Lecture 8
ELE 301: Signals and Systems
Prof. Paul Cuff
Princeton University
Fall 2011-12
Cuff (Lecture 7)
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Fall 2011-12
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Properties of the Fourier Transform
Properties of the Fourier Transform
I
I
I
I
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Linearity
Time-shift
Time Scaling
Conjugation
Duality
Parseval
Convolution and Modulation
Periodic Signals
Constant-Coefficient Differential Equations
Cuff (Lecture 7)
ELE 301: Signals and Systems
Linearity
Linear combination of two signals x1 (t) and x2 (t) is a signal of the form
ax1 (t) + bx2 (t).
Linearity Theorem: The Fourier transform is linear; that is, given two
signals x1 (t) and x2 (t) and two complex numbers a and b, then
ax1 (t) + bx2 (t) ⇔ aX1 (jω) + bX2 (jω).
This follows from linearity of integrals:
Z ∞
(ax1 (t) + bx2 (t))e −j2πft dt
−∞
Z ∞
Z
= a
x1 (t)e −j2πft dt + b
−∞
∞
x2 (t)e −j2πft dt
−∞
= aX1 (f ) + bX2 (f )
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Fall 2011-12
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Finite Sums
This easily extends to finite combinations. Given signals xk (t) with Fourier
transforms Xk (f ) and complex constants ak , k = 1, 2, . . . K , then
K
X
k=1
ak xk (t) ⇔
K
X
ak Xk (f ).
k=1
If you consider a system which has a signal x(t) as its input and the
Fourier transform X (f ) as its output, the system is linear!
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Fall 2011-12
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Linearity Example
Find the Fourier transform of the signal
1 1
2
2 ≤ |t| < 1
x(t) =
1
|t| ≤ 12
This signal can be recognized as
x(t) =
t 1
1
rect
+ rect (t)
2
2
2
and hence from linearity we have
1
1
1
2 sinc(2f ) + sinc(f ) = sinc(2f ) + sinc(f )
X (f ) =
2
2
2
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ELE 301: Signals and Systems
1.2
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1
1
rect(t/2) + rect(t)
2
2
1
0.8
0.6
0.4
0.2
0
!0.2
!2.5
!2
−2
!1.5
!1
−1
!0.5
00
0.5
11
2
sinc(ω/π) +
1.5
1.5
22
2.5
1
sinc(ω/(2π))
2
1
0.5
0
!0.5
!10
!8
−4π
!6
Cuff (Lecture 7)
!4
−2π
!2
0
0
ω
2
4
2π
6
8
4π
10
L
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Scaling Theorem
Stretch (Scaling) Theorem: Given a transform pair x(t) ⇔ X (f ), and a
real-valued nonzero constant a,
f
1
X
x(at) ⇔
|a|
a
Proof: Here consider only a > 0. (negative a left as an exercise) Change
variables τ = at
Z ∞
Z ∞
dτ
f
1
x(at)e −j2πft dt =
x(τ )e −j2πf τ /a
= X
.
a
a
a
−∞
−∞
If a = −1 ⇒ “time reversal theorem:”
X (−t) ⇔ X (−f )
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Fall 2011-12
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Scaling Examples
We have already seen that rect(t/T ) ⇔ T sinc(Tf ) by brute force
integration. The scaling theorem provides a shortcut proof given the
simpler result rect(t) ⇔ sinc(f ).
This is a good point to illustrate a property of transform pairs. Consider
this Fourier transform pair for a small T and large T , say T = 1 and
T = 5. The resulting transform pairs are shown below to a common
horizontal scale:
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Compress in time - Expand in frequency
1.2
6
rect(t)
1
sinc(ω/2π)
4
0.8
0.6
2
0.4
0.2
0
0
!0.2
!20
−10
!10
−5
0
0
t
10
5
20
10
!2
!10
!5
−10π
−5π
rect(t/5)
1
5
5π
10
10π
4
0.8
3
0.6
2
0.4
1
0.2
0
0
!1
−10
0
5
1.2
!0.2
!20
0
ω
!10
−5
00
t
Cuff (Lecture 7)
10
5
1020
!2
!10
!5
−10π −5π
ELE 301: Signals and Systems
Narrower pulse means higher bandwidth.
