Introduction

What Are High Frequency

Our familiar lumped circuit analysis is only an approximation, which is approximately valid when the circuit dimensions are less than \(\lambda/10\).

\(f\)	\(\lambda\)	\(\lambda/10\)
1 GHz	30 cm	3cm
10 GHz	3 cm	3 mm
100 GHz	3 mm	300 um

AC Analysis

For a impedance \(Z(s)\) consists of \(R, L\) and \(C\), if we apply a current \(i(t)=I \cos(\omega t + \varphi)\), then the voltage “response” can be understand as

\[v(t) \sim V \cos(\omega t + \theta)\]

with

\[V e^{j \theta} = I e^{j \varphi} \cdot Z(j\omega)\]

Average Power Consumption

Assume

\[\begin{cases} v(t) = V \cos(\omega t + \varphi)\\ i(t) = I \cos(\omega t + \theta) \end{cases}\] \[\overline{P} = \dfrac{1}{T} \int_0^{T} VI\cos(\omega t + \varphi) \cos(\omega t + \theta) dt = \dfrac{VI}{2} \cos(\theta-\varphi)\]

In phaser representation

\[\overline{P} = \dfrac{1}{2} \mathrm{Re}(V I^*)\]

Maximum Power Transfer

Assume a AC voltage source \(V_G \cos(\omega t)\) with internal impedance \(Z_G = R_G + j X_G\). What load impedance will results maximum average power on the load?

\[\overline{P_L} = \dfrac{1}{2}\mathrm{Re}\left(\dfrac{V_G(R_L+j X_L)}{R_G+jX_G + R_L + jX_L} \cdot \dfrac{V_G}{R_G-j X_G + R_L - j X_L}\right)\]

we need to maximax

\[\dfrac{R_L}{(R_G+R_L)^2+(X_G+X_L)^2}\]

so the maximum power transfer happens when

\[\begin{cases} R_L = R_G\\ X_L = -X_G \end{cases}\]

Maxwell’s Equations

\[\begin{cases} \nabla \cdot \vec{D} = \rho\\ \nabla \cdot \vec{B} = 0\\ \nabla \times \vec{E} = -\dfrac{\partial \vec{B}}{\partial t}\\ \nabla \times \vec{H} = \dfrac{\partial \vec{D}}{\partial t} + \vec{J}\\ \vec{D} = \epsilon \vec{E}\\ \vec{H} = \vec{B}/\mu\\ \nabla \cdot \vec{J} + \dfrac{\partial \rho}{\partial t} = 0\\ \vec{J} = \rho \vec{u} \end{cases}\]

In fact, the two divergence equation can be derived from the two curl equation (how?).

Lorentz’s Force

\[\vec{F} = q(\vec{E} + \vec{u}\times\vec{B})\]

Boundary Condition

\[\begin{align} E_{1t} = E_{2t}\\ \dots \end{align}\]

For boundary condition between dielectric and conductor, \(\hat{a_n}\) points fron conductor to dielectric.

Wave Equation

\[\dfrac{1}{\mu\epsilon}\nabla^2 \vec{E} = \dfrac{\partial^2 \vec{E}}{\partial t^2}\] \[\dfrac{1}{\mu\epsilon}\nabla^2 \vec{H} = \dfrac{\partial^2 \vec{H}}{\partial t^2}\] \[c^2 = \dfrac{1}{\mu\epsilon}\]

Example

\[c^2 \dfrac{\partial^2 f}{\partial x^2} = \dfrac{\partial^2 f}{\partial t^2}\]

From d’Alembert’s solution, the general solution is given by

\[f(x,t) = F(kx+\omega t) + G(kx-\omega t)\]

where \(\omega/k = c\)

Time Harmonic Field

Steady State Solution

\[\vec{E}(x,y,z;t) = \mathrm{Re}\left(\vec{E}_{0}(x,y,z) e^{j\omega t}\right)\]

Similar to the phasor in AC analysis, assume each component of the vector field is a phasor (having the same time frequency).

