signals and systems prof. h. sameti chapter 4: the continuous time fourier transform derivation of...

Signals and SystemsProf. H. Sameti

Chapter 4: The Continuous Time Fourier Transform• Derivation of the CT Fourier Transform pair• Examples of Fourier Transforms Topic three• Fourier Transforms of Periodic Signals• Properties of the CT Fourier Transform• The Convolution Property of the CTFT • Frequency Response and LTI Systems Revisited • Multiplication Property and Parseval’s Relation• The DT Fourier Transform

Book Chapter4: Section1

2

Fourier’s Derivation of the CT Fourier Transform

x(t) - an aperiodic signalview it as the limit of a periodic signal as T → ∞

For a periodic signal, the harmonic components are spaced ω0 = 2π/T apart ...

As T → ∞, ω0 → 0, and harmonic components are spaced closer and closer in frequency

Computer Engineering Department, Signals and Systems

Fourier series Fourier integral


3

Motivating Example: Square wave


increaseskept fixed

Discrete frequency points become denser in ω as T increases

0

0 1

0

1

2sin( )

2sin( )

k

k

k

k Ta

k T

TTa


4

So, on with the derivation ...


For simplicity, assumex(t) has a finite duration.

( ),2 2( )

,2

T Tx t t

x tT

periodic t

As , ( ) ( ) for all T x t x t t


5

Derivation (continued)


0

0 0

0

0

2 2

2 2

0

2( ) ( )

1 1( ) ( )

( ) ( ) in this interval

1( ) (1)

If we define

( ) ( )

then Eq.(1)

( )

jk tk

k

T T

jk t jk tk

T T

jk t

j t

k

x t a eT

a x t e dt x t e dtT T

x t x t

x t e dtT

X j x t e dt

X jka

T


6

Derivation (continued)


0

0

0

0 0

0

Thus, for 2 2

1( ) ( ) ( )

1( )

2

As , , we get the CT Fourier Transform pair

1( ) ( ) Synthesis equation

2

( ) ( ) A

k

jk t

k

a

jk t

k

j t

j t

T Tt

x t x t X jk eT

X jk e

T d

x t X j e d

X j x t e dt

nalysis equation


7

For what kinds of signals can we do this?

(1) It works also even if x(t) is infinite duration, but satisfies:a) Finite energy

In this case, there is zero energy in the error

b) Dirichlet conditions

c) By allowing impulses in x(t)or in X(jω), we can represent even more Signals

E.g. It allows us to consider FT for periodic signals


2( )x t dt

21( ) ( ) ( ) Then ( ) 0

2j te t x t X j e d e t dt

1 (i) ( ) ( ) at points of continuity

21

(ii) ( ) midpoint at discontinuity2

(iii) Gibb's phenomenon

j t

j t

X j e d x t

X j e d


8

Example #1


0

0

0

( ) ( ) ( )

( ) ( ) 1

1( ) Synthesis equation for ( )

2( ) ( ) ( )

( ) ( )

j t

j t

j t

j t

a x t t

X j t e dt

t e d t

b x t t t

X j t t e dt

e


9

Example #2: Exponential function


( )0

( )

( ) ( ), 0

( ) ( )

1 1( )

0

a j t

at

j t at j t

e

a j t

x t e u t a

X j x t e dt e e dt

ea j a j

Even symmetry Odd symmetry


10

Example #3: A square pulse in the time-domain


1

1

12sin( )

T j t

T

TX j e dt

Note the inverse relation between the two widths Uncertainty principle⇒


11

Useful facts about CTFT’s


1

(0) ( )

1(0) ( )

2

Example above: ( ) 2 (0)

1Ex. above: (0) ( )

21

(Area of the triangle)2

X x t dt

x X j d

x t dt T X

x X j d


12

Example #4:


2

A Gaussian, important in probability, optics, etc.( ) atx t e2

2 2 2

22

2

[ ( ) ] ( )2 2

( )2 4

4

( )

[ ].

at j t

j ja t j t a

a a a

ja t

a a

a

a

X j e e dt

e dt

e dt e

ea

Also a Gaussian!

(Pulse width in t)•(Pulse width in ω) ∆⇒ t•∆ω ~ (1/a1/2)•(a1/2) = 1

Uncertainty Principle! Cannot make both ∆t and ∆ω arbitrarily small.


13

CT Fourier Transforms of Periodic Signals


0

0

0

0

0

0 0

periodic in with freq

Suppose

( ) ( )

1 1( ) ( )

2 2That i

All the energy is concentrated in one fr

ue

s

2 ( )

More generall

equency

ncy

y

j tj t

j t

X j

x t t e d e

e

x

00( ) ( ) 2 ( )jk t

k kk k

t a e X j a k


14

Example #5:


0 00

0 0

1 1( ) cos

2 2

( ) ( ) ( )

j t j tx t t e e

X j

“Line Spectrum”


15

Example #6:


Sampling function( ) ( )n

x t t nT

0

0

2

2

2

1 1( ) ( )

2 2( ) ( )

k

T jk tk T

na k

x t a x t e dtT T

kX j

T T

x(t)

Same function in the frequency-domain!

