## 4 The Transfer Functions of the S&H  
In the previous section we introduced several S&H and presented the corresponding impulse responses. These have a general form  
\[\hat{x}(t)=ar(t)+bg(t), \tag{16}\]  
Figure 1: Frequency response of the extrapolations of the S&H-0, S&H-\(\alpha\) (with \(\alpha=0.3\) and \(\alpha=0.7\)), S&H-1 and S&H-exp (with \(\tau=200\)) and interpolations with S&H-\(\alpha\) (with \(\alpha=0.3\) and \(\alpha=0.7\)) and the S&H-1. Figure 2: Time response of the extrapolations of the S&H-0, S&H-\(\alpha\) (with \(\alpha=0.3\) and \(\alpha=0.7\)), S&H-1 and S&H-exp (with \(\tau=200\)) and interpolations with S&H-\(\alpha\) (with \(\alpha=0.3\) and \(\alpha=0.7\)) and the S&H-1. where \(a\) and \(b\) are real constants, \(r(t)=u(t-t_{n})-u(t-t_{n+1})\) and \(g(t)\) is a short duration signal (i.e., a pulse). The first term in (16) is a rectangle and its Laplace transform is  
\[\mathcal{L}\left[r(t)\right]=\frac{e^{-st_{n}}-e^{-st_{n+1}}}{s},\qquad s\in \mathbb{C}, \tag{17}\]  
where \(\mathcal{L}\) and \(s\) denote the Laplace operator and variable, respectively. In the case of a uniform time scale, \(t_{n}=nT\),  
\[r(t)=u(t-nT)-u(t-(n+1)T),\]  
Equation (17) gives  
\[\mathcal{L}\left[r(t)\right]=\frac{e^{-snT}-e^{-s(n+1)T}}{s},\qquad s\in \mathbb{C}. \tag{18}\]  
The second term in (16) depends on the S&H method. For the fractional case (that includes \(\alpha=1\)), that is, for relations, (13) and (15), we obtain  
\[\mathcal{L}\left[(t-t_{n})^{\alpha}r(t)\right]=e^{-st_{n}}\int_{0}^{t_{n+1}-t _{n}}t^{\alpha}e^{-st}dt,\qquad s\in\mathbb{C}. \tag{19}\]  
Let us introduce the lower incomplete gamma function defined as:  
\[\gamma(\alpha,v)=\int_{0}^{v}\tau^{\alpha-1}e^{-\tau}\mathrm{d}\tau,\quad \alpha,v>0,\]  
that verifies the relation  
\[\gamma(\alpha+1,v)=\alpha\gamma(\alpha,v)-v^{\alpha}e^{-v}.\]  
We can write  
\[\int_{0}^{t_{n+1}-t_{n}}t^{\alpha}e^{-st}dt=\alpha\frac{\gamma(\alpha,(t_{n+1} -t_{n})s)}{s^{\alpha+1}}-\frac{t_{n+1}-t_{n}}{s}e^{-(t_{n+1}-t_{n})s},\quad Re (s)>0,\]  
that leads to  
Figure 1:0  
For a uniform time scale Equation (20) yields  
Figure 1:1  
In the particular case of \(\alpha=1\), we obtain \(\gamma(1,v)=\int_{0}^{v}e^{-\tau}\mathrm{d}\tau=1-e^{-v}\), resulting  
Figure 1:2  
and, therefore, to  
Figure 1:3  
**Remark 2**. : _The choice of_  
Figure 1:4  
_is not of relevance, because it is not a function with finite support._  
Let us recall the example discussed in the previous section. Figure 2 depicts the responses in the frequency domain. We can compare the distinct items by means of the HC and MDS schemes. These clustering techniques allow the comparison and visualization of high dimensional data [26; 27; 28; 29], such as in the case of controlling nonlinear systems [30]. In our case, the \(i\)-th item corresponds to the response in the frequency domain, \(G_{i}\left(f\right)\), that is represented by a numerical array of \(m\) points. A set of \(N\) items is then compared by means of suitable distances [31]. We adopt the Euclidean and Canberra distances given by:  
Figure 1:5 Figure 1:6  
where \(i,j=1,\ldots,N\), so that \(N\) denotes the total number of items under comparison. The first distance, \(d^{E}\), follows the standard logic of Euclidean spaces, while the second, \(d^{C}\), distinguishes more clearly values close to zero and in general is more robust to data sets including large and small values. The MDS computational recursive algorithm requires the input of a matrix of all item-to-item distances (i.e., a \(N\) dimensional square and symmetric matrix with main diagonal of zeros) and tries to reproduce approximately that matrix by mean a plot in a \(m_{MDS}<m\) dimensional space where items consist of points. Therefore, similar items lead to close points and, inversely, very different items produce points very far apart. Usually, the dimensions \(m_{MDS}=2\) or \(m_{MDS}=3\) are used, because they allow a straightforward visualization [32; 33].