6.003: Signal Processing

Continuous-Time Fourier Transform

• Definition

• Examples

• Properties

• Relation to Fourier Series

September 25, 2018

Quiz 1

Thursday, October 4, from 3pm to 5pm.

No lecture on October 4.

The exam is closed book. No electronic devices. You may use one

8.5x11” sheet of notes (front and back).

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

cluding October 3.

Practice problems will be posted by the end of this week.

From Fourier Series to Fourier Transforms

Last week we represented periodic signals as sums of sinusoids.

This representation provides insights that are not obvious from

other representations (e.g., as functions of time).

• consonance and dissonance (lecture 1)

• change in pitch of a siren (lab/recitation 3b)

However, there are limitations.

• only works for periodic signals (lecture 3a)

• must know signal’s period before doing the analysis (pset 3 #7)

Today: avoid these limitation by defining the Fourier transform.

From Fourier Series to Fourier Transforms

How can we represent an aperiodic signal as a sum of sinusoids?

t

x(t)

−S 0 S

1

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

xp(t) =∞∑

i=−∞x(t− iT )

t

xp(t)

−S 0 S−T T

1

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

Take the limit as T → ∞. As xp(t) → x(t), the Fourier series will

approach the Fourier transform.

From Fourier Series to Fourier Transforms

Example:

t

xp(t)

−S 0 S−T T

1

Calculate the Fourier series coefficients Xp[k]:

Xp[k] = 1T

∫Txp(t)e−j

2πT ktdt =

2 sin 2πkT S

T 2πkT

Express Xp[k] in terms of ω = 2πkT :

Xp[k] = 2 sinωSTω

≡ 1TX(ω)

As T →∞,

TXp[k] =∫ T/2

−T/2xp(t)e−jωtdt → X(ω) =

∫ ∞−∞

x(t)e−jωtdt

X(ω) is the Fourier transform of x(t).

Relation Between Fourier Transform and Fourier Series

Fourier series coefficients are samples of continuous function of freq.

TXp[k] =2 sin 2πkS

T2πkT

= 2sinωSω

∣∣∣∣ω= 2π

T k= X(ω)

If S = 2 :

−2π 0 2π

4ω = 2π

T kω

X(ω)

If T = 8 and S = 2 :

−2π 0 2π

4ω = 2π

T k

X(ω) = TXp[k]

If T = 16 and S = 2 :

−2π 0 2π

4ω = 2π

T k

X(ω) = TXp[k]

Relation Between Fourier Transform and Fourier Series

We can reconstuct x(t) from X(ω) using Riemann sums.

xp(t) =∑k

Xp[k]e j2πT kt = 1

2π∑k

TXp[k]ej2πT kt(2π

T)→ 1

2π

∫ ∞−∞

X(ω)ejωtdω

−2π 0 2π

2Sω = 2π

T k

X(ω)

−2π 0 2π

4ω = 2π

T k

X(ω) = TXp[k] 2πT

S = 2;T = 8

−2π 0 2π

4ω = 2π

T k

X(ω) = TXp[k] 2πT

S = 2;T = 16

Fourier Transform relation: x(t) ft⇐⇒ X(ω)

Relation Between Fourier Transform and Fourier Series

Fourier series / transforms express signals by their frequency content.

Continuous-Time Fourier Series

X[k] = 1T

∫Tx(t)e−jkωotdt analysis equation

x(t) = x(t+ T ) =∞∑

k=−∞X[k]e jkωot synthesis equation

where ωo = 2πT

Continuous-Time Fourier Transform

X(ω)=∫ ∞−∞

x(t)e−jωtdt analysis equation

x(t) = 12π

∫ ∞−∞

X(ω)ejωtdω synthesis equation

Examples of Fourier Transforms

Find the Fourier Transform (FT) of a rectangular pulse:

x(t) ={

1 −1 < t < 10 otherwise

t

x(t)

−1 0 1

1

X(ω) =∫ ∞−∞

x(t)e−jωtdt =∫ 1

−1e−jωtdt = e−jωt

−jω

∣∣∣∣1−1

= 2 sinωω

−4π −2π 0 2π 4π

2X(ω)

ω

The FT is a recipe for constructing x(t) from sinusoidal components:

x(t) = 12π

∫ ∞−∞

X(ω)ejωtdω

A square pulse contains almost all frequencies ω.

Examples of Fourier Transforms

The Fourier transform of a rectangular pulse is 2 sinωω

.

t

x(t)

−1 0 1

1 ft⇐⇒

−4π −2π 0 2π 4π

2X(ω)

ω

X(ω) contains all frequencies ω except non-zero multiples of π.

Why is the transform zero at non-zero multiples of π?

What is special about those frequencies?

Why isn’t it zero at ω = 0?

Check Yourself

There are n periods of e−jnπt = cosnπt− j sinnπt in −1 < t < 1.

tx(t)−1 0 1

1

tcosπt

t− sin πt

X(ω = nπ) =∫ 1

−1e−jnπtdt =

{2 if n = 00 otherwise

No Fourier components are needed for frequencies ω = nπ.

However DC is required to offset x(t) so that x(t) ≥ 0 for all t.

