6.003: Signal Processing

Fourier Transform

Adam Hartz

hz@mit.edu

24 September 2019

Quiz 1

Thursday, 3 October, 7:30-9:30pm, room TBD

No recitation next week, no lecture on Thursday 3 Oct.

The quiz is closed book.

No electronic devices (including calculators).

You may use on 8.5x11” sheet of notes (handwritten or printed,

front and back).

Coverage: lectures, recitations, homeworks, and labs up to and in-

cluding 1 Oct.

Practice materials (old exams) are available from the web site.

Last 2 Weeks: Fourier Series

CTFS:

x(t) = x(t+ T ) =∞∑

k=−∞X[k]ej

2πktT

X[k] = 1T

∫ t0+T

t0x(t)e−j

2πktT dt

Last 2 Weeks: Fourier Series

DTFS:

x[n] = x[n+N ] =k+N−1∑k=k0

X[k]ej2πknN

X[k] = X[k +N ] = 1N

n0+N−1∑n=n0

x[n]e−j2πknN

Today: Fourier Transform

Last two weeks: representing periodic signals as sums of sinusoids.

This representation provides insights that are not obvious from other

representations.

However:

• only works for periodic signals

• must know signal’s period before doing the analysis

This is impractical (or impossible) for a large category of signals,

which we still want to be able to analyze using these methods.

Today

Fourier analysis of aperiodic signals: the Fourier transform

Fourier Transform

Consider the following (aperiodic) function of time:

t

x(t)

−S 0 S

1

Can we represent it as a sum of sinusoids?

Toward the Fourier Transform

Let’s start by considering a related signal xp(·), which we create by

summing shifted copies of x(·):

xp(t) =∞∑

m=−∞x(t−mT )

t

xp(t)

−S 0 S−T T

1

Now we can directly find the Fourier Series coefficients of this new

signal (for arbitrarily-chosen T).

However, maybe that’s not really that helpful, since this signal

doesn’t look much like our original signal x(·). How can we fix that?

Toward the Fourier Transform

t

xp(t)

−S 0 S−T T

1

This signal doesn’t really look much like our original. How can we

fix that?

Toward the Fourier Transform

Consequences of T →∞:

The frequency ω0 associated with k = 1 is defined to be 2πT . As we

increase T , ω0 gets smaller, and the spacing between the coefficients

in terms of rad/sec gets smaller and smaller.

For S = 0.5 and for different values of T :

10 5 0 5 10

0.0

0.2

0.4

0.6

0.8

1.0

X p[k

]

T=1

10 5 0 5 10

0.1

0.0

0.1

0.2

0.3

0.4

0.5

X p[k

]

T=2

10 5 0 5 100.05

0.00

0.05

0.10

0.15

0.20

X p[k

]

T=5

10 5 0 5 10

0.02

0.00

0.02

0.04

0.06

0.08

0.10

X p[k

]

T=10

10 5 0 5 10

0.01

0.00

0.01

0.02

0.03

0.04

0.05

X p[k

]

T=20

10 5 0 5 100.005

0.000

0.005

0.010

0.015

0.020

X p[k

]

T=50

Fourier Series to Fourier Transform

Once we have a periodic signal, we can find the FSC:

Xp[k] = 1T

∫Txp(t)e−jkω0tdt

where ω0 = 2πT .

Now we want to think about T → ∞ Let’s replace 1T with ω0

2π , and

explicitly pick a period to integrate over:

Xp[k] = ω02π

∫ T/2

−T/2xp(t)e−jkω0tdt

Fourier Series to Fourier Transform

Now, substitute into the synthesis equation:

xp(t) =∞∑

k=−∞Xp[k]ejkω0t

=∞∑

k=−∞

ω02π

∫ T/2

−T/2xp(t)e−jkω0tdt

ejkω0t

As we take T →∞, a few things happen:

• xp(t)→ x(t)• w0 becomes an infinitesimally small value, ω0 → dω• kω0 becomes a continuum, kω0 → ω (continuous)

• The bounds of integration approach ∞ and ∞ (respectively)

• The outer sum becomes an integral.

x(t) = 12π

∫ ∞−∞

∫ ∞−∞

x(t)e−jωtdtejωtdω

Fourier Series to Fourier Transform

x(t) = 12π

∫ ∞−∞

∫ ∞−∞

x(t)e−jωtdtejωtdω

From here, we’ll define X(ω) such that:

X(ω) =∫ ∞−∞

x(t)e−jωtdt

x(t) = 12π

∫ ∞−∞

X(ω)ejωtdω

X(·) is the Fourier Transform of x(·).

