## 3 Sampling and Reconstruction
### The S&H Reconstruction by Extrapolation  
Equation (9), is important from the mathematical point of view, but it is not of practical usefulness, since it requires the knowledge of all the signal values \(x(t_{n})\), both at the past and future. This is a consequence of the fact that \(g(t)\) is not the impulse response of a causal system. To avoid such difficulty we use causal approximations generically called S&H systems. The most known are the zero- and one-order, but we can mention also the exponential hold. Let us analyze several approaches.  
3.1.1. The Zero Order Sample and Hold  
The output signal of a zero-order hold (S&H-0) has a zero slope between two consecutive sampling instants. Therefore, the S&H-0 circuit retains the measured value for a sampling instant at that level until the next sampling instant. The impulse response is a rectangle:  
\[\hat{x}(t)=x(t_{n})\left[u(t-t_{n})-u(t-t_{n+1})\right],\]  
for \(t_{n}\leq t<t_{n+1}\), where \(u(t)\) is the Heaviside unit step.  
3.1.2. The First-Order Sample and Hold  
The first-order hold (S&H-1) exhibits an impulse response that has a constant slope between two consecutive sampling instants. This slope is the incremental ratio determined by the values of the two preceding samples. Thus, the first-order-hold predicts the response over the sampling interval from \(t_{n}\) to \(t_{n+1}\) as a ramp with the slope determined by the signal values at the two consecutive time instants. The impulse response is  
\[\hat{x}(t)=x(t_{n})\left[u(t-t_{n})-u(t-t_{n+1})\right]+(t-t_{n})\frac{x(t_{n} )-x(t_{n-1})}{(t_{n}-t_{n-1})}\left[u(t-t_{n})-u(t-t_{n+1})\right],\]  
for \(t_{n}\leq t<t_{n+1}\).  
3.1.3. The Fractional-Order Sample and Hold  
If we calculate the the arithmetic average between the S&H-0 and S&H-1, with weight factors \(1-\beta\) and \(\beta\), \(0\leq\beta\leq 1\), respectively, then we obtain the fractional-hold (S&H-\(\beta\)) [12; 13; 14].  
The impulse response of the S&H-\(\beta\) is:  
\[\hat{x}(t)=x(t_{n})\left[u(t-t_{n})-u(t-t_{n+1})\right]+\beta(t-t_{n})\frac{x( t_{n})-x(t_{n-1})}{(t_{n}-t_{n-1})}\left[u(t-t_{n})-u(t-t_{n+1})\right],\]  
for \(t_{n}\leq t<t_{n+1}\). Therefore, we obtain the S&H-0 and S&H-1 for \(\beta=0\) and \(\beta=1\), respectively.  
3.1.4. The exponential order sample and hold  
The exponential order hold S&H-exp is interesting because it has a simple electrical implementation based on an \(RC\) low-pass circuit [11]. The impulse response of the S&H-exp is:  
\[\hat{x}(t)=x(t_{n})\left[u(t-t_{n})-u(t-t_{n+1})\right]+\left[1-e^{-\frac{(t- t_{n})}{\tau}}\right]\frac{x(t_{n})-x(t_{n-1})}{(t_{n}-t_{n-1})}\left[u(t-t_{n})-u (t-t_{n+1})\right],\]  
for \(t_{n}\leq t<t_{n+1}\), where \(\tau\) is the time constant. This expression is slightly different from the one presented in [11]. However, the previous expression is preferable because gives the S&H-0 when \(\tau\rightarrow\infty\).  
3.1.5. The \(\alpha\)-Order Sample and Hold  
Given the expressions for the S&H-0 and S&H-1 we can propose a generalization as follows.  
Definition 4: The \(\alpha\)-order (S&H-\(\alpha\)) is formulated as:  
\[\hat{x}(t)=x(t_{n})+(t-t_{n})^{\alpha}\frac{x(t_{n})-x(t_{n-1})}{\left(t_{n}- t_{n-1}\right)^{\alpha}}, \tag{12}\]_for \(t_{n}\leq t<t_{n+1}\), where \(\alpha\in\mathbb{R}^{+}\)._  
The impulse response is given by:  
\[\hat{x}(t)=x(t_{n})\left[u(t-t_{n})-u(t-t_{n+1})\right]+(t-t_{n})^{\alpha}\frac {x(t_{n})-x(t_{n-1})}{\left(t_{n}-t_{n-1}\right)^{\alpha}}\left[u(t-t_{n})-u(t -t_{n+1})\right], \tag{13}\]  
for \(t_{n}\leq t<t_{n+1}\).