Which option(s) represents/ represent the dielectric loss tangent of a substrate?

The correct option representing the dielectric loss tangent of a substrate is:

(ωϵ′′ + σ)/(ωϵ′)

To understand why this is the correct representation, let us look at Maxwell's equations, specifically the Maxwell-Ampère law in a lossy dielectric medium. The curl of the magnetic field $\overrightarrow{H}$ is given by:

$$\nabla \times \stackrel{���}{H}=\overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}$$

where $\overrightarrow{J}=\sigma \overrightarrow{E}$ is the conduction current density (with $\sigma$ representing the Ohm conductivity), and $\overrightarrow{D}=ϵ\overrightarrow{E}$ is the electric displacement vector.

For time-harmonic fields with a time dependency of $e^{j\omega t}$, the time derivative $\partial /\partial t$ becomes $j\omega$. The permittivity of a lossy dielectric substrate can be represented as a complex quantity:

$$ϵ={ϵ}^{\prime }-j{ϵ}^{″}$$

Here, ${ϵ}^{\prime }$ represents the real part of the permittivity (which measures the energy stored in the material), and ${ϵ}^{″}$ represents the imaginary part of the permittivity (which accounts for the dielectric losses due to dipole relaxation/polarization damping).

Substituting these relations into the Maxwell-Ampère equation yields:

$$\nabla \times \overrightarrow{H}=\sigma \overrightarrow{E}+j\omega ({ϵ}^{\prime }-j{ϵ}^{″})\overrightarrow{E}$$

Expanding the terms on the right-hand side:

$$\nabla \times \overrightarrow{H}=\sigma \overrightarrow{E}+j\omega {ϵ}^{\prime }\overrightarrow{E}+\omega {ϵ}^{″}\overrightarrow{E}$$

Grouping the terms into real (loss-contributing) and imaginary (storage/displacement-contributing) parts:

$$\nabla \times \overrightarrow{H}=(\sigma +\omega {ϵ}^{″})\overrightarrow{E}+j\omega {ϵ}^{\prime }\overrightarrow{E}$$

In this expression, $(\sigma +\omega {ϵ}^{″})\overrightarrow{E}$ represents the total effective loss current density (in-phase with the electric field), and $\omega {ϵ}^{\prime }\overrightarrow{E}$ represents the lossless displacement current density (90 degrees out-of-phase with the electric field).

The loss tangent ($tan\delta$) of the medium is defined as the ratio of the loss current density magnitude to the displacement current density magnitude:

$$tan\delta =\frac{\omega {ϵ}^{″}+\sigma }{\omega {ϵ}^{\prime }}$$

This ratio perfectly matches the formula given in the correct option, confirming that it represents the total loss tangent of a substrate accounting for both dielectric polarization loss and ohmic conductivity loss.