6th lecture continuous-time fourier [email protected] source : hwei p hsu, ......

6th Lecture

Continuous-Time Fourier Transform

Achmad Rizal Danisya

[email protected]

Source : Hwei P Hsu, Proakis & Manolakis

mailto:[email protected]

Born: 21 March 1768 in Auxerre, Bourgogne, France

Died: 16 May 1830 in Paris, France

Jean Baptiste Joseph Fourier

Joseph’s father was a tailor in Auxerre

Joseph was the ninth of twelve children

His mother died when he was nine and

his father died the following year

Fourier demonstrated talent on math

at the age of 14.

In 1787 Fourier decided to train for

the priesthood - a religious life or a

mathematical life?

In 1793, Fourier joined the local

Revolutionary Committee

WHO IS FOURIER ?

FOURIER’S “CONTROVERSY” WORKS

Fourier did his important mathematical

work on the theory of heat (highly

regarded memoir On the Propagation of

Heat in Solid Bodies ) from 1804 to 1807

This memoir received objection from

Fourier’s mentors (Laplace and Lagrange)

and not able to be published until 1815

Napoleon awarded him a pension of 6000 francs, payable from 1 July, 1815.

However Napoleon was defeated on 1 July and Fourier did not receive any

money

Fourier Transforms are used in :

X-ray diffraction

Electron microscopy (and diffraction)

NMR spectroscopy

IR spectroscopy

Fluorescence spectroscopy

Image processing

etc. etc. etc. etc.

WHY DO WE NEED THIS ?

WHAT IS FOURIER TRANSFORM ?

A function can be described by a

summation of waves with different

amplitudes and phases.

𝑒𝑖𝑡 = cos 𝑡 + 𝑖 sin 𝑡; 𝑒−𝑖𝑡 = cos 𝑡 − 𝑖 sin 𝑡 cos(𝑠 + 𝑡) = cos 𝑠 cos 𝑡 − sin 𝑠 sin 𝑡 sin 𝑠 + 𝑡 = sin 𝑠 cos t + cos s sin t

cos 𝑡 =𝑒𝑖𝑡+𝑒−𝑖𝑡

2 ; sin 𝑡 =

𝑒𝑖𝑡−𝑒−𝑖𝑡

2𝑖

cos2 𝑡 − 𝑖2 sin2 𝑡 = cos2 𝑡 + sin2 𝑡 = 1

USEFUL MATHEMATIC TOOLS

FOURIER TRANSFORMS FAMILY

Fourier Series Representation :

SIGNAL PERIODICITY

A continuous-time signal 𝑥(𝑡) to be periodic if there is a

positive nonzero value of T for which

Fundamental Period (𝑇0) is the smallest positive T value for

which equation above satisfied.

Fundamental Frequency (𝑓0) is : 1

𝑇0

Fundamental Angular Frequency (𝜔0) is : 2𝜋/𝑇0 or 2𝜋𝑓0 Basic Examples of Periodic Signals :

𝑥 𝑡 = cos (𝜔0𝑡 + 𝜃) 𝑥 𝑡 = 𝑒𝑗𝜔0𝑡 → cos 𝜔0𝑡 + 𝑗𝑠𝑖𝑛(𝜔0𝑡)


COMPLEX EXPONENTIAL

𝑐𝑘 are known as the complex Fourier coefficients and are given by

𝑐𝑘 =1

𝑇0 𝑥 𝑡 𝑒−𝑗𝑘𝜔0𝑡

𝑇0

=1

𝑇0 𝑥 𝑡 𝑒−𝑗𝑘𝜔0𝑡𝑇0/2

−𝑇0/2

=1

𝑇0 𝑥 𝑡 𝑒−𝑗𝑘𝜔0𝑡𝑇0

0

for 𝑘 = 0 𝑐0 = 1/𝑇0 𝑥(𝑡)

𝑇0

which means 𝑐0 equals the average value of x(t) over a period.

𝑐−𝑘 = 𝑐𝑘∗

For real 𝑥(𝑡)


TRIGONOMETRIC

𝑐𝑘 =1

𝑇0 𝑥 𝑡 𝑒−𝑗𝑘𝜔0𝑡

𝑇0

If 𝑥(𝑡) is an even signal then 𝑏𝑘 = 0

If 𝑥(𝑡) is an odd signal then 𝑎𝑘 = 0


TRIGONOMETRIC

Example Problem 1

Derive the trigonometric Fourier series

𝑥 𝑡 =𝑎0

2+ (cos 𝑘𝜔0𝑡 + sin 𝑘𝜔0𝑡) ∞𝑘=1

from the complex exponential Fourier series

𝑥 𝑡 = 𝑐𝑘𝑒𝑗𝑘𝜔0𝑡

∞

𝑘=−∞

, where 𝜔0 =2𝜋

𝑇0= 2𝜋𝐹0 !

