Chapter 9, Problem 39P

(a)

To determine

Find the phasor form of the vector $A=5\sin \left(2t+\frac{\pi }{3}\right)a_x+3\cos \left(2t+30°\right)a_y$.

Answer to Problem 39P

The phasor form of the vector $A=5\sin \left(2t+\frac{\pi }{3}\right)a_x+3\cos \left(2t+30°\right)a_y$ is $\underset{\_}{5e^{-j30°}{a}_x+3e^{j30°}{a}_y}$.

Explanation of Solution

Calculation:

Write the standard form the vector field.

$A=\mathrm{Re}\left(A_s{e}^{j\omega t}\right)$        (1)

Here,

$A_s$ is the phasor form of the vector field $A$.

Rewrite the given vector field.

$\begin{array}{c}A=5\cos \left[\frac{\pi }{2}-\left(2t+\frac{\pi }{3}\right)\right]a_x+3\cos \left(2t+30°\right)a_y\\ =5\cos \left[-\left(-\frac{\pi }{6}+2t\right)\right]a_x+3\cos \left(2t+30°\right)a_y\\ =5\cos \left(2t-30°\right)a_x+3\cos \left(2t+30°\right)a_y\left\{\because \cos \left(-\theta \right)=\cos \theta \right\}\end{array}$

As $\omega =2\text{rad}/\text{s}$, rewrite the expression.

$A=5\cos \left(\omega t-30°\right)a_x+3\cos \left(\omega t+30°\right)a_y$

Rewrite the expression.

$\begin{array}{c}A=\mathrm{Re}\left[5e^{j\left(\omega t-30°\right)}{a}_x+3e^{j\left(\omega t+30°\right)}{a}_y\right]\\ =\mathrm{Re}\left[5e^{-j30°}{e}^{j\omega t}{a}_x+3e^{j30°}{e}^{j\omega t}{a}_y\right]\\ =\mathrm{Re}\left[\left(5e^{-j30°}{a}_x+3e^{j30°}{a}_y\right)e^{j\omega t}\right]\end{array}$

Compare the expression with the expression in Equation (1) and write the phasor form of the given vector field.

$A_s=5e^{-j30°}{a}_x+3e^{j30°}{a}_y$

Conclusion:

Thus, the phasor form of the vector $A=5\sin \left(2t+\frac{\pi }{3}\right)a_x+3\cos \left(2t+30°\right)a_y$ is $\underset{\_}{5e^{-j30°}{a}_x+3e^{j30°}{a}_y}$.

(b)

To determine

Find the phasor form of the vector $B=\frac{100}{\rho }\sin \left(\omega t-2\pi z\right)a_{\rho }$.

Answer to Problem 39P

The phasor form of the vector $B=\frac{100}{\rho }\sin \left(\omega t-2\pi z\right)a_{\rho }$ is $\underset{\_}{\frac{100}{\rho }{e}^{-j\left(2\pi z+90°\right)}{a}_{\rho }}$.

Explanation of Solution

Calculation:

Modify Equation (1) for the given vector field.

$B=\mathrm{Re}\left(B_s{e}^{j\omega t}\right)$        (2)

Here,

$B_s$ is the phasor form of the vector field $B$.

Rewrite the given vector field.

$\begin{array}{c}B=\frac{100}{\rho }\cos \left[\frac{\pi }{2}-\left(\omega t-2\pi z\right)\right]a_{\rho }\\ =\frac{100}{\rho }\cos \left[-\left(\omega t-2\pi z-\frac{\pi }{2}\right)\right]a_{\rho }\\ =\frac{100}{\rho }\cos \left(\omega t-2\pi z-\frac{\pi }{2}\right)a_{\rho }\end{array}$

Rewrite the expression.

$\begin{array}{c}B=\mathrm{Re}\left[\frac{100}{\rho }{e}^{j\left(\omega t-2\pi z-90°\right)}{a}_{\rho }\right]\\ =\mathrm{Re}\left[\frac{100}{\rho }{e}^{-j\left(2\pi z+90°\right)}{e}^{j\omega t}{a}_{\rho }\right]\\ =\mathrm{Re}\left[\left(\frac{100}{\rho }{e}^{-j\left(2\pi z+90°\right)}{a}_{\rho }\right)e^{j\omega t}\right]\end{array}$

Compare the expression with the expression in Equation (2) and write the phasor form of the given vector field.

