Signals and Systems Chapter 4: The Continuous Time Fourier Transform - - PowerPoint PPT Presentation

Signals and Systems

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

Fourier’s Derivation of the CT Fourier Transform

 x(t) - an aperiodic signal

view 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

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 2



Fourier series Fourierintegral 

Motivating Example: Square wave

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 3

increases kept fixed Discrete frequency points become denser in

ω as T

increases

0 1 1

2sin( ) 2sin( )

k k k

k T a k T T Ta

 

   



  

So, on with the derivation ...

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 4

For simplicity, assume x(t) has a finite duration.

( ), 2 2 ( ) , 2 T T x t t x t T periodic t            

As , ( ) ( ) for all T x t x t t   

Derivation (continued)

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 5

2 2 2 2

2 ( ) ( ) 1 1 ( ) ( ) ( ) ( )in this interval 1 ( ) (1) If we define ( ) ( ) then Eq.(1) ( )

jk t k k T T jk t jk t k T T jk t j t k

x t a e T a x t e dt x t e dt T T x t x t x t e dt T X j x t e dt X jk a T

    

   

           

         

    

Derivation (continued)

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 6

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 T t x t x t X jk e T X jk e T d x t X j e d X j x t e dt

   

         

        

           

     

nalysis equation

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

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 7 2

( ) x t dt

 

 



2

1 ( ) ( ) ( ) Then ( ) 2

j t

e t x t X j e d e t dt



  

   

  

 

1 (i) ( ) ( ) at points of continuity 2 1 (ii) ( ) midpoint at discontinuity 2 (iii) Gibb's phenomenon

j t j t

X j e d x t X j e d

 

     

   

 

 

Example #1

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 8

( ) ( ) ( ) ( ) ( ) 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

   

         

        

          

  

Example #2: Exponential function

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 9

( )

( )

( ) ( ), ( ) ( ) 1 1 ( )

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 e a j a j



  

  

 

        

         

 

Even symmetry Odd symmetry

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

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 10 1 1

1

2sin ( )

T j t T

T X j e dt



  

 

 



Note the inverse relation between the two widths ⇒ Uncertainty principle

Useful facts about CTFT’s

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 11

1

(0) ( ) 1 (0) ( ) 2 Example above: ( ) 2 (0) 1

• Ex. above: (0)

( ) 2 1 (Area of the triangle) 2 X x t dt x X j d x t dt T X x X j d       

       

      

   

Example #4:

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 12

2

A Gaussian,important in probability, optics, etc.

( )

at

x t e 

2 2 2 2 2 2 2

[ ( ) ] ( ) 2 2 ( ) 2 4 4

( ) [ ].

at j t j j a t j t a a a a j a t a a a a

X j e e dt e dt e dt e e a

       

 

               

   

  

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.

CT Fourier Transforms of Periodic Signals

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 13

periodic in with freq Suppose ( ) ( ) 1 1 ( ) ( ) 2 2 That i All the energy is concentrated in one fr ue s 2 ( ) More generall equency ncy y

j t j t j t

X j x t t e d e e x

  

              

 

          



( ) ( ) 2 ( )

jk t k k k k

t a e X j a k



    

   

   

 

Example #5:

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 14

1 1 ( ) cos 2 2 ( ) ( ) ( )

j t j t

x t t e e X j

 

       



      

“Line Spectrum”

Example #6:

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 15

Sampling function

( ) ( )

n

x t t nT 

 

  



2 2 2

1 1 ( ) ( ) 2 2 ( ) ( )

k

T jk t k T n a k

x t a x t e dt T T k X j T T

  

    

   

    

 

x(t) Same function in the frequency-domain! Note: (period in t) T ⇔ (period in ω) 2π/T Inverse relationship again!

Properties of the CT Fourier Transform

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 16

( )

1) Linearity ( ) ( ) ( ) ( ) 2) Time Shifting ( ) ( ) Proof: ( ) ( ) magnitude unchanged

j t j t j t j t t X 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

    

  

         

        

 

( ) ( ) Linear change in phase ( ( )) ( )

j t j t

e X j X j FT e X j X j t

 

    

 

    

Properties (continued)

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 17

*

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                         

The Properties Keep on Coming ...

