Unformatted text preview: ECSE303A Signals and Systems I
Wednesday, March 03, 2010 LECTURE 17
Section 4.1Continuoustime
Fourier transform
Assignment #2 due Wednesday Chapter 4: The continuousTime Fourier
transform
4.0
Introduction
4.1
The continuoustime Fourier transform
4.2
The Fourier transform for periodic signals
4.3
Properties of the continuoustime Fourier transform
4.4
The convolution property
4.5
The multiplication property
4.6
Tables of Fourier properties and of basic Fourier
transform pairs
4.7
Systems characterized by linear constantcoefficient differential equations 4.1 The ContinuousTime Fourier
Transform
At the limit of a Fourier Series where the periodicity
is infinitely large ⇒ Fourier Transform
To illustrate this, consider the Fourier series
representation of the periodic rectangular wave The spectral coefficients are ak = η sinc ( kη )
ECSE303 Chap. 4
MR  3/3/2010 2t0 ⎫
⎧
⎨η =
⎬
T⎭
⎩ 3 2t0
⎛ 2t0 ⎞
sinc ⎜ k
ak =
⎟
T
T⎠
⎝ T=2*2t0
η=0.5
Let’s increase the period T by a factor of 4… T=8*2t0
η=0.125 ECSE303 Chap. 4
MR  3/3/2010 4 The spectrum of the Fourier coefficients ak could be
plotted as a function of ω=ω0k instead than k alone 2t0
⎛ 2t0 ⎞
ak =
sinc ⎜ k
⎟
T
⎝ T⎠
2t0
⎛ k ω0 t 0 ⎞
sinc ⎜
=
⎟
T
⎝π⎠
2t0
⎛ ω t0 ⎞
sinc ⎜
=
T
π⎟
⎝
⎠
This eliminates the dependence of the sinc on the
variable k
ECSE303 Chap. 4
MR  3/3/2010 5 Now ak as a function of ω 2t0
⎛ ω t0 ⎞
sinc ⎜
ak =
⎟
T
⎝π ⎠ T=2*2t0
η=0.5 Let’s increase the period T by a factor of 4… T=8*2t0
η=0.125 ECSE303 Chap. 4
MR  3/3/2010 The sinc
envelope remain
unchanged along
the horizontal
6
axis And another way to see this solution is to plot Tak as
a function of ω 2t0
⎛ ωt0 ⎞
sinc ⎜
ak =
⎟
T
⎝π ⎠
⎛ ω t0 ⎞
Tak = 2t0sinc ⎜
⎟
⎝π ⎠
Then the sinc envelope becomes independent of the
signal period along both the amplitude and the
frequency axes ECSE303 Chap. 4
MR  3/3/2010 7 Now Tak as a function of ω ⎛ ω t0 ⎞
Tak = 2t0sinc ⎜
⎟
⎝π ⎠ T=2*2t0
η=0.5 Let’s increase the period T by a factor of 4… T=8*2t0
η=0.125 ECSE303 Chap. 4
MR  3/3/2010 8 By expressing the spectral coefficients of the square
wave function with
ωt ⎛ 0⎞
Tak = 2t0sinc ⎜
⎟
⎝π ⎠ Then the spectrum becomes a sinc envelope that
has an amplitude and width that remain unchanged
by T. This envelope is sampled at a frequency
spacing given by the periodicity of the square
function T=8*2t0
η=0.125
ECSE303 Chap. 4
MR  3/3/2010 9 Let’s define a signal x(t) equal to the periodic square
function over only one period and zero elsewhere. x (t ) x (t ) This signal with an infinite fundamental period is in
fact aperiodic ECSE303 Chap. 4
MR  3/3/2010 10 Let's define X(jkω0) to represent the envelope of the
spectral coefficients of x(t)
+∞
x (t ) = X ( jkω0 ) ≡ Tak
+∞ = ∫ k =−∞ x(t )e − jkω0t dt −∞ = t0 ∫ ∑ ak = ak e jkω0t 1
x ( t ) e − jkω0t dt
∫
TT e − jkω0t dt − t0 = 2t0sinc( t0ω π ) For a periodic square function, this spectral
envelope Tak is independent of T. Are we going to
get this for any periodic function? Let’s develop the
expression of the spectral coefficient envelope
X(jkω0) in the general case.
ECSE303 Chap. 4
MR  3/3/2010 11 ~
A periodic signal x ( t ) has the Fourier series
representation
+∞
+∞
1
jkω0t
x ( t ) = ∑ ak e
= ∑ X ( jkω0 ) e jkω0t
k =−∞
k =−∞ T
Or, equivalently with T=2π/ω0
+∞
ω0
x(t ) = ∑
X ( jkω0 )e jkω0t
k =−∞ 2π
In the limit of T⇒∞, we get
+∞
1
k ω0 = ω
x(t ) =
X ( jω )e jω t d ω
2π −∞
ω0 → d ω
It is a Tindependent
→∫
∑
spectrum
x (t ) → x (t )
representation of x(t)
ECSE303 Chap. 4 ∫ MR  3/3/2010 12 We thus have formed a set of two equations which
will be called the Fourier transform pair. 1
x(t ) =
2π
X ( jω ) = +∞ ∫ +∞ ∫ X ( jω )e jωt d ω Inverse FT −∞ x(t )e − jωt dt Fourier transform −∞ ECSE303 Chap. 4
MR  3/3/2010 13 Example
Spectrum of the impulse x (t ) = δ (t )
X ( jω ) = 1
x (t ) =
2π
X ( jω ) = +∞ +∞ ∫ X ( jω )e jωt d ω −∞
+∞ ∫ x (t )e − jωt dt −∞ δ ( t ) e − jωt dt
∫ −∞
+∞ = ∫ δ ( t ) dt −∞ =1 An impulse is made of the contribution of all
frequencies with an amplitude of 1 ECSE303 Chap. 4
MR  3/3/2010 14 Example
Spectrum of the delayed impulse X ( jω ) = x ( t ) = δ ( t − t0 )
X ( jω ) = 1
x (t ) =
2π +∞ ∫ X ( jω )e jωt d ω −∞
+∞ ∫ x (t )e − jωt dt −∞ +∞ δ ( t − t0 ) e − jωt dt
∫ −∞
+∞ = δ ( t − t0 ) e − jωt dt
∫
0 −∞ = e − jωt0 A delayed impulse is made of the contribution of all
frequencies with an amplitude of 1 and varying
phaseshift
ECSE303 Chap. 4
MR  3/3/2010 15 Example
The square function
T1 X ( jω ) = ∫e − jω t 1
x (t ) =
2π
