第4回

帯域制限された通信システム
Digital transmission through bandlimited channel
6．1 帯域制限された伝送路
Characterization of bandlimited channels and channel distortion
電話線や無線などの伝送路：帯域制限された線形フィルタ
Many communication channels, including telephone channels and radio channels are
generally characterized as bandlimited linear filter
伝送路の周波数応答C(f) Channel frequency response C(f) is expressed as
C ( f )  A( f )e j ( f )
(6.1.1)
A(f)は振幅応答，θ(f)は位相応答
A(f) is called the amplitude response, and θ(f) is called the phase response.
群遅延 Group delay
1 d  f 
(6.1.2)
f  
2 df
理想伝送路：送信信号が占有する帯域W内で，A(f)が一定かつθ(f)が周波数の線
形関数の伝送路
Non distorting channel: A(f)=constant, θ(f) is a linear function of frequency or the
group delay=constant within the bandwidth W occupied by the transmitted signal.
占有帯域幅W内でA(f)とτ(f)が一定でないと伝送路は信号を歪ませる。
If A(f) and τ(f) are not constant, channel distorts the signal.
振幅ひずみ：A(f)が一定でないひずみ Amplitude distortion: A(f) is not constant
遅延ひずみ： τ(f)が一定でないひずみ Phase distortion: τ(f) is not constant
理想でない周波数応答C(f)をもつ伝送路を通過する連続送信パルス
振幅，遅延歪みにより受信機では明瞭なパルスにならない。
As a result of amplitude and delay distortion caused by the nonideal channel
frequency response, a succession of pulse transmitted through the channel are no
longer distinguishable at the receiving terminal.
パルスが重なってシンボル間干渉(ISI)を起こす。
A sequence of successive pulses are smeared into one another.
The channel delay distortion results in intersymbol interference
シンボル間干渉の例：Example of ISI
図6.1(a)：±T，±2T，…の時間で周期的に0と
なる情報を載せたパルスが伝送される時，各
ピークは他のパルスの0にあたる位置にある。
Bandlimited pulse having zero periodically spaced
In time at point labeled ±T，±2T，…
When information is conveyed by pulse amplitude.
We can transmit a sequence of pulse, each of
which has a peak at periodic 0 of the other pulse.
理想でない伝送路を通過すると，周期的に0と
交叉しない（図6.1(b)）
Transmission of the pulse through the nonideal channel results in the received pulse having zero
crossing that are no longer periodically spaced.
連続パルス系列は干渉しあい，正しいパルスの振幅が観測できない。
A sequence of successive pulses are smeared into one another, the peak of the pulse no longer
are distinguishable.
等化器により伝送路の理想的でない周波数応答特性を補正できる。
Equalizer compensate the non ideal frequency response.
例：電話線でのシンボル間干渉 ISI on a telephone channel
図6.2：電話線の伝送路の振幅と群遅延の周波数応答モデル
Average amplitude and delay characteristics of a telephone channel
伝送路の通過帯域は300Hzから4000Hz Usable band extends from 300 to 4000Hz
インパルス応答：応答区間はパルスピーク前後10msの間に広がる
Impulse response duration is 10ms
送信シンボルレートを2500パルス/秒とすれば，20-30シンボルにわたってISIが発生。
If transmitted symbol rate is 2500 pulse per second, ISI extends over 20 to 30 symbols
6.2 シンボル間干渉 Characterization of Intersymbol Interference
送信信号：ベースバンドのパルス振幅変調（PAM）
Transmitted PAM signal is expressed as

s (t )   an g (t  nT )
(6.2.1)
g(t)：送信信号のスペクトル特性を決めるパルス形状，
g(t):basic pulse shape that controls the spectral characteristics of the transmitted signal
{an}：M点のコンスタレーションから選ばれた送信情報シンボル系列，
{an}：sequence of the transmitted information symbol selected from a signal constellation
consisting of M points
T：信号間隔(1/Tはシンボルレート) T: signal intervals(1/T is the symbol rate)
周波数応答C(f)のベースバンド伝送路を通過した受信信号
The received signal pass through the channel whose response C(f) is expressed as
n 0

r (t )   an h(t  nT )  w(t )
(6.2.2)
h(t)=g(t)*c(t)：c(t)は伝送路のインパルス応答，*は畳み込み, w(t)：加法性雑音
c(t) is the impulse response of the channel, * denotes convolution, and w(t) represents the
additive noise.
受信信号は受信フィルタを通過し1/Tサンプル/秒で標本化
The received signal passed through a receiving filter and sampled at the rate 1/T.
受信フィルタの周波数応答をH*(f)，x(t)を受信フィルタの信号パルス応答とすれば，周波数領
域で考えるとX(f)=H(f)H*(f)=|H(f)|2
The frequency response of the received filter is H*(f), x(t) is the signal pulse response.
When we consider the response in the frequency domain, X(f)=H(f)H*(f)=|H(f)|2.
n 0
これを時間領域に戻して受信信号をあらわす The filter output in the time domain is

y (t )   a n x(t  nT )  v(t )
(6.2.3)
v(t)：w(t)に対する受信フィルタ応答 response of the receiving filter to the noise w(t)
y(t)をt=kT，k=0,1,2,..で標本化する。If y(t) is sampled at times t=kT, k=0,1,2,...
n 0

y (kT )   an x(kT  nT )  v(kT )
n 0
y k   a n xk  n  vk
n 0
(6.2.4)
これを書きなおすと rewrite yk




