MIT 信号与系统 Lecture 16

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 � ∞ � ∞ X (s) = e−t u(t)e−st dt = e−(s+1)t dt =
April 6, not collected or graded. Solutions are posted. Closed book: 2 pages of notes (8 1 2 × 11 inches; front and back). Designed as 1-hour exam; two hours to complete.
Relation between Fourier and Laplace Transforms
Fourier transform “inherits” properties of Laplace transform. Property Linearity Time shift Time scale Differentiation Multiply by t Convolution x(t) ax1 (t) + bx2 (t) x(t − t0 ) x(at) dx(t) dt tx(t) x1 (t) ∗ x2 (t) X (s) aX1 (s) + bX2 (s) e−st0 X (s) 1 �s� X |a| a sX (s) − d X (s) ds − X (jω ) aX1 (jω ) + bX2 (jω ) e−jωt0 X (jω ) � � 1 jω X |a| a jωX (jω ) 1 d X (jω ) j dω
Lecture 16
Fourier Transforms
Find a general scaling rule. Let x2 (t) = x1 (at). � ∞ � ∞ X2 (jω ) = x2 (t)e−jωt dt = x1 (at)e−jωt dt
−∞ −∞
April 6, 2010
Let τ = at (a > 0). � � � ∞ 1 jω 1 X2 (jω ) = x1 (τ )e−jωτ /a dτ = X1 a a a −∞ If a < 0 the sign of dτ would change along with the limits of integra­ tion. In general, � � jω 1 . x1 (at) ↔ X1 |a| a If time is stretched (a < 1) then frequency is compressed and ampli­ tude increases (preserving area).
X1 (jω ) 2 π ω
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
3
6.003: Signals and Systems
Check Yourself
6.003: Signals and Systems
6.003: Signals and Systems
Fourier Transform
Lecture 16
Mid-term Examination #2
Tomorrow, April 7, 7:30-9:30pm. No recitations tomorrow. Coverage: Lectures 1–15 Recitations 1–15 Homeworks 1–8
T →∞ −T /2
� T /2
ω0 = 2π/T x(t)e−jωt dt =
T →∞
lim T ak = lim
� T /2
ω0 = 2π/T x(t)e−jωt dt = 2
sin ωS = E (ω )
ω
T →∞ −T /2
k=−∞
ak
k=−∞
Fourier Transform
Replacing E (ω ) by X (jω ) yields the Fourier transform relations. E (ω ) = X (s)|s=jω ≡ X (jω ) Fourier transform � ∞ x(t)e−jωt dt X (jω )= � ∞ 1 x(t)= X (jω )ejωt dω 2π −∞
t � � � � � � �X (jω)� = � 1 � � 1 + jω �
t
10
Magnitude
0
Ima 1 0 gin -1 ary (s)
1 -1 0 s) Real(
0
1
ω
Frequency plots provide intuition that is difficult to otherwise obtain.
30 20 10 0 5
x1 (t) 1 −1 1 t
� � �1 s � −s � |X (s)| = � � s (e − e )�
2. X1 (jω ) =
2 4. X1 (jω ) = sin ω ω
5. none of the above
5 0 -5 -5 0
Fourier Transform
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?
The Fourier transform is a function of real domain: frequency ω . Time representation: x1 (t) 1 −1 Frequency representation: 1 t
Check Yourself
Signal x2 (t) and its Fourier transform X2 (jω ) are shown below. x2 (t) 1 −2 Which is true? 2 t X2 (jω ) b ω0 ω
S
T
S
T
ak =
� � 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 −T /2 T −S πk T ω T ak 2 sin ωS
ω
ak =
� � 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 −T /2 T −S πk T ω T ak 2 sin ωS
X1 (jω ) × X2 (jω )
1 1 + jω
a complex-valued function of real domain.
2
6.003: Signals and Systems
Laplace Transform
The Laplace transform maps a function of time t to a complex-valued function of complex-valued domain s. x(t)
∞ � k=−∞
S x(t + kT )
t
xT (t) −S Then x(t) = lim xT (t).
T →∞
S
T
t
Fourier Transform
Represent xT (t) by its Fourier series. xT (t) −S t
Fourier Transform
Doubling period doubles # of harmonics in given frequency interval. xT (t) −S t
−∞
(“analysis” equation)
(“synthesis” equation)
Fourier transform: � ∞ x(t)e−jωt dt = H (s)|s=jω X (jω ) =
−∞
Form is similar to that of Fourier series → provides alternate view of signal.
Lecture 16
Fourier Transform
As T → ∞, synthesis sum → integral. xT (t) −S
April 6, 2010
S
T
S 2 sin ωS
ω
T ω = kω0 = k 2π T
t
ak =
� � 2π sin 2πkS 1 T /2 1 S −j 2π kt 2 sin ωS T xT (t)e−j T kt dt = e T dt = = T −T /2 T −S πk T ω T ak 2 sin ωS
−∞ 0
1 ; Re(s) > −1 1+s
a complex-valued function of complex domain. Fourier transform � ∞ � ∞ e−t u(t)e−jωt dt = e−(jω +1)t dt = X (jω ) =