Transmission Lines in the Frequency Domain

To analyze the response of a transmission line to a time-harmonic excitation, we must investigate solutions to the transmission line equations in the frequency domain. The lumped element model of a transmission line is shown in Figure 1. Phasor notation is used to represent the voltage V(z) and the current I(z) at a given frequency ω. The transmission line equations can now be written in phasor form,

$\frac{dV\left(z\right)}{dz}=-\left(R+j\omega L\right)I\left(z\right)$

$\frac{dI\left(z\right)}{dz}=-\left(G+j\omega C\right)V\left(z\right)$

Taking the derivative of (1) with respect to z and combining it with (2) yields,

$\frac{d^{2}V(z)}{d{z}^2}={\gamma }^{2}V(z)$

where,

$\gamma =\sqrt{\left(R+j\omega L\right)\left(G+j\omega C\right)}.$

Similarly, we can combine (1) and the derivative of (2) with respect to z to solve for the current,

$\frac{d^{2}I(z)}{d{z}^2}={\gamma }^{2}I(z).$

In general, γ is a complex number whose value depends only on the properties of the transmission line,

$\gamma =\alpha +j\beta$

where α is known as the attenuation constant, and β is the propagation constant of the line.

Equations (3) and (5) are ordinary differential equations with constant coefficients. They are wave equations with solutions of the form,

$V(z)=V^{+}{e}^{-\gamma z}+V^{-}{e}^{+\gamma z}$

$I(z)=I^{+}{e}^{-\gamma z}+I^{-}{e}^{+\gamma z}$

where V ⁺, V ^-, I ⁺ and I ^- are complex constants whose values are determined by the boundary conditions.

We see from Equation (7) that the voltage V(z) at any point on the transmission line has two components V ⁺e^-γz and V ^-e^γz. If we represent the complex value V ⁺ as its magnitude and phase, |V ⁺| e^jφ, and expand the first component to its full time-domain representation,

$\begin{array}{l}{V}^{+}{e}^{-\gamma z}=\mathrm{Re}\left[V^{+}{e}^{-(\alpha +j\beta )z}{e}^{j\omega t}\right]\\ =\mathrm{Re}\left[\left|V^{+}\right|e^{-\alpha z}{e}^{j(\omega t-\beta z+\varphi )}\right]\\ =\left|V^{+}\right|e^{-\alpha z}\cos (\omega t-\beta z+\varphi )\\ =\left|V^{+}\right|e^{-\alpha z}\cos \omega (t-\frac{z}{v}+\frac{\varphi }{\omega }),\end{array}$

we see that it is of the form, F(t-z/v). This indicates that V ⁺ e^-γz is a forward-traveling sinusoidal wave. Likewise, V ^- e^γz is a backward-traveling sinusoidal wave. The voltage at any position is a superposition of these two sinusoids.

The term e^-αz represents an exponential decay in the amplitude of the forward-traveling wave as it moves in the +z direction. The backward-traveling wave exhibits a similar, e^αz, decay in the ‑z direction.

From (4) and (6), it is clear that the attenuation constant is,

$\alpha =\mathrm{Re}\left[\gamma \right]=\mathrm{Re}\left[\sqrt{\left(R+j\omega L\right)\left(G+j\omega C\right)}\right].$

Because this expression includes the square root of a complex number, there is no simple general expression for α in terms of R, L, G and C. However, for most transmission lines, the loss is relatively low at the frequencies of interest (i.e., R << ωL and G << ωC). In this case, we can use the low-loss approximation for α,

$\alpha =\mathrm{Re}\left[\gamma \right]\approx \frac{1}{2}\left(\frac{R}{Z_0}+G{Z}_0\right).$

Typically, in cables and circuit board structures below 1 GHz, the conductor losses are greater than the dielectric losses, and the attenuation constant is approximately,

$\alpha \approx \frac{R}{2Z_0}.$

A wave traveling 1 meter along a transmission line is attenuated by a factor of e^α. It is common to express the attenuation of a transmission line in dB/m,

$\text{attenuation in dB/m}=20\log e^{\alpha }\simeq 8.7\alpha .$

For low-loss cables, the value obtained in (13) can be multiplied by 1000 to express the attenuation in dB/km.

The attenuation is a function of resistance, and the high-frequency resistance is a function of frequency (due to the skin effect). At frequencies where skin effect dominates, the resistance scales with the square root of the frequency. This means the attenuation expressed in dB/km scales with the square root of the frequency. So, a cable with an attenuation of 3 dB/km at 1 MHz would be expected to have an attenuation of about 30 dB/km at 100 MHz.

