6.003 Signal Processing · All of the information in a periodic signal is contained in one period....

6.003 Signal Processing

Week 4, Lecture A:Continuous Time Fourier Transform

6.003 Fall 2020

From Periodic to Aperiodic

• However, most real-world signals are not periodic.

• Previously, we have focused on Fourier representations of periodic signals: e.g., sounds, waves, music, ...

Today: generalizing Fourier representations to include aperiodic signals -> Fourier Transform

Fourier Representations of Aperiodic SignalsHow can we represent an aperiodic signal as a sum of sinusoids?

Strategy: make a periodic version of 𝑥(𝑡) by summing shifted copies:

Since 𝑥𝑝(𝑡) is periodic, it has a Fourier series (which depends on T)

Find Fourier series coefficients 𝑋𝑝[𝑘] and take the limit of 𝑋𝑝[𝑘] as T → ∞

As T → ∞, 𝑥𝑝(𝑡) → 𝑥(𝑡) and Fourier series will approach Fourier transform.

8𝜔0

Fourier Representations of Aperiodic Signals

Calculate the Fourier series coefficients 𝑋𝑝[𝑘] :

=1

𝑇න−𝑆

𝑆

1 ∙ 𝑒−𝑗2𝜋𝑇 𝑘𝑡 𝑑𝑡 =

1

𝑇∙ ቮ𝑒−𝑗

2𝜋𝑇 𝑘𝑡

(−𝑗2𝜋𝑘/𝑇)−𝑆

𝑆

=2sin(

2𝜋𝑘𝑇

𝑆)

𝑇(2𝜋𝑘𝑇

)

What happens if you double the period T?

The red samples are at new intermediate frequencies

There are twice as many samples per period of the sine function

𝑋𝑝 𝑘 =1

𝑇න−𝑇/2

𝑇/2

𝑥(𝑡) ∙ 𝑒−𝑗2𝜋𝑇 𝑘𝑡 𝑑𝑡

4𝜔0

Plot the resulting Fourier coefficients when S=1 and T=8

𝑋𝑝 𝑘 =1

𝑇න−𝑇/2

𝑇/2

𝑥𝑝(𝑡) ∙ 𝑒−𝑗

2𝜋𝑇𝑘𝑡 𝑑𝑡

Fourier Representations of Aperiodic Signals

𝑙𝑒𝑡 𝜔 =2𝜋𝑘

𝑇, Define a new function X ω = 𝑇 ∙ 𝑋𝑝 𝑘 = 2

sin(𝜔𝑆)

𝜔

𝑋𝑝 𝑘 =2sin(

2𝜋𝑘𝑇

𝑆)

𝑇(2𝜋𝑘𝑇

)

X ω =2𝜋𝑘

𝑇

If we consider 𝜔 and X ω = 2sin(𝜔𝑆)

𝜔to be continuous, 𝑇𝑋𝑝 𝑘 represents a sampled version of the

function X ω .

As T increases, the resolution in 𝜔increases.

Fourier Representations of Aperiodic SignalsWe can reconstruct 𝑥(𝑡) from X ω using Riemann sums (approximating an integral by a finite sum).

As T → ∞,

• 𝑘𝜔0 =2𝜋𝑘

𝑇becomes a continuum,

2𝜋𝑘

𝑇→ 𝜔 .

• The sum takes the from of an integral,

𝜔0 =2𝜋

𝑇→ 𝑑𝜔

• We obtain a spectrum of coefficients: 𝑋 𝜔 .

Fourier Transform

𝑋𝑝 𝑘 =1

𝑇න−𝑇/2

𝑇/2

𝑥(𝑡) ∙ 𝑒−𝑗2𝜋𝑇 𝑘𝑡 𝑑𝑡Since X ω = 𝑇 ∙ 𝑋𝑝 𝑘

𝑋 𝜔 = න−∞

∞

𝑥(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

Continuous-Time Fourier Representations

Continuous-Time Fourier Transform

Synthesis equation

Analysis equation𝑋 𝜔 = න−∞

∞


𝑥 𝑡 =1

2𝜋න−∞

∞

𝑋(𝜔) ∙ 𝑒𝑗𝜔𝑡 𝑑𝜔

Fourier series and transforms are similar: both represent signals by their frequency content.

Continuous-Time Fourier Series

𝜔0 =2𝜋

𝑇

Synthesis equation

Analysis equation𝑋 𝑘 =1

𝑇න𝑇

𝑥(𝑡)𝑒−𝑗2𝜋𝑇 𝑘𝑡𝑑𝑡

𝑥 𝑡 = 𝑥 𝑡 + 𝑇 =

𝑘=−∞

∞

𝑋[𝑘]𝑒𝑗2𝜋𝑘𝑡𝑇



Synthesis equation


∞


𝑥 𝑡 =1

2𝜋න−∞

∞


Periodic signals can be synthesized from a discrete set of harmonics. Aperiodic signals generally require all possible frequencies.


𝜔0 =2𝜋

𝑇

Synthesis equation


𝑇න𝑇


𝑥 𝑡 = 𝑥 𝑡 + 𝑇 =

𝑘=−∞

∞




Synthesis equation


∞


𝑥 𝑡 =1

2𝜋න−∞

∞


All of the information in a periodic signal is contained in one period. The information in an aperiodic signal is spread across all time.