Examples of Fourier Transforms

Find the Fourier Transform of a delayed rectangular pulse:

xd(t) ={

1 0 < t < 20 otherwise

t

xd(t)

0 2

1

Xd(ω) =∫ ∞−∞

xd(t)e−jωtdt =∫ 2

0e−jωtdt = e−jωt

−jω

∣∣∣∣20

= 1jω

(1− e−j2ω

)= 1jωe−jω

(ejω − e−jω

)= 2e−jω sinω

ω= e−jωX(ω)

Properties of Fourier Transforms

Time delays map to linear phase delay of the Fourier transform.

If x(t) ft⇐⇒ X(ω)then x(t− τ) ft⇐⇒ e−jωτX(ω)

X(ω) =∫ ∞−∞

x(t)e−jωtdt

Y (ω) =∫ ∞−∞

x(t− τ)e−jωtdt

Let u = t− τ (and therefore du = dt since τ is a constant)

Y (ω) =∫ ∞−∞

x(u)e−jω(u+τ)du = e−jωτ∫ ∞−∞

x(u)e−jωudu = e−jωτX(ω)

Examples of Fourier Transforms

Time delay.

t

x(t)

−1 0 1

1ft⇐⇒

2X(ω)

ω

−4π −2π 2π 4π

4π

−4π

∠X(ω)

ω

t

xd(t)

0 2

1ft⇐⇒

2

∣∣Xd(ω)∣∣

ω

−4π −2π 2π 4π

4π

−4π

∠Xd(ω)

ω

Just enough phase to delay every frequency component by t = 1.

Fourier Transform

Scaling time.

Consider the following signal and its Fourier transform.

Time representation:

−1 1

x1(t)

1

t

Frequency representation:

2

π

X1(ω) = 2 sinωω

ω

How would these scale if time were stretched?

Check Yourself

Signal x2(t) and its Fourier transform X2(ω) are shown below.

−2 2

x2(t)

1

t

b

ω0

X2(ω)

ω

Which of the following is true?

1. b = 2 and ω0 = π/22. b = 2 and ω0 = 2π3. b = 4 and ω0 = π/24. b = 4 and ω0 = 2π5. none of the above

Check Yourself

Find the Fourier transform.

X2(ω) =∫ 2

−2e−jωtdt = e−jωt

−jω

∣∣∣∣2−2

= 2 sin 2ωω

= 4 sin 2ω2ω

4

π/2ω

Check Yourself

Signal x2(t) and its Fourier transform X2(ω) are shown below.

−2 2

x2(t)

1

t

b

ω0

X2(ω)

ω

Which of the following is true? 3

1. b = 2 and ω0 = π/22. b = 2 and ω0 = 2π3. b = 4 and ω0 = π/24. b = 4 and ω0 = 2π5. none of the above

Fourier Transforms

Stretching time compresses frequency.

−1 1

x1(t)

1

t

2

π

X1(ω) = 2 sinωω

ω

−2 2

x2(t)

1

t

4

π/2

X2(ω) = 4 sin 2ω2ω

ω

Fourier Transforms

Find a general scaling rule.

Let x2(t) = x1(at).

X2(ω) =∫ ∞−∞

x2(t)e−jωtdt =∫ ∞−∞

x1(at)e−jωtdt

Let τ = at (a > 0, i.e., not flipped in time).

X2(ω) =∫ ∞−∞

x1(τ)e−jωτ/a 1adτ = 1

aX1(ωa

)If a < 0 the sign of dτ would change along with the limits of integra-

tion. In general,

x1(at) ↔ 1|a|X1(ωa

).

If time is stretched (0 < a < 1) then frequency is compressed and

amplitude increases (preserving area).

Moments

The value of X(ω) at ω = 0 is the integral of x(t) over time t.

X(ω)|ω=0 =∫ ∞−∞

x(t)e−jωtdt =∫ ∞−∞

x(t)e−j0tdt =∫ ∞−∞

x(t) dt

−1 1

x1(t)

1

t

area = 2 2

π

X1(ω) = 2 sinωω

ω

Moments

The value of x(0) is the integral of X(ω) divided by 2π.

x(0) = 12π

∫ ∞−∞

X(ω) e jωtdω = 12π

∫ ∞−∞

X(ω) dω

−1 1

x1(t)

1

t++

−− ++ −−

2

π

X1(ω) = 2 sinωω

ω

area

2π = 1

Moments

The value of x(0) is the integral of X(ω) divided by 2π.

x(0) = 12π

∫ ∞−∞

X(ω) e jωtdω = 12π

∫ ∞−∞

X(ω) dω

−1 1

x1(t)

1

t++

−− ++ −−

2

π

X1(ω) = 2 sinωω

ω

area

2π = 1

2

πω

equal areas !

Stretching Time

Stretching time compresses frequency and increases amplitude

(preserving area).

−1 1

x1(t)

1

t

2

π

X1(ω) = 2 sinωω

ω

−2 2

1

t

4

πω

Compressing Time to the Limit

Alternatively, we could compress time while keeping area = 1.

−12

12

x(t)

1

t

1

2πω

X(ω) = sinω/2ω/2

ω

−14

14

2

t

1

2πω

In the limit, the pulse has zero width but area 1!

We represent this limit with the delta function: δ(t).

1

t

1

ω

Summary: Continuous-Time Fourier Transform (CTFT)

Definition

• analysis and synthesis relations: analogous to CTFS

Examples

• square pulse

Properties

• time delay

• time scaling

• moment relations

Relation to Fourier Series

X(ω) =∞∑

k=−∞2πX[k] δ

(ω − k2π

T

)

Today’s Lab and Recitation

Practice with Fourier Transforms