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Signals and SystemsChapter 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 Fourier integral
Motivating Example: Square wave
Book Chapter4: Section1
Computer Engineering Department, Signals and Systems 3
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
So, on with the derivation ...
Book Chapter4: Section1
Computer Engineering Department, Signals and Systems 4
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
Derivation (continued)
Book Chapter4: Section1
Computer Engineering Department, Signals and Systems 5
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 t
k
k
T T
jk t jk t
k
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
Derivation (continued)
Book Chapter4: Section1
Computer Engineering Department, Signals and Systems 6
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
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
21( ) ( ) ( ) Then ( ) 0
2
j te 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
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
Example #2: Exponential function
Book Chapter4: Section1
Computer Engineering Department, Signals and Systems 9
( )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
Example #3: A square pulse in the time-domain
Book Chapter4: Section1
Computer Engineering Department, Signals and Systems 10
1
1
12sin( )
Tj t
T
TX 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)
1Ex. 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.( ) atx t e
2
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 makeboth ∆t and ∆ω arbitrarily small.
CT Fourier Transforms of Periodic Signals
Book Chapter4: Section1
Computer Engineering Department, Signals and Systems 13
0
0
0
0
0
0 0
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 tj t
j t
X j
x t t e d e
e
x
0
0( ) ( ) 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
0 0
0
0 0
1 1( ) cos
2 2
( ) ( ) ( )
j t j tx 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
0
0
2
2
2
1 1( ) ( )
2 2( ) ( )
k
Tjk t
kT
n
a 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!
Properties of the CT Fourier Transform
Book Chapter4: Section1
Computer Engineering Department, Signals and Systems 16
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
0
0
( ) ( )
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
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
−𝑋∗(𝑗𝜔)
The CT Fourier Transform Pair
Book Chapter4: Section2