5sinc(5ω/2π)
5
10
5π
10π
00
ω
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Scaling Example 2
As another example, find the transform of the time-reversed exponential
x(t) = e at u(−t).
This is the exponential signal y (t) = e −at u(t) with time scaled by -1, so
the Fourier transform is
X (f ) = Y (−f ) =
Cuff (Lecture 7)
1
.
a − j2πf
ELE 301: Signals and Systems
Fall 2011-12
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Scaling Example 3
As a final example which brings two Fourier theorems into use, find the
transform of
x(t) = e −a|t| .
This signal can be written as e −at u(t) + e at u(−t). Linearity and
time-reversal yield
X (f ) =
=
=
1
1
+
a + j2πf
a − j2πf
2a
a2 − (j2πf )2
2a
a2 + (2πf )2
Much easier than direct integration!
Cuff (Lecture 7)
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Complex Conjugation Theorem
Complex Conjugation Theorem: If x(t) ⇔ X (f ), then
x ∗ (t) ⇔ X ∗ (−f )
Proof: The Fourier transform of x ∗ (t) is
Z ∞
∗
Z ∞
x ∗ (t)e −j2πft dt =
x(t)e j2πft dt
−∞
−∞
Z ∞
∗
=
x(t)e −(−j2πf )t dt
= X ∗ (−f )
−∞
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Fall 2011-12
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Duality Theorem
We discussed duality in a previous lecture.
Duality Theorem: If x(t) ⇔ X (f ), then X (t) ⇔ x(−f ).
This result effectively gives us two transform pairs for every transform we
find.
Exercise What signal x(t) has a Fourier transform e −|f | ?
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Shift Theorem
The Shift Theorem:
x(t − τ ) ⇔ e −j2πf τ X (f )
Proof:
Cuff (Lecture 7)
ELE 301: Signals and Systems
Example: square pulse
Consider a causal square pulse p(t) = 1 for t ∈ [0, T ) and 0 otherwise.
We can write this as
!
t − T2
p(t) = rect
T
From shift and scaling theorems
P(f ) = Te −jπfT sinc(Tf ).
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The Derivative Theorem
The Derivative Theorem: Given a signal x(t) that is differentiable almost
everywhere with Fourier transform X (f ),
x 0 (t) ⇔ j2πfX (f )
Similarly, if x(t) is n times differentiable, then
d n x(t)
⇔ (j2πf )n X (f )
dt n
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Dual Derivative Formula
There is a dual to the derivative theorem, i.e., a result interchanging the
role of t and f . Multiplying a signal by t is related to differentiating the
spectrum with respect to f .
(−j2πt)x(t) ⇔ X 0 (f )
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The Integral Theorem
Recall that we can represent integration by a convolution with a unit step
Z t
x(τ )dτ = (x ∗ u)(t).
−∞
Using the Fourier transform of the unit step function we can solve for the
Fourier transform of the integral using the convolution theorem,
Z t
F
x(τ )dτ
= F [x(t)] F [u(t)]
−∞
1
1
= X (f )
δ(f ) +
2
j2πf
X (0)
X (f )
=
δ(f ) +
.
2
j2πf
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Fall 2011-12
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Fourier Transform of the Unit Step Function
How do we know the derivative of the unit step function?
The unit step function does not converge under the Fourier transform.
But just as we use the delta function to accommodate periodic signals, we
can handle the unit step function with some sleight-of-hand.
Use the approximation that u(t) ≈ e −at u(t) for small a.