X ( jω ) = dt =
x1 ( t ) FT t ECSE303 Chap. 4
MR  3/3/2010 +∞ ∫ x (t )e − jωt dt ω 2
⎛ ωT ⎞
sin c ⎜ 1 ⎟
T1
⎝π ⎠ T1 −∞ sin ωT1 1 −T1 ∫ X ( jω )e jωt d ω −∞ −T1 =2 +∞ ⇔ X1 ( jω ) 2
T1 Surf. area
ω −π
T1 π
T1 π/halfwidth
16 1
x (t ) =
2π Example
The Fourier transform of x ( t ) = e − at u ( t ) X ( jω ) = +∞ ∫ X ( jω )e jωt d ω −∞
+∞ ∫ x (t )e − jωt dt −∞ {a ≥ 0, real} is given by X ( jω ) = +∞ e − at u ( t ) e − jωt dt
∫ −∞
+∞ = e −( a + jω )t dt =
∫
0 = ECSE303 Chap. 4
MR  3/3/2010 1
jω + a
17 1
X ( jω ) =
jω + a The magnitude of X(jω) is X ( jω )
1/a 1 1
X ( jω ) =
ω 2 + a2 a2 a a ω The 3dB cutoff frequency of this filter is ωc=a ECSE303 Chap. 4
MR  3/3/2010 18 Phase of X(jω) 1
X ( jω ) =
jω + a ∠X ( jω ) ⎛ ω⎞
∠X ( jω ) = a tan ⎜ − ⎟
⎝ a⎠
π/2
π/4 a
−π/4
−π/2 a ω The maximum phase added to the input signal is
±π/2 for ω  >> ωc
ECSE303 Chap. 4
MR  3/3/2010 19 Remarks
For the case where eatu(t) is the impulse response
h(t) of a firstorder differential LTI system:
• The system is a lowpass filter with DC gain of 1/a
• High frequencies in the input signal x(t) are
attenuated
• The 3 dB (or 50% power, or 70.7% amplitude)
attenuation cutoff is ωc=a
• The maximum phase added to the input signal is
±π/2 for ω →±∞ ECSE303 Chap. 4
MR  3/3/2010 20 1
x (t ) =
2π +∞ ∫ X ( jω )e jωt d ω −∞ Example
+∞
X ( jω ) = ∫ x (t )e − jωt dt
The complex exponential
−∞
− at
x2 ( t ) = e u (t ) {a = α + j β , α , β ≥ 0, real}
X 2 ( jω ) = +∞ e − at u ( t ) e− jωt dt
∫ −∞
+∞ = ∫e
0 −(α + j β + jω )t 1
dt =
j (ω + β ) + α Magnitude of X2(jω) X 2 ( jω ) = 1 α 1
2
(ω + β ) + α 2 1 α2 −α−β −β α−β
ECSE303 Chap. 4
MR  3/3/2010 ω 21 Phase of X2(jω)
1
X 2 ( jω ) =
j (ω + β ) + α
⎛ (ω + β ) ⎞
∠X 2 ( jω ) = a tan ⎜ −
⎟
α⎠
⎝ π/2
π/4
α−β α −β
−β −π/4 ω −π/2
ECSE303 Chap. 4
MR  3/3/2010 22 Example
Find the Fourier transform of x3 (t ) = e −α t sin(ω0t )u (t )
−1 − (α + jω0 )t
=
(e
u (t ) − e − (α − jω0 )t u (t ))
2j
We have just seen that the Fourier transform of a
complex exponential is
x2 ( t ) = e − at u (t )
X 2 ( jω ) = ECSE303 Chap. 4
MR  3/3/2010 {a = α + j β , α , β ≥ 0, real} 1
j (ω + β ) + α 23 x2 ( t ) = e − at u (t )
X 2 ( jω ) = by linearity {a = α + j β , α , β ≥ 0, real} 1
j (ω + β ) + α ⎞
1
1
−1 ⎛
X 3 ( jω ) =
−
⎜
⎟
2 j ⎝ j (ω + ω0 ) + α − j (−ω + ω0 ) + α ⎠
⎞
1
1
−1 ⎛
=
−
⎜
⎟
2 j ⎝ j (ω + ω0 ) + α j (ω − ω0 ) + α ⎠
1 ⎛ j (ω + ω0 ) + α − [ j (ω − ω0 ) + α ] ⎞
=
⎜
⎟
2 j ⎝ [ j (ω − ω0 ) + α ][ j (ω + ω0 ) + α ] ⎠ ω0
=
2
( jω + α ) 2 + ω0 ECSE303 Chap. 4
MR  3/3/2010 24 x3 ( t ) = e −α t ω0
sin (ω0t ) u ( t ) ⇔ X 3 ( jω ) =
2
2
jω + α ) + ω0
(
FT  X 3 ( jω )
1
2α ω0
ω2 +α2
0 2
− ω0 − 2 ECSE303 Chap. 4
MR  3/3/2010 2
ω0 − α 2 ω 25 X 3 ( jω ) = ω0
2
( jω + α ) 2 + ω0 ⎡
⎤
2αω
∠X 3 ( jω ) = − atan ⎢ 2
⎥
α + ω02 − ω 2 ⎦
⎣ π
π/2
2
− ω0 + α 2
2
ω0 + α 2 −π/2 ω −π ECSE303 Chap. 4
MR  3/3/2010 26 1
x (t ) =
2π Example
Find the Fourier Transform of the
following signal X ( jω ) = X ( jω ) = ∫ x(t )e − jω t dt 0 = ∫ −T1 T1 −∞
+∞ ∫ x (t )e − jωt dt T1 T1 T1 −∞ ∫ X ( jω )e jωt d ω −∞ x (t )
+∞ +∞ (−T1 − t )e − jω t dt + ∫ (T1 − t )e − jω t dt t T1 0 ⎛ 2sin(ω T1 ) 2T1 ⎞
= j⎜
−
2
ω⎟
ω
⎝
⎠
ECSE303 Chap. 4
MR  3/3/2010 27 Spectrum of the function ⎛ 2sin(ωT1 ) 2T1 ⎞
X ( jω ) = j ⎜
−
2
ω
ω⎟
⎝
⎠ T1=1 sec
X ( jω ) = ∠X ( jω ) = 2sin(ωT1 )
− ω
π
2 π 2 ECSE303 Chap. 4
MR  3/3/2010 2 − ω>0
ω<0 28 2T1 ω Chapter 4: The continuousTime Fourier
transform
4.0
Introduction
4.1
The continuoustime Fourier transform
4.2
The Fourier transform for periodic signals
4.3
Properties of the continuoustime Fourier transform
4.4
The convolution property
4.5
The multiplication property
4.6
Tables of Fourier properties and of basic Fourier
transform pairs
4.7
Systems characterized by linear constantcoefficient differential equations x (t ) = +∞ ∑ k =−∞ ak = ak e jkω0t 1
x ( t ) e − jkω0t dt
∫
TT Question
A periodic signal x(t) is expressed in
Volts.
What are the units of ak? ECSE303 Chap. 3
MR  3/3/2010 30 1
x (t ) =
2π
X ( jω ) = +∞ ∫ X ( jω )e jωt d ω −∞
+∞ ∫ x (t )e − jωt dt −∞ Question
A non periodic signal x(t) is expressed
in Volts.
What are the units of X(jω)? ECSE303 Chap. 4
MR  3/3/2010 31 ak = 1
v(t )e − jkω0t dt
∫
TT v (t ) = = 1
δ (t )e − jkω0t dt
T −T∫/ 2
2T T ∑ δ ( t − nT ) n =−∞ 1 T /2 +∞ T 2T 3T 1
=
The spectrum of an impulse train is a real, constant
T t sequence. ak
1/T 16 8 8 16 k
ECSE303 Chap. 4
MR  3/3/2010 32 4.2 Fourier Transforms for Periodic
Signals
What happens if we use the Fourier transform to
represent periodic signals?