\[\begin{cases} \nabla \times \vec{E} = -j\omega \mu \vec{H}\\ \nabla \times \vec{H} = \vec{J} + j\omega\epsilon \vec{E}\\ \nabla \cdot \vec{E} = \dfrac{\rho}{\epsilon}\\ \nabla \cdot \vec{H} = 0 \end{cases}\]

then for source free wave equation

\[\nabla^2 \vec{E} + k^2 \vec{E} = 0\]

where \(k^2 = \omega^2 \mu\epsilon\)

In the case of wave in conductor, we can define the compex permittivity

\[\epsilon_c = \epsilon - j \dfrac{\sigma}{\omega}\]

where \(\vec{J} = \sigma \vec{E}\) (Ohms law).

Plane Wave

In Lossless Media

\[\nabla^2 \vec{E} + k^2 \vec{E} = 0\] \[\begin{cases} \left(\dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^2}{\partial z^2} + k^2\right) E_x(x,y,z) = 0\\ \left(\dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^2}{\partial z^2} + k^2\right) E_y(x,y,z) = 0\\ \left(\dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^2}{\partial z^2} + k^2\right) E_z(x,y,z) = 0 \end{cases}\]

Assume \(E_x(x,y,z) = X(x)Y(y)Z(z)\)

\[\left(\dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^2}{\partial z^2} + k^2\right) X(x)Y(y)Z(z) = 0\] \[\dfrac{X''}{X} + \dfrac{Y''}{Y} + \dfrac{Z''}{Z} + k^2 = 0\] \[k_x^2 + k_y^2 + k_z^2 = k^2\] \[E_x(x,y,z) = A e^{-j k_x x} B e^{-j k_y y} C e^{-j k_z z} = E_{x0} e^{-j \vec{k} \cdot \vec{R}}\]

Assume for \(E_y, E_z\) they have the same \(\vec{k}\) as \(E_x\).

\[\vec{E}(x,y,z) = \vec{E_0} e^{-j \vec{k}\cdot\vec{R}}\]

Does it allow different \(k_x,k_y,k_z\) for different components?

Assume source free space

\[\nabla \cdot \vec{E} = 0\] \[\vec{E} = \vec{E_0} e^{-j \vec{k}\cdot\vec{R}}\] \[\vec{k} \cdot \vec{E} = 0\]

The magnetic field can be calculated from

\[\begin{align} \vec{H} &= -\dfrac{1}{j\omega \mu} \nabla \times \left( \vec{E_0} e^{-j \vec{k}\cdot \vec{R}} \right)\\ &= \dfrac{1}{-j\omega\mu} \nabla\left(e^{-j \vec{k}\cdot\vec{R}}\right)\times \vec{E_0}\\ &= \dfrac{-j\vec{k}}{-j\omega\mu}e^{-j\vec{k}\cdot\vec{R}} \times \vec{E_0}\\ &= \dfrac{\vec{k} \times \vec{E}}{\omega\mu} = \dfrac{1}{\eta} \hat{a_n} \times \vec{E} \end{align}\]

This is typical TEM wave, where the field is orthgonal to travel direction.

Polarization

Assume

\[\begin{align} \vec{E}(z) &= \left(\hat{a_x} E_x + \hat{a_y}E_y\right) e^{-jkz}\\ &= \left(\hat{a_x} E_{x0} e^{-j\phi_{x0}} + \hat{a_y} E_{y0} e^{-j\phi_{y0}}\right) e^{-jkz} \end{align}\]

Dependent on the values of \(\phi_{x0}, \phi_{y0}\) and \(E_{x0}, E_{y0}\), it will show different polarization.