1 
y k  x0  a k   a n x k  n   v k
x0 n  0


n k


k=1,2,… (6.2.5)
x0は任意のスカラで，簡単のため1とする。
x0 is an arbitrary scale factor, for convenience set it to unity

y k  ak   an xk n  vk
k=1,2,… (6.2.6)
n 0
nk
akはk番目の標本時間の所望情報シンボル
ak represents the desired information symbol at the k-th sampling instant

(6.2.7)
a
n 0
n k
n
xk n
はISI。vkはk番目の標本時間の加法性雑音。
is ISI. and vk is the additive white noise at the k-th sampling instant.
アイパターン Eye pattern
ISIと雑音の大きさをオシロスコープで観測。
The amount of ISI and noise in a digital communication
system can be observed by an oscilloscope.
水平の掃引率を1/Tに設定して，受信信号を表示させ
る。
Display the received signal y(t) on the vertical input
with the horizontal sweep rate set at 1/T.
ISIの影響があるとアイの開きは狭くなり，加法性雑音
に対するマージンが減って，誤りが起きやすくなる。
ISI distorts the position of the zero crossings and causes
a reduction in the eye opening. It causes the system to more sensitive to a noise.
例6-1 ISI
次のインパルス応答を持つ2つの伝送路について，受信信号系列{yk}のISIの影響を考える。
Consider the effect of ISI on the received signal sequence {yk} for 2 channel
1 n  0
 0.25 n  1

xn  
0.1 n  2
0otherwise
1 n  0
0.5 n  1

xn  
 0.2 n  2
0otherwise
送信信号系列：2値{an=±1}と仮定する。 Transmitted signal sequence is binary
伝送路1：雑音のない受信信号系列{yn}(図6．6の上），
For channel1, the received signal in the absence of noise is shown in figures of upper left hand
σ2=0.25の分散を持つ白色ガウス雑音を加えると図6.6の下。
With additive white Gaussian noise with variance of σ2=0.25 is shown in figures of lower left
雑音がなければ0に設定した閾値と比較する判定器で誤りを起こさない。加法性雑音が大きく
なると誤りが発生する。
In the absent of noise, ISI alone does not cause errors at the detector that compares the
received signal sequence with the threshold set to 0. When the additive noise is sufficiently
large, errors will occur.
伝送路2：雑音がないときとある (σ2=0.1)ときの受信系列{yn}(図6.7)。ISIは雑音がなくても発生。
For channel2, the received signal in the absence of noise and with additive white Gaussian noise
with variance of σ2=0. is shown in figures of upper and lower right hand. Even in the absence of
noise , errors occur.
伝送路1 Channel 1
伝送路2 Channel 2
6.3 帯域制限された伝送路の通信システムの設計
Communication System Design for Bandlimited Channels
帯域制限された伝送路で，ゼロISIとなる送信，受信フィルタを設計
Consider the design of the transmitter and receiver filter suitable for a baseband
bandlimited channel. It results in zero ISI.
伝送路は理想(帯域内でA(f)とτ(f)は一定) Channel is ideal, that is A(f) and (f) is constant.
簡単のためA(f)=1，τ(f)=0。 For simplicity A(f)=1,τ(f)=0.
ゼロISIを持つ帯域制限された信号の設計：Signal design for zero ISI
ナイキストの第1基準：インパルス応答が符号間隔Tごとにゼロクロス
Design of bandlimited signals with zero ISI was considered by Nyquist about 80 years ago.
Its impulse response has zero crossing every symbol interval T.
レイズドコサイン周波数応答特性
Most commonly used signal in practice has a raised cosine frequency response defined as
 (1   )

