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Lecture 8ELE 301: Signals and Systems
Prof. Paul Cuff
Princeton University
Fall 2011-12
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 1 / 37
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Properties of the Fourier Transform
Properties of the Fourier TransformI LinearityI Time-shiftI Time ScalingI ConjugationI DualityI Parseval
Convolution and Modulation
Periodic Signals
Constant-Coefficient Differential Equations
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 2 / 37
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Linearity
Linear combination of two signals x1(t) and x2(t) is a signal of the formax1(t) + bx2(t).
Linearity Theorem: The Fourier transform is linear; that is, given twosignals 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:∫ ∞−∞
(ax1(t) + bx2(t))e−j2πft dt
= a
∫ ∞−∞
x1(t)e−j2πft dt + b
∫ ∞−∞
x2(t)e−j2πft dt
= aX1(f ) + bX2(f )
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 3 / 37
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Finite Sums
This easily extends to finite combinations. Given signals xk(t) with Fouriertransforms Xk(f ) and complex constants ak , k = 1, 2, . . .K , then
K∑k=1
akxk(t)⇔K∑
k=1
akXk(f ).
If you consider a system which has a signal x(t) as its input and theFourier transform X (f ) as its output, the system is linear!
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 4 / 37
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Linearity Example
Find the Fourier transform of the signal
x(t) =
{12
12 ≤ |t| < 1
1 |t| ≤ 12
This signal can be recognized as
x(t) =1
2rect
( t2
)+
1
2rect (t)
and hence from linearity we have
X (f ) =
(1
2
)2 sinc(2f ) +
1
2sinc(f ) = sinc(2f ) +
1
2sinc(f )
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 5 / 37
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!2.5 !2 !1.5 !1 !0.5 0 0.5 1 1.5 2 2.5
!0.2
0
0.2
0.4
0.6
0.8
1
1.2
!10 !8 !6 !4 !2 0 2 4 6 8 10!0.5
0
0.5
1
1.5
2
0 1 2−2 −1
0 2π−2π−4π 4πω
sinc(ω/π)+12sinc(ω/(2π))
12rect(t/2)+
12rect(t)
Linearity Example
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 6 / 37
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Scaling Theorem
Stretch (Scaling) Theorem: Given a transform pair x(t)⇔ X (f ), and areal-valued nonzero constant a,
x(at)⇔ 1|a|
X
(f
a
)
Proof: Here consider only a > 0. (negative a left as an exercise) Changevariables τ = at∫ ∞
−∞x(at)e−j2πft dt =
∫ ∞−∞
x(τ)e−j2πf τ/adτ
a=
1
aX
(f
a
).
If a = −1 ⇒ “time reversal theorem:”
X (−t)⇔ X (−f )
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 7 / 37
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Scaling Examples
We have already seen that rect(t/T )⇔ T sinc(Tf ) by brute forceintegration. The scaling theorem provides a shortcut proof given thesimpler result rect(t)⇔ sinc(f ).
This is a good point to illustrate a property of transform pairs. Considerthis Fourier transform pair for a small T and large T , say T = 1 andT = 5. The resulting transform pairs are shown below to a commonhorizontal scale:
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 8 / 37
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Compress in time - Expand in frequency
!20 !10 0 10 20
!0.2
0
0.2
0.4
0.6
0.8
1
1.2
!20 !10 0 10 20
!0.2
0
0.2
0.4
0.6
0.8
1
1.2
!10 !5 0 5 10!2
0
2
4
6
!10 !5 0 5 10!2
!1
0
1
2
3
4
5
ω
ω
−5π−10π 0 5π 10π
−5π−10π 0 5π 10π
0−5 5−10 10
0−5 5−10 10t
t
sinc(ω/2π)
5sinc(5ω/2π)
rect(t)
rect(t/5)
Narrower pulse means higher bandwidth.Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 9 / 37
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Scaling Example 2
As another example, find the transform of the time-reversed exponential
x(t) = eatu(−t).
This is the exponential signal y(t) = e−atu(t) with time scaled by -1, sothe Fourier transform is
X (f ) = Y (−f ) = 1a− j2πf
.
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 10 / 37
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Scaling Example 3
As a final example which brings two Fourier theorems into use, find thetransform of
x(t) = e−a|t|.
This signal can be written as e−atu(t) + eatu(−t). Linearity andtime-reversal yield
X (f ) =1
a + j2πf+
1
a− j2πf
=2a
a2 − (j2πf )2
=2a
a2 + (2πf )2
Much easier than direct integration!