Consider a signal x(t) with Fourier transform that is a
single impulse of area 2π at frequency kω0. This can
be represented as X ( jω ) = 2πδ (ω − kω 0 )
Taking the inverse Fourier transform yields 1
x (t ) =
2π
=e
ECSE303 Chap. 4
MR  3/3/2010 z +∞ 2πδ (ω − kω 0 )e jωt dω −∞ jkω 0t 1
x (t ) =
2π
X ( jω ) = +∞ ∫ X ( jω )e jωt d ω −∞
+∞ ∫ x (t )e − jωt dt −∞ 33 Therefore, a periodic signal for which the Fourier
transform can +∞ expressed as a sum of impulses
be X ( jω ) = ∑ 2πa δ (ω − kω
k 0 ) k =−∞ has the timedomain form x (t ) = +∞ a k e jkω 0t
∑ k =−∞ which in fact is the Fourier series representation of
x(t)
Therefore the Fourier transform of a periodic signal
with Fourier series coefficients ak, is a train of
impulses of surface area 2πak occurring at
frequencies ω=kω0.
ECSE303 Chap. 4
MR  3/3/2010 34 Example
The two impulses in the continuous frequency
domain X ( jω ) = 2π [δ (ω − ω0 ) + δ (ω + ω0 ) ]
In fact represent a periodic cosine in the timedomain
+∞
1
x(t ) =
2π [δ (ω − ω0 ) + δ (ω + ω0 ) ] e jωt d ω
∫
2π −∞
=e jω0t +e − jω0t = 2 cos (ω0t ) Similarly, 1
x (t ) =
2π
X ( jω ) = +∞ ∫ X ( jω )e jωt d ω −∞
+∞ ∫ x (t )e − jωt dt −∞ FT 2sin (ω0t ) ⇔ ( 2π j ) [δ (ω − ω0 ) − δ (ω + ω0 )]
ECSE303 Chap. 4
MR  3/3/2010 35 Fourier Transform of a Periodic Impulse Train
Consider the impulse train signal
+∞ x (t ) = ∑ δ (t − kT )
−∞ x (t ) We know that the spectral
coefficients of the impulse
train are ak =1/T. 11 2T T T 2T 3T t The Fourier spectrum is
then 2π
X ( jω ) =
T X ( jω ) +∞ 2
1π ∑ δ (ω − kω ) k =−∞ 0 −2ω0 −ω0
2T ECSE303 Chap. 4
MR  3/3/2010 T ω0
T 2T 2ω0 3T 3ω0 tω 36 Reading/exercises
Read sections 4.14.2
Problems 4.14.3, 4.21
ECSE303 Chap.4
MR  3/3/2010 42 ECSE303A Signals and Systems I
Wednesday, March 03, 2010 LECTURE 18
Section 4.3Properties of the
continuoustime Fourier
transform x(α t ) ↔ Remark
The Fouriertransform pair
1
x (t ) =
2π
X ( jω ) = +∞ ∫ 1 α X ( jω / α ) X ( jω )e jωt d ω −∞
+∞ ∫ x (t )e − jωt dt −∞ Can also be rewritten as
+∞ x (t ) = ∫ X ( jf )e j 2π ft df −∞
+∞ X ( jf ) = ∫ x (t )e − j 2π ft dt −∞
ECSE303 Chap. 4
MR  3/3/2010 44 Remark
The transform
X ( jω ) e jω0t FT ↔ 2πδ (ω − ω0 ) 2π ω ω0 Can in fact be rewritten as
X ( jω ) e jω0t FT ↔ δ ( f − f0 ) 1
f
f0 ECSE303 Chap. 4
MR  3/3/2010 45 Chapter 4: The continuousTime Fourier
transform
4.0
Introduction
4.1
The continuoustime Fourier transform
4.2
The Fourier transform for periodic signals
4.3
Properties of the continuoustime Fourier transform
4.4
The convolution property
4.5
The multiplication property
4.6
Tables of Fourier properties and of basic Fourier
transform pairs
4.7
Systems characterized by linear constantcoefficient differential equations 4.3 Properties of the Fourier Transform
We denote the relationship between a signal and its
Fourier transform as
FT x(t ) ↔ X ( jω )
The Fourier transform has the following properties Linearity
The Fourier transform is a linear operation
FT ax(t ) + by (t ) ↔ aX ( jω ) + bY ( jω )
ECSE303 Chap. 4
MR  3/3/2010 47 Time Shifting
A time shift in the time domain turns into a phase
shift in the frequency domain
FT x (t ) ↔ X ( jω )
FT x (t − t0 ) ↔ e − jωt0 X ( jω ) ECSE303 Chap. 4
MR  3/3/2010 48 1
x (t ) =
2π Example
Spectrum of the delayed impulse X ( jω ) = x (t ) x ( t ) = δ ( t − t0 ) +∞ ∫ X ( jω )e jω t dω −∞
+∞ ∫ x (t )e − jωt dt −∞ 1
t
t0 X ( jω ) = +∞ δ ( t − t0 ) e − jωt dt
∫ −∞ 1 +∞ = δ ( t − t0 ) e − jωt dt
∫
0 X ( jω ) ∠X ( jω )
ω ω −∞ = e − jωt0 A delayed impulse is made of the contribution of all
frequencies with an amplitude of 1 and varying
phaseshift
ECSE303 Chap. 4
MR  3/3/2010 49 Time/Frequency Scaling
Scaling the time variable either expands or
compresses the spectrum
FT x(t ) ↔ X ( jω )
FT x(α t ) ↔ 1 α X ( jω / α ) For α>1, the signal x(αt) is sped up (or compressed
in time), so its spectrum expands in the frequency
domain.