In Lossy Media

\[\nabla^2 \vec{E} + k_c^2 \vec{E} = 0\] \[k_c = \omega \sqrt{\mu \epsilon_c}\] \[\begin{align} \gamma &= jk_c = j\omega \sqrt{\mu\epsilon}\left(1+\dfrac{\sigma}{j\omega\epsilon}\right)^{1/2}\\ &= \alpha + j\beta \end{align}\] \[\nabla^2 \vec{E} - \gamma^2 \vec{E} = 0\] \[\vec{E} = \hat{a_x} E_0 e^{-\gamma z} = \hat{a_x} E_0 e^{-\alpha z} e^{-j\beta z}\]

Low-Loss Dielectrics

Good Conductors

Group Velocity

phase velocity

\[u_p = \dfrac{\omega}{\beta}\]

Dispersion: the situation when the phase velocity is different for different frequency.

Assume two frequency \(\omega_0 \pm \Delta \omega\) with the same amptitude.

\[\begin{align} E(z,t) &= E_0 \cos\left[(\omega_0+\Delta\omega)t - (\beta_0+\Delta\beta)z\right] + E_0\cos\left[(\omega_0-\Delta\omega)t - (\beta_0-\Delta\beta)z\right]\\ &= 2E_0\cos(\Delta \omega t - \Delta \beta z) \cos(\omega_0 t - \beta_0 z) \end{align}\]

This is a carrier at frequency \(\omega_0\) and an envelop at frequency \(\Delta\omega\). The phase velocity of carrier is

\[u = \dfrac{\omega_0}{\beta_0}\]

The phase velocity of envelop is

\[u = \dfrac{\Delta \omega}{\Delta \beta}\]

We call it group velocity

\[u_g = \dfrac{1}{d\beta/d\omega}\] \[u_g = \dfrac{u_p}{1-\dfrac{\omega}{u_p} \dfrac{du_p}{d\omega}}\]

Poynting Vector

\[\begin{align} &\nabla \cdot (\vec{E} \times \vec{H}) = \vec{H} \cdot (\nabla \times \vec{E}) - \vec{E} \cdot (\nabla \times \vec{H})\\ =& -\dfrac{\partial}{\partial t} \left(\dfrac{1}{2}\epsilon E^2 + \dfrac{1}{2}\mu H^2\right) - \sigma E^2 \end{align}\]

\(\vec{E}\times\vec{H}\) (Poynting vector) is the total power flow.

In the case of steady state sinosoidal analysis, the average power flow

\[P_{\mathrm{av}} = \dfrac{1}{2}\mathrm{Re}(\vec{E}\times \vec{H}^*)\]

Normal Incidence At A Plane Conducting Boundary

Assume the wave travel from \(\sigma=0\) to \(\sigma=\infty\). Incident wave

\[\begin{cases} \vec{E_i}(z) = \hat{a_x} E_{i0} e^{-j\beta_{1}z}\\ \vec{H_i}(z) = \hat{a_y} \dfrac{E_{i0}}{\eta_1} e^{-j\beta_1 z} \end{cases}\]

Reflected wave

\[\vec{E_r} = \hat{a_x} E_{r0} e^{+j\beta_1 z}\]

Total \(\vec{E}\) in medium 1

\[\vec{E_i} + \vec{E_r} = \hat{a_x} E_{i0} e^{-j\beta_{1}z} + \hat{a_x} E_{r0} e^{+j\beta_1 z}\]

using boundary condition for perfect conductor

\[E_{r0} = - E_{i0}\]

\(\vec{H_r}\) can get from \(\vec{E_r}\).

Oblique Incidence At A Plane Conducting Boundary

Perpendicular Polarization

For the components that \(\vec{E_i}\) is perpendicular to plane of incidence.

We guess the reflected wave \(\vec{E_r}\) is also perpendicular to the plane of incidence.

\[\hat{a_{ni}} = \hat{a_x}\sin\theta_i + \hat{a_z} \cos\theta_i\] \[\vec{E_i} = \hat{a_y} E_{i0} e^{-j\beta_1 (x\sin\theta_i + z\cos\theta_i)}\]

Incidence At A Plane Dielectric

Waveguides

Uniform Guiding Structure

\[\vec{E}(x,y,z,t) = \mathrm{Re}\left\{ \vec{E_0}(x,y) e^{-\gamma z} e^{j\omega t}\right\}\] \[(\nabla^2 + k^2) \vec{E} = 0\] \[\nabla^2 = \nabla_{xy}^2 + \gamma^2 = \nabla_{t}^2 + \gamma^2\] \[\left(\nabla_{xy}^2 + \gamma^2 + k^2\right) \vec{E} = 0\]

where \(t\) means transverse (TE,TM,TEM), which means orthogonal to something.

