《信号与系统》第七章课件(英文版)

n = −∞
π ωc (t − nT )
7.3 the effect of undersampling : aliasing (p527)（欠采样的效果：混叠现象）
Example：x(t) = cosω0t
spectrum of original signal
ω0
=
ωs
6
xr (t ) = cosω0t
k =−∞
∑ =
1 T
∞ k =−∞
X(
j(ω
−
kωs
))
X p ( jω ) is a periodic function of ω consisting of a superposition of shifted replicas of X ( jω ), scaled by 1
T
x(t)
x p (t )
7.0 Introduction
9What is sampling? A continuous-time signal is represented by its values, or samples, at points equally spaced in time.
9The importance of the sampling
x(t)
xp(t) = x(t ) ⋅ p(t )
p(t )
∑ P(
jω )
=
2π
T
∞
δ (ω
n = −∞
−
kωs )
X p ( jω ) =
1
2π
X ( jω ) ∗ P ( jω )ωs=2π
T
∑ In frequency
domain:
=
1
2π
X ( jω ) ∗ 2π
T
∞
δ (ω − kωs )
Under certain conditions a c-t signal can be completely recovered from a sequence of its samples .
Due to the dramatic development of digital technology, we can use d-t system technology to implement c-t systems and process c-t signals:
ω0
=
2ωs
6
xr (t ) = cosω0t
ωs > 2ω0
spectrum of original signal
ω0
=
4ωs
6
xr (t ) = cos(ωs − ω0 )t ≠ x(t)
ωs < 2ω0
ω0
=
5ωs
6
xr (t ) = cos(ωs − ω0 )t ≠ x(t)
problems：
We exploit sampling to convert a c-t signal to a d-t signal, process the d-t signal using a d-t system and then convert back to continuous time.
7.1 representation of a continuous-time signal by its samples: the sampling theorem (p516)（用信号样本表示连续时间 信号：采样定理）
Consider a c-t signal x ( t ) whose Fourier
transform has the property that X ( jf ) = 0 for f > 200 H z
The signal lasted for 2 minutes. Determine the minimum sample points numberN
¾Interpolation（内插）: a process in which the fitting（拟合） of a continuous-time signal to a set of sample values is performed.
¾band-limited interpolation（带限内插）: interpolation using the impulse response of an ideal low-pass filter
7.2 7.6 7.9 7.22
Ch7 Sampling
第七章 采样
V Main content
1. The sampling theorem (抽样定理); 2. reconstruction of a signal from its
samples using interpolation(利用内插 从采样信号中重建信号); 3. The effect of undersampling : aliasing(欠采样的效果：混叠现象）
∞
∑ xr (t) = xp (t) ∗ h(t) = x(nT )δ (t − nT ) ∗ h(t) n−∞
∞
= ∑ x(nT )h(t − nT ) n = −∞
h(t ) = Tωc Sinωct = T ⋅ Sinωct
π ωct
πt
∴ x(t)
∑ = ∞ x(nT ) ⋅ ωcT Sinωc (t − nT )
ωs > 2ωM Nyquist rate（奈奎斯特频率）
where
ωs
=
2π
T
9Exact recovery of a c-t signal from its samples using an ideal lowpass filter
ωM < ωc < (ωs − ωM )
Example 1(supplementary):
Solution: fM = 100Hz
fsMin = 2 fM = 200Hz
TsMax =
1 f sMin
= 1s 200
N Min
=
τ
TsMax
= (2× 60)
1 = 24000 200
7.2 reconstruction of a signal from its samples using interpolation (p522)（利用 内插由样本重建信号）
T?
7.1.1 Impulse-train sampling (冲激串采样)
In time domain:
x(t)
∞
p(t) = ∑ δ (t − nT ) n= −∞
xp(t) = x(t ) ⋅ p(t )
∞
p(t) = ∑ x(nT )δ (t − nT ) −∞
T :Sampling period Sampling function
In the absence of any additional conditions , could a signal be specified by a sequence of equally spaced samples?
NO!
9In addition, we can get different sequences if a signal is sampled at different regular intervals .
p(t )
ωs > 2ωM
ωM
ωs < 2ωM
9Sampling theorem
Let x ( t ) be a band-limited signal with X ( jω ) = 0 for ω > ω M
Then x ( t )is uniquely determined by its samples x ( n T ), n = 0 , ± 1, ± 2 ,K , if