6.003: Signals and Systems
Fourier Transform
November 3, 2011
1
Last Time: Fourier Series
Representing periodic signals as sums of sinusoids.
→ new representations for systems as filters.
Today: generalize for aperiodic signals.
2
Fourier Transform
An aperiodic signal can be thought of as periodic with infinite period.
Let x(t) represent an aperiodic signal.
x(t)
−S
“Periodic extension”: xT (t) =
t
S
∞
0
x(t + kT )
k=−∞
xT (t)
−S
S
Then x(t) = lim xT (t).
T →∞
3
t
T
Fourier Transform
Represent xT (t) by its Fourier series.
xT (t)
−S
ak =
t
S
T
2π
sin 2πkS
2 sin ωS
1 T /2
1 S −j 2π kt
T
=
xT (t)e−j T kt dt =
e T dt =
T ω
T −T /2
πk
T −S
T ak
2 sin ωS
ω
ω = kω0 = k
2π
T
k
ω
ω0 = 2π/T
4
Fourier Transform
Doubling period doubles # of harmonics in given frequency interval.
xT (t)
−S
t
S
T
Z
Z
sin 2πkS
2 sin ωS
1 T /2
1 S −j 2π kt
−j 2π
kt
T
=
ak =
xT (t)e T dt =
e T dt =
T ω
T −T /2
πk
T −S
T ak
2 sin ωS
ω
ω = kω0 = k
2π
T
k
ω
ω0 = 2π/T
5
Fourier Transform
As T → ∞, discrete harmonic amplitudes → a continuum E(ω).
xT (t)
−S
t
S
T
Z
Z
sin 2πkS
2 sin ωS
1 T /2
1 S −j 2π kt
−j 2π
kt
T
=
ak =
xT (t)e T dt =
e T dt =
T ω
T −T /2
πk
T −S
T ak
2 sin ωS
ω
ω = kω0 = k
2π
T
k
ω
ω0 = 2π/T
2
lim T ak = lim
x(t)e−jωt dt = sin ωS = E(ω)
ω
T →∞
T →∞ −T /2
Z T /2
6
Fourier Transform
As T → ∞, synthesis sum → integral.
xT (t)
−S
t
S
T ak
T
2 sin ωS
ω
ω = kω0 = k
2π
T
k
ω
ω0 = 2π/T
2
lim T ak = lim
x(t)e−jωt dt = sin ωS = E(ω)
ω
T →∞
T →∞ −T /2
Z
∞
∞
0 1
0 ω0
1
∞
j 2π
kt
jωt
T
x(t) =
E(ω)
e
=
E(ω)e
→
E(ω)e
jωt dω
T
2π
2π
−∞
k=−∞ � �� �
k=−∞
Z T /2
ak
7
Fourier Transform
Replacing E(ω) by X(jω) yields the Fourier transform relations.
E(ω) = X(jω)
Fourier transform
Z ∞
X(jω)=
x(t)e−jωt dt
(“analysis” equation)
−∞
Z ∞
1
x(t)=
X(jω)ejωt dω
2π −∞
(“synthesis” equation)
Form is similar to that of Fourier series
→ provides alternate view of signal.
8
Relation between Fourier and Laplace Transforms
If the Laplace transform of a signal exists and if the ROC includes the
jω axis, then the Fourier transform is equal to the Laplace transform
evaluated on the jω axis.
Laplace transform:
Z ∞
X(s) =
x(t)e−st dt
−∞
Fourier transform:
Z ∞
X(jω) =
x(t)e−jωt dt = X(s)|s=jω
−∞
9
Relation between Fourier and Laplace Transforms
Fourier transform “inherits” properties of Laplace transform.
Property
x(t)
X(s)
X(jω)
Linearity
ax1 (t) + bx2 (t)
aX1 (s) + bX2 (s)
aX1 (jω) + bX2 (jω)
Time shift
x(t − t0 )
e−st0 X(s)
e−jωt0 X(jω)
Time scale
x(at)
1
s
X
|a|
a
Differentiation
dx(t)
dt
sX(s)
Multiply by t
tx(t)
Convolution
x1 (t) ∗ x2 (t)
−
d
X(s)
ds
X1 (s) × X2 (s)
10
1
X
|a|
jω
a
jωX(jω)
−
1 d
X(jω)
j dω
X1 (jω) × X2 (jω)
Relation between Fourier and Laplace Transforms
There are also important differences.
Compare Fourier and Laplace transforms of x(t) = e−t u(t).
x(t)
t
Laplace transform
Z ∞
Z ∞
−t
−st
X(s) =
e u(t)e dt =
e−(s+1)t dt =
−∞
0
1
; Re(s) > −1
1+s
a complex-valued function of complex domain.
Fourier transform
Z ∞
Z ∞
−t
−jωt
X(jω) =
e u(t)e
dt =
e−(jω+1)t dt =
−∞
0
a complex-valued function of real domain.