1
e u (t ) ↔
jω + a
FT
1
at
e u (−t ) ↔
jω − a
− at Example
eatu(t) and eatu(t), a>0
ECSE303 Chap. 4
MR  3/3/2010 FT 50 Example
The square function
T1 X ( jω ) = ∫ e − jωt dt =2 1
x (t ) =
2π =
x1 ( t ) −∞
+∞ ∫ ω 2
⎛ ωT ⎞
sin c ⎜ 1 ⎟
T1
⎝π ⎠ FT t T1 x (t )e − jωt dt −∞ 1 ECSE303 Chap. 4
MR  3/3/2010 ∫ X ( jω )e jωt d ω sin ωT1 X ( jω ) = −T1 −T1 +∞ ↔ X1 ( jω ) 2T1 Surf. area
ω −π
T1 π
T1 π/halfwidth
51 Conjugation and Conjugate Symmetry
Taking the general case of a complex signal, the
conjugate of the signal has a Fourier transform
FT x(t ) ↔ X ( jω )
FT x (t ) ↔ X ∗ (− jω )
∗ Proof ⎡∞
⎤
∗
− jωt
[ X ( jω )] = ⎢ ∫ x ( t ) e dt ⎥
⎣ −∞
⎦
X ∗ ( jω ) = ∞ ∫ ∗ x∗ ( t ) e jωt dt −∞ X ∗ (− jω ) =
ECSE303 Chap. 4
MR  3/3/2010 ∞ ∫ −∞ x∗ ( t ) e − jωt dt
52 FT x(t ) ↔ X ( jω )
Then for a real signal with x(t) = x*(t),
Considering a cartesian notation, FT x (t ) ↔ X ∗ (− jω )
∗ X ( jω ) = Re { X ( jω )} + j Im { X ( jω )} the spectrum takes the form of:
an even function of ω (represented by cosines), Re{ X ( jω )} = Re{ X ( − jω )} or an odd function of ω (represented by sinuses), Im{ X ( jω )} = − Im{ X ( − jω )} Considering a phasor notation X ( jω ) = X ( jω ) e j∠X ( jω )
The magnitude is an even function of ω  X ( jω ) = X ( − jω )
The phase, an odd even function of ω ∠X ( jω ) = −∠X ( − jω ) ECSE303 Chap. 4
MR  3/3/2010 53 FT x(t ) ↔ X ( jω )
FT x (t ) ↔ X ∗ (− jω )
∗ If x(t) is real and even (combination of cosines), then
the spectrum is real and even X ( jω ) = X ( − jω )
If x(t) is real and odd (sum of sinus), then the
spectrum is purely imaginary and odd X ( jω ) = − X ( − jω ) = − X ∗ ( jω ) ECSE303 Chap. 4
MR  3/3/2010 54 Differentiation
Differentiating a signal in the time domain is
equivalent to a multiplication of the Fourier transform
by jω
FT x(t ) ↔ X ( jω )
FT
d
x (t ) ↔ jω X ( jω )
dt 1
x (t ) =
2π
X ( jω ) = +∞ ∫ X ( jω )e jωt d ω −∞
+∞ ∫ x (t )e − jωt dt −∞ ECSE303 Chap. 4
MR  3/3/2010 55 Integration
Integrating a signal in the time domain is equivalent
to a division of the Fourier transform by jω.
Also, an infinite DC term πX(0)δ(ω) must be added to
the Fourier transform.
FT x(t ) ↔ X ( jω )
t 1
∫ x(τ )dτ ↔ jω X ( jω ) + π X (0)δ (ω )
−∞
FT 1
x (t ) =
2π
X ( jω ) = +∞ ∫ X ( jω )e jωt d ω −∞
+∞ ∫ x (t )e − jωt dt −∞ ECSE303 Chap. 4
MR  3/3/2010 56 t 1
∫ x(τ )dτ ↔ jω X ( jω ) + π X (0)δ (ω )
−∞
FT Example
Find the Fourier transform of the running integral of
x(t)=δ(t)
FT δ (t ) ↔ 1 t 1
u (t ) ↔
+ πδ (ω )
jω
FT 1
∫ δ (τ )dτ ↔ jω + πδ (ω )
−∞
FT Note that FT{u(t)} cannot be found from the Fourier
transform integral alone because
X ( jω ) = +∞ ∫ e − jωt dt = undefined 0 u(t) is not integrable. Even then, the spectrum found
for u(t) is often used.
ECSE303 Chap. 4
MR  3/3/2010 57 FT u (t ) ↔
t 1
+ πδ (ω )
jω 1
X ( jω ) + π X (0)δ (ω )
Example
∫
jω
Find the Fourier transform of the running integral of
x(t)=u(t)u(tt0)
FT x(τ )dτ ↔ −∞ FT u (t ) − u (t − t0 ) ↔ ⎡1
⎤
1
+ πδ (ω ) − e − jωt0 ⎢
+ πδ (ω ) ⎥
jω
⎣ jω
⎦ (1 − e
u (t ) − u (t − t ) ↔
FT − jωt0 ) jω 0 FT u (t ) − u (t − t0 ) ↔ e − jωt0 2 sin (ωt0 2 ) ω2 FT u (t ) − u (t − t0 ) ↔ t0 e − jωt0 2 sinc (ωt0 2π )
t FT ∫ u (τ ) − u (τ − t0 )dτ ↔ −∞ ECSE303 Chap. 4
MR  3/3/2010 1
⎡t0 e − jωt0 2 sinc (ωt0 2π ) ⎤ + π t0δ (ω )
⎦
jω ⎣
59 Parseval's Relation
Just like the total average power of a periodic signal
is equal to the sum of the powers of all its
harmonics,
∞
2
1
P = ∫ x ( t ) dt = ∑ ak
TT
k =−∞ 2 the total energy in an aperiodic signal is equal to the
total energy in its spectrum.
+∞ ∫ x (t ) −∞
ECSE303 Chap. 4
MR  3/3/2010 2 1
dt =
2π +∞ ∫ X ( jω ) 2 dω −∞ 60 +∞ ∫ x (t ) Example
Find the energy contained in 2 −∞ 1
dt =
2π +∞ ∫ X ( jω ) 2 −∞ 1
e u (t ) ↔
jω + a
FT − at In time
+∞ ∫ x ( t ) dt =
2 −∞ +∞ ∫ e − at u ( t ) dt −∞
+∞ = e −2 at dt
∫
0 1
=
2a
ECSE303 Chap. 4
MR  3/3/2010 In frequency
2 1
2π +∞ ∫ X ( jω ) −∞ 2 1
dω =
2π
1
=
2π
1
=
2π
= 1
2a +∞ ∫ −∞ 2 1
dω
jω + a +∞ 1
∫ ω 2 + a 2 dω
−∞
+∞ ⎡1
⎛ ω ⎞⎤
⎢ a atan ⎜ a ⎟ ⎥
⎝ ⎠ ⎦ −∞
⎣
61 dω Energydensity spectrum
The energy density spectrum of an aperiodic signal
is defined as the magnitude squared of its spectrum,
i.e.,
2 X ( jω ) We can find the energy of a signal in a given
frequency band by integrating E[ωa ,ωb ] ECSE303 Chap. 4
MR  3/3/2010 1
=
2π ωb ∫
ω X ( jω ) dω
2 a 62 Note that it is customary to include the negative
frequency band as well.
For example, to estimate the energy contained in a
signal between, say, f=5 kHz and f=10 kHz E[10000π ,20000π ] −10000π
⎡ 20000π
⎤
1
2
2
=
⎢ ∫ X ( jω ) d ω + ∫ X ( jω ) dω ⎥
2π ⎣10000π
−20000π
⎦ And if the signal is real  X ( jω ) = X ( − jω ) E[10000π ,20000π ] = ECSE303 Chap. 4
MR  3/3/2010 1 π 20000π ∫π X ( jω ) d ω
2 10000 63 Duality
The Fourier transform pair is symmetric. This results
in a duality between the time domain and the
frequency domain. For example,
x1 ( t ) X1 ( jω ) FT x1 ( t ) ↔ X 1 ( jω ) 1 2T1 Surf. area t −T1 ω
−π
T1 T1 ωc
π x2 ( t ) π/halfwidth T1 X 2 ( jω ) FT x2 ( t ) ↔ X 2 ( jω ) π 1 ω t −π
ECSE303 Chap. 4
MR  3/3/2010 ωc π
ωc −ωc ωc
64 Example
Find the frequency derivative of X(jω)=1/(jω+a).
What does it represents in the time domain?
