Chapter 7, Problem 7.1P

To determine

The discrete time signal of given analog voltage by using different time samples. Also, plot each series as a function of time and state the difference between the discrete representations of analog signal.

Explanation of Solution

Given:

  $\begin{array}{l}\begin{array}{l}\text{Analog voltage }=2sin4\pi t\text{ mV}\hfill \\ \text{Time increment }=1/8\text{s},1/5\text{s},1/3\text{s},1/21\text{s}\hfill \end{array}\\ \text{The data set comprised of }N\text{ Points},N=128\end{array}$

Concept Used:

At any sampling interval$\delta t$ ,$E\left(t\right)=2sin4\pi \times \delta \left(t\right)$

Calculation:

For$\delta t=1/8\text{s}\text{=0}\text{.125}\text{s}$.

At time$\delta t=0$ ,

  $\begin{array}{l}E\left(t\right)=2sin4\pi \times 0\\ E\left(t\right)=0\end{array}$

At time$\delta t=0.125\text{s}$ ,

  $\begin{array}{l}E\left(t\right)=2sin(4\pi \times 0.125)\\ E\left(t\right)=2\text{ mV}\end{array}$

To compute the remaining points ofE(t) using an increment of 0.125s, we use a spreadsheet tool.

The graph plotted between the amplitude and time interval is:

The analog signal was converted into a discrete time signal using a spreadsheet tool and a plot of amplitude vs time has been plotted for the time increment of 1/8 s.

For$\delta t=1/5\text{s}\text{=0}\text{.2}\text{s}$.

At time$\delta t=0$ ,

  $\begin{array}{l}E\left(t\right)=2sin4\pi \times 0\\ E\left(t\right)=0\end{array}$

At time$\delta t=0.2\text{s}$ ,

  $\begin{array}{l}E\left(t\right)=2sin(4\pi \times 0.2)\\ E\left(t\right)=\text{ 1}\text{.175 mV}\end{array}$

To compute the remaining points ofE(t) using an increment of 0.2s, we use a spreadsheet tool.

The graph plotted between the amplitude and time interval is:

The analog signal was converted discrete time signal using a spreadsheet tool and a plot of amplitude vs time has been plotted for the time increment of 1/5 s.

For$\delta t=1/3\text{s}\text{=}\text{0}\text{.33}\text{s}$.

At time$\delta t=0$ ,

  $\begin{array}{l}E\left(t\right)=2sin4\pi \times 0\\ E\left(t\right)=0\end{array}$

At time$\delta t=0.33\text{s}$ ,

  $\begin{array}{l}E\left(t\right)=2sin(4\pi \times 0.33)\\ E\left(t\right)=-1.688\text{ mV}\end{array}$

To compute the remaining points ofE(t) using an increment of 0.333 s, we use a spreadsheet tool.

The graph plotted between the amplitude and time interval is:

The analog signal was converted discrete time signal using a spreadsheet tool and a plot of amplitude vs time has been plotted for the time increment of 1/3 s.

At time$\delta t=0$,

  $\begin{array}{l}E\left(t\right)=2sin4\pi \times 0\\ E\left(t\right)=0\end{array}$

At time$\delta t=0.047\text{s}$ ,

  $\begin{array}{l}E\left(t\right)=2sin(4\pi \times 0.047)\\ E\left(t\right)=\text{ 1}\text{.126 mV}\end{array}$

To compute the remaining points ofE(t) using an increment of 0.0476s, we use a spreadsheet tool, The graph plotted between the amplitude and time interval is:

The data set is given below:

The analog signal was converted discretized time signal using a spreadsheet tool and a plot of amplitude vs time has been plotted for the time increment of 1/21 s.

The nature of the plot for$\delta t=1/3\text{s}$ is different when compared to the other sampling frequencies. Hence, it possesses different sampling and does not meet the required criterion.