Chapter A8, Problem 1P

To determine

Plot the magnitude of the impedance $Z\left(s\right)=2+5s$ as a function of both $\sigma$ and $j\omega$.

Answer to Problem 1P

The magnitude of the impedance as a function of $\sigma$ and $j\omega$ is sketched as shown in Figure 2 and Figure 3.

Explanation of Solution

Given data:

The admittance is,

$Z\left(s\right)=2+5s$

Calculation:

The admittance is given by,

$Z\left(s\right)=2+5s$

Substitute $\sigma +j\omega$ for $s$ in the above equation as follows,

$Z\left(s\right)=2+5\left(\sigma +j\omega \right)$

$Z\left(s\right)=2+5\sigma +5j\omega$        (1)

Finding the magnitude of the above function,

$\begin{array}{c}\left|Z\left(s\right)\right|=\left|2+5\sigma +5j\omega \right|\\ =\sqrt{{\left(2+5\sigma \right)}^2+{\omega }^2}\end{array}$

Matlab code for the Impedance model:

sigma = linspace(-10,10,21);

omega = linspace(-10,10,21);

[X, Y] = meshgrid(sigma,omega);

Z = abs(2 + 5*X + j*5*Y);

colormap(hsv);

s = [-5 3 8];

surfl(X,Y,Z,s);

view (-20,5)

Matlab output:

The plot for the magnitude of given impedance is shown in Figure 1.

Sketch for $j\omega$:

Consider the admittance,

$Z\left(s\right)=2+5s$

Replace $s=j\omega$ in the above equation,

$Z\left(j\omega \right)=2+5\left(j\omega \right)$

Taking magnitude,

$\left|Z\right|=\sqrt{2^2+{\left(5\omega \right)}^2}$

$\left|Z\right|=\sqrt{4+25{\omega }^2}$

For various values of $\omega$ in the above equation, the sketch will be as follows,

Sketch for $\sigma$:

Consider the admittance,

$Z\left(s\right)=2+5s$

Replace $s=\sigma$ in the above equation,

$Z\left(\sigma \right)=2+5\sigma$

For various values of $\sigma$ in the above equation, the sketch will be as follows,

Conclusion:

Thus, the magnitude of the impedance as a function of $\sigma$ and $j\omega$ is sketched as shown in Figure 2 and Figure 3.