# Relationship of input-outputs in Hartley transformation based on few elements in a vector

This question is related also to that question HERE, but with further details and another issue I need to solve.

Assuming we have a vector $$X = [x_0,x_1,\ldots,x_N]$$, its Hartley transformation is equal to $$H(X) = \Re(FFT(X)) - \Im(FFT(X))$$. where $$H$$ denotes the Hartley transformation, and $$FFT$$ is the discrete Fourier transformation. Which means that $$y = H(X)$$. My question : Is it possible to fix some values in the vector $$X$$, for example, $$X_{1:4:N}$$ to be any values in order to have their correspondent values $$y_{1:4:N}$$ equal to $$[1,1,\ldots,1]?$$. What supposed to be the valued of $$X_{1:4:N}$$ to have the results of $$y_{1:4:N}=[1,1,\ldots,1]?$$

$$NP:$$ The values in both vectors $$x$$ and $$y$$ in the other index of $$1:4:N$$ are random data.

• Since each entry in the input makes some contribution to each entry in the output, I don't think there is a clean relationship of the kind that I suspect you want. That is, one needs very specific cancellations to guarantee 1s in specific entries of the output. There might be some special values of N that have some simpler relations. What kinds of N are under consideration? May 20 '20 at 20:31
• @JoeMac Thank you for you comment, $N = 1024$ is under consideration. Second, you said "There might be some special values of $N$ that have some simpler relations." Which $N$ have simpler relations?
– Gze
May 20 '20 at 22:17
• For N = 1024, for example, if x = (1,0,0,0,...) with that pattern repeated 1024/4 = 256 times, and if y = DHTx, then y is almost all 0, except (with MATLAB indexing) y(1) = y(257) = y(513) = y(769) = 256. May 21 '20 at 0:38
• @JoeMac what you mean by x = (1,0,0....) is it all the vector $X$ of size $N$ or the vector of size 256 representing $X_{1:4:N}?$
– Gze
May 21 '20 at 2:50