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A symmetric construction for approximating u(t)
Example: Find the Fourier transform of the signum or sign signal

 1 t>0
0 t=0 .
f (t) = sgn(t) =

−1 t < 0
We can approximate f (t) by the signal
fa (t) = e −at u(t) − e at u(−t)
as a → 0.
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This looks like
1.5
sgn(t)
e−t/5
e−t
1
0.5
0
!0.5
!1
!1.5
!2
!1.5
!1
!0.5
0
0.5
1
1.5
2
t
As a → 0, fa (t) → sgn(t).
The Fourier transform of fa (t) is
Fa (f ) = F [fa (t)]
= F e −at u(t) − e at u(−t)
−at
at
= F e u(t) − F e u(−t)
1
1
=
−
a + j2πf
a − j2πf
−j4πf
=
a2 + (2πf )2
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Therefore,
lim Fa (f ) =
a→0
=
=
−j4πf
a2 + (2πf )2
−j4πf
(2πf )2
1
.
jπf
lim
a→0
This suggests we define the Fourier transform of sgn(t) as
2
f 6= 0
j2πf
sgn(t) ⇔
.
0
f =0
Cuff (Lecture 7)
ELE 301: Signals and Systems
With this, we can find the Fourier transform of the unit step,
u(t) =
1 1
+ sgn(t)
2 2
as can be seen from the plots
sgn(t)
1
t
0
u(t)
1
0
−1
t
−1
The Fourier transform of the unit step is then
1 1
F [u(t)] = F
+ sgn(t)
2 2
1
1
1
=
δ(f ) +
.
2
2 jπf
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The transform pair is then
1
1
u(t) ⇔ δ(f ) +
.
2
j2πf
πδ(ω) +
1
jω
π
1
jω
Cuff (Lecture 7)
ELE 301: Signals and Systems
ω
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Parseval’s Theorem
(Parseval proved for Fourier series, Rayleigh for Fourier transforms. Also
called Plancherel’s theorem)
Recall signal energy of x(t) is
Z
∞
|x(t)|2 dt
Ex =
−∞
Interpretation: energy dissipated in a one ohm resistor if x(t) is a voltage.
Can also be viewed as a measure of the size of a signal.
Theorem:
Z
∞
Ex =
|x(t)|2 dt =
Z
−∞
Cuff (Lecture 7)
∞
|X (f )|2 df
−∞
ELE 301: Signals and Systems
Fall 2011-12
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Example of Parseval’s Theorem
Parseval’s theorem provides many simple integral evaluations. For
example, evaluate
Z ∞
sinc2 (t) dt
−∞
We have seen that sinc(t) ⇔ rect(f ).
Parseval’s theorem yields
Z ∞
Z
sinc2 (t) dt =
−∞
∞
rect2 (f ) df
−∞
Z 1/2
=
1 df
−1/2
= 1.
Try to evaluate this integral directly and you will appreciate Parseval’s
shortcut.
Cuff (Lecture 7)
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? The Convolution Theorem ?
Convolution in the time domain ⇔ multiplication in the frequency domain
This can simplify evaluating convolutions, especially when cascaded.
This is how most simulation programs (e.g., Matlab) compute
convolutions, using the FFT.
The Convolution Theorem: Given two signals x1 (t) and x2 (t) with Fourier
transforms X1 (f ) and X2 (f ),
(x1 ∗ x2 )(t) ⇔ X1 (f )X2 (f )
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Proof: The Fourier transform of (x1 ∗ x2 )(t) is


Z∞ Z∞

x1 (τ )x2 (t − τ ) dτ  e −j2πft dt
−∞
−∞
Z∞
=

Z∞
x1 (τ ) 
−∞

x2 (t − τ )e −j2πft dt  dτ.
−∞
Using the shift theorem, this is
Z∞
=
x1 (τ ) e −j2πf τ X2 (f ) dτ
−∞
Z∞
= X2 (f )
x1 (τ )e −j2πf τ dτ
−∞
= X2 (f )X1 (f ).
