Chapter 11, Problem 58P

To determine

Calculate the length of the shorted shunt-stub tuner in terms of $\lambda$.

Answer to Problem 58P

The length of the shorted shunt-stub tuner is $\mathcal{l}=0.0867\lambda$ and $\mathcal{l}=0.0819\lambda$.

Explanation of Solution

Calculation:

Consider the general formula for transformed load impedance at stub position $z=-d$.

$Z\left(z=-d\right)=Z_{\text{o}}\left(\frac{Z_L+j{Z}_{\text{o}}t}{Z_{\text{o}}+j{Z}_{L}t}\right)$        (1)

Where,

$t=\{\begin{array}{l}\frac{X_L\pm \sqrt{\left(\frac{R_L}{Z_{\text{o}}}\right)\left[{\left(Z_{\text{o}}-R_L\right)}^2+X_L^2\right]}}{R_L-Z_{\text{o}}}{R}_L\ne Z_{\text{o}}\\ \frac{-X_L}{2Z_{\text{o}}}{R}_L=Z_{\text{o}}\end{array}$        (2)

Consider the general formula for load admittance.

$Y=\frac{1}{Z}=G+jB$        (3)

Consider the general formula for distance.

$\frac{d}{\lambda }=\{\begin{array}{l}\frac{1}{2\pi }{\tan }^{-1}\left(t\right)t\ge Z_{\text{o}}\\ \frac{1}{2\pi }\left[\pi +{\tan }^{-1}\left(t\right)\right]t<Z_{\text{o}}\end{array}$        (4)

Consider the general formula for length of the short-circuit stub.

$\frac{\mathcal{l}}{\lambda }=\frac{1}{2\pi }{\tan }^{-1}\left(\frac{Y_{\text{o}}}{B}\right)$        (5)

Given that, the load impedance is,

$Z_L=75+j100$        (6)

Consider the general formula for load impedance.

$Z_L=R_L+j{X}_L$        (7)

s

On comparing Equation (6) and (7),

$R_L=75$

$X_L=100$

Substitute 75 for $R_L$, 50 for $Z_{\text{o}}$, and 100 for $X_L$ in Equation (2).

$\begin{array}{c}t=\frac{100\pm \sqrt{\left(\frac{75}{50}\right)\left[{\left(50-75\right)}^2+{100}^2\right]}}{75-50}{R}_L\ne Z_{\text{o}}\\ =\frac{100\pm 126.2}{25}\end{array}$

Therefore,

$t=9.048\text{s}$ and $t=-1.048\text{s}$

Find the length of the stub for $t=9.048\text{s}$ and $t=-1.048\text{s}$.

Consider $t=9.048\text{s}$.

Substitute 50 for $Z_{\text{o}}$, 9.048 for t, and $75+j100$ for $Z_L$ in Equation (1).

$\begin{array}{c}Z=\left(50\right)\left[\frac{\left(75+j100\right)+j\left(50\right)\left(9.048\right)}{\left(50\right)+j\left(75+j100\right)\left(9.048\right)}\right]\\ =\left(50\right)\left[\frac{75+j100+j452.4}{50-904.8+j678.6}\right]\\ =\left(50\right)\left[\frac{75+j552.4}{-854.8+j678.6}\right]\\ =13.04-j21.95\Omega \end{array}$

Substitute $13.04-j21.95$ for Z in Equation (3).

$Y=\frac{1}{13.04-j21.95}$

$Y=0.02+j0.033$        (8)

Rewrite Equation (3) as follows,

$Y=Y_{\text{o}}+jB\left\{\because G=Y_{\text{o}}\right\}$        (9)

On comparing Equation (8) and (9),

$Y_{\text{o}}=0.02$

$B=0.033$

Substitute 0.02 for $Y_{\text{o}}$ and 0.033 for B in Equation (5).

$\begin{array}{l}\frac{\mathcal{l}}{\lambda }=\frac{1}{2\pi }{\tan }^{-1}\left(\frac{0.02}{0.033}\right)\\ \frac{\mathcal{l}}{\lambda }=0.0867\\ \mathcal{l}=0.0867\lambda \end{array}$

Consider $t=-1.048\text{s}$.

Substitute 50 for $Z_{\text{o}}$, $-1.048$ for t, and $75+j100$ for $Z_L$ in Equation (1).

$\begin{array}{c}Z=\left(50\right)\left[\frac{\left(75+j100\right)+j\left(50\right)\left(-1.048\right)}{\left(50\right)+j\left(75+j100\right)\left(-1.048\right)}\right]\\ =\left(50\right)\left[\frac{75+j100-j52.4}{50+104.8-j78.6}\right]\\ =\left(50\right)\left[\frac{75+j47.6}{154.8-j78.6}\right]\\ =13.053-j22\Omega \end{array}$

Substitute $13.053-j22$ for Z in Equation (3).

$Y=\frac{1}{13.053-j22}$

$Y=0.019+j0.0336$        (10)

On comparing Equation (9) and (10),

$Y_{\text{o}}=0.019$

$B=0.0336$

Substitute 0.019 for $Y_{\text{o}}$ and 0.0336 for B in Equation (5).

$\begin{array}{l}\frac{\mathcal{l}}{\lambda }=\frac{1}{2\pi }{\tan }^{-1}\left(\frac{0.019}{0.0336}\right)\\ \frac{\mathcal{l}}{\lambda }=0.0819\\ \mathcal{l}=0.0819\lambda \end{array}$

Conclusion:

Thus, the length of the shorted shunt-stub tuner is $\mathcal{l}=0.0867\lambda$ and $\mathcal{l}=0.0819\lambda$.