T

0




T
 
T 
   (1   )
 (1   )
T
(6.3.1)
X rc ( f )   1  sin
 
    
2 
T 
T
T
2 
 (1   )

0




T
α：ロールオフ率(0≤α≤1) 1/T：シンボルレート
α is called the roll-off factor in the range 0≤α≤1, and 1/T is the symbol rate.
周波数応答Xrc(f)(α=0，α=0.5，α=1）の例：図6.8
The frequency response for α=0，α=0.5，α=1 is illustrated in next slide.
α=0の時，Xrc(f)は物理的に実現できない矩形応答 for α=0, it is physically nonrealized.
1/(2T)：ナイキスト周波数 Nyquist frequency
超過帯域幅：α>0では所望信号Xrc(f)の帯域幅はナイ 1
キスト周波数1/(2T)を超える。超過帯域幅はナイキス 0.9
ト周波数の百分率で表される。α=0.5のとき超過帯域 0.8
0.7
幅は50%，α=1のとき超過帯域幅は100%
Excess bandwidth: For α>0, bandwidth occupied by 0.6
Xrc(f) beyond Nyquist frequency. Excess bandwidth is 0.5
expressed by percentage of Nyquist frequency. when 0.4
0.3
α=0.5, excess bandwidth is 50%.
レイズドコサイン周波数応答を持つ信号パルスxrc(t)は0.2
0.1
The signal pulse xrc(t) having the raised cosine
0
0
spectrum is
sin( t / T ) cost / T 
(6.3.2)
xrc (t ) 
2 2
2
(t / T ) 1  4 t / T
alpha=0
alpha=0.5
alpha=1
5
10
15
fN=1/(2T)
チャネルひずみがなければ標本時間t=kT，k≠0で隣接シンボル間でISIが生じない。伝送路歪
が発生するとISIは0でなく，等化器が必要となる。
At sampling instance t=kT，k≠0, there is no ISI when there is no channel distortion.
In the presence of channel distortion, the ISI given by 6.2.7 is no longer 0. In this case,
equalizer is required.
ゼロISIとするには以下の式を満足させる送信フィルタGT(f)と受信フィルタGR(f)を適用する
To yield 0 ISI, transmitter and receiver filter with frequency response of GT(f) and GR(f) are
designed so as to satisfy
GT ( f )GR ( f )  X rc ( f )
(6.3.3)
と選ばれると，標本時間t=nTでISIは0となる。Then, ISI at t=nT is 0
ここでXrc(f)はレイズドコサイン周波数応答を持つ
X (f) is the raised cosine frequency response.
例6-2 コサインロールオフフィルタ Optimum transmitter and receiver filter
式(6.3.3)を満足するGR(f)とGT(f)を設計せよ。
Design transmitter and receiver filter GR(f) and GT(f) such that their product satisfies 6.3.3
解
所望の振幅応答は以下のとおりである。The desired magnitude response is
GT ( f )  GR ( f )  X rc ( f )
(6.3.4)
Xrc(f)は式(6.3.1)で与えられる。 Xrc(f) is given by 6.3.1
デジタルフィルタの周波数応答はインパルス応答ｇT(n)によって表せる。
The frequency response is related to the impulse response of the digital filter by
GT ( f ) 
( N 1) / 2
 j 2fnTs
g
(
n
)
e
 T
(6.3.5)
Tsは標本間隔，Nはフィルタの長さ(奇数)。標本周波数Fsは
Ts is the sampling interval and N is the length of the filter(N is odd). Sampling frequency sets
Fs 
( N 1) / 2
1 4

Ts T
(6.3.6)
このとき折り返し周波数はFs/2=2/T。 The folding frequency is Fs/2=2/T
GT ( f )  X rc ( f ) から Xrc(f)を等間隔周波数Δf=Fs/Nで標本化
Sample Xrc(f) at equally spaced points in frequency, with frequency separation Δf=Fs/N
( N 1) / 2
 mFs 
 j 2mn / N
X rc (mf )  X rc 
   gT (n)e
 N  n N 1 / 2
(6.3.7)
逆フーリエ変換の関係から From the inverse transform relation
g T ( n) 
( N 1) / 2

n    N 1 / 2
 4m  j 2mn / N
X rc 
e
 NT 
n  0,1,  ,
N 1
2
gT(n)は対称で，所望の線形位相送信フィルタはgT(n)を(N-1)/2サンプル遅らせる。
gT(n) is symmetric, the impulse response of the filter is delaying gT(n) by (N-1)/2 sample.
フィルタのインパルス応答
Impulse response
送信FIRフィルタの周波数応答
Frequency response of FIR filter at transmitter
カスケード接続したフィルタのインパルス応答
Impulse response of the cascade of the transmitter and receiver filter
アイパターン
Eye pattern
Xrc関数
T
T 
   (1   )
 (1   )
1

sin
|

|







2 
2 
T 
T
T
T
T   1     
 (1   )
X

1

sin
 


→
rc

2
2  T
T 
T
X rc 


 (1   )
T
→
T
T    T 
  
1

sin


1

sin


    T


2
2  T  2 
 2 
T
T 
1   
1

cos
f




2
 
2T 
T
T    T 
  
 1  sin

  1  sin    0
2
2  T  2 
 2 
X rc 
T
T 
1      (1   )  f   (1   )
X rc  1  cos
 f 

2T
2T
2
 
2T 
T
T  1   1   
(1   ) →
X rc  1  cos


  T
f 
2

2
T
2
T
2T



f 
(1   )
2T
→
→
1
1 
 f 
2T
2T
T
T  1   1    T 
T  2 
1

cos


1

cos







2
  2T
2T  2 
  2T 
T
 1  cos    0
2
X rc 

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