Note that the solution may not be plane wave, which means the \(\vec{E}, \vec{H}\) may not be orthogonal to travelling direction. So we need to solve all the components \(E_x, E_y, E_z, H_x, H_y, H_z\).

Assuming source free in the waveguide (not on the surface)

\[\begin{cases} \nabla \times \vec{E} = -j\omega\mu\vec{H}\\ \nabla \times \vec{H} = -j\omega\epsilon \vec{E} \end{cases}\]

Once we solved \(E_z, H_z\), the other four components

\[\begin{cases} H_x = -\dfrac{1}{h^2}\left(\gamma \dfrac{\partial H_z}{\partial x} - j\omega\epsilon \dfrac{\partial E_z}{\partial y}\right)\\ \vdots \end{cases}\]

where \(h^2 = \gamma^2 + k^2\)

TEM waves: \(E_z=H_z=0\).
TM waves: \(H_z=0\).
TE waves: \(E_z=0\).

Is it possible \(E_z\ne 0, H_z\ne 0\)?

Parallel-Plate Waveguide

Assume \(w \gg b\).

TEM

Assume TEM mode, and assume a plane wave

\[\begin{cases} \vec{E} = \hat{a_y} E_0 e^{-\gamma z}\\ \vec{H} = -\hat{a_x} \dfrac{E_0}{\eta} e^{-\gamma z} \end{cases}\]

where

\[\begin{cases} \gamma = j\beta = j\omega\sqrt{\mu\epsilon}\\ \eta = \sqrt{\dfrac{\mu}{\epsilon}} \end{cases}\]

Since we already know such wave satisfy Maxwell’s equation,we need to further check the boundary condition \(y=0, y=b\) for the plane wave.

\[\begin{cases} E_t = 0\\ H_n = 0 \end{cases}\]

and the boundary condition is satisfied. So we find a simple solution. The frequency \(\omega\) can be any frequency.

The parallel plate will have surface charge and surface current.

\[\begin{cases} \rho = \hat{a_n} \cdot \vec{D}\\ \vec{J} = \hat{a_n} \times \vec{H} \end{cases}\]

for \(y=0\), \(\hat{a_n} = +\hat{a_y}\).

\[\begin{cases} \rho = \epsilon E_0 e^{-j\beta z}\\ \vec{J} = \hat{a_z} \dfrac{E_0}{\eta} e^{-j\beta z} \end{cases}\]

TM

Assume \(E_z(x,y,z)\) doesn’t depends on \(x\), since \(x\) is wide enough.

\[E_z(x,y,z) = E_z^0(y)e^{-\gamma z}\] \[\left(\nabla_{xy}^2 + h^2\right) \vec{E} = 0\] \[\dfrac{d^2}{dy^2} E_{z}^0 (y) + h^2 E_z^0 (y) = 0\]

with boundary condition

\[E_z^0(y) \vert_{y=0,y=b} = 0\] \[E_z^0(y) = A_n \sin \dfrac{n\pi y}{b}\] \[h = \dfrac{n\pi}{b}\]

then all the other components can be calculated by the equations.

\[\gamma = \sqrt{h^2-k^2} = \sqrt{\left(\dfrac{n\pi}{b}\right)^2 - \omega^2\mu\epsilon}\]

Dielectric Waveguide

Antennas

For static field (static \(\rho, \vec{J}\)).

\[\begin{cases} \nabla \times \vec{E} = 0 \quad \implies \quad \vec{E} = -\nabla V\\ \nabla \cdot \vec{B} = 0 \quad \implies \quad \vec{B} = \nabla \times \vec{A} \end{cases}\] \[\begin{cases} V = \int \dfrac{1}{4\pi\epsilon} \dfrac{\rho dv}{R}\\ \vec{A} = \int \dfrac{\mu}{4\pi} \dfrac{\vec{J} dv}{R} \end{cases}\]