We know that e − at 1
u (t ) ↔
jω + a
FT From the definition of the Fourier transform
X ( jω ) = +∞ ∫ x (t )e − jωt dt ⇒ −∞ +∞ dX ( jω )
= ∫ − jtx (t )e − jωt dt
dω
−∞ Taking the derivative of X(jω) thus leads to
⎛1⎞
− jte u ( t ) ↔− j ⎜
⎟
jω + a ⎠
⎝
− at ECSE303 Chap. 4
MR  3/3/2010 FT ⎛1⎞
te u ( t ) ↔ ⎜
⎟
jω + a ⎠
⎝
− at FT 2 2 First derivative in
frequency is similar to
a multiplication by –jt
in the time domain
65 Example
The impulse response of a system for which
H(jω)=eaω {Re(a)>0} is
1
h (t ) = 1
2π +∞ ∫ e − aω e jωt d ω + 0 1
2π 0 ∫ x (t ) = e aω e jωt d ω −∞ +∞ 0 2π X ( jω ) = ⎡ eω ( jt − a ) ⎤
1 ⎡ eω ( jt + a ) ⎤
⎢
⎥+
⎢
⎥
2π ⎣ jt + a ⎦ −∞
jt − a ⎦ 0
⎣
FT
1
2a
1
2a
−a ω
↔e
=
2π ( t 2 + a 2 )
2π t 2 + a 2 +∞ ∫ X ( jω )e jωt d ω −∞
+∞ ∫ x (t )e − jωt dt −∞ 1
=
2π ( Compare with ) e −a t FT ↔ 2a
ω 2 + a2 {Re ( a ) > 0} {Re ( a ) > 0} Duality leads to equivalent profiles when converting
from one domain (t, ω) to the opposite one (ω, t) with
ECSE303 Chap. 4
a factor of 2π
MR  3/3/2010 66 Suggested exercises
4.1, 4.2, 4.4, 4.6, 4.8, 4.12, 4.17 Note: for those attending the Thursday
tutorial session, please contact Vahid at
[email protected] for an
appointment ECSE303A Signals and Systems I
Wednesday, March 03, 2010 LECTURE 19
Sections 4.44.7
Convolution, multiplication, tables
and systems characterized by
linear differential equations
Mid term #2 Friday Feb 5th Chapter 4: The continuousTime Fourier
transform
4.0
Introduction
4.1
The continuoustime Fourier transform
4.2
The Fourier transform for periodic signals
4.3
Properties of the continuoustime Fourier transform
4.4
The convolution property
4.5
The multiplication property
4.6
Tables of Fourier properties and of basic Fourier
transform pairs
4.7
Systems characterized by linear constantcoefficient differential equations 4.4 Convolution
The convolution of two signals in the time domain is
equivalent to a multiplication of their Fourier
transforms in the frequency domain.
FT a (t ) ∗ b(t ) ↔ A( jω ) B( jω )
A direct application is the calculation of the response
of an LTI system to an arbitrary input signal: Y ( jω ) = H ( jω ) X ( jω )
The output signal in the timedomain is obtained by
taking the inverse Fourier transform of its spectrum.
ECSE303 Chap. 4
MR  3/3/2010 70 Proof ∞ y (t ) = ∫ x(τ )h(t − τ )dτ −∞ ⎡∞
⎤ − jωt
Y ( jω ) = ∫ ⎢ ∫ x (τ ) h(t − τ ) dτ ⎥ e dt
−∞ ⎣ −∞
⎦
∞
⎡∞
⎤
− jω t
= ∫ x (τ ) ⎢ ∫ h(t − τ )e dt ⎥ dτ
−∞
⎣ −∞
⎦
∞ ∞ = ∫ −∞ x (τ ) ⎡ e − jωτ H ( jω ) ⎤ dτ
⎣
⎦ = H ( jω ) X ( jω )
ECSE303 Chap. 4
MR  3/3/2010 71 Example
Find the spectrum of a system with impulse
response h(t)=eatsin(ω0t)u(t) to an input
x(t)=rectangle (width=2t0,center=t0) H ( jω ) = 2
( jω + a ) 2 + ω0 X ( jω ) = e ECSE303 Chap. 4
MR  3/3/2010 ω0 − jωt0 ⎛ t0ω ⎞
2t0 sinc ⎜
π⎟
⎝
⎠ Y ( jω ) = X ( jω ) H ( jω )
⎛t ω ⎞
2ω0t0 e − jωt0 sinc ⎜ 0 ⎟
⎝π ⎠
=
2
( jω + a ) 2 + ω0 72 Chapter 4: The continuousTime Fourier
transform
4.0
Introduction
4.1
The continuoustime Fourier transform
4.2
The Fourier transform for periodic signals
4.3
Properties of the continuoustime Fourier transform
4.4
The convolution property
4.5
The multiplication property
4.6
Tables of Fourier properties and of basic Fourier
transform pairs
4.7
Systems characterized by linear constantcoefficient differential equations 4.5 Multiplication
This property is the dual of the convolution property.
The multiplication of two signals in the time domain
is equivalent to a convolution of their spectra. 1
a (t )b(t ) ↔
A( jω ) ∗ B ( jω )
2π
FT Modulation is based on this property. ECSE303 Chap. 4
MR  3/3/2010 74 Example
Consider the amplitudemodulated signal x(t)
described by y (t ) = c(t ) x(t )
The Fourier transform of the carrier c(t)=cos(ω0t) is C ( jω ) = πδ (ω − ω0 ) + πδ (ω + ω0 )
Spectrally, it is represented by two impulses of
surface area π, one at ω0 and the other at +ω0. ECSE303 Chap. 4
MR  3/3/2010 75 Suppose that the spectra of x(t) and c(t) look like this Then the Fourier transform of the output signal
FT 1
x(t)c(t) looks like this
c(t ) x(t ) ↔
C ( jω ) ∗ X ( jω )
2π Hence multiplication of a signal by a sinusoid
duplicates the spectrum at the sinusoid’s frequency
band.