11
1
1 + jω
Laplace Transform
The Laplace transform maps a function of time t to a complex-valued
function of complex-valued domain s.
x(t)
t
Magnitude
1 |X(s)| = 1 + s
10
0
Ima 1 0
gin -1
ary
(s)
1
-1 0eal(s)
R
12
Fourier Transform
The Fourier transform maps a function of time t to a complex-valued
function of real-valued domain ω.
x(t)
t
X(j ω) = 1 1 + jω ω
0
1
Frequency plots provide intuition that is difficult to otherwise obtain.
13
Check Yourself
Find the Fourier transform of the following square pulse.
x1 (t)
1
−1
1
t
1. X1 (jω) =
1 ω
e − e−ω
ω
2. X1 (jω) =
1
sin ω
ω
3. X1 (jω) =
2 ω
e − e−ω
ω
4. X1 (jω) =
2
sin ω
ω
5. none of the above
14
Fourier Transform
Compare the Laplace and Fourier transforms of a square pulse.
x1 (t)
1
−1
1
Laplace transform:
Z 1
1
1
e−st
=
e
s − e−s
e
−st dt =
X1 (s) =
s
−s −1
−1
Fourier transform
Z 1
X1 (jω) =
e
−jωt dt =
−1
t
[function of s = σ + jω]
1
2 sin ω
e−jωt
=
−jω
−1
ω
15
[function of ω]
Check Yourself
Find the Fourier transform of the following square pulse. 4
x1 (t)
1
−1
1
t
1. X1 (jω) =
1 ω
e − e−ω
ω
2. X1 (jω) =
1
sin ω
ω
3. X1 (jω) =
2 ω
e − e−ω
ω
4. X1 (jω) =
2
sin ω
ω
5. none of the above
16
Laplace Transform
Laplace transform: complex-valued function of complex domain.
x1 (t)
1
−1
t
1
1 s
−s |X(s)| = (e − e )
s
30
20
10
0
5
5
0
0
-5 -5
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Fourier Transform
The Fourier transform is a function of real domain: frequency ω.
Time representation:
x1 (t)
1
−1
1
t
Frequency representation:
X1 (jω) =
2 sin ω
ω
2
ω
π
18
Check Yourself
Signal x2 (t) and its Fourier transform X2 (jω) are shown below.
x2 (t)
X2 (jω)
1
−2
2
b
ω
t
ω0
Which is true?
1.
2.
3.
4.
5.
b = 2 and ω0
b = 2 and ω0
b = 4 and ω0
b = 4 and ω0
none of the
= π/2
= 2π
= π/2
= 2π
above
19
Check Yourself
Find the Fourier transform.
Z 2
X2 (jω) =
e
−2
−jωt
2
e−jωt 2 sin 2ω
4 sin 2ω
dt =
=
=
−jω −2
ω
2ω
4
ω
π/2
20
Check Yourself
Signal x2 (t) and its Fourier transform X2 (jω) are shown below.
x2 (t)
X2 (jω)
1
−2
2
Which is true?
1.
2.
3.
4.
5.
b
ω
t
ω0
3
b = 2 and ω0
b = 2 and ω0
b = 4 and ω0
b = 4 and ω0
none of the
= π/2
= 2π
= π/2
= 2π
above
21
Fourier Transforms
Stretching time compresses frequency.
X1 (jω) =
x1 (t)
2
1
−1
2 sin ω
ω
ω
t
1
π
X2 (jω) =
4 sin 2ω
2ω
4
x2 (t)
1
−2
2
ω
t
π/2
22
Check Yourself
Stretching time compresses frequency.
Find a general scaling rule.
Let x2 (t) = x1 (at).
If time is stretched in going from x1 to x2 , is a > 1 or a < 1?
23
Check Yourself
Stretching time compresses frequency.
Find a general scaling rule.
Let x2 (t) = x1 (at).
If time is stretched in going from x1 to x2 , is a > 1 or a < 1?
x2 (2) = x1 (1)
x2 (t) = x1 (at)
Therefore a = 1/2, or more generally, a < 1.
24
Check Yourself
Stretching time compresses frequency.
Find a general scaling rule.
Let x2 (t) = x1 (at).
If time is stretched in going from x1 to x2 , is a > 1 or a < 1?
a<1
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Fourier Transforms
Find a general scaling rule.
Let x2 (t) = x1Z(at).
∞
X2 (jω) =
−∞
−jωt
x2 (t)e
Z ∞
dt =
−∞
x1 (at)e−jωt dt
Let τ = at (aZ> 0).
∞
1
jω
−jωτ /a 1
x1 (τ )e
dτ = X1
X2 (jω) =
a
a
a
−∞
If a < 0 the sign of dτ would change along with the limits of integra­
tion. In general, 1
jω
x1 (at) ↔
X1
.
|a|
a
If time is stretched (a < 1) then frequency is compressed and ampli­
tude increases (preserving area).
26
Moments
The value of X(jω) at ω = 0 is the integral of x(t) over time t.
Z ∞
X(jω)|ω=0 =
x(t)e
−jωt
Z ∞
dt =
−∞
x(t)e
Z ∞
dt =
−∞
area = 2
1
1
x(t) dt
−∞
X1 (jω) =
x1 (t)
−1
j0t
2 sin ω
ω
2
ω
t
π
27
Moments
The value of x(0) is the integral of X(jω) divided by 2π.