Book Chapter4: Section1 Computer Engineering Department, Signals and Systems 18

1 Time-Scaling ( ) ( ) 1 E.g. 1 ( ) ( ) compressed in time stretched in frequency a) ( )real and even x at X j a 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

−𝑌∗(𝑘𝜕)

The CT Fourier Transform Pair

Book Chapter4: Section2 Computer Engineering Department, Signals and Systems 19

) 𝑦(𝑢) ↔ 𝑌(𝑘𝜕 𝑌(𝑘𝜕) = න

−∞ ∞

𝑦(𝑢)𝑓−𝑘𝜕𝑢 𝑒𝑢 𝑦(𝑢) = 1 2𝜌 න

−∞ ∞

𝑌(𝑘𝜕)𝑓𝑘𝜕𝑢𝑒𝜕

─ FT

(Analysis Equation)

─ Inverse FT

(Synthesis Equation) Last lecture: some properties Today: further exploration

) 𝑧(𝑢) = ℎ(𝑢) ∗ 𝑦(𝑢) ↔ 𝑍(𝑘𝜕) = 𝐼(𝑘𝜕)𝑌(𝑘𝜕 ) 𝑥ℎ𝑓𝑠𝑓 ℎ(𝑢) ↔ 𝐼(𝑘𝜕

𝑦(𝑢) = ඲

−∞ ∞

1 2𝜌 𝑌(𝑘𝜕)𝑒𝜕 𝑓𝑘𝜕𝑢 Coefficient a 𝑏𝑓𝑘𝜕𝑢 → → 𝐼(𝑘𝜕)𝑏𝑓𝑘𝜕𝑢 𝑧(𝑢) = ඲

−∞ ∞

𝐼(𝑘𝜕) 1 2𝜌 𝑌(𝑘𝜕)𝑒𝜕 𝑓𝑘𝜕𝑢 = 1 2𝜌 න

−∞ ∞

𝐼(𝑘𝜕)𝑌(𝑘𝜕)𝑓𝑘𝜕𝑢𝑒𝜕 Y(jω)

Book Chapter4: Section2 Computer Engineering Department, Signal and Systems 20

Convolution Property

A consequence of the eigenfunction property :

h(t)

H(jω).a

Synthesis equation for y(t)

The Frequency Response Revisited

Book Chapter4: Section2 Computer Engineering Department, Signal and Systems 21

h(t)

) 𝑦(𝑢) → → 𝑧(𝑢) = ℎ(𝑢) ∗ 𝑦(𝑢

impulse response

) 𝑍(𝑘𝜕) = 𝐼(𝑘𝜕)𝑌(𝑘𝜕

frequency response

The frequency response of a CT LTI system is simply the Fourier transform of its impulse response Example #1:

൯ 𝑦 𝑢 = 𝑓𝑘𝜕0𝑢 → → 𝑧(𝑢 H(jω) ൯ 𝑓𝑘𝜕0𝑢 ↔ 2𝜌𝜀(𝜕 − 𝜕0

Recall

) 𝑍(𝑘𝜕) = 𝐼(𝑘𝜕)𝑌(𝑘𝜕) = 𝐼(𝑘𝜕)2𝜌𝜀(𝜕 − 𝜕0) = 2𝜌𝐼(𝑘𝜕0)𝜀(𝜕 − 𝜕0 ⇓ 𝑧(𝑢) = 𝐼(𝑘𝜕0)𝑓𝑘𝜕0𝑢

inverse FT

Example #2 A differentiator 𝑒 𝑒𝑢 sin𝜕0𝑢 = 𝜕0cos𝜕0𝑢 = 𝜕0sin(𝜕0𝑢 + 𝜌 2) 𝑒 𝑒𝑢 cos𝜕0𝑢 = − 𝜕0sin𝜕0𝑢 = 𝜕0cos(𝜕0𝑢 + 𝜌 2)

Book Chapter4: Section2 Computer Engineering Department, Signal and Systems 22

𝑧(𝑢) = ) 𝑒𝑦(𝑢 𝑒𝑢

• an LTI system

Differentiation property:

) 𝑍(𝑘𝜕) = 𝑘𝜕𝑌(𝑘𝜕 ⇓ 𝐼(𝑘𝜕) = 𝑘𝜕

1) Amplifies high frequencies (enhances sharp edges)

Larger at high ωo phase shift

2) +π/2 phase shift ( j = ejπ/2)

Example #3: Impulse Response of an Ideal Lowpass Filter

Book Chapter4: Section2 Computer Engineering Department, Signal and Systems 23

ℎ(𝑢) = 1 2𝜌 න

−𝜕𝑑 𝜕𝑑

𝑓𝑘𝜕𝑢𝑒𝜕 = sin𝜕𝑑𝑢 𝜌𝑢 = 𝜕𝑑 𝜌 sin𝑑 𝜕𝑑𝑢 𝜌

Define: sinc(θ) = sin𝜌𝜄

𝜌𝜄

Questions: 1) Is this a causal system? No. 2) What is h(0)?

