[PDF] Working out Fourier Transforms Pairs PDF ft

Fourier Transform of Gaussian Let f(t) be a Gaussian: f(t) = e −π t 2 By the definition of Fourier transform we see that: F(s) = / ∞ −∞ e −πt 2 e −j2πst dt

The Fourier transform of e −ax 2 Introduction Let a > 0 be constant We define a function fa(x) by fa(x)=e−ax2 and denote by ˆ fa(w) the Fourier transform of

[PDF] Table of Fourier Transform Pairs

Definition of Inverse Fourier Transform Р ¥ ¥- = w w p w de F tf tj )( 2 1 )( Definition of Fourier Transform Р ¥ ¥- - = dt etf F tjw w )( )( ) ( 0 ttf- 0 )( tj e F w

[PDF] EE2 Mathematics Solutions to Example Sheet 4: Fourier Transforms

e−t(1+iω)dt = 2 1 + ω2 2) (i) Designate J1f(t)l = f(ω) with a a real constant of either 3) To find the Fourier transform of the non-normalized Gaussian f(t) = e− t2

[PDF] Chapter 1 The Fourier Transform - Math User Home Pages

1 mar 2010 · cos(λt)dt = 2 sin(πλ) λ = 2π sinc λ Thus sinc λ is the Fourier transform of the box function The inverse Fourier transform is ∫ ∞ −

[PDF] Working out Fourier Transforms Pairs

Fourier Transform of Gaussian Let f(t) be a Gaussian: f(t) = e −π t 2 By the definition of Fourier transform we see that: F(s) = / ∞ −∞ e −πt 2 e −j2πst dt

[PDF] 5 Fourier transform

2 ω sin(ω/2) It is common to define the sinc function as follows: sinc(x) = sin(πx) πx Thus the following Fourier transform pair has been established: e−btu(t) F

[PDF] Lecture 3 - Fourier Transform

2 Before we consider Fourier Transform, it is important to understand the relationship between sinusoidal signals and exponential functions So far we have

[PDF] Integral Transformation Methods 1 Fourier transforms 11

We call this transform the Fourier transform Definition 1 (Fourier transform) The Fourier transform ofa function f(x) is F( f)( ξ) = 1 2 π √ ∫− ∞ ∞ e− iξx f(x) dx

Fourier Transform Pairs

The Fourier transform transforms a function of

time,f(t), into a function of frequency,F(s):

F{f(t)}(s) =F(s) =Z

-¥f(t)e-j2pstdt. tion of frequency,F(s), into a function of time, f(t): F -1{F(s)}(t) =f(t) =Z -¥F(s)ej2pstds. form is the identity transform: f(t) =Z Z¥ -¥f(t)e-j2pstdt? e j2pstds.

Fourier Transform Pairs (contd).

Because the Fourier transform and the inverse

Fourier transform differ only in the sign of the

exponential's argument, the following recipro- cal relation holds betweenf(t)andF(s): f(t)F-→F(s) is equivalent to

F(t)F-→f(-s).

This relationship is often written more econom-

ically as follows: f(t)F←→F(s) wheref(t)andF(s)are said to be aFourier transform pair.

Fourier Transform of Gaussian

Letf(t)be a Gaussian:

f(t) =e-pt2.

By the definition of Fourier transform we see

that:

F(s) =Z

-¥e-pt2e-j2pstdt Z -¥e-p(t2+j2st)dt.

Now we can multiply the right hand side by

e -ps2eps2=1:

F(s) =e-ps2Z¥

-¥e-p(t2+j2st)+ps2dt =e-ps2Z¥ -¥e-p(t2+j2st-s2)dt =e-ps2Z¥ -¥e-p(t+js)(t+js)dt =e-ps2Z¥ -¥e-p(t+js)2dt

Fourier Transform of Gaussian (contd.)

F(s) =e-ps2Z¥

-¥e-p(t+js)2dt

After substitutingufort+jsanddufordtwe

see that:

F(s) =e-ps2Z¥

-¥e-pu2du 1.

It follows that the Gaussian is its own Fourier

transform: e -pt2F←→e-ps2.

Fourier Transform of Dirac Delta Function

To compute the Fourier transform of an impulse

we apply the definition of Fourier transform:

F{d(t-t0)}(s) =F(s) =Z

-¥d(t-t0)e-j2pstdt which, by the sifting property of the impulse, is just: e -j2pst0.

It follows that:

d(t-t0)F-→e-j2pst0.

