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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
)