目錄

• DCT
• • 定義
  • • c程序
    • c結(jié)果
    • matlab結(jié)果
  • 測試的問題
  • • • 得出的數(shù)據(jù)
  • 修改成功后

DCT

定義

給定序列x(n),n=0,1…N-1,其離散余弦變換定義為

$X_c(0)=\frac{1}{\sqrt{N}}{\sum }_{n=0}^{N?1}x(n)$

$X_c(k)=\sqrt{\frac{2}{N}}{\sum }_{n=0}^{N?1}x(n)cos\frac{(2n+1)?k?\pi }{2N},n=0,1...,N?1$

其變換的核函數(shù)

$C_{k,n}=\sqrt{\frac{2}{N}}{g}_{k}cos\frac{(2n+1)k\pi }{2N},k,n=0,1....,N?1$

是實數(shù)，式中系數(shù)

c程序

#include<stdio.h> #include<math.h> #define N 64 #define pi 3.14159 int main() {int n,k;float y[100],xn[100];//輸出，輸入信號for(k=0; k<N; k++)//循環(huán)求X(k){if(k==0){for(n=0; n<N; n++){xn[n]=cos(n*pi/6.0);y[n]=sqrt(1.0/N)*xn[n];//公式y(tǒng)[k]+=y[n];}}else{for(n=0; n<N; n++){xn[n]=cos(n*pi/6.0);y[n]=sqrt(2.0/N)*xn[n]*cos((2.0*n+1)*k*pi/(2.0*N));//公式y(tǒng)[k]+=y[n];}}printf(" %d %lf\n",k,y[k]);}}

matlab

>> N=64; >> n=0:1:N; >> xn=cos(n*pi/6); >> figure >> y=dct(xn); >> stem(n,y)

c結(jié)果

matlab結(jié)果

對比有點差異，跟FFT的差異相似。
有大佬知道怎么修改的請評論下，感謝

測試的問題

#include <stdio.h>//標(biāo)準(zhǔn)輸入輸出頭文 #include<math.h>//聲明了常用的一些數(shù)學(xué)運(yùn)算 #define pi 3.1415 #define N 64//64個點 int main() {int n,k;double xn;//輸入信號double y[100];//輸出，輸入信號for(k=0; k<N; k++)//循環(huán)求X(k){for(n=0; n<N; n++){xn=1.0;y[k] += xn;}printf("%d %f\n",k,y[k]);}}

得出的數(shù)據(jù)

0
-1.#QNAN0
1 64.000000
2
-1.#QNAN0
3 64.000000
4 64.000000
5 64.000000
6 64.000000
7
-1.#QNAN0
8 64.000000
9 64.000000
10 3072570228342306800000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
11 64.000000
12 64.000000
13 64.000000
14 64.000000
15 3072569644077048600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
16 64.000000
17 64.000000
18 64.000000
19 64.000000
20 64.000000
21 64.000000
22 64.000000
23 64.000000
24 64.000000
25 64.000000
26 64.000000
27 64.000000
28 64.000000
29 64.000000
30 64.000000
31 2833760479163969400000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
32 64.000000
33 64.000000
34 64.000000
35 64.000000
36 64.000000
37 64.000000
38 64.000000
39 64.000000
40 64.000000
41 3488645024638201700000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
42
-1.#QNAN0
43 3581851895918701500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000
44 64.000000
45 64.000000
46 64.000000
47 64.000000
48 64.000000
49 64.000000
50 64.000000
51 64.000000
52 64.000000
53 64.000000
54 64.000000
55 64.000000
56 64.000000
57 64.000000
58 64.000000
59 64.000000
60 64.000000
61 64.000000
62 64.000000
63 64.000000

發(fā)現(xiàn) 0，2，7，10，15，31，41，42，43是錯誤的。

修改成功后

發(fā)現(xiàn)是定義的范圍大小不夠,以及增添一個下x[n]函數(shù)，改了如下

double y[1000]，x[100]

#include<stdio.h> #include<math.h> #define N 64 #define pi 3.14159 int main() {int n,k;double xn;double y[1000],x[100];//輸出，輸入信號for(k=0; k<N; k++)//循環(huán)求X(k){if(k==0){for(n=0; n<N; n++){xn=cos(n*pi/6.0);x[n]=sqrt(1.0/N)*xn;//公式y(tǒng)[k]+=x[n];}}else{for(n=0; n<N; n++){xn=cos(n*pi/6.0);x[n]=sqrt(2.0/N)*xn*cos((2.0*n+1)*k*pi/(2.0*N));//公式y(tǒng)[k]+=x[n];}}printf(" %d %lf\n",k,y[k]);}}

然后我發(fā)現(xiàn)matlab的好像程序錯了，修改了下

> n=0:1:63; >> x=cos(n*pi/6); >> y=dct(x); >> figure >> stem(n,y)

c 的結(jié)果圖

matlab的結(jié)果圖

結(jié)果圖跟matlab相差不多

總結(jié)

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