Science I Tutorial

Fourier and Laplace Transforms

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Kartik Dutta, kartik.dutta@research.iiit.ac.in

Center for Visual Information Technology (CVIT), IIIT Hyderabad

Transform based approach

A problem is defined in one setting
E.g. denoising a given rectangular image

Transform the problem to a new domain (to a different basis)

Where it is more easily solvable

Solve the problem in transformed setting

Transform the solution back to original domain

Vineet Gandhi
Center for Visual Information Technology (CVIT), IIIT Hyderabad

Motivation
RECIPE

Transform

Banana juice (100ml)

Strawberry juice (50ml)
Raspberry juice (50ml)
Mint syrup (5 ml)
Vanilla syrup (5 ml)
Water (50 ml)

Adjust and remix

https://betterexplained.com/articles/an-interactive-guide-to-the-fourier-transform/

Motivation

Motivation

Courtesy: floridalinguistics.com

Motivation

Courtesy: floridalinguistics.com

Motivation

Courtesy: floridalinguistics.com

Motivation

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Idea
the Laplace transform converts integral and differential equations into
algebraic equations
this is like phasors, but
applies to general signals, not just sinusoids
handles non-steady-state conditions
allows us to analyze
LCCODEs
complicated circuits with sources, Ls, Rs, and Cs
complicated systems with integrators, differentiators, gains

The Laplace transform

Slides taken from https://web.stanford.edu/~boyd/ee102/laplace.pdf

Complex numbers
complex number in Cartesian form: z = x + jy
x = <z, the real part of z

y = =z, the imaginary part of z

j = 1 (engineering notation); i = 1 is polite term in mixed

company
complex number in polar form: z = rej
r is the modulus or magnitude of z

is the angle or phase of z

exp(j) = cos + j sin

complex exponential of z = x + jy:

ez = ex+jy = exejy = ex(cos y + j sin y)

The Laplace transform

The Laplace transform

well be interested in signals defined for t 0
the Laplace transform of a signal (function) f is the function F = L(f )
defined by
Z
f (t)est dt
F (s) =
0

for those s C for which the integral makes sense

F is a complex-valued function of complex numbers

s is called the (complex) frequency variable, with units sec1; t is called

the time variable (in sec); st is unitless
for now, we assume f contains no impulses at t = 0
common notation convention: lower case letter denotes signal; capital
letter denotes its Laplace transform, e.g., U denotes L(u), Vin denotes
L(vin), etc.
The Laplace transform

Three definitions of the Fourier Transform

1. Definition 1: (used in these notes)
f () =

f (t)eit dt ,

with the inverse Fourier transform written as

Z
1
f ()eit d .
f (t) =
2
2. Definition 2: This definition is sometimes used in signal processing :
Z
f (s) =
f (t)e2ist dt ,

with the inverse Fourier transform written as

Z
f (t) =
f (s)e2ist ds .

Hence s acts like the frequency with = 2s.

3. Definition 3: This next definition is used more in mathematical physics because of the
symmetry in the coefficients of both the transform and its inverse :
Z
1
f () =
f (t)eit dt ,
2
with the inverse Fourier transform written as
Z
1

f ()eit d .
f (t) =
2

Notes are taken from

http://wwwf.imperial.ac.uk/~jdg/ee2maft.pdf

Fourier Transforms

1.1

Introduction

The definition used here is

f () =

f (t)eit dt .

(1.1)

where the frequency is real. Another common notation is to write F () or F() for f ().
Given the spectrum f () the function f (t) can be recovered through the inverse transform
Z
1
f (t) =
f ()eit d .
(1.2)
2
Note the factor of 1/2 in the coefficient. The interplay between the function of time f (t)
(or a sampled time series) and the FT f () is subtle. Clearly, it is possible that functions f (t)
could be chosen for which the integral (1.1) is infinite which means that this transform does
not exist. There are two conditions that must be satisfied for the FT to exist :
(i) f (t) must be absolutely integrable : that is
Z
|f (t)| dt < .

(1.3)

(ii) If f (t) has discontinuities then it must be finite at these.