For time varying field

\[\nabla \times \vec{E} = -\dfrac{\partial \vec{B}}{\partial t} = -\dfrac{\partial}{\partial t} \left(\nabla \times \vec{A}\right)\] \[\vec{E} = \dfrac{1}{j\omega\epsilon} \nabla \times \vec{H}\] \[\nabla \times \left(\vec{E} + \dfrac{\partial \vec{A}}{\partial t}\right) = 0\] \[\vec{E} + \dfrac{\partial \vec{A}}{\partial t} = -\nabla V\] \[\vec{E} = -\nabla V - \dfrac{\partial \vec{A}}{\partial t}\] \[\nabla \times \left(\dfrac{\nabla \times \vec{A}}{\mu}\right) = \vec{J} + \dfrac{\partial}{\partial t} \left[\epsilon \left(-\nabla V - \dfrac{\partial \vec{A}}{\partial t}\right)\right]\] \[\begin{align} \nabla \times \left(\nabla \times \vec{A}\right) &= \mu \vec{J} + \mu\epsilon \dfrac{\partial}{\partial t}(-\nabla V - \dfrac{\partial \vec{A}}{\partial t})\\ &= \nabla \left(\nabla \cdot \vec{A}\right) - \nabla^2 \vec{A} \end{align}\] \[\nabla^2 \vec{A} - \mu\epsilon \dfrac{\partial^2 \vec{A}}{\partial t^2} = -\mu \vec{J} + \nabla\left(\nabla\cdot \vec{A} + \mu\epsilon\dfrac{\partial V}{\partial t}\right)\]

Let’s define

\[\begin{cases} \nabla \times \vec{A} = \vec{B}\\ \nabla \cdot \vec{A} + \mu\epsilon\dfrac{\partial V}{\partial t} = 0 \quad \text{(Lorentz condition)} \end{cases}\] \[\nabla^2 \vec{A} - \mu\epsilon\dfrac{\partial^2 \vec{A}}{\partial t^2} = -\mu \vec{J}\] \[\begin{align} - \nabla \cdot \left(\nabla V\right) &= \nabla \cdot \left(\vec{E}+\dfrac{\partial \vec{A}}{\partial t}\right)\\ &= \dfrac{\rho}{\epsilon} + \dfrac{\partial}{\partial t} \left(\nabla \cdot \vec{A}\right)\\ &= \dfrac{\rho}{\epsilon} + \dfrac{\partial}{\partial t} \left(-\mu\epsilon \dfrac{\partial V}{\partial t}\right)\\ &= -\nabla^2 V \end{align}\] \[\nabla^2 V - \mu\epsilon \dfrac{\partial^2 V}{\partial t^2} = -\dfrac{\rho}{\epsilon}\]

the solution is given by

\[V(\vec{R},t) = \dfrac{1}{4\pi\epsilon} \int \dfrac{\rho \left(\vec{R'}, t-\dfrac{\vert \vec{R} - \vec{R}' \vert}{u}\right)}{\vert \vec{R}-\vec{R'} \vert} dv\]

where \(u = \dfrac{1}{\sqrt{\mu\epsilon}}\).

For sinosoidal steady state

\[\vec{A}(\vec{R}, t) = \dfrac{\mu}{4\pi} \int \dfrac{\mathrm{Re}\left\{\vec{J}(\vec{R'}) e^{j\omega(t-\Delta t)}\right\}}{\vert \vec{R'} - \vec{R} \vert} dv\]

In phasor form

\[\vec{A} = \dfrac{\mu}{4\pi}\int \dfrac{\vec{J} \cdot e^{-jkR}}{R} dv\] \[V = \dfrac{1}{4\pi\epsilon}\int \dfrac{\rho e^{-jkR}}{R} dv\]