ℎ(0) = 1 2𝜌 න

−∞ ∞

𝐼(𝑘𝜕)𝑒𝜕 = 2𝜕𝑑 2𝜌 = 𝜕𝑑 𝜌

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

𝑡(𝑢) = න

−∞ 𝑢

ℎ(𝑢)𝑒𝑢 𝑡(∞) = න

−∞ ∞

ℎ(𝑢)𝑒𝑢 = 𝐼(𝑘0) = 1

Example #4:

Cascading filtering operations

Book Chapter4: Section2 Computer Engineering Department, Signal and Systems 24

) 𝑓. 𝑕. 𝐼1(𝑘𝜕) = 𝐼2(𝑘𝜕 𝐼(𝑘𝜕) = 𝐼1

2(𝑘𝜕) ℎ𝑏𝑡 𝑏

𝑡ℎ𝑏𝑠𝑞𝑓𝑠 𝑔𝑠𝑓𝑟𝑣𝑓𝑜𝑑𝑧 𝑡𝑓𝑚𝑓𝑑𝑢𝑗𝑤𝑗𝑢𝑧

⇒

Book Chapter4: Section2 Computer Engineering Department, Signal and Systems 25

Example #5:

⇓

⇓ sin4𝜌𝑢 𝜌𝑢 ∗ sin8𝜌𝑢 𝜌𝑢 = ?

h(t) x(t)

) 𝑍(𝑘𝜕) = 𝑌(𝑘𝜕 ) ⇒ 𝑧(𝑢) = 𝑦(𝑢 Example #6: 𝑓−𝑏𝑢2 ∗ 𝑓−𝑐𝑢2 = ? 𝜌 𝑏 + 𝑐 . 𝑓−

𝑏𝑐 𝑏+𝑐 𝑢2

𝜌 𝑏 𝑓− 𝜕2

4𝑏 ×

𝜌 𝑐 𝑓− −𝜕2

4𝑐 =

𝜌 𝑏𝑐 𝑓− 𝜕2

4 1 𝑏+1 𝑐

Gaussian × Gaussian = Gaussian ⇒ Gaussian ∗ Gaussian = Gaussian

Example #2 from last lecture

Book Chapter4: Section2 Computer Engineering Department, Signal and Systems 26

𝑦(𝑢) = 𝑓−𝑏𝑢𝑣(𝑢) , 𝑏 > 0

𝑌(𝑘𝜕) = ඲

−∞ ∞

𝑦(𝑢)𝑓−𝑘𝜕𝑢𝑒𝑢 = න

∞

𝑓−𝑏𝑢𝑓−j𝜕𝑢𝑒𝑢 𝑓

) −(𝑏+𝑘𝜕

= − 1 𝑏 + 𝑘𝜕 𝑓− 𝑏+𝑘𝜕 𝑢]∞ 0 = 1 𝑏 + 𝑘𝜕

Book Chapter4: Section2 Computer Engineering Department, Signal and Systems 27

Example #7:

⇓ ⇓

) 𝑧(𝑢) = [𝑓−𝑢 − 𝑓−2𝑢]𝑣(𝑢 ) ℎ(𝑢) = 𝑓−𝑢𝑣(𝑢) , 𝑦(𝑢) = 𝑓−2𝑢𝑣(𝑢 ) 𝑧(𝑢) = ℎ(𝑢) ∗ 𝑦(𝑢 ⇓ 𝑍(𝑘𝜕) = 𝐼(𝑘𝜕)𝑌(𝑘𝜕) = 1 1 + 𝑘𝜕 . 1 2 + 𝑘𝜕