The following is a list of common functions :
1. The sign-function
sgn(t) =
2. The triangle or tent function :

3. The rectangle function :

1 t 0
+1 t 0

1t 0t1
(t) =
1 + t 1 t 0

0
otherwise
(t) =

4. The filtering function :

(1.5)

1 12 t 12
0 otherwise

(1.6)

sin(t/2)
.
t/2

(1.7)

sinc (t) =

The 12 -factor is unusual but is the natural definition for this definition of the FT.
2

(1.4)

The overbar notation f should not be confused with complex conjugate.

5. The Heaviside step function :

H(t) =

1 t0
0 t0

(1.8)

6. The error function :

1
erf (t) =

e
t

x2

2
dx =

ex dx .

7. The normalized autocorrelation function :

R
f (u) f (t u) du
(t) = R
.
2 du
|f
(u)|

1.2

(1.9)

(1.10)

The Dirac -function

Observing the list of functions in the previous section, it is clear that there is one missing. How
can a spike be represented? For instance, it is intuitive that the spectrum f () of a single
sine-wave f (t) = sin 0 t should be a spike at 0 but the condition of absolute integrability
(1.3) is not satisfied because of the infinite range of the integral. How can such an improper
function be represented? One way of formaizing a spiky function is to introduce the Dirac
Delta function (t t0 ) by considering the properties of a box of unit area under a limiting
process, as in the figure below :

A box of unit area : width h & height h1 at a

point t0 on the t-axis which limits to a spike as
h 0 but retains unit area. The curve f (t) is
some other function : The product of the two
is non-zero only within the range of the box.

f (t)
h1
h
t0

From the picture we represent (t t0 ) as

(t t0 ) = lim

h0

h1
0

t 0 t t0 + h
otherwise

(1.11)

with the property of unit area

Area =

(t t0 ) dt = 1 .

(1.12)

In the limit h 0 the -function3 acts as a spike at t0 : of course it is not a proper function
at all but it possesses a the powerful property
Z

f (t)(t t0 ) dt =

N
X
i=1

f (ti )(ti t0 ) ti



= lim f (t0 )h1 h
h0

= f (t0 ) .

(1.13)

To express this in words, when multiplied on a function f (t) and integrated, the -function
simply picks out the value of f (t) at the point of the spike t0 . This result can be expressed in
a more general way :
Z
f (t0 )(t0 t) dt0 = f (t) .
(1.14)

2. As in 1.1, the tent function (t) is defined as

1t 0t1
(t) =
1 + t 1 t 0

0
otherwise
Thus we have

() =
Now we know that

te
a

it

(1 + t)e

it

dt +

(1 t)eit dt

Z
i b  it 
dt =
td e
a


Z b
i  it b
it

e
dt
=
te
a

a

b
i
i it
it
=
te
e
,

(1.45)

(1.46)

(1.47)

and

Using these in (1.46)

eit dt =
a

i  it b
.
e
a

(1.48)


i 
1 eit + eit 1






i
i
i i
i
i i
i
i
i
e e
e

e
+
+

i i
2
i
1
= e ei + 2 +
ei ei 2 ei + ei

2(1 cos )
=
2
2 1
4 sin 2
=
= sinc2 .
(1.49)
2

() =

Example : Consider a function of time with one frequency an exact sine-wave in the form
f (t) = f0 ei 0 t
and so
f () = f0

ei(0)t dt = 2f0 ( 0 ) .

(1.23)

(1.24)

Thus the spectrum is just a single frequency a spike at = 0 . The inverse transform is
Z
f0
f (t) =
(1.25)
2( 0 )eit d = f0 ei0 t .
2

Laplace Transforms

2.1

Introduction

For a function f (t) uniquely defined on 0 t , its Laplace transform (LT) is defined as
Z
L [f (t)] = f (s) =
est f (t) dt ,
(2.1)
0

where s may be complex. The LT may not exist if f (t) becomes singular in [0 ]. The LT is
a one-sided transform in that it operates on [0 ] and not, like the FT, on [ ]. For
this reason, LTs are useful for initial value problems, such as circuit theory, where a function
switches on at t = 0 and where f (0) has been specified.
Because s is a complex variable the inverse transform
I


1
f (s) =
est f (s) ds
f (t) = L

(2.2)

is more difficult to handle because the contour C is a tricky infinite rectangle in the righthand-half of the s-plane. Referred to as Bromwich integrals the evaluation of these is beyond
our present course. To circumvent this difficulty we resort firstly to a library of transforms
(see Handout 7) for the standard functions and secondly to ways of piecing combinations of
these together for those not in the list.