Fourier Transform of Harmonic Signal

What is the inverse Fourier transform of an im-

pulse located ats0? Applying the definition of inverse Fourier transform yields: F -1{d(s-s0)}(t)=f(t)=Z -¥d(s-s0)ej2pstds which, by the sifting property of the impulse, is just: e j2ps0t.

It follows that:

e j2ps0tF-→d(s-s0).

Fourier Transform of Sine and Cosine

We can compute the Fourier transforms of the

sine and cosine by exploiting the sifting prop- erty of the impulse:Z¥ -¥f(x)d(x-x0)dx=f(x0). •QuestionWhat is the inverse Fourier trans- form of a pair of impulses spaced symmetri- cally about the origin? F -1{d(s+s0)+d(s-s0)} form: f(t) =Z -¥[d(s+s0)+d(s-s0)]ej2pstds.

Fourier Transform of Sine and Cosine (contd.)

Expanding the above yields the following ex-

pression forf(t):Z¥ -¥d(s+s0)ej2pstds+Z -¥d(s-s0)ej2pstds

Which by the sifting property is just:

f(t) =ej2ps0t+e-j2ps0t =2cos(2ps0t).

Fourier Transform of Sine and Cosine (contd.)

It follows that

cos(2ps0t)F←→1

2[d(s+s0)+d(s-s0)].

A similar argument can be used to show:

sin(2ps0t)F←→j

2[d(s+s0)-d(s-s0)].

Fourier Transform of the Pulse

To compute the Fourier transform of a pulse we

apply the definition of Fourier transform:

F(s) =Z

-¥P(t)e-j2pstdt Z 1 2 1

2e-j2pstdt

1 -j2pse-j2pst????1 2 1 2 =1 -j2ps?e-jps-ejps? 1 ps? ejps-e-jps?2j

Using the fact that sin(x) =(ejx-e-jx)

2jwe see

that:

F(s) =sin(ps)

ps.

Fourier Transform of the Shah Function

Recall the Fourier series for the Shah function:

2pIII?t2p?

=12p¥å w=-¥ejwt.

By the sifting property,

III?t 2p? w=-¥Z -¥d(s-w)ejstds.

Changing the order of the summation and the

integral yields III?t 2p? =Z w=-¥d(s-w)ejstds.

Factoring outejstfrom the summation

III?t 2p? =Z -¥ejst¥å w=-¥d(s-w)ds Z -¥ejstIII(s)ds.

Fourier Transform of the Shah Function

Substituting 2ptfortyields

III(t) =Z

-¥III(s)ej2pstds =F-1{III(s)}(t).

Consequently we see that

F{III}=III.

quotesdbs_dbs14.pdfusesText_20

[PDF] [PDF] Working out Fourier Transforms Pairs

Fourier Transform Pairs

The Fourier transform transforms a function of

F{f(t)}(s) =F(s) =Z

Fourier Transform Pairs (contd).

Because the Fourier transform and the inverse

Fourier transform differ only in the sign of the

F(t)F-→f(-s).

This relationship is often written more econom-

Fourier Transform of Gaussian

Letf(t)be a Gaussian:

By the definition of Fourier transform we see

F(s) =Z

Now we can multiply the right hand side by

F(s) =e-ps2Z¥

Fourier Transform of Gaussian (contd.)

F(s) =e-ps2Z¥

After substitutingufort+jsanddufordtwe

F(s) =e-ps2Z¥

It follows that the Gaussian is its own Fourier

Fourier Transform of Dirac Delta Function

To compute the Fourier transform of an impulse

F{d(t-t0)}(s) =F(s) =Z

It follows that:

Fourier Transform of Harmonic Signal

What is the inverse Fourier transform of an im-

It follows that:

Fourier Transform of Sine and Cosine

We can compute the Fourier transforms of the

Fourier Transform of Sine and Cosine (contd.)

Expanding the above yields the following ex-

Which by the sifting property is just:

Fourier Transform of Sine and Cosine (contd.)

It follows that

2[d(s+s0)+d(s-s0)].

A similar argument can be used to show:

2[d(s+s0)-d(s-s0)].

Fourier Transform of the Pulse

To compute the Fourier transform of a pulse we

F(s) =Z

2e-j2pstdt

Using the fact that sin(x) =(ejx-e-jx)

2jwe see

F(s) =sin(ps)

Fourier Transform of the Shah Function

Recall the Fourier series for the Shah function:

2pIII?t2p?

By the sifting property,

Changing the order of the summation and the

Factoring outejstfrom the summation

Fourier Transform of the Shah Function

Substituting 2ptfortyields

III(t) =Z

Consequently we see that

F{III}=III.