The propagation constant of a transmission line is 0.15×10–3 + j1.5×10–3. The wavelength of the traveling wave

Question

The propagation constant of a transmission line is 0.15×10–3 + j1.5×10–3. The wavelength of the traveling wave is

Examoogle · Accepted Answer

The correct answer is Option B: $\frac{2\pi}{1.5 	imes 10^{-3}}$. The propagation constant of a transmission line is defined as $\gamma = \alpha + j\beta$, where $\alpha$ is the attenuation constant and $\beta$ is the phase constant. The wavelength $\lambda$ of the traveling wave is related to the phase constant $\beta$ by the expression $\lambda = \frac{2\pi}{\beta}$.

Step-by-Step Solution & Concepts:
The propagation constant of a transmission line is defined as $\gamma = \alpha + j\beta$, where $\alpha$ is the attenuation constant and $\beta$ is the phase constant. The wavelength $\lambda$ of the traveling wave is related to the phase constant $\beta$ by the expression $\lambda = \frac{2\pi}{\beta}$.

Working Principle: As an electromagnetic wave propagates down a transmission line, it undergoes a phase shift determined by the phase constant $\beta$. The distance required for the wave to complete one full cycle (a phase shift of $2\pi$ radians) is defined as the wavelength $\lambda$. Since $\beta$ represents the phase shift per unit length (radians/meter), $\lambda$ is simply the inverse proportion of $\beta$.

Key Formulas: $\gamma = \alpha + j\beta$ — Propagation constant definition; $\lambda = \frac{2\pi}{\beta}$ — Relationship between wavelength and phase constant

Key Points: The propagation constant is a complex quantity where the imaginary part $\beta$ dictates the wave's phase velocity.; The units of $\beta$ are radians per meter (rad/m) and $\alpha$ are nepers per meter (Np/m).; Wavelength is inversely proportional to the phase constant: higher frequency signals generally correspond to higher $\beta$ values and shorter wavelengths.

Advantages: Provides a simple mathematical relation for wave analysis; Essential for calculating line impedance matching

Disadvantages: Assumes a linear, time-invariant transmission line; Does not account for non-uniform line geometry

Applications: Designing impedance matching networks; Determining signal propagation delays in power and communication lines

Additional Info: Given $\gamma = 0.15 	imes 10^{-3} + j(1.5 	imes 10^{-3})$, the imaginary part $\beta = 1.5 	imes 10^{-3}$.. Substituting $\beta$ into the formula $\lambda = \frac{2\pi}{\beta}$ yields $\frac{2\pi}{1.5 	imes 10^{-3}}$.

Answer: B is correct — The wavelength is defined by $\lambda = \frac{2\pi}{\beta}$, where $\beta$ is the imaginary part of the propagation constant.

Exam Tip: Always ensure you extract the imaginary part ($\beta$) for phase calculations, as the real part ($\alpha$) only affects signal amplitude/attenuation.

Examoogle · Answer

Option A: $\frac{15 	imes 10^{-3}}{2\pi}$

Examoogle · Answer

Option C: $\frac{15 	imes 10^{-3}}{\pi}$

Examoogle · Answer

Option D: $\frac{\pi}{15 	imes 10^{-3}}$

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