𝜔0 =2𝜋

𝑇

Synthesis equation


𝑇න𝑇


𝑥 𝑡 = 𝑥 𝑡 + 𝑇 =

𝑘=−∞

∞




Synthesis equation


∞


𝑥 𝑡 =1

2𝜋න−∞

∞


Harmonic frequencies 𝑘𝜔0 are samples of continuous frequency ω


𝜔0 =2𝜋

𝑇

Synthesis equation


𝑇න𝑇


𝑥 𝑡 = 𝑥 𝑡 + 𝑇 =

𝑘=−∞

∞

𝑋[𝑘]𝑒𝑗2𝜋𝑘𝑇 𝑡

Fourier Transform of a Rectangular Pulse

𝑋 𝜔 = න−∞

∞

𝑥(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡 = න−1

1

1 ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡 = อ𝑒−𝑗𝜔𝑡

−𝑗𝜔−1

1

= 2sin(𝜔)

𝜔


By definition, the value of 𝑋 𝜔 = 0 is the integral of 𝑥 𝑡 over all time

𝑋 𝜔


By definition, the value of 𝑥 𝑡 = 0 is the integral of 𝑋 𝜔 over all frequencies, divided by 2π

𝑥 𝑡𝑋 𝜔

Check yourself!

𝑋2 𝜔 = න−∞

∞

𝑥2(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡 = න

−2

2

1 ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡 = อ𝑒−𝑗𝜔𝑡

−𝑗𝜔−2

2

= 2sin(2𝜔)

𝜔=4sin(2𝜔)

2𝜔

= 2sin(𝜔𝑆)

𝜔, (S=2)

Stretching In Time

Stretching in time compresses in frequency.

How would 𝑋 𝜔 scale if time were stretched?

Compressing Time to the LimitAlternatively, compress time while keeping area = 1:

In the limit, the pulse has zero width but area 1! We represent this limit with the delta function: δ(t).

Math With Impulses

δ(t) only has a nonzero value at t = 0, but it has finite area: it is most easily described as an integral:

Importantly, it has the following property (the “sifting property”):

𝑙𝑒𝑡 𝜏 = 𝑡 − 𝑎, ∞−∞

𝛿 𝜏 𝑓 𝜏 + 𝑎 𝑑𝜏 = −00+ 𝛿 𝜏 𝑓 𝑎 𝑑𝜏 = 𝑓(𝑎) ∙ −0

0+ 𝛿 𝜏 𝑑𝜏 = 𝑓(𝑎)

The Fourier Transform of δ(t):

Although physically unrealizable, the impulse (a.k.a. Dirac delta) function δ(t) is useful as a mathematically tractable approximation to a very brief signal.

𝑋 𝜔 = න−∞

∞

𝛿(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡 = න0−

0+

𝛿(𝑡) ∙ 𝑒−𝑗𝜔0 𝑑𝑡 = 1

𝛿(𝑡) 𝑋 𝜔

න−∞

∞

𝛿(𝑡 − 𝑎) 𝑑𝑡 = න𝑎−

𝑎+

𝛿(𝑡) 𝑑𝑡 = 1

Math With ImpulsesFind the function whose Fourier transform is a unit impulse.

𝑥 𝑡 =1

2𝜋න−∞

∞

𝛿(𝜔) ∙ 𝑒𝑗𝜔𝑡 𝑑𝜔 =1

2𝜋න0−

0+

𝛿(𝜔) ∙ 𝑒𝑗0𝑡 𝑑𝜔 =1

2𝜋

Notice the similarity to the previous result:

These relations are duals of each other: • A constant in time consists of a single frequency at ω = 0. • An impulse in time contains components at all frequencies.

DualityThe continuous-time Fourier transform and its inverse are symmetric except for the minus sign in the exponential and the factor of 2π.

𝑥 𝑡 =1

2𝜋න−∞

∞


𝑋 𝜔 = න−∞

∞

𝑥(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡 𝑥 𝑡 =1

2𝜋න−∞

∞


If 𝑥 𝑡𝐶𝑇𝐹𝑇

𝑋 𝜔 then 𝑋 𝑡𝐶𝑇𝐹𝑇

2𝜋𝑥 −𝜔

How do we show this?

FT synthesis

Swapping 𝜔 and 𝑡 𝑥 𝜔 =1

2𝜋න−∞

∞

𝑋(𝑡) ∙ 𝑒𝑗𝜔𝑡 𝑑𝑡

Change sign of 𝜔 𝑥 −𝜔 =1

2𝜋න−∞

∞

𝑋(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

Multiply by 2π 2𝜋𝑥 −𝜔 = න−∞

∞

𝑋(𝑡) ∙ 𝑒−𝑗𝜔𝑡 𝑑𝑡

DualityDuality can be used to simplify math for transform pairs.

Summary

• Continuous Time Fourier Transform: Fourier representation to all CT signals!

• Very useful signals:• Rectangular pulse and its Fourier Transform (sinc)

• Delta function (Unit impulse) and its Fourier Transform

Synthesis equation


∞


𝑥 𝑡 =1

2𝜋න−∞

∞


𝛿(𝑡) 𝑋 𝜔

• Duality:• Make it easier to find the FT of new signals

If 𝑥 𝑡𝐶𝑇𝐹𝑇

𝑋 𝜔 then 𝑋 𝑡𝐶𝑇𝐹𝑇

2𝜋𝑥 −𝜔

6.003 Signal Processing · All of the information in a periodic signal is contained in one period....

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Transcript of 6.003 Signal Processing · All of the information in a periodic signal is contained in one period....