帮帮忙，能不能给我 基于C语言的FIR滤波器设计的程序代码（包括CMD，C,ASM)，谢谢了 真的很急！！！

#include"math.h"

void firwin(n,band,fln,fhn,wn,h)

int n,band,wn

double fln,fhn,h[]

{int i,n2,mid

double s,pi,wc1,wc2,beta,delay

double window()

beta=0.0

if(wn==7)

{printf("input beta parameter of Kaiser window(2<beta<10)\n")

scanf("%1f",&beta)

}

pi=4.0*atan(1.0)

if((n%2)==0)/*如果n是偶数*/

{n2=n/2+1/*这行什么意思*/

mid=1

}

else

{n2=n/2

mid=0

}

delay=n/2.0

wc1=2.0*pi*fln

if(band>=3) wc2=2.0*pi*fhn/*先判断用户输入的数据，如果band参数大于3*/

switch(band)

{case 1:

{for(i=0i<=n2i++)

{s=i-delay

h[i]=(sin(wc1*s)/(pi*s))*window(wn,n+1,i,beta)

h[n-i]=h[i]

}

if(mid==1) h[n/2]=wc1/pi

break

}

case 2:

{for(i=0i<=n2i++)

{s=i-delay

h[i]=(sin(pi*s)-sin(wc1*s))/(pi*s)

h[i]=h[i]*window(wn,n+1,i,beta)

h[n-i]=h[i]

}

if(mid==1) h[n/2]=1.0-wc1/pi

break

}

case 3:

{for(i=0i<n2i++)

{s=i-delay

h[i]=(sin(wc2*s)-sin(wc1*s))/(pi*s)

h[i]=h[i]*window(wn,n+1,i,beta)

h[n-i]=h[i]

}

if(mid==1)h[n/2]=(wc2-wc1)/pi

break

}

case 4:

{for(i=0i<=n2i++)

{s=i-delay

h[i]=(sin(wc1*s)+sin(pi*s)-sin(wc2*s))/(pi*s)

h[i]=h[i]*window(wn,n+1,i,beta)

h[n-i]=h[i]

}

if(mid==1)h[n/2]=(wc1+pi-wc2)/pi

break

}

}

}

static double window(type,n,i,beta)

int i,n,type

double beta

{int k

double pi,w

double kaiser()

pi=4.0*atan(1.0)

w=1.0

switch(type)

{case 1:

{w=1.0

break

}

case 2:

{k=(n-2)/10

if(i<=k)

w=0.5*(1.0-cos(i*pi/(k+1)))

break

}

case 3:

{w=1.0-fabs(1.0-2*i/(n-1.0))

break

}

case 4:

{w=0.5*(1.0-cos(2*i*pi/(n-1)))

break

}

case 5:

{w=0.54-0.46*cos(2*i*pi/(n-1))

break

}

case 6:

{w=0.42-0.5*cos(2*i*pi/(n-1))+0.08*cos(4*i*pi/(n-1))

break

}

case 7:

{w=kaiser(i,n,beta)

break

}

}

return(w)

}

static double kaiser(i,n,beta)

int i,n

double beta

{

double a,w,a2,b1,b2,beta1

double bessel0()

b1=bessel0(beta)

a=2.0*i/(double)(n-1)-1.0

a2=a*a

beta1=beta*sqrt(1.0-a2)

b2=bessel0(beta1)

w=b2/b1

return(w)

}

static double bessel0(x)

double x

{int i

double d,y,d2,sum

y=x/2.0

d=1.0

sum=1.0

for(i=1i<=25i++)

{d=d*y/i

d2=d*d

sum=sum+d2

if(d2<sum*(1.0e-8)) break

}

return(sum)

}

这是窗函数法的，当然还有其他的比如切比雪夫，零相位滤波什么的，我也在研究，不是很懂哈

#include<stdio.h>

void main()

{

char ch

int num1,num2,num3,num4,m

num1=0

num2=0

num3=0

num4=0

puts("Type in a sentence,then press <ENTER>\n")

while ((ch=getchar())!='\n')

{

if(ch>=48&&ch<=57) num1++

else if(ch>=65&&ch<=90) num2++

else if(ch>=97&&ch<=122) num3++

else num4++

}

printf("%4d%4d%4d%4d",num1,num2,num3,num4)

}

帮你改了一下 你的太复杂了 可以运行

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