2.2

Library of Laplace Transforms

1. The constant function f (t) = 1 :

f (t) = 1 ;
Proof :
f (s) =
provided Re s > 0.

f (s) =

st

1
s


est
dt =
s

Re s > 0


(2.3)

1
,
s

(2.4)

Re s > a

(2.5)

1
,
sa

(2.6)

2. The exponential-function f (t) = eat :

f (t) = exp(at) ;
Proof :
f (s) =
provided Re(s a) > 0.

(sa)t

f (s) =


1
;
sa

e(sa)t
dt =
sa


0

3. The sine function :

f (t) = sin(at) ;

f (s) =

s2

a
;
+ a2

Re s > 0

(2.7)

Proof : Take both the sine and cosine functions in combination : cos at + i sin at = eiat
Z


1
s + ia
iat
= 2
L e
=
e(sia)t dt =
(2.8)
s ia
s + a2
0
provided Re(s) > 0. Then the imaginary (real) part gives the result for sine (cosine).
4. The cosine function
f (t) = cos(at) ;

f (s) =

s2

s
;
+ a2

Re s > 0

(2.9)

5. The polynomial function f (t) = tn :

f (t) = tn ;

f (s) =

n!
sn+1

(n 0) ;

Re s > 0

Proof : Define the LT as f (s) = In as

Z
Z
1 n  st 
st n
In =
e t dt =
t d e
s 0
0
Z
n st n1
n
e t
dt = In1
=
s 0
s

(2.10)

(2.11)

provided Re(s) > 0. With n = 0 and L[1] = s|1 we obtain I1 = s2 and end up with
n!

f (s) = In =

sn+1

(2.12)

6. The Heaviside function :

f (t) = H(t t0 ) ;

f (s) =

Proof : For Re s > 0



L H(t t0 )

=
=

exp(st0 )
;
s

Z0
t0

Re s > 0

(2.13)

est H(t t0 ) dt
est dt =

est0
.
s

(2.14)

7. The Dirac -function :

f (t) = (t t0 ) ;

f (s) = exp(st0 ) ;

Proof : t0 needs to reside within the positive range of t

 st0
Z
t0 0 ,
e
st
e (t t0 ) dt =
0
t0 < 0 .
0

t0 0

(2.15)

(2.16)

Solving ODEs using Laplace Transforms

x
+ x + 02x = f(t)


s2x(s) sx0 x 0 + (sx(s) x0 ) + 20 x(s) = f(s) .
This re-organizes into
where x0 = x(0) and x 0 = x(0).

s2 + s + 02 x(s) = f (s) + (s + )x0 + x 0 .

(2.47)

Note that the final expression for x(s) divides conveniently into two parts corresponding to
the Complementary Function and the Particular Integral
x(s) =

f (s)
(s + )x0 + x 0
+ 2
2
+ s + 0
s + s + 02
|
{z
} |
{z
}
s2

P.I.

(2.48)

C.F.

The initial conditions appear in x0 and x 0 as part of the Complementary Function. How to
take the inverse depends on whether the denominator has real or complex roots. These we
consider by example.
Example 1 : Solve x + x 2x = et with x0 = 3 and x 0 = 0.

(2.47) becomes

s2 + s 2 x(s) =

1
+ 3(s + 1) .
s1

(2.49)

Noting that s2 + s 2 = (s 1)(s + 2) we have

x(s) =

3(s + 1)
1
+
2
(s 1) (s + 2) (s 1)(s + 2)
|
{z
} |
{z
}
P.I.

