ELEC 3360 RF & Microwave Engineering

Practice Questions for Week 6: Transmission Lines

Question 1. a)

Calculate the resistance of a $1 m$ length of copper wire at DC and at $20 GHz$ - the wire is $1 mm$ in diameter - assume for this example that at $20 GHz$ all the current is restricted to 1 skin depth from the surface.

Solution 1. a)

\begin{eqnarray*} l &=& 1 \,m \\ \rho_{copper} &=& 1.7\times10^{-6} \, \Omega cm \end{eqnarray*}

Remember $$R = \frac{\rho l}{A},$$ so if units of $R$ are $\Omega$, then units of $$\rho = \frac{\Omega m^2}{m} = \Omega m$$
Therefore:



In [31]:

    
import math  # get access to most mathematical functions
l = 100  # length converted to cm
rho_copper = 1.7e-6
r_wire = 0.05  # radius converted to cm
R = (rho_copper * l) / (math.pi * r_wire**2)
print("Copper wire resistance at DC is {:.3f} Ohms".format(R))









    



Copper wire resistance at DC is 0.022 Ohms

But at $20 GHz$, skin depth is $$\sigma_s=\sqrt{\frac{2\rho}{\omega\mu}}$$



In [32]:

    
omega = 2 * math.pi * 20e9  # get angular frequency for 20 GHz
mu = 4 * math.pi * 1e-9  # equal to mu_0, but converted to V.s/A.cm units
skin_depth = math.sqrt((2 * rho_copper) / (omega * mu))
print("Skin depth at 20 GHz is {:.3e} cm".format(skin_depth))









    



Skin depth at 20 GHz is 4.640e-05 cm

Around $63\%$ of the current is contained in 1 skin depth from surface. We will assume that all the current is restricted to 1 skin depth for the purpose of this excercise in calculating worst case resistance at $20 GHz$.
$$R_{20 GHz} = \frac{\rho l}{A_{Cylinder}}$$



In [33]:

    
# Area of ring around edge of conductor
A_cylinder = math.pi * (r_wire**2 - (r_wire-skin_depth)**2)
R_20GHz = (rho_copper * l) / A_cylinder
print("Copper wire resistance at 20 GHz is {:.3f} Ohms".format(R_20GHz))









    



Copper wire resistance at 20 GHz is 11.667 Ohms

Question 1. b)

Calculate the impedance of a parallel plate capacitor at $20 GHz$, where the separation of the plates is 1 $\mu m$ and the area is $1000\mu m^2$, and where the dielectric constant is $\epsilon_r = 2.5-j0.025$.

Solution 1. b)

\begin{eqnarray*} X_c &=& -j\frac{1}{\omega C} \\ C &=& \frac{A\epsilon_r \epsilon_0}{d} = 8.85\times10^{-15}\,\epsilon_r = 8.85\times10^{-15}(2.5 - j0.025) \, F \end{eqnarray*}



In [34]:

    
e_r = 2.5 - 0.025j
c = 8.85e-15
x_c = (-1j) / (omega * c * e_r)
print("The impedance of the parallel plate capacitor at 20 GHz is {:.3f}{:.3f}j Ohms".
      format(x_c.real, x_c.imag))









    



The impedance of the parallel plate capacitor at 20 GHz is 3.596-359.636j Ohms

Question 1. c)

What do you notice about this impedance and how does this compare with what you know of the properties of an ideal capacitor?

Solution 1. c)

No longer an ideal capacitor, i.e. energy is dissipated in lossy dielectric.

Question 1. d)

What is the $tan\delta$ for the dielectric material used in this example?

Solution 1. d)

\begin{equation*} tan\delta = -\frac{\epsilon''}{\epsilon'} = \frac{0.025}{2.5} = 0.01 \end{equation*}

Question 1. e)

If I now squeeze the capacitor plates together the dielectric constant changes to $\epsilon_r = 2.7 - j0.03$. What is going on and what is the effect on the impedance of the capacitor?

Solution 1. e)



In [35]:

    
e_r_new = 2.7 - 0.03j
x_c_new = (-1j) / (omega * c * e_r_new)
print("The new impedance of the parallel plate capacitor at 20 GHz is {:.3f}{:.3f}j Ohms".
      format(x_c_new.real, x_c_new.imag))









    



The new impedance of the parallel plate capacitor at 20 GHz is 3.700-332.989j Ohms

As a result of the squeeze the dielectric becomes slightly more lossy, and the reactance drops a bit.

Question 2. a)

Find and write down an expression for the impedance of free space and the speed of light in free space (i.e. using the values of permeability and permittivity) and thus calculate values for the impedance of and the speed of light in free space.

Solution 2. a)

\begin{equation*} Z_{0_{free space}} = \eta = \sqrt{\frac{\mu}{\epsilon}} = \sqrt{\frac{\mu_0 \mu_r}{\epsilon_0 \epsilon_r}} \end{equation*}

Assuming $\epsilon_r = \mu_r = 1$:



In [36]:

    
mu_0 = 4 * math.pi * 1e-7
epsilon_0 = 8.85e-12
z0 = math.sqrt(mu_0 / epsilon_0)
print("The impedance of free space is {:.3f} Ohms".format(z0))









    



The impedance of free space is 376.819 Ohms

Note the similarity to $Z = \sqrt{\frac{L}{C}}$. For ideal lossless line, by similar analog $V_p = \frac{1}{\sqrt{LC}}$. What if $V_p = \frac{1}{\sqrt{\epsilon\mu}}$, and again, $\epsilon_r = \mu_r = 1$:



In [37]:

    
v_p = 1 / math.sqrt(mu_0 * epsilon_0)
print("Phase velocity in free space is {:.3f} m/s".format(v_p))









    



Phase velocity in free space is 299863380.466 m/s

Question 2. b)

If you now had a lens composed of a dielectric with a relative permittivity of 2.5 what would the impedance be for a plane wave (TEM) propagaint within this lens?

Solution 2. b)

\begin{eqnarray*} \epsilon_r &=& 2.5 \\ Z_{dielectric} &=& \eta_{dielectric} = \sqrt{\frac{\mu_0}{\epsilon_r \epsilon_0}} = \frac{Z_0}{\sqrt{\epsilon_r}} \end{eqnarray*}



In [38]:

    
e_r_dielectric = 2.5
z_dielectric = z0 / math.sqrt(e_r_dielectric)
print("The impedance of the dielectric is {:.3f} Ohms".format(z_dielectric))









    



The impedance of the dielectric is 238.322 Ohms

Question 3. c)

What is the speed of light in this material, i.e. slower by a factor of what? What is the refractive index of the dialectric material? Compare this to the permittivity and comment on your finding.

Solution 3. c)

\begin{equation*} V_{p_{dielectric}} = \frac{1}{\sqrt{\mu \epsilon}} => V_{p_{dielectric}} = \frac{V_p}{\sqrt{\epsilon_r}} \end{equation*}



In [39]:

    
v_p_dielectric = v_p / math.sqrt(e_r_dielectric)
print("Velocity in the dielectric is {:.3f} m/s and is slower by a factor of {:.3f}".
      format(v_p_dielectric, v_p_dielectric/v_p))









    



Velocity in the dielectric is 189650253.830 m/s and is slower by a factor of 0.632

If $n = \frac{V_p}{V_{material}}$, then $n = 1.581 = \sqrt{\epsilon_r}$.

ELEC 3360 RF & Microwave Engineering Practice Questions for Week 6: Transmission Lines