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# radix 2 disspation in frequency

Asked by ayman osama on 5 Dec 2012
```for K=0:7
for N=0:3
if mod(K,2)==0
v(N+1)=(x(N+1)+x(N+5))*exp(-1i*N*K*pi/2);
elseif  mod(K,2)==1
v(N+1)=((x(N+1)-x(N+5))*exp(-1i*N*pi/4))*exp(-1i*N*K*pi/2);
end
```
```      end
Y(K+1)=sum(v);
end```

i assume that is radix 2 disspation in frequency what's wron in it

Walter Roberson on 5 Dec 2012

Are you seeing an error message? If not then what difference do you see between what you are getting and what you expected?

ayman osama on 5 Dec 2012

no i didn'y got an error msg i should have fft of x what i see that the first term in fft is right and other terms are wrong i was using dissipation in frequency teq.

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Answer by Azzi Abdelmalek on 6 Dec 2012
Edited by Azzi Abdelmalek on 6 Dec 2012
```x=rand(1,8);    % Example
xo=x(2:2:end);  % odd part
xe=x(1:2:end);  % even part
for N=0:3
for k=0:3
Xo(k+1)=xo(k+1)*exp(-j*2*k*N*pi/4)
Xe(k+1)=xe(k+1)*exp(-j*2*k*N*pi/4)
end
Yo(N+1)=sum(Xo)
Ye(N+1)=sum(Xe)
Y(N+1)=Ye(N+1)+exp(-j*2*N*pi/n)*Yo(N+1)
Y(N+5)=Ye(N+1)-exp(-j*2*N*pi/n)*Yo(N+1)
end
```