Very similar to the Fourier series, except:

• x(·) need not be periodic

• x(·) can contain all possible frequencies

Example: FT of a Square Pulse

Find the Fourier Transform of a rectangular pulse:

t

x(t)

−1 0 1

1

Check Yourself!

The FT of a rectangular pulse contains almost all frequencies ω.

t

x(t)

−1 0 1

1

−4π −2π 0 2π 4π

2X(ω)

ω

• Why is X(mπ) = 0?

• What is special about those frequencies?

• Why isn’t it zero at ω = 0?

Check Yourself!

A signal and its Fourier transform are shown below:

−2 2

x2(t)

1

t

b

ω1

X2(ω)

ω

Which of the following is true?

1. b = 2 and ω1 = π/22. b = 2 and ω1 = π

3. b = 4 and ω1 = π/24. b = 4 and ω0 = π

5. none of the above

Recap: CT Fourier Relations

Fourier series/transforms express signals by frequency content.

CTFS

x(t) = x(t+ T ) =∞∑

k=−∞X[k]ej

2πktT

X[k] = 1T

∫ t0+T

t0x(t)e−j

2πktT dt

CTFT

x(t) = 12π

∫ ∞−∞

X(ω)ejωtdω

X(ω) =∫ ∞−∞

x(t)e−jωtdt

Recap: CT Fourier Relations

All information in a periodic signal is contained in one period. Infor-

mation in an aperiodic signal is spread across all time.

CTFS

x(t) = x(t+ T ) =∞∑

k=−∞X[k]ej

2πktT

X[k] = 1T

∫ t0+T

t0x(t)e−j

2πktT dt

CTFT

x(t) = 12π

∫ ∞−∞

X(ω)ejωtdω

X(ω) =∫ ∞−∞

x(t)e−jωtdt

Recap: CT Fourier Relations

Harmonic frequencies kω0 are samples of a continuous frequency ω.

CTFS

x(t) = x(t+ T ) =∞∑

k=−∞X[k]ej

2πktT

X[k] = 1T

∫ t0+T

t0x(t)e−j

2πktT dt

CTFT

x(t) = 12π

∫ ∞−∞

X(ω)ejωtdω

X(ω) =∫ ∞−∞

x(t)e−jωtdt

Recap: CT Fourier Relations

Periodic signals can be synthesized from a discrete set of harmonics.

Aperiodic signals generally require all frequencies.

CTFS

x(t) = x(t+ T ) =∞∑

k=−∞X[k]ej

2πktT

X[k] = 1T

∫ t0+T

t0x(t)e−j

2πktT dt

CTFT

x(t) = 12π

∫ ∞−∞

X(ω)ejωtdω

X(ω) =∫ ∞−∞

x(t)e−jωtdt

DT Fourier Transform

We can apply this same idea to DT signals. If x[·] is an arbitrary DT

signal:

X(Ω) =∞∑

n=−∞x[n]e−jΩn

x[n] = 12π

∫2πX(Ω)ejΩndΩ

DT Sinusoids: Aliasing

Because n is an integer, we can only faithfully represent frequencies

in the base band: 0 ≤ Ω ≤ π.

Frequencies outside that range alias to frequencies in that range.

For example, the following graphs are of cosines with Ω = 0.2π, 2.2π,

and 4.2π, respectively:

0 2 4 6 8 10 12 14

1.00

0.75

0.50

0.25

0.00

0.25

0.50

0.75

1.00

0 2 4 6 8 10 12 14

1.00

0.75

0.50

0.25

0.00

0.25

0.50

0.75

1.00

0 2 4 6 8 10 12 14

1.00

0.75

0.50

0.25

0.00

0.25

0.50

0.75

1.00

They all produce exactly the same samples!

Aliasing and DTFT

DT frequencies alias, which has consequences for DTFT analysis.

In particular, where m is an integer:

X(Ω + 2mπ) =∞∑

n=−∞x[n]ej(Ω+2πm)n

=∞∑

n=−∞x[n]ejΩn ej2πmn︸ ︷︷ ︸

=1

=∞∑

n=−∞x[n]ejΩn

= X(Ω)

X(Ω) is periodic in 2π, so all information is contained one period.

Only need to integrate over 2π.

Example: DTFT of Square Pulse

Find the DTFT of a DT rectangular pulse, which is zero outside

the range shown:

0 2-2

Summary

Today, extended Fourier analysis to aperiodic signals by introducing

CTFT and DTFT:

CTFT

x(t) = 12π

∫ ∞−∞

X(ω)ejωtdω X(ω) =∫ ∞−∞

x(t)e−jωtdt

DTFT

x[n] = 12π

∫2πX(Ω)ejΩndΩ X(Ω) =

∞∑n=−∞

x[n]e−jΩn

Next time, properties of Fourier transforms.