Hertizen Dipoles

\[i(t) = I\cos\omega t = \dfrac{d q(t)}{dt}\] \[I=j\omega Q\] \[\vec{P} = \hat{a_z} \cdot dl \cdot Q = \hat{a_z} \dfrac{I \cdot dl}{j\omega}\] \[\begin{align} \vec{A} &= \dfrac{\mu}{4\pi} \dfrac{\vec{J} e^{-jkR} dv}{R}\\ &= \hat{a_z} \dfrac{\mu}{4\pi} \dfrac{I dl \cdot e^{-j\beta R}}{R}\\ &= \hat{a_R} A_R + \hat{a_\theta} A_\theta \end{align}\] \[\vec{H} = \dfrac{1}{\mu} \nabla \times \vec{A}\]

Transmission Lines

\[-\dfrac{\partial v(z,t)}{\partial z} = R i(z,t) + L \dfrac{\partial i(z,t)}{\partial t}\] \[-\dfrac{\partial i(z,t)}{\partial z} = G v(z,t) + C \dfrac{\partial v(z,t)}{\partial t}\] \[\begin{cases} -\dfrac{d V(z)}{dz} = (R+j\omega L) I(z)\\ -\dfrac{d I(z)}{dz} = (G + j\omega C) V(z) \end{cases}\] \[\begin{cases} \dfrac{d^2 V(z)}{dz^2} = \gamma^2 V(z)\\ \dfrac{d^2 I(z)}{dz^2} = \gamma^2 V(z) \end{cases}\] \[\gamma = \alpha + j\beta = \sqrt{(R+j\omega L)(G+j\omega C)}\] \[Z_0 = \dfrac{R+j\omega L}{\gamma} = \dfrac{\gamma}{G+j\omega C} = \sqrt{\dfrac{R+j\omega L}{G+j\omega C}}\]

Finite Transmission Lines

\[\begin{cases} V(z) = V_0^+ e^{-\gamma z} + V_0^- e^{+\gamma z}\\ I(z) = I_0^+ e^{-\gamma z} + I_0^- e^{+\gamma z} \end{cases}\] \[\dfrac{V_0^+}{I_0^+} = - \dfrac{V_0^-}{I_0^-} = Z_0\]

Boundary condition at \(z = l\)

\[\dfrac{V(l)}{I(l)} = \dfrac{V_L}{I_L} = Z_L\]

we can imagine if \(Z_L = Z_0\), then with only \(V_0^+\), it can satisfy the boundary condition, such that there is no refection wave.

\[\begin{cases} V(l) = V_L = V_0^+ e^{-\gamma l} + V_0^- e^{+\gamma l}\\ I(l) = I_L = \dfrac{V_0^+}{Z_0} e^{-\gamma l} - \dfrac{V_0^-}{Z_0} e^{+\gamma l} \end{cases}\] \[\begin{cases} V(z) = \dfrac{I_L}{2} \left[(Z_L+Z_0)e^{\gamma (l-z)} + (Z_L-Z_0)e^{-\gamma(l-z)}\right]\\ I(z) = \dfrac{I_L}{2Z_0} \left[(Z_L+Z_0)e^{\gamma (l-z)} - (Z_L-Z_0)e^{-\gamma(l-z)}\right] \end{cases}\] \[\begin{cases} V(z') = I_L (Z_L \cosh \gamma z' + Z_0 \sinh \gamma z')\\ I(z') = \dfrac{I_L}{Z_0} (Z_L \sinh \gamma z' + Z_0 \cosh \gamma z') \end{cases}\] \[Z_i = Z_0 \dfrac{Z_L + Z_0 \tanh \gamma l}{Z_0 + Z_L \tanh \gamma l}\]

Transmission Line As Circuit Elements

For lossless transmission line

\[Z_i = Z_0 \dfrac{Z_L + j Z_0 \tan\beta l}{Z_0 + j Z_L \tan\beta l}\]

If \(Z_L \to \infty\)

\[Z_i = -j Z_0 \cot \beta l\]

If \(Z_L = 0\)

\[Z_i = j Z_0 \tan\beta l\]

Quarter-wave section: \(l = \lambda/4, \beta l = \pi/2\).

\[Z_i = \dfrac{Z_0^2}{Z_L}\]

also called quarter wave transformer.

