# Signals and Systems – Zero-Order Hold and its Transfer Function (Practical Reconstruction)

Signals and SystemsElectronics & ElectricalDigital Electronics

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## 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.

Updated on 05-Jan-2022 11:15:38