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 11 / 37
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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∫ ∞−∞
x∗(t)e−j2πft dt =
(∫ ∞−∞
x(t)e j2πft dt
)∗=
(∫ ∞−∞
x(t)e−(−j2πf )t dt
)∗= X ∗(−f )
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 12 / 37
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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 wefind.
Exercise What signal x(t) has a Fourier transform e−|f |?
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 13 / 37
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Shift Theorem
The Shift Theorem:x(t − τ)⇔ e−j2πf τX (f )
Proof:
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 14 / 37
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Example: square pulse
Consider a causal square pulse p(t) = 1 for t ∈ [0,T ) and 0 otherwise.We can write this as
p(t) = rect
(t − T2
T
)From shift and scaling theorems
P(f ) = Te−jπfT sinc(Tf ).
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 15 / 37
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The Derivative Theorem
The Derivative Theorem: Given a signal x(t) that is differentiable almosteverywhere with Fourier transform X (f ),
x ′(t)⇔ j2πfX (f )
Similarly, if x(t) is n times differentiable, then
dnx(t)
dtn⇔ (j2πf )nX (f )
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 16 / 37
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Dual Derivative Formula
There is a dual to the derivative theorem, i.e., a result interchanging therole of t and f . Multiplying a signal by t is related to differentiating thespectrum with respect to f .
(−j2πt)x(t)⇔ X ′(f )
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 17 / 37
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The Integral Theorem
Recall that we can represent integration by a convolution with a unit step∫ t−∞
x(τ)dτ = (x ∗ u)(t).
Using the Fourier transform of the unit step function we can solve for theFourier transform of the integral using the convolution theorem,
F[∫ t−∞
x(τ)dτ
]= F [x(t)]F [u(t)]
= X (f )
(1
2δ(f ) +
1
j2πf
)=
X (0)
2δ(f ) +
X (f )
j2πf.
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 18 / 37
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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, wecan handle the unit step function with some sleight-of-hand.
Use the approximation that u(t) ≈ e−atu(t) for small a.
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 19 / 37
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A symmetric construction for approximating u(t)
Example: Find the Fourier transform of the signum or sign signal
f (t) = sgn(t) =
1 t > 00 t = 0−1 t < 0
.
We can approximate f (t) by the signal
fa(t) = e−atu(t)− eatu(−t)
as a→ 0.
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 20 / 37
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This looks like
!2 !1.5 !1 !0.5 0 0.5 1 1.5 2!1.5
!1
!0.5
0
0.5
1
1.5
t
e−te−t/5sgn(t)
As a→ 0, fa(t)→ sgn(t).
The Fourier transform of fa(t) is
Fa(f ) = F [fa(t)]= F
[e−atu(t)− eatu(−t)
]= F
[e−atu(t)
]−F
[eatu(−t)
]=
1
a + j2πf− 1
a− j2πf
=−j4πf
a2 + (2πf )2
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 21 / 37
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Therefore,
lima→0
Fa(f ) = lima→0
−j4πfa2 + (2πf )2
=−j4πf(2πf )2
=1
jπf.
This suggests we define the Fourier transform of sgn(t) as
sgn(t)⇔{ 2
j2πf f 6= 00 f = 0
.
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 22 / 37
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With this, we can find the Fourier transform of the unit step,
u(t) =1
2+
1
2sgn(t)
as can be seen from the plots
t0
1
−1t0
1
−1
sgn(t) u(t)
The Fourier transform of the unit step is then
F [u(t)] = F[
1
2+
1
2sgn(t)
]=
1
2δ(f ) +
1
2
(1
jπf
).
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 23 / 37
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The transform pair is then
u(t)⇔ 12δ(f ) +
1
j2πf.
1jω
ω
πδ(ω)+1jω π
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 24 / 37
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Parseval’s Theorem
(Parseval proved for Fourier series, Rayleigh for Fourier transforms. Alsocalled Plancherel’s theorem)
Recall signal energy of x(t) is
Ex =∫ ∞−∞|x(t)|2 dt
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:
Ex =∫ ∞−∞|x(t)|2 dt =
∫ ∞−∞|X (f )|2 df
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 25 / 37
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Example of Parseval’s Theorem
Parseval’s theorem provides many simple integral evaluations. Forexample, evaluate ∫ ∞
−∞sinc2(t) dt
We have seen that sinc(t)⇔ rect(f ).
Parseval’s theorem yields∫ ∞−∞
sinc2(t) dt =
∫ ∞−∞
rect2(f ) df
=
∫ 1/2−1/2
1 df
= 1.