• a rational function of jω, ratio of polynomials of jω

Partial fraction expansion

𝑍(𝑘𝜕) = 1 1 + 𝑘𝜕 − 1 2 + 𝑘𝜕

Inverse FT

Example #8: LTI Systems Described by LCCDE’s

(Linear-constant-coefficient differential equations)

Book Chapter4: Section2 Computer Engineering Department, Signal and Systems 28

෎

𝑙=0 𝑂

𝑏𝑙 ൯ 𝑒𝑙𝑧(𝑢 𝑒𝑢𝑙 = ෎

𝑙=0 𝑁

𝑐𝑙 ൯ 𝑒𝑙𝑦(𝑢 𝑒𝑢𝑙

Using the Differentiation Property

൯ 𝑒𝑙𝑦(𝑢 𝑒𝑢𝑙 ↔ 𝑘𝜕 𝑙𝑌 𝑘𝜕

⇓ Transform both sides of the equation

෍

𝑙=0 𝑂

𝑏𝑙. 𝑘𝜕 𝑙𝑍 𝑘𝜕 = ෍

𝑙=0 𝑁

൯ 𝑐𝑙. 𝑘𝜕 𝑙𝑌(𝑘𝜕 1) Rational, can use PFE to get h(t) 2) If X(jω) is rational e.g. x(t)=Σcie-at u(t) then Y(jω) is also rational 𝑍(𝑘𝜕) = ෌𝑙=0

𝑁

𝑐𝑙 𝑘𝜕 𝑙 σ𝑙=0

𝑂

𝑏𝑙 𝑘𝜕 𝑙 𝑌 𝑘𝜕 H(jω)

⇓

Parseval’s Relation

Book Chapter4: Section2 Computer Engineering Department, Signal and Systems 29

න

−∞ ∞

|𝑦(𝑢)|2𝑒𝑢 = 1 2𝜌 න

−∞ ∞

|𝑌(𝑘𝜕)|2𝑒𝜕

Total energy in the time-domain Total energy in the time-domain

1 2𝜌 |𝑌(𝑘𝜕)|2

• Spectral density

Multiplication Property

FT is highly symmetric,

𝑦(𝑢)𝐺−1 1 2𝜌 න

−∞ ∞

𝑌(𝑘𝜕)𝑓𝑘𝜕𝑢𝑒𝜕, 𝑌(𝑘𝜕)ഫ ഫ 𝐺 න

−∞ ∞

𝑦(𝑢)𝑓−𝑘𝜕𝑢𝑒𝑢

We already know that: Then it isn’t a surprise that:

) 𝑦(𝑢) ∗ 𝑧(𝑢) ↔ 𝑌(𝑘𝜕). 𝑍(𝑘𝜕 𝑦(𝑢). 𝑧(𝑢) ↔ 1 2𝜌 𝑌(𝑘𝜕) ∗ 𝑍 𝑘𝜕 Convolution in ω = 1 2𝜌 න

−∞ ∞

𝑌(𝑘𝜄)𝑍 𝑘 𝜕 − 𝜄 𝑒𝜄

— A consequence of Duality

Examples of the Multiplication Property

Book Chapter4: Section2 Computer Engineering Department, Signal and Systems 30

𝑠(𝑢) = 𝑡(𝑢). 𝑞(𝑢) ↔ 𝑆 𝑘𝜕 = 1 2𝜌 𝑇(𝑘𝜕) ∗ 𝑄 𝑘𝜕 𝐺𝑝𝑠 𝑞(𝑢) = cos𝜕0𝑢 ↔ 𝑄 𝑘𝜕 = 𝜌𝜀 𝜕 − 𝜕0 + 𝜌𝜀 𝜕 + 𝜕0 𝑆 𝑘𝜕 = 1 2 𝑇 𝑘 𝜕 − 𝜕0 + 1 2 𝑇 𝑘 𝜕 + 𝜕0

For any s(t) ...

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Example (continued)

Book Chapter4: Section2 Computer Engineering Department, Signal and Systems 31

𝑠(𝑢) = 𝑡(𝑢). cos 𝜕0𝑢 Amplitude modulation (AM) 𝑆(𝑘𝜕) = 1 2 ൣ𝑇 𝑘 𝜕 − 𝜕0 Drawn assuming: 𝜕0 − 𝜕1 > 0 𝑗. 𝑓. 𝜕0 > 𝜕1