Using PFs
x(s) =
and so the Library gives us

(2.50)

C.F.

1
17
10
+
+
2
3(s 1)
9(s 1) 9(s + 2)

1
17
10
x(t) = tet + et + e2t .
3
9
9

(2.51)

(2.52)

Example 2 : Solve x + 16x = sin 2t with x0 = 0 and x 0 = 1.

(2.47) becomes

s2 + 16 x(s) = 1 +

and so

s2

2
,
+4

1
2
+ 2
+ 16 (s + 4)(s2 + 16)
5
1
=
+
2
2
6(s + 16) 6(s + 4)




5
4
1
2
+
.
=
24 s2 + 42
12 s2 + 22

x(s) =

(2.53)

s2

x(t) =

5
1
sin 4t +
sin 2t .
24
12

(2.54)

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Multidimensional transform
From Wikipedia, the free encyclopedia
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In mathematical analysis and applications, multidimensional transforms are used to analyze the frequency content of signals in a domain of two or

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1 Multidimensional Fourier transform

1.1 Properties of Fourier transform

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1.1.1 Linearity
1.1.2 Shift

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1.1.3 Modulation

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1.1.4 Multiplication

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1.1.5 Differentiation

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1.1.6 Transposition

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1.1.7 Reflection

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1.1.8 Complex conjugation

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1.1.9 Parseval's theorem (MD)

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1.1.10 Separability

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1.2 MD FFT
1.3 MD DFT
2 Multidimensional discrete cosine transform
3 Multidimensional Laplace Transform
4 Multidimensional Z Transform[6]
4.1 Region of Convergence

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5 Applications
5.1 Image processing
5.2 Spectral analysis
5.3 Partial differential equations

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5.4 Image processing for arts surface analysis by FFT

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5.5 Application to weakly nonlinear circuit simulation[14]

6 See also
7 References

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Multidimensional Fourier transform

[edit]

One of the more popular multidimensional transforms is the Fourier transform, which converts a signal from a time/space domain representation to a
frequency domain representation.[1] The discrete-domain multidimensional Fourier transform (FT) can be computed as follows:

where F stands for the multidimensional Fourier transform, m stands for multidimensional dimension. Define f as a multidimensional discrete-domain
signal. The inverse multidimensional Fourier transform is given by

The multidimensional Fourier transform for continuous-domain signals is defined as follows:[1]

Properties of Fourier transform

[edit]

Similar properties of the 1-D FT transform apply, but instead of the input parameter being just a single entry, its a Multi-dimensional (MD) array or
vector. Hence, its x(n1,,nM) instead of x(n).
Linearity [edit]
if

, and

then,

Shift [edit]
if

, then

Modulation [edit]
if

, then

Multiplication [edit]

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for 0 n1, n2, ... , nm N(1, 2, ... , m) 1.

Multidimensional discrete cosine transform

[edit]

The discrete cosine transform (DCT) is used in a wide range of applications such as data compression, feature extraction, Image reconstruction, multiframe detection and so on. The multidimensional DCT is given by:

for ki = 0, 1, ..., Ni 1, i = 1, 2, ..., r.

Multidimensional Laplace Transform

[edit]

The multidimensional Laplace transform is useful for the solution of boundary value problems. Boundary value problems in two or more variables
characterized by partial differential equations can be solved by a direct use of the Laplace transform.[3] The Laplace transform for an M-dimensional
case is defined[3] as

where F stands for the s-domain representation of the signal f(t).

A special case (along 2 dimensions) of the multi-dimensional Laplace transform of function f(x,y) is defined[4] as

F(x,y) is called the image of f(x,y) and f(x,y) is known as the original of F(x,y).[5] This special case can be used to solve the Telegrapher's equations.[5]

Multidimensional Z Transform[6]

[edit]

The multidimensional Z transform is used to map the discrete time domain multidimensional signal to the Z domain. This can be used to check the
stability of filters. The equation of the multidimensional Z transform is given by

where F stands for the z-domain representation of the signal f(n).

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