Answer :

Example Problem 1(contd)

So that,

From

Representation of real periodic signal 𝑥 𝑡

𝐶0= dc component

𝐶𝑘𝑐𝑜𝑠(𝑘𝜔0𝑡 − 𝜃𝑘) = kth Harmonic component of 𝑥 𝑡

𝐶1𝑐𝑜𝑠(𝜔0𝑡 − 𝜃1)= Fundamental component

𝐶𝑘= Harmonic Amplitude

𝜃𝑘= Harmonic Phase Angles


HARMONIC FORM


CONVERGENCE a periodic signal x(t) has a Fourier series representation if it satisfies the following Dirichlet conditions:

1. 𝑥(𝑡) is absolutely integrable over any period, that is

𝑥 𝑡 𝑑𝑡 < ∞

𝑇0

2. 𝑥(𝑡) has a finite number of maxima and minima

within any finite interval of 𝑡.

3. 𝑥(𝑡) has a finite number of discontinuities within any finite interval of 𝑡 , and each of these discontinuities is finite.

From

We can express it like this :

𝑐𝑘 = 𝑐𝑘 ejk

Amplitude Spectra 𝑥(𝑡) : Plot on 𝜔 vs |𝑐𝑘|

Phase Spectra of 𝑥(𝑡) : Plot on 𝜔 vs 𝑘

since 𝑘 is only integer so the curves appear discrete frequency, therefore refered as : discrete frequency spectra or line spectra.


AMPLITUDE & PHASE SPECTRA

For a real periodic signal 𝑥(𝑡) we have

𝑐−𝑘 = 𝑐𝑘∗

thus,


AMPLITUDE & PHASE SPECTRA

The average power of a periodic signal 𝑥(𝑡) over any period as

Parseval Identity :


POWER CONTENT OF PERIODIC SIGNAL

A non periodic 𝑥(𝑡)

where 𝑥 𝑡 = 0 when 𝑡 > 𝑇1

Let 𝑥𝑇0(𝑡) be a periodic signal formed by repeating 𝑥 𝑡

with the fundamental period 𝑇0 → ∞

lim𝑇0→∞𝑥𝑇0(𝑡) = 𝑥(𝑡)

Fourier Transform :

FROM SERIES TO TRANSFORMATION

Fourier Transform :


Let us define :

, So that 𝑐𝑘 will become

Fourier Transform :


𝜔0 =2𝜋

𝑇0→1

𝑇0=𝜔02𝜋

As 𝑇0 → ∞ , 𝜔0 → 0 , thus let 𝜔0 = ∆𝜔

Fourier Transform :


Example Problem 2

Determine the complex exponential Fourier series representation for each of the following signals: a) 𝑥 𝑡 = cos𝜔0𝑡 b) 𝑥 𝑡 = cos 4𝑡 + sin 6𝑡 c) 𝑥 𝑡 = sin2 𝑡 Answers :

a)

b)

Find the fundamental period of 𝑥(𝑡) first :

𝜔0𝑡1 = 6𝑡 → 𝑇01 =𝜋

3

𝜔0𝑡2 = 4𝑡 → 𝑇02 =𝜋

2

The 𝑇0 is the least common multiple between 𝑇01 and 𝑇02, that

is 𝑇0 = 𝜋, so 𝜔0 =2𝜋

𝜋= 2

A

A

C

C

a

Example Problem 2 (Contd)

Example Problem 3

Answers :

(a)


(b)

FREQUENCY RESPONSE

𝑦(𝑡) ↔ 𝑌(𝜔) 𝑥 𝑡 ∗ ℎ(𝑡) ↔ 𝑋 𝜔 𝐻(𝜔)

𝐻(𝜔) is called frequency response of the

system

𝐻 𝜔 =𝑌 𝜔

𝑋 𝜔

𝐻 𝜔 = 𝐻 𝜔 𝑒𝑗𝜃𝐻(𝜔)

|𝐻(𝜔)|=Magnitude response

𝜃𝐻(𝜔)=Phase Response of the system

|𝑌 𝜔 | = |𝑋 𝜔 ||𝐻 𝜔 |

With

𝑌 𝜔 = 𝑌 𝜔 𝑒𝑗𝜃𝑋(𝜔) and 𝑋 𝜔 = 𝑋 𝜔 𝑒𝑗𝜃𝑌(𝜔)

We have :

𝜃𝑌 𝜔 = 𝜃𝑋 𝜔 + 𝜃𝐻(𝜔)

FREQUENCY RESPONSE

Duality or Symmetry From basic transform pair :

We have a special characteristic

𝑋 𝑡 ↔ 2𝜋𝑥(−𝜔)

This property allows us to obtain both of these dual Fourier transform pairs from one evaluation

Problem Solved 1

Find the Fourier transform of 𝑥 𝑡 = 𝑒−𝑎 𝑡

for 𝑎 > 0 !

