Understanding Doppler Shift in Cellular Systems
This question asks about the change in received carrier frequency for a mobile user in a digital cellular system due to their movement.
The phenomenon that causes a change in the received frequency due to relative motion between the transmitter and the receiver is known as the Doppler effect.
Doppler Effect Formula
The Doppler shift ($\Delta f$) can be calculated using the formula:
$ \Delta f = f_c \frac{v}{c} \cos(\theta) $Where:
- $f_c$ is the original carrier frequency.
- $v$ is the relative velocity between the transmitter and the receiver (mobile user).
- $c$ is the speed of light in a vacuum (approximately $3 \times 10^8$ m/s).
- $\theta$ is the angle between the direction of the transmitted signal's propagation and the direction of the mobile user's velocity vector.
Applying the Formula to the Problem
Let's break down the given information:
- Transmitter carrier frequency, $f_c = 900$ MHz $= 900 \times 10^6$ Hz.
- Mobile user's velocity, $v = 70$ km/hr.
- The mobile is moving in a direction perpendicular to the direction of arrival of the transmitted signal. This is a key piece of information.
Converting Velocity Units
First, we need to convert the mobile's velocity from km/hr to m/s:
$ v = 70 \frac{\text{km}}{\text{hr}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ hr}}{3600 \text{ s}} $ $ v = \frac{70 \times 1000}{3600} \text{ m/s} \approx 19.44 \text{ m/s} $Determining the Angle ($\theta$)
The problem states the mobile moves perpendicular to the direction of signal arrival. In the context of the Doppler formula, the direction of signal arrival is the direction the signal is traveling from the transmitter to the receiver. The velocity vector describes the mobile's motion. If the motion is perpendicular to the signal path, the angle $\theta$ between the velocity vector and the direction of signal propagation is $90^\circ$.
$ \theta = 90^\circ $Calculating the Doppler Shift ($\Delta f$)
Now, we can calculate the Doppler shift using the formula:
$ \Delta f = f_c \frac{v}{c} \cos(\theta) $Substitute the values:
$ \Delta f = (900 \times 10^6 \text{ Hz}) \times \frac{19.44 \text{ m/s}}{3 \times 10^8 \text{ m/s}} \times \cos(90^\circ) $Since $\cos(90^\circ) = 0$, the entire term for the Doppler shift becomes zero:
$ \Delta f = (900 \times 10^6 \text{ Hz}) \times (\approx 6.48 \times 10^{-8}) \times 0 $ $ \Delta f = 0 \text{ Hz} $Calculating the Received Carrier Frequency
The received carrier frequency ($f_r$) is the original carrier frequency plus the Doppler shift:
$ f_r = f_c + \Delta f $Substituting the values:
$ f_r = 900 \text{ MHz} + 0 \text{ Hz} $ $ f_r = 900 \text{ MHz} $Conclusion
When the mobile user is moving perpendicularly to the direction of the transmitted signal, there is no Doppler shift. Therefore, the received carrier frequency is the same as the transmitted carrier frequency.
The received carrier frequency is 900 MHz.
Important Questions from Electromagnetic Wave Propagation
Arrange the formulas of $\beta$ in the order where plane wave is travelling in lossy dielectric, free space, good conductor, loss-less dielectric respectively.
A. $\beta = \omega \sqrt{\frac{\mu \epsilon}{2} [\sqrt{1 + (\frac{\sigma}{\omega \epsilon})^2} +1]}$
B. $\beta = \frac{\omega}{C}$
C. $\beta = \omega \sqrt{\mu \epsilon}$
D. $\beta = \sqrt{\frac{\omega \mu \sigma}{2}}$
Choose the correct answer from the options given below:
A plane wave is propagating in a uniform medium with propagation constant \((0.6\pi + j\,0.4\pi)\) its wavelength is
Assertion (A) : The ionosphere is that region of the earth’s atmosphere in which the constituent gases are ionised by radiations from outer space.
Reason (R) : Throughout the ionosphere, there are several layers in which ionization density either reaches maximum or remains almost constant. It depend on the intensity of sun. It also depends upon the atmospheric pressure.
Ultra High Frequency (UHF) spectrum is defined as :
Assertion (A) : Light is electromagnetic wave in nature, these may be x-rays, radio waves, microwaves etc.
Reason (R) : The amount of energy depends on the intensity of the light rays. The energy associated with each photon is proportional to the frequency.
Select your answer using the codes given below :