Computer Engineering Department, Signals and Systems 19
)𝑥(𝑡) ↔ 𝑋(𝑗𝜔
𝑋(𝑗𝜔) = න
−∞
∞
𝑥(𝑡)𝑒−𝑗𝜔𝑡 𝑑𝑡
𝑥(𝑡) =1
2𝜋න
−∞
∞
𝑋(𝑗𝜔)𝑒𝑗𝜔𝑡𝑑𝜔
─ FT(Analysis Equation)
─ Inverse FT(Synthesis Equation)
Last lecture: some propertiesToday: further exploration
)𝑦(𝑡) = ℎ(𝑡) ∗ 𝑥(𝑡) ↔ 𝑌(𝑗𝜔) = 𝐻(𝑗𝜔)𝑋(𝑗𝜔)𝑤ℎ𝑒𝑟𝑒 ℎ(𝑡) ↔ 𝐻(𝑗𝜔
𝑥(𝑡) =
−∞
∞1
2𝜋𝑋(𝑗𝜔)𝑑𝜔 𝑒𝑗𝜔𝑡
Coefficient a
𝑎𝑒𝑗𝜔𝑡 → → 𝐻(𝑗𝜔)𝑎𝑒𝑗𝜔𝑡
𝑦(𝑡) =
−∞
∞
𝐻(𝑗𝜔)1
2𝜋𝑋(𝑗𝜔)𝑑𝜔 𝑒𝑗𝜔𝑡
=1
2𝜋න−∞
∞
𝐻(𝑗𝜔)𝑋(𝑗𝜔)𝑒𝑗𝜔𝑡𝑑𝜔
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) )𝑥(𝑡) → → 𝑦(𝑡) = ℎ(𝑡) ∗ 𝑥(𝑡
impulse response
)𝑌(𝑗𝜔) = 𝐻(𝑗𝜔)𝑋(𝑗𝜔
frequency response
The frequency response of a CT LTI system is simply the Fourier transform of its impulse response
Example #1: ൯𝑥 𝑡 = 𝑒𝑗𝜔0𝑡 → → 𝑦(𝑡H(jω)
൯𝑒𝑗𝜔0𝑡 ↔ 2𝜋𝛿(𝜔 − 𝜔0Recall
)𝑌(𝑗𝜔) = 𝐻(𝑗𝜔)𝑋(𝑗𝜔) = 𝐻(𝑗𝜔)2𝜋𝛿(𝜔 − 𝜔0) = 2𝜋𝐻(𝑗𝜔0)𝛿(𝜔 − 𝜔0
⇓𝑦(𝑡) = 𝐻(𝑗𝜔0)𝑒
𝑗𝜔0𝑡
inverse FT
Example #2 A differentiator
𝑑
𝑑𝑡sin𝜔0𝑡 = 𝜔0cos𝜔0𝑡 = 𝜔0sin(𝜔0𝑡 +
𝜋
2)
𝑑
𝑑𝑡cos𝜔0𝑡 = − 𝜔0sin𝜔0𝑡 = 𝜔0cos(𝜔0𝑡 +
𝜋
2)
Book Chapter4: Section2
Computer Engineering Department, Signal and Systems 22
𝑦(𝑡) =)𝑑𝑥(𝑡
𝑑𝑡- an LTI system
Differentiation property: )𝑌(𝑗𝜔) = 𝑗𝜔𝑋(𝑗𝜔
⇓
𝐻(𝑗𝜔) = 𝑗𝜔
1) Amplifies high frequencies (enhances sharp edges)
Larger at high ωo phase shift2) +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(𝑡) =1
2𝜋න−𝜔𝑐
𝜔𝑐
𝑒𝑗𝜔𝑡𝑑𝜔
=sin𝜔𝑐𝑡
𝜋𝑡
=𝜔𝑐𝜋sin𝑐
𝜔𝑐𝑡
𝜋
Define: sinc(θ) =sin𝜋𝜃
𝜋𝜃
Questions:1) Is this a causal system? No.
2) What is h(0)?
ℎ(0) =1
2𝜋න−∞
∞
𝐻(𝑗𝜔)𝑑𝜔 =2𝜔𝑐2𝜋
=𝜔𝑐𝜋
3) What is the steady-state value of the step response, i.e. s(∞)?
𝑠(𝑡) = න−∞
𝑡
ℎ(𝑡)𝑑𝑡
𝑠(∞) = න−∞
∞
ℎ(𝑡)𝑑𝑡 = 𝐻(𝑗0) = 1
Example #4: Cascading filtering operations
Book Chapter4: Section2
Computer Engineering Department, Signal and Systems 24
)𝑒. 𝑔. 𝐻1(𝑗𝜔) = 𝐻2(𝑗𝜔
𝐻(𝑗𝜔) = 𝐻12(𝑗𝜔) ℎ𝑎𝑠 𝑎
𝑠ℎ𝑎𝑟𝑝𝑒𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑣𝑖𝑡𝑦
⇒
Book Chapter4: Section2
Computer Engineering Department, Signal and Systems 25
Example #5:
⇓
⇓
sin4𝜋𝑡
𝜋𝑡∗sin8𝜋𝑡
𝜋𝑡= ?
h(t)x(t)
)𝑌(𝑗𝜔) = 𝑋(𝑗𝜔)⇒ 𝑦(𝑡) = 𝑥(𝑡
Example #6: 𝑒−𝑎𝑡2∗ 𝑒−𝑏𝑡
2= ?
𝜋
𝑎 + 𝑏. 𝑒
−𝑎𝑏𝑎+𝑏 𝑡2
𝜋
𝑎𝑒−
𝜔2
4𝑎 ×𝜋
𝑏𝑒−
−𝜔2
4𝑏 =𝜋
𝑎𝑏𝑒−𝜔2
41𝑎+1𝑏
Gaussian × Gaussian = Gaussian ⇒ Gaussian ∗ Gaussian = Gaussian
Example #2 from last lecture
Book Chapter4: Section2
Computer Engineering Department, Signal and Systems 26
𝑥(𝑡) = 𝑒−𝑎𝑡𝑢(𝑡) , 𝑎 > 0
𝑋(𝑗𝜔) =
−∞
∞
𝑥(𝑡)𝑒−𝑗𝜔𝑡𝑑𝑡 = න0
∞
𝑒−𝑎𝑡𝑒−j𝜔𝑡𝑑𝑡
𝑒 )−(𝑎+𝑗𝜔
= −1
𝑎 + 𝑗𝜔𝑒− 𝑎+𝑗𝜔 𝑡]
∞0=
1
𝑎 + 𝑗𝜔
Book Chapter4: Section2
Computer Engineering Department, Signal and Systems 27
Example #7:
⇓
⇓)𝑦(𝑡) = [𝑒−𝑡 − 𝑒−2𝑡]𝑢(𝑡
)ℎ(𝑡) = 𝑒−𝑡𝑢(𝑡) , 𝑥(𝑡) = 𝑒−2𝑡𝑢(𝑡)𝑦(𝑡) = ℎ(𝑡) ∗ 𝑥(𝑡
⇓
𝑌(𝑗𝜔) = 𝐻(𝑗𝜔)𝑋(𝑗𝜔) =1
1 + 𝑗𝜔.