Note: (period in t) T ⇔ (period in ω) 2π/T Inverse relationship again!


16

Properties of the CT Fourier Transform


0

0

0

0

( )

1) Linearity ( ) ( ) ( ) ( )

2) Time Shifting ( ) ( )

Proof: ( ) ( )

magnitude unchanged

j t

j tj t j t

tX j

ax t by t aX j bY j

x t t e X j

x t t e dt e x t e dt

FT

0

00

( ) ( )

Linear change in phase

( ( )) ( )

j t

j t

e X j X j

FT

e X j X j t


17

Properties (continued)


*

Conjugate Symmetry

( ) real ( ) ( )

( ) ( )

( ) ( )

{ ( )} { ( )}

{ ( )} { ( )}

Even

x t X j X j

X j X j

X j X j

Re X j Re X j

Im X j Im X j

Odd

Od

n

d

Eve


18

The Properties Keep on Coming ...


1Time-Scaling ( ) ( )

1 E.g. 1

( ) ( ) compressed in time stretched in frequency

a) ( ) real and even

x at X ja a

a a at t

x t X j

x t

*

*

( ) ( )

( ) ( ) ( ) Real & even

b) ( ) real and odd ( ) ( )

( ) ( ) ( ) Purely imaginary &:

c) ( ) { ( )}+ { ( )}

x t x t

X j X j X j

x t x t x t

X j X j X j

X j Re X j jIm X j

( ) { ( )} { ( )}x t Ev x t Od x t

For real


19

The CT Fourier Transform Pair


─ FT(Analysis Equation)

─ Inverse FT(Synthesis Equation)

Last lecture: some propertiesToday: further exploration


20

Coefficient a

Y(jω)

Computer Engineering Department, Signal and Systems

Convolution Property

A consequence of the eigenfunction property :

h(t)

H(jω).a

Synthesis equation for y(t)


21

The Frequency Response Revisited


h(t)

¿impulse response

¿frequency response

The frequency response of a CT LTI system is simply the Fourier transform of its impulse response

Example #1: ¿H(jω)

¿Recall

¿⇓𝑦 (𝑡)=𝐻 ( 𝑗 𝜔0)𝑒

𝑗 𝜔0𝑡inverse FT


22

Example #2 A differentiator


𝑦 (𝑡)=¿¿ - an LTI system

Differentiation property:¿⇓

𝐻 ( 𝑗 𝜔)= 𝑗 𝜔1) Amplifies high frequencies (enhances sharp edges)

Larger at high ωo phase shift 2) +π/2 phase shift ( j = ejπ/2)


23

Example #3: Impulse Response of an Ideal Lowpass Filter


h (𝑡)=1

2𝜋 −𝜔𝑐

𝜔𝑐

𝑒 𝑗 𝜔𝑡 𝑑𝜔

¿sin𝜔𝑐𝑡𝜋 𝑡

¿𝜔𝑐

𝜋sin𝑐 (𝜔𝑐𝑡

𝜋 ) Define: sinc(θ) ¿

sin 𝜋𝜃𝜋𝜃

Questions:1) Is this a causal system? No.

2) What is h(0)?

h (0)= 12𝜋 −∞

∞

𝐻 ( 𝑗 𝜔)𝑑𝜔=2𝜔𝑐

2𝜋=𝜔𝑐

𝜋3) What is the steady-state value of the step response, i.e. s(∞)?

𝑠(𝑡)=− ∞

𝑡

h (𝑡)𝑑𝑡

𝑠(∞)=− ∞

∞

h (𝑡)𝑑𝑡=𝐻 ( 𝑗 0)=1


24

Example #4: Cascading filtering operations


¿

𝐻 ( 𝑗 𝜔)=𝐻12( 𝑗𝜔 ) h𝑎𝑠 𝑎

𝑠h𝑎𝑟𝑝𝑒𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑣𝑖𝑡𝑦

⇒


25Computer Engineering Department, Signal and Systems

Example #5:

sin 4 𝜋𝑡𝜋 𝑡

∗sin 8𝜋𝑡𝜋 𝑡

=?

h(t)x(t)

¿

Example #6: 𝑒−𝑎𝑡 2

∗𝑒−𝑏𝑡 2

= ? √ 𝜋𝑎+𝑏

.𝑒−( 𝑎𝑏𝑎+𝑏 )𝑡2

√ 𝜋𝑎 𝑒− 𝜔2

4𝑎×√ 𝜋𝑏 𝑒− −𝜔2

4𝑏 = 𝜋√𝑎𝑏

𝑒− 𝜔2

4 ( 1𝑎+ 1𝑏)

