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Question:
Published on: 22 December, 2024

Distinguish between group velocity and phase velocity.

Answer:

Group velocity:

If there is modulation in the carrier, the modulation envelope actually travels at velocity slower than that of carrier alone and of course slower than speed of light. The velocity of modulation envelope is called the group velocity Vg. This happens when a modulation signal travels in a waveguide; the modulation goes on slipping backward with respect to the carrier. It is defined as the rate at which the wave propagates through the waveguide and is given

\(V_g=\frac{d\omega}{d\beta}\).

Where \(\omega=2\pi f\) and \(\beta=\frac{2\pi}{\lambda_g}\), w is the wave’s angular frequency and \beta is the phase constant.

\(V_g=c\sqrt{1-\left(\frac{\lambda_0}{\lambda_c}\right)^2}\)

Phase velocity:

We know that wave propagates in the waveguide when guide wavelength λg is greater than the free space wavelength λ0. Since the velocity of propagation is the product of λ and f, it follows that in a waveguide, Vpgf where Vp is the phase velocity. But the speed of light is equal to product of λ0 and f. This Vp is greater than the speed of light since λg0. This is contradicting since no signal can travel faster than the speed of light. However, the wavelength in the guide is the length of the cycle and Vp represents the velocity of the phase. In fact it is defined as the rate at which the wave changes its phase in terms of the guide wavelength. The phase velocity is the velocity with which the wave changes phase at a plane boundary and not the velocity with which it travels along the boundary.

\(V_p=\frac{\omega}{\beta}\)

Where \(\omega=2\pi f\) and \(\beta=\frac{2\pi}{\lambda_g}\), w is the wave’s angular frequency and \beta is the phase constant.

The example of phase velocity is the velocity of a waves train corresponding to sea waves approaching a beach at an angle rather than straight in. The phenomenon which accompanies this is that the edge of the wave appears to sweep along the beach much faster than the wave is really travelling. It is the phase velocity that provides this effect.

\(V_p=\frac{c}{\sqrt{1-\left(\frac{f_c}{f}\right)^2}}\)

\(V_p=\frac{c}{\sqrt{1-\left(\frac{\lambda_0}{\lambda_c}\right)^2}}\)

Where fc, λ0 and λc are the cut – off frequency, signal wavelength in free space and cut – off wave length.

\(V_pV_g=\frac{c}{\sqrt{1-\left(\frac{\lambda_0}{\lambda_c}\right)^2}}\times c\sqrt{1-\left(\frac{\lambda_0}{\lambda_c}\right)^2}=c^2\)

Hence, the product of phase and group velocities is constant and it is equal to the square of the velocity of light in free space.

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