1
2 + 𝑗𝜔
- a rational function of jω, ratio of polynomials of jω
Partial fraction expansion
𝑌(𝑗𝜔) =1
1 + 𝑗𝜔−
1
2 + 𝑗𝜔
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
𝑘=0
𝑁
𝑎𝑘൯𝑑𝑘𝑦(𝑡
𝑑𝑡𝑘=
𝑘=0
𝑀
𝑏𝑘൯𝑑𝑘𝑥(𝑡
𝑑𝑡𝑘
Using the Differentiation Property
൯𝑑𝑘𝑥(𝑡
𝑑𝑡𝑘↔ 𝑗𝜔 𝑘𝑋 𝑗𝜔
⇓ Transform both sides of the equation
𝑘=0
𝑁
𝑎𝑘 . 𝑗𝜔𝑘𝑌 𝑗𝜔 =
𝑘=0
𝑀
൯𝑏𝑘 . 𝑗𝜔𝑘𝑋(𝑗𝜔
1) Rational, can usePFE to get h(t)
2) If X(jω) is rationale.g. x(t)=Σcie-at u(t)
then Y(jω) is also rational
𝑌(𝑗𝜔) =
𝑘=0
𝑀𝑏𝑘 𝑗𝜔 𝑘
σ𝑘=0𝑁 𝑎𝑘 𝑗𝜔 𝑘
𝑋 𝑗𝜔
H(jω)
⇓
Parseval’s Relation
Book Chapter4: Section2
Computer Engineering Department, Signal and Systems 29
න−∞
∞
|𝑥(𝑡)|2𝑑𝑡 =1
2𝜋න−∞
∞
|𝑋(𝑗𝜔)|2𝑑𝜔
Total energyin the time-domain
Total energyin the time-domain
1
2𝜋|𝑋(𝑗𝜔)|2
- Spectral density
Multiplication PropertyFT is highly symmetric,
𝑥(𝑡)𝐹−11
2𝜋න−∞
∞
𝑋(𝑗𝜔)𝑒𝑗𝜔𝑡𝑑𝜔, 𝑋(𝑗𝜔)ഫഫ𝐹න−∞
∞
𝑥(𝑡)𝑒−𝑗𝜔𝑡𝑑𝑡
We already know that:
Then it isn’t asurprise that:
)𝑥(𝑡) ∗ 𝑦(𝑡) ↔ 𝑋(𝑗𝜔). 𝑌(𝑗𝜔
𝑥(𝑡). 𝑦(𝑡) ↔1
2𝜋𝑋(𝑗𝜔) ∗ 𝑌 𝑗𝜔
Convolution in ω=
1
2𝜋න−∞
∞
𝑋(𝑗𝜃)𝑌 𝑗 𝜔 − 𝜃 𝑑𝜃
— A consequence of Duality
Examples of the Multiplication Property
Book Chapter4: Section2
Computer Engineering Department, Signal and Systems 30
𝑟(𝑡) = 𝑠(𝑡). 𝑝(𝑡) ↔ 𝑅 𝑗𝜔 =1
2𝜋𝑆(𝑗𝜔) ∗ 𝑃 𝑗𝜔
𝐹𝑜𝑟 𝑝(𝑡) = cos𝜔0𝑡 ↔ 𝑃 𝑗𝜔 = 𝜋𝛿 𝜔 − 𝜔0 + 𝜋𝛿 𝜔 + 𝜔0
𝑅 𝑗𝜔 =1
2𝑆 𝑗 𝜔 − 𝜔0 +
1
2𝑆 𝑗 𝜔 + 𝜔0
For any s(t) ...
⇓
Example (continued)
Book Chapter4: Section2
Computer Engineering Department, Signal and Systems 31
𝑟(𝑡) = 𝑠(𝑡). cos 𝜔0𝑡
Amplitude modulation(AM)
𝑅(𝑗𝜔)
=1
2ൣ𝑆 𝑗 𝜔 − 𝜔0
Drawn assuming:
𝜔0 − 𝜔1 > 0𝑖. 𝑒. 𝜔0 > 𝜔1