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Signals and Systems – Zero-Order Hold and its Transfer Function (Practical Reconstruction)
Data Reconstruction
The data reconstruction is defined as the process of obtaining the analog signal $\mathrm{\mathit{x\left ( t \right )}}$ from the sampled signal $\mathrm{\mathit{x_{s}\left ( t \right )}}$. The data reconstruction is also known as interpolation.
The sampled signal is given by,
$$\mathrm{\mathit{x_{s}\left ( t \right )\mathrm{=}x\left ( t \right )\sum_{n\mathrm{=}-\infty }^{\infty }\delta \left ( t-nT \right )}}$$
$$\mathrm{\Rightarrow \mathit{x_{s}\left ( t \right )\mathrm{=}\sum_{n\mathrm{=}-\infty }^{\infty }x\left ( nT \right )\delta \left ( t-nT \right )}}$$
Where, $\mathrm{\mathit{\delta \left ( t-nT \right )}}$ is zero except at the instants $\mathrm{\mathit{t\mathrm{=}nT}}$. A reconstruction filter which is assumed to be linear and time invariant has unit impulse response ℎ(𝑡). The output of the reconstruction filter is given by the convolution as,
$$\mathrm{ \mathit{y\left ( t \right )\mathrm{=}\int_{-\infty }^{\infty }\sum_{n\mathrm{=}-\infty }^{\infty }x\left ( nT \right )\delta \left ( k-nT \right )h\left ( t-k \right )dk}}$$
By rearranging the order of integration and summation, we get,
$$\mathrm{ \mathit{y\left ( t \right )\mathrm{=}\sum_{n\mathrm{=}-\infty }^{\infty }x\left ( nT \right )\int_{-\infty }^{\infty }\delta \left ( k-nT \right )h\left ( t-k \right )dk}}$$
$$\mathrm{ \therefore \mathit{y\left ( t \right )\mathrm{=}\sum_{n\mathrm{=}-\infty }^{\infty }x\left ( nT \right )h\left ( t-nT \right )}}$$
What is Zero-Order Hold?
The zero-order hold is a method which is widely used to reconstruct the signals in real time. In the zero-order hold reconstruction method, the continuous signal is reconstructed from its samples by holding the given sample for an interval until the next sample is received. Therefore, the zero-order hold generates the step approximations.
The process of reconstruction by zero-order hold is shown in the figure.
Mathematically, the output of the zero-order hold is given by,
$$\mathrm{\mathit{\tilde{x}_{a}\left ( t \right )\mathrm{=}x\left ( n \right );\; \; \mathrm{for}\: nT\leq n\leq \left ( n\mathrm{\: +\: }\mathrm{1} \right )T }}$$
Therefore,
$$\mathrm{\mathit{\tilde{x}_{a}\left ( t \right )\mathrm{=}x\left ( \mathrm{0} \right );\; \; \mathrm{for}\: \mathrm{0}\leq t\leq T }}$$
$$\mathrm{\mathit{\tilde{x}_{a}\left ( t \right )\mathrm{=}x\left ( T \right );\; \; \mathrm{for}\: T\leq t\leq \mathrm{2}T}}$$
$$\mathrm{\mathit{\tilde{x}_{a}\left ( t \right )\mathrm{=}x\left ( \mathrm{2}T \right );\; \; \mathrm{for}\: \mathrm{2}T\leq t\leq \mathrm{3}T\: \: \mathrm{and\: so\: on}}}$$
Also, the impulse response of the zero-order hold is given by,
$$\mathrm{\mathit{h(t)\mathrm{=}\left\{\begin{matrix} \mathrm{1} &\mathrm{for\: 0}\leq t\leq T \ \mathrm{0}& \mathrm{otherwise} \ \end{matrix}\right.}}$$
Transfer Function of Zero-Order Hold
The output $\mathrm{\mathit{\tilde{x}_{a}\left ( t \right )}}$ of a zero-order hold is given by the convolution of its impulse response ℎ(𝑡) and its input $\mathrm{\mathit{x\left ( nT \right )}} $, i.e.,
$$\mathrm{\mathit{\tilde{x}_{a}\left ( t \right )\mathrm{=}x\left ( nT \right )\ast h\left ( t \right )}}$$
$$\mathrm{\Rightarrow \mathit{\tilde{x}_{a}\left ( t \right )\mathrm{=}\sum_{n\mathrm{=}-\infty }^{\infty }x\left ( nT \right ) h\left ( t -nT\right )}}$$
Since the impulse response of the zero-order hold is given by,
$$\mathrm{ \mathit{h\left ( t \right )\mathrm{=}u\left ( t \right )-u\left ( t-T \right )}}$$
$$\mathrm{\Rightarrow \mathit{h\left ( t-nT \right )\mathrm{=}u\left ( t -nT\right )-u\left [ t-\left ( n\mathrm{\: +\: }\mathrm{1} \right )T \right ]}} $$
$$\mathrm{\mathit{\therefore \tilde{x}_{a}\left ( t \right )\mathrm{=}\sum_{n\mathrm{=}-\infty }^{\infty }x\left ( nT \right )\left\{ u\left ( t -nT\right )-u\left [ t-\left ( n\mathrm{\: +\: }\mathrm{1} \right )T \right ]\right\}}}$$
By taking Laplace transform on both sides, we get,
$$\mathrm{\mathit{ L\left [ \tilde{x}_{a}\left ( t \right ) \right ]\mathrm{=}L\left [ \sum_{n\mathrm{=}-\infty }^{\infty }x\left ( nT \right )\left\{ u\left ( t -nT\right )-u\left [ t-\left ( n\mathrm{\: +\: }\mathrm{1} \right )T \right ]\right\} \right ]}}$$
$$\mathrm{\mathit{\tilde{X}_{a}\left ( s \right )\mathrm{=}\sum_{n\mathrm{=}-\infty }^{\infty }x\left ( nT \right )\left [ \frac{e^{-nTs}}{s}-\frac{e^{-\left ( n\mathrm{\: +\: }\mathrm{1} \right )Ts}}{s} \right ]\mathrm{=}\left ( \frac{\mathrm{1}-e^{-Ts}}{s} \right )\sum_{n\mathrm{=}-\infty }^{\infty }x\left ( nT \right )e^{-nTs}}}$$
$$\mathrm{\Rightarrow \mathit{\tilde{X}_{a}\left ( s \right )\mathrm{=}\left ( \frac{\mathrm{1}-e^{-Ts}}{s} \right )X^{\ast }\left ( s \right )}}$$
Therefore, the transfer function of the zero-order hold is given by,
$$\mathrm{ \mathit{TF\mathrm{=}\frac{\tilde{X}_{a}\left ( s \right )}{X^{\ast }\left ( s \right )}\mathrm{=}\left ( \frac{\mathrm{1}-e^{-Ts}}{s} \right )}}$$
The output of the zero order hold consists of higher order harmonics because it consists of steps. These harmonics can be removed by applying the output of ZOH to a low pass filter. This LPF tends to smooth the corners on the step approximations generated by the zero order hold. Thus, this LPF is also known as smoothing filter.
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