5 Thenthe rect slides out and the area decreases linearly back to zero at t=1.5.

The total width of the trapezoid is 3, and each segment is of width 1. Since thetwo inputs are centered at t = 0, the output is also centered at the sum of the twoinputs, so is also centered at t = 0.

b(t) =

ˆ ∞

−∞Π

[t

2

]Π(t− τ )dτ =

ˆ 1

−1Π(t− τ )dτ

=

0 0 < t−1.5´ t+.5

−1 dτ = τ∣∣t+.5−1 = t + 1.5 −1.5 < t < −.5

´ t+.5

t−.5 dτ = 1 −.5 < t < .5´ 1

t−.5 dτ = τ∣∣1t−.5 = 1.5− t .5 < t < 1.5

0 t > 1.5


1-D Convolution Movie


1-D Convolution Movie


CT Table of Convolutions

f(t) h(t) f(t) ∗ h(t) F (ω)H(ω)

f(t) δ(t− T ) f(t− T ) F (ω)e−iωt

Π(t) Π(t) Λ(t) sinc(ω2π

)sinc

(ω2π

)

e−λtu(t) u(t) 1−e−λtλ u(t) 1

iω+λ

(1iω + πδ(ω)

)

u(t) u(t) tu(t)(1iω + πδ(ω)

) (1iω + πδ(ω)

)

e−λtu(t) e−λtu(t) te−λtu(t) 1iω+λ

1iω+λ

e−αtu(t) e−βtu(t) e−αt−e−βtβ−α u(t) 1

iω+α1

iω+β

te−λtu(t) e−λtu(t) 12t2e−λtu(t) 1

(λ+iω)21

iω+λ

tme−λtu(t) tne−λtu(t) m!n!(m+n+1)!

tm+n+1eλtu(t) m!(λ+iω)m

n!(λ+iω)n

e−αt cos(βt + θ)u(t) e−λtu(t) cos(θ−φ)eλt−e−αt cos(βt+θ−φ)√(α+λ)2+β2

u(t) (α+iω) cos θ+β sin θ(α+iω)2+β2

1iω+λ

φ = − tan−1 βα+λ


Linear Space Invariant (LSI) Systemseg Shift Invariant

Space Variant Linear Transform

g(x) =

ˆ ∞

−∞f(x′)h(x′, x)dx′ = Lf(x) g = H f

impulses at different positions of the input x′ yield different outputs h(x, x′)

When the transfom is space invariant

g(x− a) = Lf(x− a) ⇒ h(x′, x) = h(x− x′)

⇒ g(x) =

ˆ ∞

−∞f(x′)h(x− x′)dx′ g = h ∗ f = F−1x HF

1-D impulse response h(x)

Eigenfunctions of any LSI operator are the complex exponentials with eigenvaluesgiven by the Transfer function H(f) = Ff(x)

Lei2πf0x =ˆ

ei2πf0x′h(x− x′)dx′ =

ˆ

ei2πf0(x−x”)h(x”)dx”

= ei2πf0xˆ

e−i2πf0x”h(x”)dx” = ei2πf0xH(f0)


Auto Convolution Movie:e−tu(t) ∗ e−tu(t) = te−tu(t)


Auto Convolution Movie:e−2tu(t) ∗ e−2tu(t) = te−2tu(t)


Cross Convolution Movie:e−2tu(t) ∗ e−tu(t) = e−t−e−2t

2−1 u(t)



4−1 u(t)



8−1 u(t)


Convolution Movie:te−2tu(t) ∗ e−2tu(t) = 1

2!t2e−2tu(t)


Convolution Movie:te−2tu(t) ∗ te−2tu(t) = 1

3!t3e−2tu(t)


Convolution Movie:t2e−2tu(t) ∗ t2e−2tu(t) = 2!2!

5! t5e−2tu(t)


Convolution Movie:t2e−4tu(t) ∗ t2e−4tu(t) = 2!2!

5! t5e−4tu(t)
