OFDM 十六讲9 How to Avoid ISI in Digital Communications

发布于:2023-01-20 ⋅ 阅读:(409) ⋅ 点赞:(0)

前言:

  

 前面讲过Nyquist Zero ISI Theorem,主要是讲采样频率 必须大于2倍主信息频率。

 f_s>2f_w

直观上跟数据发送的速率也有关系,这篇就是把它跟数据发送的速率结合起来了

这篇主要结合Nyquist Theorem讨论一下数字通讯系统中最大的发送数据速率和W关系。


一  整体流程

      

       这个 线性系统(linear system),是3个过滤器的卷积。

       x(t)=h_T(t)\star h_c(T)\star h_R(t)

     h_T: a filter in your send,impulse response

    h_c:  a filter in your channel, impulse response.这个是难以改变的,比如多径传输,直接取决于信道本身,一般都是设计其它两个过滤器

    h_R: a filter in your receiver.

  

 

 我们就是要设计这种过滤器

  x_n=\left\{\begin{matrix} A,n=0\\ 0, n \neq 0 \end{matrix}\right.


二  NYQUIST ZERO ISI therorem

      x(t) 傅里叶变换结果为常数

  定理1:   \sum_{m=-\infty}^{\infty}X(f-\frac{m}{T})=T,

     m =0

           

m>0时候:

2.1:     \frac{1}{T}>2w

       

           

          从图中我们可以看到,其和不是一个常数。   

           不能满足上面定理1,会产生ISI

  2.2      \frac{1}{T}=2w

 

如果设计发送接收的滤波器,使得 频域信号为square,  ISI也能为0

 each one of these elements is one of 
the elements of this summation and the 
overall function is the overall function 
that i've drawn and it has to equal the same value 
for all values of frequency and the way i've 
drawn it here ,it does not do that .
  however there is a special shape that would achieve 
that , so instead of this shape that i've drawn hre , if you 
could design your transmit filter and receive filter so 
the over all square was a square ,then i think you can see that if
these shapes were replaced by a square , when they added up 
they would be absolutely flat ,acroos all the values of frequency 
and they would be exactly the same because they can each one of these 
shape is contained between -w and w
 

 1.3  发送速率小于2倍采样频率

 从图上可以看到频域中,降低的部分由ISI 部分补上去了,其最终结果是一个常数

满足定理1


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