[交流] 【求助】在matlab中如何用希尔伯特变换定义相位 已有3人参与

Originally posted by lijinfeng042 at 2010-05-28 18:58:48:

根据定义啊
例子
[code]
%余弦信号
n=0:512;
Ts=0.0125;
xn=cos(40*pi*n*Ts);
subplot(2,1,1);plot(n,xn);
xlabel('采样点数n');ylabel('原始图像x(n)');title('余弦信号')

xn1=hilbert(xn)
xr=re ...

Originally posted by lijinfeng042 at 2010-05-29 11:42:44:

按定义来啊 呵呵

Originally posted by lijinfeng042 at 2010-05-29 16:29:30:

呵呵 看到错误 你就知道hilbert的实质计算了吧 就是实部用fft 虚部iff计算的 但是你那样的话也不对啊 直接算不了的

Originally posted by lijinfeng042 at 2010-05-29 19:15:42:

http://www.docin.com/p-3374693.html
看看定义 呵呵并不是那么好积出来

Originally posted by lijinfeng042 at 2010-05-29 22:17:21:

看一下帮助
In detail, hilbert uses a four-step algorithm:

   1.

      It calculates the FFT of the input sequence, storing the result in a vector x.
   2.

      It creates a vector ...

Originally posted by lijinfeng042 at 2010-05-30 14:26:53:

无限接近事实啊 呵呵
>> A=[1 2 3 4]
A =
     1     2     3     4
>> xn=hilbert(A)
xn =
   1.0000 + 1.0000i   2.0000 - 1.0000i   3.0000 - 1.0000i   4.0000 + 1.0000i
>>

Originally posted by lijinfeng042 at 2010-05-30 14:26:53:

无限接近事实啊 呵呵
>> A=[1 2 3 4]
A =
     1     2     3     4
>> xn=hilbert(A)
xn =
   1.0000 + 1.0000i   2.0000 - 1.0000i   3.0000 - 1.0000i   4.0000 + 1.0000i
>>