The correct answer is:
- A. $$\left( {\frac{1}{s}} \right)\left( {1 + {e^{ – sT}}} \right)$$
A zero-order hold (ZOH) is a type of data acquisition system that samples a continuous-time signal at regular intervals and then holds the samples constant between samples. The transfer function of a ZOH is given by:
$$H(s) = \frac{1}{s}\left(1 + e^{-sT}\right)$$
where $T$ is the sampling period.
The transfer function of a ZOH can be derived by considering the impulse response of a ZOH. The impulse response of a ZOH is given by:
$$h(t) = u(t) – u(t-T)$$
where $u(t)$ is the unit step function.
The transfer function of a ZOH can then be found by taking the Laplace transform of the impulse response:
$$H(s) = \int_{0}^{\infty} h(t) e^{-st} dt = \int_{0}^{\infty} u(t) – u(t-T) e^{-st} dt = \frac{1}{s}\left(1 + e^{-sT}\right)$$
Option B is incorrect because it does not include the factor of $1/s$. Option C is incorrect because it does not include the factor of $e^{-sT}$. Option D is incorrect because it does not include the factor of $1$.