Substituting (7) and (8) into (1), we can show that,

$-\gamma \left[V^{+}{e}^{-\gamma z}-V^{-}{e}^{\gamma z}\right]=-\left(R+j\omega L\right)\left[I^{+}{e}^{-\gamma z}+I^{-}{e}^{\gamma z}\right].$

Using (4), we get,

$V^{+}{e}^{-\gamma z}-V^{-}{e}^{\gamma z}=\sqrt{\frac{R+j\omega L}{G+j\omega C}}\left[I^{+}{e}^{-\gamma z}+I^{-}{e}^{\gamma z}\right].$

Since this must be true for all z,

$\frac{V^{+}}{I^{+}}=Z_0=\sqrt{\frac{R+j\omega L}{G+j\omega C}}$

$\frac{V^{-}}{I^{-}}=-Z_0=-\sqrt{\frac{R+j\omega L}{G+j\omega C}}.$

Z₀ gives the ratio of voltage to current in the forward-traveling component of the wave, or the negative of that ratio in the backward-traveling component. In general, however, the ratio of the total voltage to the total current on the line is a function of position.

Once again, for convenience, we will define z = 0 to be the location of the load. Thus, the load impedance, Z_L, is equal to the ratio of the total voltage at z = 0 to the total current at z = 0,

$Z_L=\frac{V(0)}{I(0)}=\frac{V^{+}+V^{-}}{I^{+}+I^{-}}=\frac{V^{+}+V^{-}}{\raisebox{1ex}{$V^{+}$}\!\left/ \!\raisebox{-1ex}{$Z_0$}\right.-\raisebox{1ex}{$V^{-}$}\!\left/ \!\raisebox{-1ex}{$Z_0$}\right.}=Z_0\frac{V^{+}+V^{-}}{V^{+}-V^{-}}.$

Rearranging the terms in (18) yields,

$\frac{V^{-}}{V^{+}}=\frac{Z_L-Z_0}{Z_L+Z_0}.$

As in the time-domain case, we define the voltage reflection coefficient to be the ratio of the reflected wave to the incident wave,

${\text{Γ}}_{\text{L}}=\frac{V^{-}}{V^{+}}=\frac{Z_L-Z_0}{Z_L+Z_0}.$

Comparing this result with the time-domain expression in the previous section, we see that Γ_L is calculated from the load impedance and the characteristic impedance in the same way in both the time and frequency domains. In the frequency domain, however, the quantities Z₀ and Z_L are complex; therefore, Γ_L may also be a complex quantity.

Lossless Transmission Lines in the Frequency Domain

Many practical transmission lines have relatively low loss and can be modeled as lossless (i.e., R ≈ G ≈ 0). Lossless lines have a real characteristic impedance,

$Z_0=\sqrt{\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$C$}\right.}$

and an imaginary propagation constant,

$\gamma =j\beta =j\text{}\omega \text{}\sqrt{LC}.$

The voltage along a lossless line is given by,

$V(z)=V^{+}\left[e^{-j\beta z}+{\Gamma }_{\text{L}}{e}^{j\beta z}\text{}\right].$

The current along a lossless line is given by,

$I(z)=I^{+}\left[e^{-j\beta z}-{\Gamma }_{\text{L}}{e}^{j\beta z}\text{}\right].$

The impedance at z = -ℓ is found by combining Equations (23) and (24) with Equations (16) and (20),

$Z(-\text{ℓ})=\frac{V(-\text{ℓ})}{I(-\text{ℓ})}=Z_0\frac{Z_L\cos \beta \text{ℓ}+j\text{}{Z}_0\sin \beta \text{ℓ}}{Z_0\cos \beta \text{ℓ}+j\text{}{Z}_L\sin \beta \text{ℓ}}=Z_0\frac{Z_L+j\text{}{Z}_0\tan \beta \text{ℓ}}{Z_0+j\text{}{Z}_L\tan \beta \text{ℓ}}.$

From Equation (25), it is clear that Z(-ℓ) will equal Z_L when the imaginary terms are zero. This occurs when tan βℓ = 0, or $\text{ℓ}=n\frac{\lambda }{2}.$ In other words, Z(‑ℓ) = Z_L whenever the length of the transmission line is an integer multiple of a half wavelength.