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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Vimium has been updated to 1.29.x

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])







    

    

    

    

    

    

    

    

    

    


 

word 第23,24,3,4课

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 hospitality

float
redundant
overall
Panther
economic
squeeze
 
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barge
gage??
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??
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===
24
tyranny
simple???
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enclosure

 
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stall
 

==4==
robust
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correctly
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vegetation
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latent
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bull
ther??
mild
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gorve
 

burial
pam aite
disastrous
symmestry

 

dictate
dose
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distress
dim
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disposition
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disposal
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eliminate
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asset
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bare
accordingly

 

 


 

bbb


 

ccc


 

ddd


 

eee


 

fff


 

 

山寨机就是山寨机 哎

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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!