Zero-Order Hold and its Transfer Function (Practical Reconstruction)

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Data Reconstruction

The data reconstruction is defined as the process of obtaining the analog signal x(t) from the sampled signal $\mathrm{x_{s}(t)}$. The data reconstruction is also known as interpolation.

The sampled signal is given by,

$$\mathrm{x_{s}(t) \:=\: x(t)\sum_{n=-\infty}^{\infty}\:\delta(t\:-\:nT)}$$

$$\mathrm{\Rightarrow\: x_{s}(t) \:=\: \sum_{n=-\infty}^{\infty}\:x(nT)\delta(t\:-\:nT)}$$

Where, $\mathrm{\delta(t\:-\:nT)}$ is zero except at the instants $\mathrm{t\:=\:nT}$. 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,

$$\mathrm{y(t) \:=\: \int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}\:x(nT)\delta(k\:-\:nT)h(t\:-\:k)dk}$$

By rearranging the order of integration and summation, we get,

$$\mathrm{y(t) \:=\: \sum_{n =-\infty}^{\infty}\:x(nT)\int_{-\infty}^{\infty}\:\delta(k\:-\:nT)h(t\:-\:k)dk}$$

$$\mathrm{\therefore\: y(t) \:=\: \sum_{n=-\infty}^{\infty}\:x(nT)h(t\:-\:nT)}$$

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,

$$\mathrm{\tilde{x}_{a}(t)\:=\:x(n)\:;\: \: for \: nT\:\leq\: n\:\leq \: ( n \: +\: 1 )T }$$

Therefore,

$$\mathrm{\tilde{x}_{a}(t) \:=\: x(0)\: ; \:\: for \: 0\:\leq\: t\:\leq\: T }$$

$$\mathrm{\tilde{x}_{a}(t) \:=\: x(T)\: ; \:\: for \: T\:\leq\: t\:\leq\: 2T}$$

$$\mathrm{\tilde{x}_{a}(t) \:=\: x(2T)\:; \:\: for \: 2T\:\leq\: t\:\leq\: 3T\: \:and\: so\: on}$$

Also, the impulse response of the zero-order hold is given by,

$$\mathrm{h(t)\:=\: \begin{cases} 1\:\: ;\:\:for\:\: 0\:\leq\: t\:\leq\: T \\\\ 0\:\:; otherwise \ \end{cases} }$$

Transfer Function of Zero-Order Hold

The output $\mathrm{\tilde{x}_{a}(t)}$ of a zero-order hold is given by the convolution of its impulse response h(t) and its input $\mathrm{x(nT)}$, i.e.,

$$\mathrm{\tilde{x}_{a}(t) \:=\: x(nT)\:\cdot\: h(t)}$$

$$\mathrm{\Rightarrow\: \tilde{x}_{a}(t) \:=\: \sum_{n=-\infty}^{\infty}\:x(nT)h(t \:-\: nT)}$$

Since the impulse response of the zero-order hold is given by,

$$\mathrm{h(t) \:=\: u(t) \:-\: u(t \:-\: T)}$$

$$\mathrm{\Rightarrow\: h(t\:-\:nT) \:=\: u(t \:-\:nT)\:-\:u\left [ t \:-\: (n \:+\: 1)T \right ]}$$

$$\mathrm{\therefore\: \tilde{x}_{a}(t) \:=\: \sum_{n=-\infty}^{\infty}\:x(nT)\left\{ u(t\:-\:nT)\:-\:u[t\:-\:(n \: +\: 1 )T]\right\}}$$

By taking Laplace transform on both sides, we get,

$$\mathrm{L[\tilde{x}_{a}(t)]\:=\: L\left[\sum_{n=-\infty}^{\infty}\:x(nT)\{u(t\:-\:nT)\:-\:u[t\:-\:(n\:+\:1)T] \} \right ]}$$

$$\mathrm{\tilde{X}_{a}(s)\:=\:\sum_{n=-\infty}^{\infty}\:x(nT)\left [ \frac{e^{-nTs}}{s}\:-\:\frac{e^{-(n\:+\:1)Ts}}{s} \right ]\:=\:\left(\frac{1\:-\:e^{-Ts}}{s}\right)\sum_{n=-\infty}^{\infty}\:x(nT)e^{-nTs}}$$

$$\mathrm{\Rightarrow\: \tilde{X}_{a}(s)\:=\:\left (\frac{1\:-\:e^{-Ts}}{s} \right)X^{\ast}(s)}$$

Therefore, the transfer function of the zero-order hold is given by,

$$\mathrm{TF\:=\:\frac{\tilde{X}_{a}(s)}{X^{\ast}(s)}\:=\:\left(\frac{1\:-\:e^{-Ts}}{s} \right)}$$

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.