Matlab - how to reverse a function?

Author

how to reverse a function?

2006-09-30, 7:07 pm

like x[n] = [1+2i, 3+4i, 5+6i], find out x[-n]

Duncan Po

2006-09-30, 7:07 pm

PZ wrote:
>
>
> like x[n] = [1+2i, 3+4i, 5+6i], find out x[-n]
>
>

Do you mean flip an array? If so, there is the function FLIPLR:
[color=darkred]

x =

1.0000 + 2.0000i 3.0000 + 4.0000i 5.0000 + 6.0000i
[color=darkred]

y =

5.0000 + 6.0000i 3.0000 + 4.0000i 1.0000 + 2.0000i

2006-09-30, 7:07 pm

thanks. Close how to specify index 0?

I am supposed to find out fft(conj(array) = conj(fliplr (fft(array))).
basically the fft of a conjugate of one sequence equals to the
conjugate of the -k/N of the fft of the sequence, (messy, guess you
know what I wrote).

The problem is it is -k, it is supposed to start from -5, -4 -3, -2...
If I just use fliplr, it is indexing from zero, right? how to express
this?

pz
Duncan Po wrote:
> PZ wrote:
>
> Do you mean flip an array? If so, there is the function FLIPLR:
>
>
> x =
>
> 1.0000 + 2.0000i 3.0000 + 4.0000i 5.0000 + 6.0000i
>
>
> y =
>
> 5.0000 + 6.0000i 3.0000 + 4.0000i 1.0000 + 2.0000i

2006-09-30, 7:07 pm

OK, i am wrong. it is supposed to be mod, N-K, not fliplr.

let's say [1+2i, 3+4i, 5+6i, 7+8i],

how to find out
[1+2i, 7+8i, 5+6i, 3+4i ]

basically, zero postion wont' chanage, the rest reversed, 1 to final
postion, the final one goes to 1 position?

PZ wrote:[color=darkred]
> thanks. Close how to specify index 0?
>
> I am supposed to find out fft(conj(array) = conj(fliplr (fft(array))).
> basically the fft of a conjugate of one sequence equals to the
> conjugate of the -k/N of the fft of the sequence, (messy, guess you
> know what I wrote).
>
> The problem is it is -k, it is supposed to start from -5, -4 -3, -2...
> If I just use fliplr, it is indexing from zero, right? how to express
> this?
>
>
> pz
> Duncan Po wrote:

Greg Heath

2006-09-30, 7:07 pm

CORRECTED FOR TOPPOSTING
PZ wrote:
> Duncan Po wrote:
>
> thanks. Close how to specify index 0?
>
> I am supposed to find out fft(conj(array) = conj(fliplr (fft(array))).
> basically the fft of a conjugate of one sequence equals to the
> conjugate of the -k/N of the fft of the sequence, (messy, guess you
> know what I wrote).
>
> The problem is it is -k, it is supposed to start from -5, -4 -3, -2...
> If I just use fliplr, it is indexing from zero, right? how to express
> this?

MATLAB only uses positive indices. Negative frequency fft
components are moved to the beginning of the array using
FFTSHIFT. However, the zero frequency component at
index = 1 is only moved to the exact middle of the array when
N is odd. For example

fftshift( [1 2 3 4 5] ) = [4 5 1 2 3]
whereas
fftshift( [1 2 3 4 5 6] ) = [4 5 6 1 2 3]

similarly, when FLIPLR is used next, the zero frequency
component only remains in the exact middle if N is odd.

fliplr( [4 5 1 2 3] ) = [3 2 1 5 4]
whereas
fliplr( [4 5 6 1 2 3]) = [ 3 2 1 6 5 4]

Therefore (FLIPUD used for columns):

Y1 = fft(conj(x))
Y2 = conj(fft(x))
Y3 = fftshift(Y1)
y4 = fftshift(Y2)

%For arbitrary N

Y1 = [ Y2(1); flipud(Y2(2:end)) ]

%For odd N

Y3 = flipud(Y4);

%For even N

Y3 = f[Y4(1); flipud(Y4(2:end))]

Running the following code may help:

clear all, clc
for N = 5:6
N=N
x = randn(N,1)+i*randn(N,1)
Y1 = fft(conj(x));
Y2 = conj(fft(x));
Y1Y2 = [Y1 Y2]
E12 = Y1-[Y2(1);flipud(Y2(2:end))];
E21 = Y2-[Y1(1);flipud(Y1(2:end))];
E12E21 = [E12 E21]'

Y3 = fftshift(Y1);
Y4 = fftshift(Y2);
Y34 =[ Y3 Y4 ]

if floor(N/2) < N/2 % N odd
E34 = Y3-flipud(Y4);
E43 = Y4-flipud(Y3);
else % N even
E34 = Y3-[Y4(1); flipud(Y4(2:end))];
E43 = Y4-[Y3(1); flipud(Y3(2:end))];
end
E34E43 = [E34 E43]'

end

Hope this helps.

Greg

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