华中科技大学电信系信号与系统课件-chapter 5

n = ∞
∑ x[n] < ∞
+∞
In contrast to the situation for the analysis equation, there are generally no convergence issues associated with the synthesis equation.
5.2 THE FOURIER TRANSFORM FOR PERIODIC SIGNALS
First consider the Fourier transform of the sequence x[n] = e jω 0 n
X (e ) =
X (e ) =
jω
jω
l = ∞
∑ 2πδ (ω ω
,
1 = N
x[n]
n = N1
~[n]e jk ( 2π / N ) n ∑x
N2
N →∞
n
–N1
0
N2
1 ak = N
n = N1
∑
N2
x[n]e jk ( 2π / N ) n
1 = N
n = ∞
∑
+∞
x[n]e jk ( 2π / N ) n
x[n]e jωn ∑
+∞
Defining a function: Thus, Consequently,
x1[n] x2[n] n
n
0
0
X 1 (e
jω
)
X 2 ( e jω )
–2π–π 0 π 2π
ω
–2π –π 0 π 2π
ω
In discrete time, Low frequencies are the values of ω near even multiple of π ; high frequencies are those values of ω near odd multiples of π.
1/(1+α)
–2π –π
0
π
2π ω
–2π –π
arg X (e jω )
tan 1 (α / 1 α 2 )
π arg X (e jω )
0
2π ω
tan 1 ( α / 1 α 2 )
–2π –π
0
π
2π
ω
–2π
tan 1 (α / 1 α 2 )
tan ( α / 1 α )
1 2
5.1 REPRESENTATION OF APERIODIC SIGNALS: THE DISCRETE-TIME FOURIER TRANSFORM
~[ n] x
… –N –N1 …
~[n] = x
n
k= N
∑
a k e jk ( 2π / N ) n
0
N2
N
1 ak = N
As
n= N
~[n]e jk ( 2π / N ) n ∑x
1 X (e jω )e jωndω = e j(ω0 +2π r)n = e jω0n 2π ∫2π
Now consider an arbitrary periodic sequence x[n] with period N and with the Fourier series representation x[n] = ∑ a e jk ( 2π / N ) n .
–2π
–π
0
π
2π
ω
Convergence Issues of the Discrete-Time Fourier Transform
If x[n] is an infinite duration signal, we must consider the question of convergence of the infinite summation in the analysis equation. The analysis equation will converge if x[n] is absolutely summable; that is
2π jω X (e ) = N
2πk ∑ δ ω N k =∞
+∞
2π/N
X (e jω )
… …
2π N
ω
5.3 PROPERTIES OF THE DISCRETE-TIME FOURIER TRANSFORM
5.3.1 Periodicity of the Discrete-Time Fourier Transform
2π ω0 = 5
That is, X (e jω ) = πδ (ω 2π ) + πδ (ω + 2π ), π ≤ ω < π, 5 5 X ( e jω )
…
π
…
Example 5.5
–2π –ω0 0 ω0 2π (–2π–ω0) (–2π+ω0) (2π–ω0) (2π+ω0)
ω
Consider the periodic sample train x[n] =
X (e j (ω + 2π ) ) = X (e jω )
5.3.2 Linearity of the Fourier Transform FT x1[n] ←→ X 1 (e jω ) If FT x2 [n] ←→ X 2 (e jω ) and
FT ax1[n] + bx2 [n] ←→ aX1 (e jω ) + bX 2 (e jω ) then 5.3.3 Time Shifting and Frequency Shifting FT If x[n] ←→ X (e jω ) FT then x[n n0 ] ←→ e jωn0 X (e jω )
n
+∞ jω n n =0
+∞
X (e ) = ∑ (αe
) + ∑ (αe jω ) m
m =1
∞
α e jω 1 + X ( e jω ) = jω 1 αe 1 αe jω 1α 2 = 1 2α cos ω + α 2 for 0 < α < 1,
x[n]
X (e
jω
)
(1+α)/(1–α)
jω
∑
+∞
Example 5.4
Consider the periodic signal
jω +∞
x[n] = cosω0n = 1 e jω0n + 1 e jω0n , 2 2
+∞ 2π 2π X (e ) = ∑πδ (ω 2πl ) + ∑πδ (ω + 2πl ) 5 5 l =∞ l =∞
Consider the unit sample x[n] = δ[n]
X (e ) = 1
Approximating δ[n] by an integral of complex exponentials with frequencies taken over the interval |ω| ≤ W , i.e.,
+∞
n = ∞
∑e
+∞
jω 0 n jω n
e
=
n = ∞
0
∑e
+∞
j (ω 0 ω ) n
X ( e jω )
2π
2πl )
…
…
ω
(ω0–4π) ω0 (ω0+4π) (ω0–2π) (ω0+2π)
To check the validity of this expression,
+∞ 1 1 X (e jω )e jωndω = ∫ ∑2πδ(ω ω0 2π l)e jωndω 2π ∫2π 2π 2π l=∞
(1–α)/(1+α)
0
n
–2π
0
2π
ω
Example 5.3
X (e ) =
jω
1, Consider the rectangular pulse x[n] = 0,
n = N1
n ≤ N1 n > N1
x[n]
1
∑e
N1
j ωn
X (e ) =
jω
e
jωN1
(1 e 1 e jω
jω( 2 N1 +1)
X ( e jω ) =
1 a k = X (e jkω 0 ) N
n = ∞
discrete-time Fourier transform
~[n] = x
As
k= N
∑
1 1 jkω 0 jkω 0 n X (e ) e = N 2π
,
k= N
∑ X (e
jωn
jkω 0)ejkω源自0 nω0N →∞
–π
0
π
2πω
1 >α > 0
0 >α > -1
Example 5.2
Consider the signal
x[n] = α ,
n
α < 1.
+∞ 1 n =0 n = ∞
X ( e jω ) =
n = ∞
jω
α e j ωn = ∑ α n e j ω n + ∑ α n e j ωn ∑
Example 5.1
Consider the signal
x[n] = α n u[n],
α < 1.
+∞ jω n =0
X (e ) =
X (e jω )