ECSE303 Chap. 4
MR  3/3/2010 76 Application
Voice has a bandwidth of less than 5kHz . It is
transmitted over typical AM radio at carrier
modulation frequencies in the range 500kHz to 1.5
MHz. ECSE303 Chap. 4
MR  3/3/2010 77 Example
Knowing that a triangular function (=convolution
between two identical square functions) in the
frequency domain is represented as a sinc2 in the
time domain, and using the properties of Fourier
transforms, evaluate the following integral
∞ ∫ ⎡ sinc ( t )⎤ ⎡3 sinc ( 3t )⎤ dt
⎣
⎦⎣
⎦ −∞ We first modify the argument of the integral
∞ ∞ −∞ −∞ ⎣
⎦⎣
⎦
∫ ⎡ sin c ( t )⎤ ⎡3 sin c ( 3t )⎤ dt = ∫ x ( t ) dt x ( t ) = ⎡ sin c ( t ) ⎤ ⎡3 sin c ( 3t ) ⎤
⎣
⎦⎣
⎦
ECSE303 Chap. 4
MR  3/3/2010 78 ωc
π x2 ( t ) X 2 ( jω ) FT x2 ( t ) ↔ X 2 ( jω ) 1 ω t −π ωc π
ωc −ωc ωc ⎡1, ω < π ⎤ ⎡1, ω < 3π ⎤
⎢
⎥* ⎢
⎥
0 , ω > π ⎦ ⎣0 , ω > 3π ⎦
⎣
⎡ω + 4π −4π < ω < −2π ⎤
1⎢
2π
=
−2π < ω < 2π ⎥
⎥
2π ⎢
⎢ 4π − ω
2π < ω < 4π ⎥
⎣
⎦ 1
X ( jω ) =
2π 1 ⎡ − ω + 4π ω < 4π ⎤ 1 ⎡ − ω + 2π ω < 2π ⎤
=
⎢
⎥−
⎢
⎥
0
0
ω > 4π ⎦ 2π ⎣
ω > 2π ⎦
2π ⎣
1 ⎡1 ω < 2π ⎤ ⎡1 ω < 2π ⎤ 1 ⎡1 ω < π ⎤ ⎡1
=
⎢
⎥* ⎢
⎥−
⎢
⎥* ⎢
2π ⎣0 ω > 2π ⎦ ⎣0 ω > 2π ⎦ 2π ⎣0 ω > π ⎦ ⎣0
ECSE303 Chap. 4
MR  3/3/2010 ω <π⎤
⎥
ω >π⎦
79 x ( t ) = ⎡ 2 sin c ( 2t ) ⎤ − ⎡ sin c ( t ) ⎤
⎣
⎦⎣
⎦
2 ∞ ∞ −∞ −∞ ∫ x ( t ) dt = ∫ ⎡2 sin c ( 2t )⎤
⎣
⎦ 2 2 − ⎡ sin c ( t ) ⎤ dt
⎣
⎦
2 This integral can easily be performed using
Parseval’s relation
1 ⎡1
⎣
⎦⎣
⎦
∫ ⎡2 sin c ( 2t )⎤ − ⎡ sin c ( t )⎤ dt = 2π −∞ ⎢0
∫⎣
−∞
∞ 2 ∞ 2 1
=
2π 2 2 ω < 2π ⎤ ⎡1 ω < π ⎤
⎥ −⎢
⎥ dω
2π ⎦ ⎣0 ω > π ⎦
ω> 2π 1
12 d ω −
∫
2π
−2π π 12 d ω
∫ −π 1
1
4π −
2π
2π
2π
=1
= ECSE303 Chap. 4
MR  3/3/2010 80 Example
Problem 4.21f:
Calculate the Fourier transform of the signal ⎡ sin (π t ) ⎤ ⎡ sin ( 2π ( t − 1) ) ⎤
⎥
⎢
⎥⎢
⎣ π t ⎦ ⎢ π ( t − 1) ⎥
⎣
⎦ ECSE303 Chap. 4
MR  3/3/2010 81 ωc
π x2 ( t ) X 2 ( jω ) FT x2 ( t ) ↔ X 2 ( jω ) 1 ω t −π ωc π
ωc −ωc ωc ⎡ sin (π t ) ⎤ ⎡ sin ( 2π ( t − 1) ) ⎤
⎥ = ⎡ sinc ( t ) ⎤ ⎡ 2 sin c ( 2t ) ∗ δ ( t − 1) ⎤
⎢
⎥⎢
⎦⎣
⎦
π t ⎦ ⎢ π ( t − 1) ⎥ ⎣
⎣
⎣
⎦
FT ⎧ 1, ω < π
δ ( t − 1) ↔ e − jω
sinc ( t ) ↔ ⎨
⎩0 , otherwise
FT ⎧ 1,
ω < 2π
2 sinc ( 2t ) ↔ ⎨
⎩0 , otherwise
FT ECSE303 Chap. 4
MR  3/3/2010 82 1
⎡
⎤⎡
⎤
⎣ sinc ( t ) ⎦ ⎣ 2 sin c ( 2t ) ∗ δ ( t − 1) ⎦ ↔ 2π
FT Sq4π ( Ω ) ⎡ e − jΩ , Ω < 2π ⎤
Sq4π ( Ω ) = ⎢
⎥
0 , otherwise ⎦
⎣ 1 2π π π 2π Sq2π (ω − Ω )
1 ωπ
ECSE303 Chap. 4
MR  3/3/2010 ω+π ⎡ 1, ω < π ⎤ ⎡ e − jω , ω < 2π ⎤
⎥
⎢
⎥* ⎢
0 , otherwise ⎦ ⎣ 0 , otherwise ⎦
⎣ Ω
⎡1, ω − Ω < π ⎤
Sq2π (ω − Ω ) = ⎢
⎥
0 , otherwise ⎦
⎣ Ω 83 Pattern 1: No overlap
Pattern 2: Partial overlap
ωπ ω+π 2π
ω +π ∫π −2 3π<ω<π π π Sq4π ( Ω ) Sq2π (ω − Ω )d Ω = Ω 2π
ω +π ∫π e − jΩ d Ω −2 e − j ( ω +π ) − e j 2π
=
−j ( ) = j −e − jω − 1 ECSE303 Chap. 4
MR  3/3/2010 84 Pattern 3: Full overlap
ωπ 2π
ω +π ∫π −2 π π<ω<π
ω+π π 2π Sq4π ( Ω ) Sq2π (ω − Ω )d Ω = ω +π ∫π
ω Ω e − jΩ d Ω − e − j ( ω + π ) − e − j ( ω −π )
=
−j ( = j −e − jω + e − jω ) =0
ECSE303 Chap. 4
MR  3/3/2010 85 Pattern 4: Partial overlap
Pattern 5: No overlap π<ω<3π
ωπ 2π
ω +π ∫π −2 π π Sq4π ( Ω ) Sq2π (ω − Ω )d Ω = ω+π
Ω 2π
2π ∫π
ω e − jΩ d Ω − e − j ( 2 π ) − e − j ( ω −π )
=
−j ( = j 1 + e − jω ECSE303 Chap. 4
MR  3/3/2010 )
86 1
⎡ sinc ( t ) ⎤ ⎡ 2 sin c ( 2t ) ∗ δ ( t − 1) ⎤ ↔
⎣
⎦⎣
⎦ 2π
FT ⎡ 1, ω < π ⎤ ⎡ e − jω , ω < 2π ⎤
⎥
⎢
⎥* ⎢
0 , otherwise ⎦ ⎣ 0 , otherwise ⎦
⎣ ( ) ( ) ⎡ − j 1 + e − jω
FT 1 ⎢
⎡ sinc ( t ) ⎤ ⎡ 2 sin c ( 2t ) ∗ δ ( t − 1) ⎤ ↔
0
⎣
⎦⎣
⎦ 2π ⎢
⎢
j 1 + e − jω
⎢
⎣ ECSE303 Chap. 4
MR  3/3/2010 −3π < ω < −π ⎤
⎥
−π < ω < π ⎥
⎥
π < ω < 3π ⎥
⎦ 87 Chapter 4: The continuousTime Fourier
transform
4.0
Introduction
4.1
The continuoustime Fourier transform
4.2
The Fourier transform for periodic signals
4.3
Properties of the continuoustime Fourier transform
4.4
The convolution property
4.5
The multiplication property
4.6
Tables of Fourier properties and of basic Fourier
transform pairs
4.7
Systems characterized by linear constantcoefficient differential equations ECSE303 Chap. 4
MR  3/3/2010 89 +Duality: but be careful as ω and t can be interchanged by
including a multiplication factor of 2π
ECSE303 Chap. 4
MR  3/3/2010 90 ECSE303 Chap. 4
MR  3/3/2010 91 ECSE303 Chap. 4
MR  3/3/2010 92 The Inverse Fourier Transform
1) By direct calculation of the integral
Example
Consider the ideal lowpass filter with cutoff
frequency ωc and given the frequency spectrum R1,
H ( jω ) = S
T0, ω < ωc
ω > ωc The corresponding impulse response is calculated
from the frequency spectrum ECSE303 Chap. 4
MR  3/3/2010 •93 R1,
H ( jω ) = S
T0, 1
x (t ) =
2π ω < ωc
ω > ωc X ( jω ) = h(t ) = 1
2π 1
=
2π +∞ ∫ +∞ ∫ X ( jω )e jωt d ω −∞
+∞ ∫ x (t )e − jωt dt −∞ H ( jω )e jωt d ω −∞
+ ωc ∫
ω − e jωt d ω c 1
jωt ωc
⎡e ⎤
=
⎦ −ωc
2π ( jt ) ⎣ ωc
ωc sin( π tπ )
=
ωcπ t
π
π
ωc
⎛ ωc ⎞
= sinc ⎜ t ⎟
π
⎝π ⎠
ECSE303 Chap. 4
MR  3/3/2010 94 Thus the impulse response of an ideal lowpass filter
is a sinc function extending from t=∞ to t=+∞ ωc
π ωc
⎛ ωc ⎞
h(t ) = sinc ⎜ t ⎟
π
⎝π ⎠ t
π
−
ωc π
ωc Take note that this function is real but noncausal
ECSE303 Chap. 4
MR  3/3/2010 95 2) Using transforms pairs from tables
The Fourier transform is often in the form of a
rational function of jω (a ratio of polynomials). 2 − jω
Y ( jω ) = 2
ω − 5 jω − 6
To determine the inverse Fourier transform:
• Perform a partial fraction expansion to simplify the
expression as a sum of first order polynomial
terms.
