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untitled2.asv
% d=[1,1,0,1,1,0,0,0,0,1,1,0,0,1,0,0,1,1]
% for n=1:2:16
% if d(n)==0&d(n+1)==0;select=1
% elseif d(n)==0&d(n+1)==1;select=2
% elseif d(n)==1&d(n+1)==0;select=3
% else d(n)==1&d(n+1)==1;select=4
untitled2.m
d=[1,1,0,1,1,0,0,0,0,1,1,0,0,1,0,0,1,1]
for n=1:2:16
if d(n)==0&d(n+1)==0;select=1
elseif d(n)==0&d(n+1)==1;select=2
elseif d(n)==1&d(n+1)==0;select=3
else d(n)==1&d(n+1)==1;select=4
end
sw
qpsk_mod.m
function [d] = qpsk_mod(b1,numbits) % qpsk modulation
%
% pairs of bits,b(i),b(i+1)are mapped to complex-valued modulation symbols d_symb=Re+j*Im
% modulation scheme: b(i) b(i+1)| Re |
untitled2.asv
% d=[1,1,0,1,1,0,0,0,0,1,1,0,0,1,0,0,1,1]
% for n=1:2:16
% if d(n)==0&d(n+1)==0;select=1
% elseif d(n)==0&d(n+1)==1;select=2
% elseif d(n)==1&d(n+1)==0;select=3
% else d(n)==1&d(n+1)==1;select=4
untitled2.m
d=[1,1,0,1,1,0,0,0,0,1,1,0,0,1,0,0,1,1]
for n=1:2:16
if d(n)==0&d(n+1)==0;select=1
elseif d(n)==0&d(n+1)==1;select=2
elseif d(n)==1&d(n+1)==0;select=3
else d(n)==1&d(n+1)==1;select=4
end
sw
vowels.c
#include
void main()
{
char letter;
int vowel_count = 0;
for (letter = 'A'; letter
vowels.c
#include
void main()
{
char letter;
int vowel_count = 0;
for (letter = 'A'; letter
习题－38.c
//本程序只给出了算法思想
//读者可以自己完善本程序
typedef enum {RED,WHITE,BLUE} color; //三种颜色
void Flag_Arrange(color a[ ],int n)
//把由三种颜色组成的序列重排为按照红,白,蓝的顺序排列
{
i=0;j=0;k=n-1;
while(j
vowels.c
#include
void main()
{
char letter;
int vowel_count = 0;
for (letter = 'A'; letter
mtd_print.m
function []=mtd_print(res,file_sal)
% PURPOSE: Save output of multivariate temporal disaggregation methods
% ------------------------------------------------------------
% SYNTAX: []=mtd_print(res,