Try to evaluate this integral directly and you will appreciate Parseval’sshortcut.
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 26 / 37
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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) computeconvolutions, using the FFT.
The Convolution Theorem: Given two signals x1(t) and x2(t) with Fouriertransforms X1(f ) and X2(f ),
(x1 ∗ x2)(t)⇔ X1(f )X2(f )
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 27 / 37
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Proof: The Fourier transform of (x1 ∗ x2)(t) is
∞∫−∞
∞∫−∞
x1(τ)x2(t − τ) dτ
e−j2πft dt=
∞∫−∞
x1(τ)
∞∫−∞
x2(t − τ)e−j2πft dt
dτ.Using the shift theorem, this is
=
∞∫−∞
x1(τ)(
e−j2πf τX2(f ))
dτ
= X2(f )
∞∫−∞
x1(τ)e−j2πf τ dτ
= X2(f )X1(f ).
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 28 / 37
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Examples of Convolution Theorem
Unit Triangle Signal ∆(t){1− |t| if |t| < 10 otherwise.
-1 10
1
t
Δ(t)
Easy to show ∆(t) = rect(t) ∗ rect(t).
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 29 / 37
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Sincerect(t)⇔ sinc(f )
then∆(t)⇔ sinc2(f )
!10 !8 !6 !4 !2 0 2 4 6 8 100
0.05
0.1
0.15
0.2
0.25
0 4π−4πω
0
1.0
2π−2π
sinc2(ω/2π)
Transform of Unit Triangle Signal
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 30 / 37
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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.
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 31 / 37
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Multiplication Example - Bandpass Filter
A bandpass filter can be implemented using a low-pass filter andmultiplication by a complex exponential.
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 32 / 37
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Modulation
The Modulation Theorem: Given a signal x(t) with spectrum x(f ), then
x(t)e j2πf0t ⇔ X (f − f0),
x(t) cos(2πf0t)⇔1
2(X (f − f0) + X (f + f0)) ,
x(t) sin(2πf0t)⇔1
2j(X (f − f0)− X (f + f0)) .
Modulating a signal by an exponential shifts the spectrum in the frequencydomain. This is a dual to the shift theorem. It results from interchangingthe roles of t and f .
Modulation by a cosine causes replicas of X (f ) to be placed at plus andminus the carrier frequency.
Replicas are called sidebands.
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 33 / 37
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Amplitude Modulation (AM)
Modulation of complex exponential (carrier) by signal x(t):
xm(t) = x(t)ej2πf0t
Variations:
fc(t) = f (t) cos(ω0t) (DSB-SC)fs(t) = f (t) sin(ω0t) (DSB-SC)fa(t) = A[1 + mf (t)] cos(ω0t) (DSB, commercial AM radio)
I m is the modulation indexI Typically m and f (t) are chosen so that |mf (t)| < 1 for all t
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 34 / 37
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Examples of Modulation Theorem
!2 !1 0 1 2
!0.2
0
0.2
0.4
0.6
0.8
1
1.2
!20 !10 0 10 20
!0.2
0
0.2
0.4
0.6
0.8
1
1.2
!2 !1 0 1 2
!1
!0.5
0
0.5
1
!20 !10 0 10 20
!0.2
0
0.2
0.4
0.6
0.8
1
1.2
0−10π 10π−20π 20π
0−10π 10π−20π 20π
0 1−1−2 2
0 1−1−2 2t
t
ω
ω
rect(t) sinc(ω/2π)
rect(t)cos(10πt)12sinc
(ω−10π2π
)+12sinc
(ω+10π2π
)
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 35 / 37
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Periodic Signals
Suppose x(t) is periodic with fundamental period T and frequencyf0 = 1/T . Then the Fourier series representation is,
x(t) =∞∑
k=−∞ake
j2πkf0t .
Let’s substitute in some δ functions using the sifting property:
x(t) =∞∑
k=−∞ak
∫ ∞−∞
δ(f − kf0)e j2πftdf ,
=
∫ ∞−∞
( ∞∑k=−∞
akδ(f − kf0)
)e j2πftdf .
This implies the Fourier transform: x(t)⇔∑∞
k=−∞ akδ(f − kf0).
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 36 / 37
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Constant-Coefficient Differential Equations
n∑k=0
akdky(t)
dtk=
M∑k=0
bkdkx(t)
dtk.
Find the Fourier Transform of the impulse response (the transfer functionof the system, H(f )) in the frequency domain.
Cuff (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 37 / 37