• Apply the Inverse Fourier transform to the
simplified terms and use the linear property of FT.
ECSE303 Chap. 4
MR  3/3/2010 96 Example
Consider the response of an LTI system with
impulse response h(t)=e2tu(t) to the input
x(t)=e3tu(t) using Fourier transforms.
The solution is given from multiplying the Fourier
transforms of the input and the impulse response. 1
H ( jω ) =
jω + 2 1
X ( jω ) =
jω + 3 1
Y ( jω ) = X ( jω ) H ( jω ) =
( jω + 3)( jω + 2)
ECSE303 Chap. 4
MR  3/3/2010 97 The partial fraction expansion consists in expressing
the function as a sum of firstorder terms.
Y ( jω ) = A
B
1
=
+
( jω + 2)( jω + 3) ( jω + 2) ( jω + 3) To solve for A and B:
(1) Equate the transform with its sum of partial fractions,
and let s=jω 1
A
B
=
+
( s + 2)( s + 3) ( s + 2) ( s + 3)
(2) To obtain the coefficient A, multiply both sides of the
equation by (s+2) and evaluate at s=2 1
( s + 2) B
= A+
( s + 3) s = −2
( s + 3)
ECSE303 Chap. 4
MR  3/3/2010 ⇒ 1
A=
=1
−2 + 3 s =−2 98 Applying again step (2) for the constant B, we obtain 1
A
B
=
+
( s + 2)( s + 3) ( s + 2) ( s + 3)
1
( s + 3) A
=
+B
( s + 2) s =−3 ( s + 2) s =−3
1
⇒ B=
= −1
−3 + 2
The coefficients A and B are called residues at poles ECSE303 Chap. 4
MR  3/3/2010 99 FT x (t ) = e u (t ) ↔
− at X ( jω ) = 1
jω + a Finally, the partial fraction expansion of the Fourier
transform of the output is given by 1
1
−
Y ( jω ) =
( jω + 2) ( jω + 3)
Which leads to y (t ) = e −2t u (t ) − e −3t u (t ) ECSE303 Chap. 4
MR  3/3/2010 100 Chapter 4: The continuousTime Fourier
transform
4.0
Introduction
4.1
The continuoustime Fourier transform
4.2
The Fourier transform for periodic signals
4.3
Properties of the continuoustime Fourier transform
4.4
The convolution property
4.5
The multiplication property
4.6
Tables of Fourier properties and of basic Fourier
transform pairs
4.7
Systems characterized by linear constantcoefficient differential equations 4.7 Systems characterized by linear
differential equations
The output y(t) of a stable LTI system with impulse
response h(t) and subjected to an input x(t) is given
by
+∞
y ( t ) = x ( t ) * h ( t ) = ∫ x (τ ) h ( t − τ ) dτ
−∞ We can also calculate the output by taking the
inverse Fourier transform of
Y ( jω ) = H ( jω ) X ( jω )
Represented in a block diagram as
X ( jω ) ECSE303 Chap. 4
MR  3/3/2010 H ( jω ) Y ( jω ) 102 For a cascade of two stable LTI systems with
impulse responses h1(t), h2(t), we have X ( jω ) H1 ( jω ) H 2 ( jω ) Y ( jω ) y ( t ) = x ( t ) * h1 ( t ) * h2 ( t ) Y ( jω ) = H2 ( jω ) H1 ( jω ) X ( jω ) ECSE303 Chap. 4
MR  3/3/2010 103 For a parallel connection of two stable LTI systems
with impulse responses h1(t), h2(t), we have
H1 ( jω )
X ( jω ) + Y ( jω ) +
H 2 ( jω ) y ( t ) = ⎡ h1 ( t ) + h2 ( t ) ⎤ * x ( t )
⎣
⎦ Y ( jω ) = [ H1 ( jω ) + H2 ( jω )] X ( jω )
ECSE303 Chap. 4
MR  3/3/2010 104 For a feedback interconnection of two stable LTI
systems with impulse responses h1(t), h2(t), we have E ( jω ) = X ( jω ) − Y ( jω ) H 2 ( jω )
Y ( jω ) = E ( jω ) H1 ( jω ) H1 ( jω )
X ( jω )
Y ( jω ) =
1 + H1 ( jω ) H 2 ( jω )
ECSE303 Chap. 4
MR  3/3/2010 105 LTI Differential Systems
Consider the stable LTI system defined by an Nthorder linear constantcoefficient differential equation
initially at rest: d k y (t ) M
d k x (t )
∑0 ak dt k = ∑0 bk dt k
k=
k=
N Assume that
FT FT x ( t ) ↔ X ( jω )
y ( t ) ↔ Y ( jω )
Recall that differentiation in the time domain is
equivalent to a multiplication in the Fourier domain
by jω. Thus,
. N M k =0 k =0 a k ( jω ) k Y ( jω ) = ∑ bk ( jω ) k X ( jω )
∑
ECSE303 Chap. 4
MR  3/3/2010 106 N M k =0 k =0 a k ( jω ) k Y ( jω ) = ∑ bk ( jω ) k X ( jω )
∑
Because Y ( jω )
H ( jω ) =
X ( jω )
the frequency response of the system is given by
M Y ( jω )
H ( jω ) =
=
X ( jω ) bk ( jω ) k
∑
k =0
N a k ( jω ) k
∑
k =0 ECSE303 Chap. 4
MR  3/3/2010 107 Example
Consider the LTI differential system
dy ( t )
dt + ay ( t ) = Ax ( t ) Applying the Fourier transform operation leads to ( jω + a ) Y ( jω ) = AX ( jω )
The frequency response of the system is
Y ( jω )
A
H ( jω ) =
=
X ( jω ) jω + a
And thus it has an impulse response
h ( t ) = Ae − at u ( t )
ECSE303 Chap. 4
MR  3/3/2010 108 Example
The frequency response of the secondorder LTI
differential system
d 2 y (t )
dy (t )
dx (t )
+3
+ 2 y (t ) =
− x (t )
2
dt
dt
dt is calculated as follows:
[( jω ) 2 + 3 jω + 2]Y ( jω ) = ( jω − 1) X ( jω )
Y ( jω )
jω − 1
H ( jω ) =
=
X ( jω ) ( jω ) 2 + 3 jω + 2 ECSE303 Chap. 4
MR  3/3/2010 109 H ( jω ) = jω − 1 ( jω ) 2 + 3 jω + 2 Suppose we want to obtain the step response of this
system.