Cuff (Lecture 7)
ELE 301: Signals and Systems
Examples of Convolution Theorem
Unit Triangle Signal ∆(t)
1 − |t|
0
if |t| < 1
otherwise.
1
Δ(t)
-1
0
t
1
Easy to show ∆(t) = rect(t) ∗ rect(t).
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Since
rect(t) ⇔ sinc(f )
then
∆(t) ⇔ sinc2 (f )
sinc2(ω/2π)
0.25
1.0
0.2
0.15
0.1
0.05
0
0 !10
!8
−4π
!6
!4
−2π
Cuff (Lecture 7)
!2
0
0
ω
2
4
2π
6
8
4π
ELE 301: Signals and Systems
10
Transform
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Multiplication Property
If x1 (t) ⇔ X1 (f ) and x2 (t) ⇔ X2 (f ),
x1 (t)x2 (t) ⇔ (X1 ∗ X2 )(f ).
This is the dual property of the convolution property.
Note: If ω is used instead of f , then a 1/2π term must be included.
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Multiplication Example - Bandpass Filter
A bandpass filter can be implemented using a low-pass filter and
multiplication by a complex exponential.
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Modulation
The Modulation Theorem: Given a signal x(t) with spectrum x(f ), then
x(t)e j2πf0 t ⇔ X (f − f0 ),
1
(X (f − f0 ) + X (f + f0 )) ,
2
1
x(t) sin(2πf0 t) ⇔
(X (f − f0 ) − X (f + f0 )) .
2j
x(t) cos(2πf0 t) ⇔
Modulating a signal by an exponential shifts the spectrum in the frequency
domain. This is a dual to the shift theorem. It results from interchanging
the roles of t and f .
Modulation by a cosine causes replicas of X (f ) to be placed at plus and
minus the carrier frequency.
Replicas are called sidebands.
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Amplitude Modulation (AM)
Modulation of complex exponential (carrier) by signal x(t):
xm (t) = x(t)e j2πf0 t
Variations:
fc (t) = f (t) cos(ω0 t)
fs (t) = f (t) sin(ω0 t)
fa (t) = A[1 + mf (t)] cos(ω0 t)
I
I
(DSB-SC)
(DSB-SC)
(DSB, commercial AM radio)
m is the modulation index
Typically m and f (t) are chosen so that |mf (t)| < 1 for all t
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Examples of Modulation Theorem
rect(t)
sinc(ω/2π)
1.2
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
!0.2
!2
−2
−1
!1
0
11
0
t
!0.2
!20
!10
−20π −10π
22
00
ω
10
10π
20
20π
0
10
10π
20
20π
1.2
1
1
0.5
0.8
0.6
0
0.4
!0.5
0.2
0
!1
!2
−2
!1
−1
0
1
!0.2
!20
!10
−20π −10π
2
0
1
2
t
rect(t) cos(10πt)
Cuff (Lecture 7)
0
!
!
"ω
"
1
ω − 10π
ω + 10π
1
sinc
+ sinc
2
2π
2
2π
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Periodic Signals
Suppose x(t) is periodic with fundamental period T and frequency
f0 = 1/T . Then the Fourier series representation is,
∞
X
x(t) =
ak e j2πkf0 t .
k=−∞
Let’s substitute in some δ functions using the sifting property:
x(t) =
∞
X
k=−∞
Z
∞
X
−∞
k=−∞
=
δ(f − kf0 )e j2πft df ,
−∞
∞
Z
This implies the Fourier transform:
Cuff (Lecture 7)
∞
ak
!
ak δ(f − kf0 ) e j2πft df .
x(t) ⇔
P∞
ELE 301: Signals and Systems
k=−∞ ak δ(f
− kf0 ).
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Constant-Coefficient Differential Equations
n
X
k=0
ak
M
X
d k y (t)
d k x(t)
=
bk
.
k
dt
dt k
k=0
Find the Fourier Transform of the impulse response (the transfer function
of the system, H(f )) in the frequency domain.
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