Z ∞
Z ∞
1
1
jωt
x(0) =
X(jω) e dω =
X(jω) dω
2π −∞
2π −∞
X1 (jω) =
x1 (t)
area
=1
2π
1
2
+
−1
1
t
+
−
28
−
2 sin ω
ω
+
π
−
+
−
ω
Moments
The value of x(0) is the integral of X(jω) divided by 2π.
Z ∞
Z ∞
1
1
x(0) =
X(jω) e jωt dω =
X(jω) dω
2π −∞
2π −∞
X1 (jω) =
x1 (t)
area
=1
2π
1
2
+
−1
1
t
+
−
2 sin ω
ω
+
−
π
−
+
−
ω
equal areas !
2
ω
π
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Stretching to the Limit
Stretching time compresses frequency and increases amplitude
(preserving area).
2 sin ω
X1 (jω) =
x1 (t)
ω
2
1
−1
ω
t
1
4
π
1
−2
2
ω
t
π
1
2π
ω
t
New way to think about an impulse!
30
Fourier Transform
One of the most useful features of the Fourier transform (and Fourier
series) is the simple “inverse” Fourier transform.
Z ∞
X(jω)=
x(t)e−jωt dt
(Fourier transform)
−∞
Z ∞
1
x(t)=
X(jω)ejωt dω
2π −∞
(“inverse” Fourier transform)
31
Inverse Fourier Transform
Find the impulse reponse of an “ideal” low pass filter.
H(jω)
1
ω
ω0
−ω0
ω
Z ∞
Z ω0
1
1 ejωt 0
sin ω0 t
1
jωt
jωt
=
h(t) =
H(jω)e dω =
e dω =
πt
2π −∞
2π −ω0
2π jt −ω0
h(t)
ω0 /π
t
π
ω0
This result is not so easily obtained without inverse relation.
32
Fourier Transform
The Fourier transform and its inverse have very similar forms.
Z ∞
X(jω)=
x(t)e−jωt dt
(Fourier transform)
−∞
Z ∞
1
x(t)=
X(jω)ejωt dω
2π −∞
(“inverse” Fourier transform)
Convert one to the other by
• t→ω
• ω → −t
• scale by 2π
33
Duality
The Fourier transform and its inverse have very similar forms.
Z ∞
X(jω) =
x(t)e−jωt dt
Z−∞
∞
x(t) =
1
X(jω)ejωt dω
2π −∞
Two differences:
•
•
minus sign: flips time axis (or equivalently, frequency axis)
divide by 2π (or multiply in the other direction)
x1 (t) = f (t) ↔ X1 (jω) = g(ω)
t → ω ; flip ; ×2π
ω→t
x2 (t) = g(t) ↔ X2 (jω) = 2πf (−ω)
34
Duality
Using duality to find new transform pairs.
x1 (t) = f (t) ↔ X1 (jω) = g(ω)
t → ω ; flip ; ×2π
ω→t
x2 (t) = g(t) ↔ X2 (jω) = 2πf (−ω)
f (t) = δ(t)
g(ω) = 1
1
↔
ω
t
↓
g(t) = 1
↔
2πf (−ω) = 2πδ(ω)
2π
ω
t
The function g(t) = 1 does not have a Laplace transform!
35
More Impulses
Fourier transform of delayed impulse: δ(t − T ) ↔ e−jωT .
x(t) = δ(t − T )
1
T
t
R∞
X(jω) = −∞ δ(t − T )e−jωt dt = e−jωT
X(jω) = 1
1
ω
∠X(jω) = −ωT
ω
−T
36
Eternal Sinusoids
Using duality to find the Fourier transform of an eternal sinusoid.
↔
δ(t − T )
e−jωT
t → ω ; flip ; ×2π
ω→t
e−jtT
↔
2πδ(ω + T )
e−jω0 t
↔
2πδ(ω + ω0 )
T → ω0 :
x(t) = x(t + T ) =
∞
0
2π kt
T
CTFS
←→
j 2Tπ kt
CTFT
←→
a k ej
k=−∞
x(t) = x(t + T ) =
∞
0
ak e
k=−∞
37
{ak }
∞
0
k=−∞
2πak δ ω −
2π
k
T
Relation between Fourier Transform and Fourier Series
Each term in the Fourier series is replaced by an impulse.
∞
X
x(t) =
xp (t − kT )
k=−∞
···
···
t
0
ak
T
···
···
k
X(jω) =
∞
X
2π ak δ(ω − k
k=−∞
···
2π
)
T
···
ω
0
38
2π
T
Summary
Fourier transform generalizes ideas from Fourier series to aperiodic
signals.
Fourier transform is strikingly similar to Laplace transform
•
•
similar properties (linearity, differentiation, ...)
but has a simple inverse (great for computation!)
Next time – applications (demos) of Fourier transforms
39
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6.003 Signals and Systems
Fall 2011
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