Fast fading, Slow fading and Multipath effect in wireless communicationsPei-Che Chang
Fast fading, Slow fading and multipath effect in wireless communications
QPSK in AWGN channel
QPSK in AWGN + Rayleigh fading channel
using GNU Octave simulation
Fast fading, Slow fading and Multipath effect in wireless communicationsPei-Che Chang
Fast fading, Slow fading and multipath effect in wireless communications
QPSK in AWGN channel
QPSK in AWGN + Rayleigh fading channel
using GNU Octave simulation
C2 discrete time signals and systems in the frequency-domainPei-Che Chang
Discrete-Time Signals and Systems in the Frequency-Domain
Discrete-Time Fourier Transform
time domain convolution theorem
frequency domain convolution theorem
Z transform
1. 1. 平面简谐波 Plane Harmonic Waves
为了定量地描述波在空间的传播,可将处在任意位置的
质元在任意时刻的振动状态表示为
)( tryy ,
r
=
)( try ,
r
多元函数 称为波函数
返回 退出
x ty
x
v
r
O P
时刻
⎥
⎦
⎤
⎢
⎣
⎡
+⎟
⎠
⎞
⎜
⎝
⎛
−= ϕω
v
x
tAt,xy cos)(
)cos()0( ϕω += tAt,y
v
x
v
x
t −
2. x
ty
x
v
r
O P
时刻
π
ωλ
2
=v)cos()0( ϕω += tAt,y
⎥
⎦
⎤
⎢
⎣
⎡
+⎟
⎠
⎞
⎜
⎝
⎛
−= ϕω
v
x
tAt,xy cos)( ⎥
⎦
⎤
⎢
⎣
⎡
−+=
v
x
tA ωϕω )(cos
λ
π2
=k
角波数 angular wave number k
k
r
波矢
( )ϕω +−= kxtAt,xy cos)(
退出返回
4. ( )ϕω +−= kxtAt,xy cos)(
x∆ t tt ∆+
y
x
v
r
O P
( ) ( ) ϕ∆∆ω ++−+= xxkttϕω +−kxt
k
ω
= λ
λ
π
π
f
f
v ==
2
2
t
x
v
∆
∆
=xkt ∆∆ω =
v Phase velocity相速度
退出返回
6. T T
T
T
2α
1α
dxx+x x
dxdm µ=
( )1212 tgtgsinsin αααα −≈− TTT
dx
x
y
T
x
y
x
y
T
xdxx
2
2
∂
∂
∂
∂
∂
∂
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=
+
2
2
t
y
dx
∂
∂
µ=
2
2
2
2
x
yT
t
y
∂
∂
µ∂
∂
=
退出返回
12. 已知:向右, ( )αω += tAy,v a cos
如图建立坐标,求波动式及b点振动式
y
xO
v
a b
ax l
[ ]a
kxtAy ++= αωcos0
[ ]kxkxtAy a −++= αωcos
[ ] [ ]kltAkxkxtAy bab −+=−++= αωαω coscos
退出返回
13. 已知:向右, ( )αω += tAy,v a cos
如图建立坐标,求波动式及b点振动式
x
y
O
v
a b
ax
l
[ ]a
kxtAy −+= αωcos0
[ ]kxkxtAy a +−+= αωcos
[ ] [ ]kltAkxkxtAy bab −+=+−+= αωαω coscos
退出返回
14. [例3] 已知 t = 0 时的波形曲线为Ⅰ,波沿 ox 方向传播,经
t=1/2 s 后波形变为曲线Ⅱ。已知波的周期 T >1s,试根据图
中绘出的条件求出波的表达式,并求 A点的振动式。
m010.A =
y(cm)
x(cm)1 2 3 4 5 6
Ⅰ Ⅱ
A
1
-1
0
m040.=λ
101
sm020
21
010 −
⋅==
−
= .
.
t
xx
v
s2
020
040
===
.
.
v
T
λ 1
sπ
π2 −
==
T
ω
返回 退出
17. y′
)
20
(π4cos3
x
ty +=
y
B点振动式:
波动式: ]π)
20
(π4cos[3 −+=
x
ty
[例4] 一平面简谐波在介质中以速度 ,沿 x
轴的负向传播。已知A点的振动式为 ,则
(1)以A点为坐标原点求波动式;(2)以距A点5m处的B
为坐标原点求波动式。
m/s20=v
ty π4cos3=
返回 退出
π)π4cos(3 −= t]
20
)5(
[π4cos3
−
+= tyB
A xB
v
18. x
入射波波动式:
)(cos
v
x
tAy −= ω
)(cos
v
L
tAyB −= ω
v
反射波波动式:
)
2
(cos
v
L
tAy −= ω
)
2
(cos
v
x
v
L
tAy +−= ω
O B x
L
v
y
[例5] 有一平面简谐波沿 x 轴方向传播,在距反射面 B 为
L处的振动规律为 ,设波速为v ,反射时无
半波损失,求入射波和反射波的波动式。
tAy ωcos=
反射波在原点的振动式:
返回 退出
19. §13.3 波的能量和能流
1.平面简谐波的能量密度 µ线密度为弦上的横波
2
)(
2
1
t
y
dxdK
∂
∂
µ=
( ) ( )[ ]
2
1
2
2
1
22
1
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+=+→
x
y
dxdydxdx
∂
∂
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
−⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+= dx
x
y
dxTdU
2
1
2
1
∂
∂
a
b
dx
dy
v
O x
x
考虑弦上距原点O 为 处
的原长为 的线元xd ab
x
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+⎟
⎠
⎞
⎜
⎝
⎛
+= dx
x
y
dxT L
2
2
1
1
∂
∂
返回 退出
在 t 时刻,线元 ab 偏
离平衡位置到 y 处,
其振动速度为
t
y
∂
∂
21. ( )kxtAdxdE −= ωµω 222
sin
S
1
设弦的横截面积为 S
Sρµ =
dxS
dE
w = ( )kxtA
S
−= ωω
µ 222
sin
( )kxtA −= ωρω 222
sin
22
0
2
11
Awdt
T
w
T
ρω=>=< ∫
退出返回
22. snapshot
The oscillating string element thus has both its maximum kinetic
energy and its maximum elastic potential energy at y = 0
When the string element is at its y = ym position, its length has its
normal undisturbed value dx, so its elastic potential energy is
zero. However, when the element is rushing through its y = 0
position, it is stretched to its maximum extent, and its elastic
potential energy then is a maximum.
退出返回
23. snapshot
The regions of the string at maximum displacement have no
energy, and the regions at zero displacement have maximum
energy.
As the wave travels along the string, forces due to the tension in
the string continuously do work to transfer energy from regions
with energy to regions with no energy.
退出返回