# On the Impulse Response of Singular Discrete LTI Systems and Three Fourier Transform Pairs
## 1 Introduction  
A linear time-invariant (LTI) system is sufficiently described by the unit impulse response function (IRF). Given an LTI system with an unknown IRF, we can use any input function to obtain the IRF. A conceptually simple way to determine a system's IRF is to use a unit impulse and unit step function as input. However, as a unit impulse has an infinitely short duration and a unit step function has an infinitely long duration, they are not always practical to produce in reality. A more practical and convenient input test function is the rectangular function. For a discrete system, this corresponds to a sequence of uniform pulses. For example, let us have the input to a discrete LTI system as \(x_{1}[n]=\left\{ \begin{array}{l}1,\ n=0,\ 1\\ 0,\ otherwise\end{array} \right.\) and the corresponding output is \(y_{1}[n]=\left\{ \begin{array}{l}1,\ n=0,\ 1,\ 2\\ 0,\ otherwise\end{array} \right.\). It can be seen that the IRF is \(h[n]=[1,0,\ 1,-1,\ldots]\), which is an infinite sequence with the repeating pattern underlined. For an LTI system, the standard method to find the system's IRF is to inverse Fourier transform the frequency transfer function \(H[k]=Y[k]/X[k]\), where \(X[k]\) and \(Y[k]\) are the discrete Fourier transform (DFT) of the input and output sequences, respectively. In this example, however, inverse discrete Fourier Transform (IDFT) cannot be applied since a singularity (pole) exists. In this article, we are largely concerned only with the singular frequency transfer function (SFTF) of marginally stable systems. In general, deconvolution via IDFT is not doable for SFTF and part of the information is considered lost [1]. To be more general, let the input sequence \(x[n]\) have \(L\) consecutive unit impulses, i.e., \(x[n]=\left\{ \begin{array}{l}1,\ 0\leq n\leq L-1\\ 0,\ n\geq L\end{array} \right.\). The corresponding output of an LTI system is a sequence of real values, \(y[n]\). The last non-zero value in \(y[n]\) occurs at index \(L_{y}-1\), i.e., \(y[n]=0,\ \forall n\geq L_{y}\) and \(y[n]\
eq 0\) for \(n=L_{y}-1\). In this study, we are confined to finite \(L\) and \(L_{y}\). To find the IRF, \(h[n]\), we can carry out a time-domain deconvolution via polynomial division or arecursive sum (e.g., [2,3]). Alternatively, the IRF for the above defined \(x[n]\) can be found through the convolution sum \(y[n]=\sum_{m=0}^{\infty}h[m]x[n-m]\) as  
\[h[n]=\left\{ \begin{array}{l}y[n]-y[n-1],\qquad 0\leq n\leq L-1\\ y[n]-y[n-1]+h[n-L],\quad n\geq L\end{array} \right.. \tag{1}\]  
Since \(y[n]=0\) for \(n\geq L_{y}\), \(h\big{[}L_{y}+1\big{]}=h\big{[}L_{y}-L+1\big{]}\). \(h[n]\) becomes periodic with a period of \(L\) starting at \(max\big{(}L_{y}-L+1,\ 0\big{)}\) or sooner, where \(max\) is the maximum operator. The number of independent points in \(h[n]\) is \(M=max\big{(}L_{y}+1,\ L\big{)}\). \(h[n]\) can alternatively be expressed as  
\[h[n]=\sum_{m=0}^{L_{y}-1}y[m]q[n-m]+h[n-L]u[n-L], \tag{2}\]  
where \(q[m]=\left\{ \begin{array}{l}1,\ m=0\\ -1,\ m=1\\ 0,\ otherwise\end{array} \right.\), and \(u[m]=\left\{ \begin{array}{l}0\ m<0\\ 1\ m\geq 0\end{array} \right.\) is the unit step function. In general, such an LTI system has an infinite IRF. Fundamental to LTI systems is that they are sufficiently described by either \(h[n]\) or its Fourier transform, the frequency transfer function \(H[k]\) (e.g., [3,4]). If \(h[n]\) and \(H[k]\) are indeed equivalent in characterizing an LTI system, a logical conclusion is that we can _always_ obtain one from the other. The fact that we cannot find \(h[n]\) from a singular \(H[k]\) via IDFT may not be because \(H[k]\) has incomplete information but because we have not found the right way to carry it out. Otherwise, we have to admit \(H[k]\) and \(h[n]\) are not always equivalent. The problem appears to be simple and yet fundamental to LTI systems. Although LTI theory is very mature, obtaining IRF from an SFTF has not received much attention.