ELEC 3360 RF & Microwave Engineering

Practice Questions for Week 7: Transmission Lines II

Question 1. a)

Use the expressions from your notes fo $L$ and $C$ to derive the expressions for the characteristic impedance $Z_0$ and the phase constant $\beta$ for a length of coax and thus show that $V_p = \frac{c}{\sqrt{\epsilon'}}$.

Solution 1. a)

\begin{eqnarray*} L &=& \frac{\mu}{2\pi}\ln\frac{b}{a} \, H/m\\ C &=& \frac{2\pi\epsilon'}{\ln\frac{b}{a}} \, F/m\\ Z_0 = \sqrt{\frac{L}{C}} \Rightarrow Z_0^2 &=& \frac{\frac{\mu}{2\pi}\ln\frac{b}{a}}{2\pi\epsilon'}\times\ln\frac{b}{a} \Rightarrow \\ \Rightarrow Z_0^2 &=& \frac{\mu\ln^2\frac{b}{a}}{(2\pi)^2\epsilon'} \Rightarrow \\ \Rightarrow Z_0 &=& \frac{1}{2\pi}\ln\frac{b}{a}\sqrt{\frac{\mu}{\epsilon'}} \, \Omega \end{eqnarray*}

The above is the analytical formula for $Z_0$ for a coaxial transmission line. Now, \begin{eqnarray*} \eta &=& \sqrt{\frac{\mu}{\epsilon'\epsilon_0}} \Rightarrow \\ \Rightarrow Z_0 &=& \frac{\eta}{2\pi}\ln\frac{b}{a} \, \Omega \end{eqnarray*}

In lossless coaxial transmission line $\beta = \omega\sqrt{LC}$, so: \begin{eqnarray*} \beta^2 &=& \omega^2\frac{\mu}{2\pi}\ln\frac{b}{a}\times\frac{2\pi\epsilon'}{\ln\frac{b}{a}} \Rightarrow \\ \Rightarrow \beta^2 &=& \omega^2\mu\epsilon' \Rightarrow \\ \Rightarrow \beta &=& \omega\sqrt{\mu\epsilon'} \end{eqnarray*}

As a check we know that $V_p = \frac{\omega}{\beta}$ so then: \begin{eqnarray*} V_p &=& \frac{\omega}{\omega\sqrt{\mu\epsilon'}} \Rightarrow \\ \Rightarrow V_p &=& \frac{1}{\sqrt{\mu\epsilon'}} \end{eqnarray*} However, $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$, so $V_p = \frac{c}{\sqrt{\epsilon'}}$.

Question 1. b)

Calculate the phase constant and overall loss ($Np/m$ and $dB/m$) for a $1m$ length of coaxial cable at $10GHz$ if $R_s = 2.5$, the inner conductor radius is $1mm$, the outer is $2.5mm$, and the dielectric constant is $\epsilon_r = \epsilon_r' - j\epsilon_r'' = 2.5 - j0.0025$.

Solution 1. b)

\begin{eqnarray*} R &=& \frac{R_s}{2\pi}(\frac{1}{a} + \frac{1}{b}) \, \Omega/m \\ L &=& \frac{\mu_0}{2\pi}\ln\frac{b}{a} \, H/m \\ C &=& \frac{2\pi\epsilon'}{\ln\frac{b}{a}} \, F/m \\ G &=& \frac{2\pi\omega\epsilon''}{\ln\frac{b}{a}} \, S/m \end{eqnarray*}

These four values can be used to describe the coaxial cable. They are also used to calculate the phase constant $\beta$, the characteristic impedance $Z_0$, and the losses per unit length.

Now we can find the following values: \begin{eqnarray*} \beta &=& \omega\sqrt{LC} = \frac{\omega\sqrt{\epsilon_r'}}{c} \, rad/m\\ Z_0 &=& \frac{\sqrt{\frac{\mu}{\epsilon'}}\ln\frac{b}{a}}{2\pi} = \frac{377\ln\frac{b}{a}}{2\pi\sqrt{\epsilon_r'}} \, \Omega\\ \alpha &=& \frac{R}{2Z_0} + \frac{GZ_0}{2} \, Np/m \end{eqnarray*}

Question 2. c)

Use the expression for effective dielectric constant of a microstrip line and derive the expressions where $w = 10d$, $w = d$, and $w = \frac{d}{4}$. What do you notice?

