8 3

z变换求响应,离散时间系统分析

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从这个

http://jpkc.wyu.cn/xhyxt/kejian/chapter5/%E4%BE%8B%E9%A2%98/%E4%BE%8B5-27.htm

拷贝来的,备忘!!

【例5-27】已知离散时间系统的差分方程为

激励信号为，起始状态为，求该系统的系统函数、单位取样响应、零输入响应、零状态响应和完全响应。

解：（1）可以直接写出系统函数为

（2）对进行逆z变换就得系统的单位取样响应为

（3）系统的特征根为，所以，系统的零输入响应为，代入起始状态，得

          解得       

所以，系统的零输入响应为

（4）由得系统零状态响应的z变换为

对上式进行逆z变换，得系统的零状态响应为

（5）系统完全响应为零输入响应与零状态响应之和，即

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5 16

不知道能不能高亮显示，频偏仿真。

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%**************************************************************************
 
%The simulation adopted the model in paper "ML estimation of time and
 
%frequency offset" written by J.J van de Beek e.t.Programmed by binyue-Liu
 
%**************************************************************************
 
close all; clear all;
 
N=1024;                                     %length of one symbol
 
L=128;                                      %length of CP
 
SNR=15;
 
Symbol_num=6;
 
Bit_sym=4;                                  %16_QAM
 
s=randint(1,N*Symbol_num*Bit_sym);
 
s_map=qammod(2.^(Bit_sym-1:-1:0)*reshape(s,Bit_sym,N*Symbol_num),16);
 
%Reflecting to the 16_QAM constallation
 
t=ifft(reshape(s_map,N,Symbol_num))*N;        %OFDM modulation
 
t_add_CP=[t(N-L+1:N,:);t];                  %Addition with CP
 
SNR_line=10^(SNR/10);                   
 
transmit=reshape(t_add_CP,1,(N+L)*Symbol_num);
 
sigmma=(1^2+3^2)/2/SNR_line;
 
Noise=(randn(1,(N+L)*Symbol_num)+j*randn(1,(N+L)*Symbol_num))*sqrt(sigmma/2);
 
recieve=zeros(1,Symbol_num*(N+L));
 
erro_rate=zeros(1,10);
 
index=1;
 
for e=0.02:0.01:0.12;                     %The offset of norminal carrier frequency
 
    for i=1:Symbol_num;   
 
        k=1:N+L;
 
        recieve((i-1)*(N+L)+1:(N+L)*i)=transmit((i-1)*(N+L)+1:(N+L)*i).*exp(j*2*pi*e*k/N);
 
    end
 
    r=reshape(recieve+Noise,N+L,Symbol_num);
 
    r1=reshape(fft(r(L+1:end,:))/N,1,N*Symbol_num);
 
    Out_symbol=qamdemod(r1,16);
 
     
 
    Out_bit=zeros(4,N*Symbol_num);
 
    for i=1:N*Symbol_num;
 
      for k=1:Bit_sym
 
        Out_bit(k,i)=floor(Out_symbol(i)/(2^(Bit_sym-k)));
 
        if Out_bit(k,i)==1
 
           Out_symbol(i)=Out_symbol(i)-2^(Bit_sym-k);
 
        else
 
        end
 
      end
 
    end
 
    Out_bit=reshape(Out_bit,1,Bit_sym*N*Symbol_num);
 
erro_rate(index)=length(find(Out_bit~=s))/length(s);
 
index=index+1;
 
end
 
semilogy(0.02:0.01:0.12,erro_rate,'-*')
 
axis([0.01 0.12 10^(-5) 1])
 
 
 
 
 
 
 
     
 
     
 
     
 
     
 
     
 
     
 
     
 
     
 
     
 
    

 

4 14

word 第23,24,3,4课

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 hospitality

float
redundant
overall
Panther
economic
squeeze
 
editor？？
 
quest
barge
gage？？
dove
elegant
immerse
？
confirmation
cloudy
resident
commute
recur
prope？？？
erase
reflect？？？
embed
？？
advocate
 
===
24
tyranny
simple？？？
porch
velvet
propeller
enclosure
 
==3==
disorder
fell
shameful
settlement
toad 蟾蜍
board
fixture
obstinate
grumble
booklet
booth电话亭
kindle
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bourgeois中产阶级
bowel
confidential
catalogue 按目录
dissolved
pluck
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nourishment
bribe
bridegroom
marsh 湿地
stall
 
==4==
robust
contrive
footpath
overseas
viscous
instabilty
correctly
aviation航空
vegetation
denial
instructor
budget
challenge
latent
gigantic
bull
ther??
mild
commence
gorve
 
burial
pam aite
disastrous
symmestry

 

dictate
dose
discount
distress
dim
dispose
disposition
earnest
distant
disposal
dramatic
elastic
eliminate
dominate
asset
assess
bare
accordingly

 

 

 

bbb

 

ccc

 

ddd

 

eee

 

fff

 

 

4 13

山寨机就是山寨机 哎

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4 10

google要以手机和上网本打败传统PC了

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 三年时间虽然夸张了一点,不过未来趋势肯定是这样的。把计算复杂度都集中到云端,这样上网本就能有更大的优势.当终端只依赖于网速而不依赖于处理速度的时候,传统PC就要被淘汰了.当然这在中国还不现实，因为中国的网速还是这么慢！