Answers :

Evaluate the value of 𝑡 𝑡 > 0 → 𝑥 𝑡 = 𝑒−𝑎𝑡 𝑡 < 0 → 𝑥 𝑡 = 𝑒𝑎𝑡

Problem Solved 1

Problem Solved II

Find the Fourier transform of

𝑥 𝑡 =1

𝑎2 + 𝑡2

Answer :

We exploit the result we have in previous problem and symmetry/duality property :

𝑒−𝑎 𝑡 ↔2𝑎

𝑎2 + 𝜔2

2𝑎

𝑎2 +𝜔2→2𝑎

𝑎2 + 𝑡2

Remember duality property : 𝑋 𝑡 ↔ 2𝜋𝑥(−𝜔)

Then, 2𝑎

𝑎2+𝑡2↔ 2𝜋𝑒−𝑎 −𝜔 = 2𝜋𝑒−𝑎 𝜔

Problem Solved II

Then to solve our problem we just need

to divide each side with 2𝑎 2𝑎

𝑎2 + 𝑡2×1

2𝑎↔ 2𝜋𝑒−𝑎 𝜔 ×

1

2𝑎

1

𝑎2 + 𝑡2↔𝜋

𝑎𝑒−𝑎 𝜔

Distortionless Transmission

exact input signal shape be reproduced at the output although its amplitude may be different and it may be delayed in time

𝑦 𝑡 = 𝐾𝑥(𝑡 − 𝑡𝑑) 𝑡𝑑= time delay

𝐾= Gain constant (𝐾 > 0) 𝑦 𝑡 = 𝐾𝑥 𝑡 − 𝑡𝑑 ↔ 𝑌 𝜔 = 𝐾𝑒

−𝑗𝜔𝑡𝑑𝑋(𝜔) Thus 𝐻 𝜔 = 𝐾𝑒−𝑗𝜔𝑡𝑑 From 𝐻 𝜔 = 𝐻 𝜔 ejθH(𝜔) We have 𝐾 = |𝐻 𝜔 | and 𝑒−𝑗𝜃𝐻(𝜔) = 𝑒−𝑗𝜔𝑡𝑑 or 𝜃𝐻 𝜔 = 𝜔𝑡𝑑

So the the magnitude of frequency response must be constant over the entire frequency range, and the phase must be linear with the frequency.

Distortionless Transmission

Inconsistent magnitude (𝐾) Different Gain/Attenuation = Amplitude distortion

Not linear Phase Spectrum (𝜃𝐻 𝜔 ) Different waveform = phase distortion

LTI System Differential Equation

FILTERS Ideal Low Pass Filter (LPF)

Absolute Bandwidth : 𝑊𝐵 = 𝜔𝑐

FILTERS

Ideal High Pass Filter

FILTERS

Ideal Bandpass Filter

Absolute Bandwidth : 𝑊𝐵 = 𝜔2 − 𝜔1

Narrowband if : 𝑊𝐵 ≪ 𝜔0 , where

𝜔0 =1

2(𝜔1 + 𝜔2)

FILTERS

Ideal Bandstop Filter

FILTERS

−𝑥 𝑡 + 𝑖 𝑡 𝑅 + 𝑉𝑐 = 0

−𝑥 𝑡 +𝐶 𝑑𝑉𝑐𝑑𝑡𝑅 + 𝑦 𝑡 = 0

Because 𝑉𝑐 = 𝑦(𝑡) , then

𝑅𝐶𝑗𝜔𝑌 𝜔 + 𝑌 𝜔 = 𝑋(𝜔) 𝑌 𝜔 𝑗𝜔𝑅𝐶 + 1 = 𝑋(𝜔)

3 dB BANDWIDTH

𝑊3𝑑𝐵 also known as half power bandwidth

For LPF : positive frequency at which

the amplitude spectrum |𝐻 𝜔 | drops to a

value equal to 𝐻 0

2

For BPF : the difference between the

frequencies at which the amplitude

spectrum |𝐻 𝜔 | drops to a value equal to 𝐻 𝜔𝑚

2

3 dB BANDWIDTH

6th lecture continuous-time fourier [email protected] source : hwei p hsu, ......

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