Reconstruction of Signal by Interpolation

In time domain, the reconstruction of the continuous signal $x(t)$ from its sampled version $x_s(t)$ can be considered as an interpolation process of filling the gaps between neighboring samples. The interpolation can be considered as convolution of $x_s(t)$ with a certain function $h(t)$:

$$x(t)_{\mbox{reconstructed}}=h(t)*x_s(t)$$

In frequency domain, the interpolation can be considered as a filtering process:

$$X(\omega)_{\mbox{reconstructed}}=H(\omega) X_s(\omega)$$

with the general effect of reserving the central portion of the periodic spectrum $X_x(\omega)$ while suppressing all its replica at higher frequencies.

• Zero-order hold
  
  $$h_0(t)=\left\{\begin{array}{ll} 1 & 0 \le t < T_s 0 & \mbox{otherwise} \end{array} \right.$$
  
  
  A continuous signal $x_0(t)$ can be recovered by
  
  $$x_0(t)=h_0(t) * x_s(t)$$
  
  
  which is a series of square pulses with their heights modulated by $x(mT_s)$. The interpolation corresponds a low-pass filtering in frequency domain by
  
  $$H_0(\omega)={\cal F}[h_0(t)]=\frac{2}{\omega}sin(\frac{\omega T_s}{2})e^{-j\omega T_s/2}$$
  
  

• First-order hold
  
  $$h_1(t)=\left\{\begin{array}{ll} 1-\vert t\vert/T_s & 0 \le \vert t\vert < T_s 0 & \mbox{otherwise} \end{array} \right.$$
  
  
  A continuous signal $x_1(t)$ can be recovered by
  
  $$x_1(t)=h_1(t) * x_s(t)$$
  
  
  which is the linear interpolation of the sample train $x(mT_s)$ (connecting every two consecutive samples by a straight line). This interpolation corresponds a low-pass filtering in frequency domain by
  
  $$H_1(\omega)={\cal F}[h_1(t)]=\frac{4}{\omega^2 T_s}sin^2(\frac{\omega T_s}{2})$$
  
  

  

• Ideal reconstruction

  The reconstructed signals $x_0(t)$ and $x_1(t)$ using 0th or 1st order hold interpolation are certainly different from the original signal $x(t)$, for the reason that the low-pass filter is non-ideal. To find the interpolation function for a perfect reconstruction of the original signal $x(t)$, consider an ideal low-pass filter in frequency domain:
  

  $$H_{lp}(\omega)=\left\{ \begin{array}{ll} T_s & -\omega_c < \omega < \omega_c 0 & else \end{array} \right.$$
  
  
  with time domain impulse response
  
  $$h_{lp}(t)={\cal F}[H_{lp}(\omega)]=T_s\;\frac{sin(\omega_c t)}{\pi t}$$
  
  
  The reconstruction of the continuous signal $X(\omega)$ can now be realized by applying this ideal low-pass filter to the sampled signal $X_s(\omega)$:
  
  $$X_r(\omega)=H_{lp}(\omega)\;X_s(\omega)$$
  
  
  If the cutoff frequency $\omega_c$ is higher than $\omega_m$ but lower than $\omega_s-\omega_m$, then the central portion of $X_s(\omega)$ can be extracted and scaled by factor $T$, while all other replicas beyond the cutoff frequency are suppressed, i.e., the original signal is perfectly reconstructed:
  
  $$X_r(\omega)=X(\omega)$$
  
  
  In time domain, this perfect reconstruction is
  

  $$x(t) = h_{lp}(t) * x_s(t) = T_s\;\frac{sin(\omega_c t)}{\pi t}*\sum_{m=-\infty}^\infty x(mT_s)\delta(t-mT_s)$$

  $$= T_s\;\sum_{m=-\infty}^\infty x(mT_s) [\delta(t-mT_s)*\frac{sin(\omega_c t)}{\pi t}]$$

  $$= T_s\;\sum_{m=-\infty}^\infty x(mT_s) \frac{sin(\omega_c (t-mT_s))}{\pi (t-mT_s)}$$