Để giải quyết vấn đề này chúng ta cần xử lý biên độ phổ một chút bằng hàm log. Hàm logarit sẽ sửa độ khuếch đại, và thay thế cho hiển thị phổ |H(u,v)| chúng ta hiển thị: D(u,v) = log10(1+|H(u,v)|) | xr j xr j Tr xi j xi j Ti if iter 0 for k 1 k ip k i k kk-1 for j k j N j incr i j ip Tr xr i wr l -xi i wi l Ti xr i wi l xi i wr l xr i xr j -Tr xi i xi j -Ti xr j xr j Tr xi j xi j Ti kk 1 ip l incr l This procedure is for obtaining the transpose of a large file of complex numbers stored on secondary memory. We assume that the file is large to the point that it would not fit in active memory. struct COMPLEX float REAL 126 float IMAG void transpose FILE fptr int N int n Algorithm int N1 inc int iter i k int k1 inc1 int k2 j k3 k4 NS struct COMPLEX buff1 buff2 tmp long loc NT NS sizeof struct COMPLEX NT N NS buff1 struct COMPLEX malloc NT buff2 struct COMPLEX malloc NT N1 N 2 inc 1 inc1 2 for iter 0 iter n iter gotoxy 1 6 printf Iteration number 4c. iter 1 k1 0 for k 0 k N1 k for i k1 i k1 inc i loc NT long i if fseek fptr loc SEEK_SET 0 perror fseek failed exit 1 else fread buff1 NT 1 fptr loc NT long j inc if fseek fptr loc SEEK_SET 0 127 perror fseek failed exit 1 else fread buff2 NT 1 fptr k3 0 for k2 0 k2 N1 k2 for j k3 j k3 inc j tmp buff1 j inc buff1 j inc buff2 j buff2 j tmp k3 inc1 loc NT long i if fseek fptr loc SEEK_SET 0 perror fseek failed exit 1 else fwrite buff1 NT 1 fptr loc NT long i inc if fseek fptr loc SEEK_SET 0 perror fseek failed exit 1 else fwrite buff2 NT 1 fptr k1 inc1 inc 2 inc1 2 N1 2 .
