Relationship between input and output sequence in Hartley transformation

Question

As you know that Discrete Hartley transformation is related to the discrete Fourier transformation, $i.e$, assuming we have a vector $X = [x_0,x_1,\ldots,x_N]$, its Hartley transformation is equal to $H(X) = Real(FFT(X)) - Imag(FFT(X))$. where $H$ denotes the Hartley transformation.

I wonder, if I want the output of the Hartley transformation equals to a vector of length $N$ whose all elements are $1$, it means $H(X) = [1,1,\ldots,1] $, what supposed to be the input $X$? I means I need to formulate the relationship between the inputs and outputs.

Answer 1

Wikipedia's entry for the discrete Hartley transform shows states that the $\mathsf{DHT}$ is, up to a scaling, its own inverse. If $x$ is a vector with $N$ entries and $y$ is its discrete Hartley transform, \begin{equation} y = \mathsf{DHT}x, \end{equation} then \begin{equation} x = \frac{1}{N}\mathsf{DHT}y. \end{equation}

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If $x$ is a vector with $N$ entries such that \begin{equation} \mathsf{DHT}x = \underbrace{(1,1,\ldots,1)^{\mathsf{T}}}_{\textrm{$N$ entries}}, \end{equation} then we recover $x$ with \begin{equation} x = \frac{1}{N}\mathsf{DHT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right). \end{equation} This means that $x$ is \begin{equation} \begin{split} x &=~ \frac{1}{N}\left(\mathsf{Re}\left[\mathsf{DFT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right)\right] - \mathsf{Im}\left[\mathsf{DFT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right)\right]\right), \end{split} \end{equation} where $\mathsf{DFT}$ is the discrete Fourier transform, which we usually compute with a FFT algorithm. The $\ell^{\textrm{th}}$ entry of the $\mathsf{DFT}$ of the all-1 vector is \begin{equation} \begin{split} \sum_{n=0}^{N-1}1\times e^{-2\pi j \ell n/N} &=~ 1 + e^{-2\pi j \ell/N} + \left(e^{-2\pi j \ell/N}\right)^2 + \cdots + \left(e^{-2\pi j \ell/N}\right)^{N-1}\\ &=~ \left\{\begin{array}{rl}N & \textrm{if}~\ell=0,\\0&\textrm{if}~\ell\neq 0.\end{array}\right.\\ &=~ N\delta_{\ell,0}, \end{split} \end{equation} where $\delta_{p,q}$ is the Kronecker delta. One way to show this is to note that if $\ell = 0$, then each exponent is $0$, so each term in the sum is $1$. On the other hand, if $\ell\neq 0$, then $\exp(-2\pi j\ell/N) \neq 1$,and \begin{equation} \begin{split} 1 + e^{-2\pi j \ell/N} + \left(e^{-2\pi j \ell/N}\right)^2 + \cdots + \left(e^{-2\pi j \ell/N}\right)^{N-1} &=~ \frac{1 - \left(e^{-2\pi j \ell/N}\right)^N}{1 - e^{-2\pi j \ell/N}}\\ &=~ \frac{1 - e^{-2N\pi j \ell/N}}{1 - e^{-2\pi j \ell/N}}\\ &=~ \frac{1 - 1}{1 - e^{-2\pi j \ell/N}} ~=~ 0. \end{split} \end{equation}

That shows that the $\mathsf{DFT}$ of the all-1 vector has no imaginary part, and its real part is $(N,0,0,\ldots,0)^{\mathsf{T}}$. Hence \begin{equation} x ~=~ \frac{1}{N}\left(\begin{array}{c} N\\0\\0\\\vdots\\0 \end{array}\right) ~=~ \left(\begin{array}{c} 1\\0\\0\\\vdots\\0 \end{array}\right). \end{equation}

Answer 2

The Hartley transform is an involution: it is (up to a scale factor) its own inverse. The classical discrete Hartley transform of order $N$ is such that $H_N^{-1} = \frac{1}{N}H_N$. Be careful with your notation, the vector $x$ has $N+1$ entries, so maybe you are after an $N+1$-order DHT!

If $\mathbf{1}$ denotes the all-ones vector, then in matrix-vector product $H_Nx = \mathbf{1} $ is equivalent to $H_N^{-1}H_Nx = H_N^{-1}\mathbf{1} $, and you get more directly the mysterious $X$:

$$x = \frac{1}{N}H_N\mathbf{1} $$

and you get the result given by Joe Mac by a direct computation. A little interpretation: the discrete "Dirac" signal at the origin yields a flat "Hartley" spectrum, just like for Fourier.