Solution 2. c)

\begin{eqnarray*} \epsilon_{eff} &=& \frac{\epsilon_r+1}{2}+\frac{\epsilon_r-1}{2}\frac{1}{\sqrt{1+\frac{12d}{w}}} \\ w = 10d \Rightarrow \epsilon_{eff} &=& \frac{\epsilon_r+1}{2}+\frac{\epsilon_r-1}{2}\frac{1}{\sqrt{2.2}} = \frac{\epsilon_r+1}{2}+0.674\times\frac{\epsilon_r-1}{2} \\ w = d \Rightarrow \epsilon_{eff} &=& \frac{\epsilon_r+1}{2}+\frac{\epsilon_r-1}{2}\frac{1}{\sqrt{13}} = \frac{\epsilon_r+1}{2}+0.277\times\frac{\epsilon_r-1}{2} \\ w = \frac{d}{4} \Rightarrow \epsilon_{eff} &=& \frac{\epsilon_r+1}{2}+\frac{\epsilon_r-1}{2}\frac{1}{\sqrt{49}} = \frac{\epsilon_r+1}{2}+0.143\times\frac{\epsilon_r-1}{2} \end{eqnarray*}

The above show that $\epsilon_{eff}$ is an average of two dielectric constants plus a term which implies a disproportianate ammount of the field is in the dielectric and as the line width goes up more of the field is in the dielectric than you might expect - further increasing dielectric losses.

Question 2. e)

Design a microstrip transmission line with a characteristic impedance of $Z_0 = 60\Omega$, where the substrate thickness is $d=1.58mm$, the loss tangent $\tan\delta$ is given in Question 1. b), as is the relative permittivity $\epsilon_r$. The surface resistance of copper at $10 GHz$ is $R_S = 2.5\Omega$. Assume the ratio of $\frac{w}{d}<2$. What is the width of the microstrip line in this case?

Solution 2. e)

We know from Microwave Office excercise that for a $Z_0 = 50 \Omega$ the line width is $~2.4mm$. Therefore, if $Z_0 = 60\Omega$ is higher we might expect a somewhat narrower line that is perhaps $~1.5-2mm$. Then, if $d = 1.588mm$, the ratio $\frac{w}{d}$ will be $~1-1.2$, so the assumption for $<2$ is probably correct.
From Pozar: \begin{eqnarray*} \frac{w}{d} &=& \frac{8{e}^A}{{e}^{2A}-2} \\ A &=& \frac{Z_0}{60}\sqrt{\frac{\epsilon_r+1}{2}}+\frac{\epsilon_r-1}{\epsilon_r+1}(0.23+\frac{0.11}{\epsilon_r}) \end{eqnarray*}

Question 2. f)

Calculate the guide wavelength at $10 GHz$ and calculate the total loss per cm at $10 GHz$.

Solution 2. f)

Recall that \begin{eqnarray*} \lambda_g &=& \frac{\lambda_0}{\sqrt{\epsilon_{eff}}} \\ \lambda_0 &=& \frac{c}{f} \\ \epsilon_{eff} &=& \frac{\epsilon_r+1}{2}+\frac{\epsilon_r-1}{2}\frac{1}{\sqrt{1+\frac{12d}{w}}} \end{eqnarray*}

Now, to find the total loss per cm at $10 GHz$ we can use the following formulas: \begin{eqnarray*} \alpha_t &=& \alpha_c + \alpha_d \, Np/m \\ \alpha_c &=& \frac{R_s}{Z_0w} \, Np/m \\ \alpha_d &=& \frac{k_0\epsilon_r(\epsilon_{eff}-1)\tan\delta}{2\sqrt{\epsilon_{eff}}(\epsilon_r-1)} \, Np/m \\ k_0 &=& \frac{2\pi}{\lambda_0} \end{eqnarray*}