Gaussian × Gaussian = Gaussian ⇒ Gaussian ∗ Gaussian = Gaussian


26

Example #2 from last lecture


𝑥 (𝑡)=𝑒−𝑎𝑡𝑢 (𝑡) , 𝑎>0

𝑋 ( 𝑗 𝜔)= −∞

∞

𝑥 (𝑡)𝑒− 𝑗 𝜔𝑡 𝑑𝑡 =0

∞

𝑒−𝑎𝑡𝑒− j 𝜔𝑡𝑑𝑡𝑒¿ ¿

¿ −( 1𝑎+ 𝑗 𝜔 )𝑒− (𝑎+ 𝑗 𝜔 ) 𝑡 ¿∞

0= 1𝑎+ 𝑗𝜔


27Computer Engineering Department, Signal and Systems

Example #7:

¿- a rational function of jω, ratio of polynomials of jω

Partial fraction expansion

𝑌 ( 𝑗 𝜔)=1

1+ 𝑗 𝜔−

12+ 𝑗 𝜔

Inverse FT


28

Example #8: LTI Systems Described by LCCDE’s (Linear-constant-coefficient differential equations)


∑𝑘=0

𝑁

𝑎𝑘¿¿¿

Using the Differentiation Property¿¿

⇓Transform both sides of the equation

∑𝑘=0

𝑁

𝑎𝑘 . ( 𝑗 𝜔 )𝑘𝑌 ( 𝑗 𝜔 )=∑𝑘=0

𝑀

¿¿1) Rational, can use

PFE to get h(t)

2) If X(jω) is rationale.g. x(t)=Σcie-at u(t)then Y(jω) is also rational

𝑌 ( 𝑗 𝜔)=[∑𝑘=0

𝑀

𝑏𝑘 ( 𝑗𝜔 )𝑘

∑𝑘=0

𝑁

𝑎𝑘 ( 𝑗 𝜔 )𝑘 ]𝑋 ( 𝑗𝜔 )

H(jω)

⇓


29

Parseval’s Relation


− ∞

∞

¿ 𝑥 (𝑡)¿2𝑑𝑡= 12𝜋 −∞

∞

¿ 𝑋 ( 𝑗𝜔)¿2𝑑𝜔

Total energyin the time-domain

Total energyin the time-domain

12𝜋∨𝑋 ( 𝑗 𝜔)¿2

- Spectral density

Multiplication Property FT is highly symmetric,

𝑥 (𝑡)´́𝐹 −1 1

2𝜋 −∞

∞

𝑋 ( 𝑗𝜔)𝑒 𝑗 𝜔𝑡 𝑑𝜔 , 𝑋 ( 𝑗 𝜔) ´́𝐹−∞

∞

𝑥(𝑡)𝑒− 𝑗 𝜔𝑡 𝑑𝑡

We already know that:

Then it isn’t asurprise that:

¿𝑥 (𝑡). 𝑦 (𝑡)↔

12𝜋

𝑋 ( 𝑗 𝜔)∗𝑌 ( 𝑗 𝜔 )Convolution in ω

¿ 12𝜋 − ∞

∞

𝑋 ( 𝑗 𝜃)𝑌 ( 𝑗 (𝜔−𝜃 ) )𝑑𝜃

— A consequence of Duality


30

Examples of the Multiplication Property


𝑟 (𝑡)=𝑠(𝑡 ).𝑝 (𝑡)↔𝑅 ( 𝑗 𝜔 )= 12𝜋

[𝑆 ( 𝑗𝜔 )∗𝑃 ( 𝑗 𝜔 ) ]

𝐹𝑜𝑟 𝑝 (𝑡)=cos𝜔0 𝑡↔𝑃 ( 𝑗𝜔 )=𝜋𝛿 (𝜔−𝜔0 )+𝜋𝛿 (𝜔+𝜔0 )

𝑅 ( 𝑗 𝜔 )=12𝑆 ( 𝑗 (𝜔−𝜔0 ))+

12𝑆 ( 𝑗 (𝜔+𝜔0 ))

For any s(t) ...

⇓


31

Example (continued)


𝑟 (𝑡)=𝑠(𝑡 ).cos (𝜔0𝑡 )Amplitude modulation(AM)

𝑅 ( 𝑗 𝜔)=12 [𝑆 ( 𝑗 (𝜔−𝜔0 ) )+𝑆 ( 𝑗 (𝜔+𝜔0 )) ]

Drawn assuming:𝜔0 −𝜔1>0𝑖 .𝑒 . 𝜔0>𝜔1

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