Half-wave section: \(l = \lambda/2, \beta l = \pi\).

\[Z_i = Z_L\]

Lines With Resistive Termination

\[\begin{align} V(z') &= \dfrac{I_L}{2} (Z_L + Z_0) e^{\gamma z'} \left[1 + \dfrac{Z_L - Z_0}{Z_L + Z_0}e^{-2\gamma z'}\right]\\ &= \dfrac{I_L}{2}(Z_L + Z_0) e^{\gamma z'} [1 + \Gamma e^{-2\gamma z'}] \end{align}\] \[\Gamma = \dfrac{Z_L - Z_0}{Z_L + Z_0}\]

Transmission Line Circuits

\[V(z') = \dfrac{Z_0 V_g}{Z_0 + Z_g} e^{-\gamma z} \left(\dfrac{1+\Gamma e^{-2\gamma z'}}{1-\Gamma_g \Gamma e^{-2\gamma l}}\right)\]

The Smith Chart

Lossless Transmission Line.

\[\Gamma = \dfrac{z_L - 1}{z_L + 1}\] \[z_L = \dfrac{1+\Gamma}{1-\Gamma}\]

plot \(\Gamma\) on the complex plane.

Now 9.6(1/2)

LC Resonance and Matching Networks

LC Resonance

For series-connected LC circuit, when the reactance magnitudes of \(L\) and \(C\) are equal, the pair resonates.

\[\omega_0 = \dfrac{1}{\sqrt{LC}}\]

Series Circuit Quality Factors

Impedance \(Z\), resistance \(R\) and reactance \(X\).

Q of Inductors and Capacitors

We can model real inductor as the combination of resistance and reactance

\[Z_L = R_L + j X_L\] \[Z_C = R_C + j X_C\]

For series \(L\) and \(R_L\)

\[Q = \dfrac{X_L}{R_L} = \dfrac{\omega L}{R_L}\]

For series \(C\) and \(R_C\)

\[Q = \dfrac{X_C}{R_C} = \dfrac{1}{\omega C R_C}\]

Unloaded Q

The above apply to inductor or capacitor only. If we connect such real inductor and capacitor in series, it seems we get two different \(Q\)

\[Q_1 = \dfrac{\omega L}{R_L + R_C}\] \[Q_2 = \dfrac{1}{\omega C (R_L+R_C)}\]

Which one should we use? Actually at frequency \(\omega_0 = 1/\sqrt{LC}\)

\[Q_U = \dfrac{\omega_0 L}{R_L+R_C} = \dfrac{1}{\omega_0 C(R_L+R_C)}\]

the \(Q_U\) is called as unloaded Q.

Loaded Q

Then, the series connected real inductor and capacitor, and connected to some resistors and sources (including their internal resistance). Naturally, the loaded Q is defined as

\[Q_L = \dfrac{\omega_0 L}{Z_{\text{others}} + R_L + R_C} = \dfrac{1}{\omega_0 C (Z_{\text{others}} + R_L + R_C)}\]

Parallel Circuit Quality Factors

Admittance \(Y\), conductance \(G\) and susceptance \(B\).

\[Y_L = G_L + jB_L\] \[Y_C = G_C + jB_C\]

For parallel \(L\) and \(G_L\)

\[Q = \dfrac{B_L}{G_L} = \dfrac{1}{\omega L G_L}\]

For parallel \(C\) and \(G_C\)

\[Q = \dfrac{B_C}{G_C} = \dfrac{\omega C}{G_C}\]

Similarly

\[Q_U = \dfrac{1}{\omega L (G_L+G_C)} = \dfrac{\omega C}{G_L + G_C}\] \[Q_L = \dfrac{1}{\omega_L (G_{\text{others}} + G_L + G_C)} = \dfrac{\omega C}{G_{\text{\others}} + G_L + G_C}\]

High Frequency Design Technique