The Fourier transform of the step function is
X ( jω ) = 1
+ πδ (ω )
jω Then the output Y ( jω ) = H ( jω ) X ( jω )
⎡
⎤⎡ 1
⎤
jω − 1
=⎢
+ πδ (ω ) ⎥
⎥⎢
2
⎢ ( jω ) + 3 jω + 2 ⎥ ⎣ jω
⎦
⎣
⎦
jω − 1
1
=
− πδ (ω ) Partial fraction
2
⎡( jω ) + 3 jω + 2 ⎤ jω 2
expansion
⎣
⎦
ECSE303 Chap. 4
MR  3/3/2010 110 Y ( jω ) = jω − 1 1
− πδ ( jω )
⎡( jω )2 + 3 jω + 2 ⎤ jω 2
⎣
⎦ Expanding the rational function on the righthand
side into partial fractions, we get (s=jω)
s −1
s −1
A
B
C
=
=+
+
s 2 + 3s + 2 s ( s + 1)( s + 2) s s s + 1 s + 2 and the coefficients are calculated as follows
A= s −1
1
=−
( s + 1)( s + 2 ) s =0 2 s −1
B=
=2
s ( s + 2 ) s =−1
C=
ECSE303 Chap. 4
MR  3/3/2010 s −1
3
=−
2
s ( s + 1) s =−2 s −1
=
2
⎡ s + 3s + 2 ⎤ s
⎣
⎦
−1
2
3
+
−
2s s + 1 2 ( s + 2 )
111 Hence, jω − 1 1
Y ( jω ) =
− πδ (ω )
2
⎡( jω ) + 3 jω + 2 ⎤ jω 2
⎣
⎦
⎛ 1 ⎞ 3⎛ 1 ⎞ 1
11
2⎜
=−
+
⎟− ⎜
⎟ − πδ (ω )
2 jω
⎝ jω + 1 ⎠ 2 ⎝ jω + 2 ⎠ 2
⎤
⎛ 1 ⎞ 3⎛ 1 ⎞
1⎡ 1
=− ⎢
+ πδ (ω ) ⎥ + 2 ⎜
⎟− ⎜
⎟
2 ⎣ jω
jω + 1 ⎠ 2 ⎝ jω + 2 ⎠
⎦
⎝ We find the output by inspection LM− 1 + 2e
y (t ) =
N2
ECSE303 Chap. 4
MR  3/3/2010 −t OP
Q 3 −2 t
− e u( t )
2
112 Example
Find h(t) for the following stable secondorder LTI
differential system d 2 y (t ) dy (t )
+
+ y (t ) = x (t )
2
dt
dt
The frequency response of this system is given by H ( jω ) =
= ECSE303 Chap. 4
MR  3/3/2010 Y ( jω ) X ( jω )
1 ( jω ) 2 + jω + 1
113 Letting s=jω, and expanding the righthand side into
partial fractions, we get
1
1
=
s 2 + 1s + 1 ⎡ ⎛ 1
3 ⎞⎤ ⎡ ⎛ 1
3 ⎞⎤
⎢s + ⎜ + j
⎟⎥ ⎢ s + ⎜ − j
⎟⎥
2
2 ⎠⎥ ⎢ ⎝ 2
2 ⎠⎥
⎢⎝
⎣
⎦⎣
⎦
A
B
=
+
⎛1
⎛1
3⎞
3⎞
s+⎜ + j
⎟ s+⎜ − j
⎟
2⎠
2⎠
⎝2
⎝2 the residues at poles are then
A= 1
⎛1
3⎞
s+⎜ − j
⎟
2
2 ⎠ ⎛1
⎝
s =−⎜ + j
⎜
⎝2 ECSE303 Chap. 4
MR  3/3/2010 =j
3⎞
⎟
2⎟
⎠ 1
3 B= 1
⎛1
3⎞
s+⎜ + j
⎟
2
2 ⎠ ⎛1
⎝
s =−⎜ − j
⎜
⎝2 =−j 1
3 3⎞
⎟
2⎟
⎠ 114 Hence, 1
⎡
⎛1
⎛1
3 ⎞⎤ ⎡
3 ⎞⎤
⎢ jω + ⎜ + j
⎟ ⎥ ⎢ jω + ⎜ − j
⎟⎥
2
2 ⎠⎥ ⎢
2
2 ⎠⎥
⎢
⎝
⎝
⎣
⎦⎣
⎦
j
1
j
1
=
−
⎛1
⎛1
3
3
3⎞
3⎞
jω + ⎜ + j
jω + ⎜ − j
⎟
⎟
2
2⎠
2
2⎠
⎝
⎝ H ( jω ) = we find h(t)
⎡ j −1t− j 3t
j − 1 t + j 23 t ⎤
h(t ) = ⎢
e2 2 −
e2
⎥ u (t )
3
⎢3
⎥
⎣
⎦
3
j t⎤
j − 1 t ⎡ − j 23 t
=
− e 2 ⎥ u (t )
e 2 ⎢e
3
⎢
⎥
⎣
⎦ ECSE303 Chap. 4
MR  3/3/2010 2 −1t
3
=
e 2 sin( t )u (t )
2
3
115 Example
Find the differential equation that leads to ( jω ) 2 + ωc 2 jω
H hp ( jω ) =
( jω ) 2 + ωc 2 jω + ωc2
Y ( jω )
H hp ( jω ) =
X ( jω )
⎡( jω ) 2 + ωc 2 jω + ωc2 ⎤ Y ( jω ) = ⎡( jω ) 2 + ωc 2 jω ⎤ X ( jω )
⎣
⎦
⎣
⎦
d 2 y (t )
dy (t )
d 2 x(t )
dx(t )
2
+ ωc 2
+ ω c y (t ) =
+ ωc 2
2
2
dt
dt
dt
dt
ECSE303 Chap. 4
MR  3/3/2010 116 Suggested exercises
4.184.20, 4.25, 4.26, 4.33 ECSE303 Chap. 4
MR  3/3/2010 121 ...
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This note was uploaded on 04/07/2010 for the course ELEC ecse 303 taught by Professor Rochette during the Winter '10 term at McGill.
 Winter '10
 Rochette

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