$B_s=\frac{100}{\rho }{e}^{-j\left(2\pi z+90°\right)}{a}_{\rho }$

Conclusion:

Thus, the phasor form of the vector $B=\frac{100}{\rho }\sin \left(\omega t-2\pi z\right)a_{\rho }$ is $\underset{\_}{\frac{100}{\rho }{e}^{-j\left(2\pi z+90°\right)}{a}_{\rho }}$.

(c)

To determine

Find the phasor form of the vector $C=\frac{\cos \theta }{r}\sin \left(\omega t-3r\right)a_{\theta }$.

Answer to Problem 39P

The phasor form of the vector $C=\frac{\cos \theta }{r}\sin \left(\omega t-3r\right)a_{\theta }$ is $\underset{\_}{\frac{\cos \theta }{r}{e}^{-j\left(3r+90°\right)}{a}_{\theta }}$.

Explanation of Solution

Calculation:

Modify Equation (1) for the given vector field.

$C=\mathrm{Re}\left(C_s{e}^{j\omega t}\right)$        (3)

Here,

$C_s$ is the phasor form of the vector field $C$.

Rewrite the given vector field.

$\begin{array}{c}C=\frac{\cos \theta }{r}\cos \left[\frac{\pi }{2}-\left(\omega t-3r\right)\right]a_{\theta }\\ =\frac{\cos \theta }{r}\cos \left[-\left(\omega t-3r-90°\right)\right]a_{\theta }\\ =\frac{\cos \theta }{r}\cos \left(\omega t-3r-90°\right)a_{\theta }\end{array}$

Rewrite the expression.

$\begin{array}{c}C=\mathrm{Re}\left[\frac{\cos \theta }{r}{e}^{j\left(\omega t-3r-90°\right)}{a}_{\theta }\right]\\ =\mathrm{Re}\left[\frac{\cos \theta }{r}{e}^{-j\left(3r+90°\right)}{e}^{j\omega t}{a}_{\theta }\right]\\ =\mathrm{Re}\left[\left(\frac{\cos \theta }{r}{e}^{-j\left(3r+90°\right)}{a}_{\theta }\right)e^{j\omega t}\right]\end{array}$

Compare the expression with the expression in Equation (3) and write the phasor form of the given vector field.

$C_s=\frac{\cos \theta }{r}{e}^{-j\left(3r+90°\right)}{a}_{\theta }$

Conclusion:

Thus, the phasor form of the vector $C=\frac{\cos \theta }{r}\sin \left(\omega t-3r\right)a_{\theta }$ is $\underset{\_}{\frac{\cos \theta }{r}{e}^{-j\left(3r+90°\right)}{a}_{\theta }}$.

(d)

To determine

Find the phasor form of the vector $D=10\cos \left(k_{1}x\right)\cos \left(\omega t-k_{2}z\right)a_y$.

Answer to Problem 39P

The phasor form of the vector $D=10\cos \left(k_{1}x\right)\cos \left(\omega t-k_{2}z\right)a_y$ is $\underset{\_}{10\cos \left(k_{1}x\right)e^{-j{k}_{2}z}{a}_y}$.

Explanation of Solution

Calculation:

Modify Equation (1) for the given vector field.

$D=\mathrm{Re}\left(D_s{e}^{j\omega t}\right)$        (4)

Here,

$D_s$ is the phasor form of the vector field $D$.

Rewrite the given vector field.

$\begin{array}{c}D=\mathrm{Re}\left[10\cos \left(k_{1}x\right)e^{j\left(\omega t-k_{2}z\right)}{a}_y\right]\\ =\mathrm{Re}\left[10\cos \left(k_{1}x\right)e^{-j{k}_{2}z}{e}^{j\omega t}{a}_y\right]\\ =\mathrm{Re}\left[\left(10\cos \left(k_{1}x\right)e^{-j{k}_{2}z}{a}_y\right)e^{j\omega t}\right]\end{array}$

Compare the expression with the expression in Equation (4) and write the phasor form of the given vector field.

$D_s=10\cos \left(k_{1}x\right)e^{-j{k}_{2}z}{a}_y$

Conclusion:

Thus, the phasor form of the vector $D=10\cos \left(k_{1}x\right)\cos \left(\omega t-k_{2}z\right)a_y$ is $\underset{\_}{10\cos \left(k_{1}x\right)e^{-j{k}_{2}z}{a}_y}$.