(a) Explain any two factors that justify the need of modulating a low-frequency signal.
(b) Write two advantages of frequency modulation over amplitude modulation.
Modulating low-frequency signals are required because:
i) Size of the antenna: Antenna is required for both transmission and reception of the signal whose size (at least ) should be comparable to the wavelength of the signal so that time variation of the signal can be properly sensed by the antenna. For a low-frequency signal, the wavelength is large. Therefore, a large antenna of such a huge height is practically impossible to construct and operate.
Hence, the need for high-frequency transmission arises.
ii) Overlapping signals from different transmitters: When an information is transmitted using different transmitters, the signals get overlapped and the information is lost. Hence, high-frequency transmission is required. Each band should be allocated to each frequency range in order to avoid mixing of signals.
iii) Higher energy: High-frequency signals have high energy and therefore, even after loss due to attenuation, signals can be carried over longer distances.
Advantage of frequency modulation over amplitude modulation are:
i) FM is more efficient than AM because the amplitude of an FM wave is constant, irrespective of its modulation index. Thus, the transmitted power is constant. Also, in AM the transmitted power goes waste in the transmitted carrier.
ii) FM reception is quite immune to noise in comparison to AM reception. Noise is a form of amplitude variation in the transmitted signal. Thus, using amplitude limiters in the FM receivers, the noise is eliminated.
Explain the terms (i) Attenuation and (ii) Demodulation used in Communication System.
The B.E. of the nucleus of mass number 240, B1 = 7.6 x 240 = 1824 MeV
The B.E of each product nucleus, B2 = 8.5 x 120 - 1020 MeV
Then, the energy released as the nucleus breaks is given by,
E = 2B2 - B1 = 2 x 1020 - 1824 = 216 MeV
Energy in the fusion reaction is given by,
The three characteristics of photoelectric effect, which cannot be explained on the basis of wave theory of light are:
1. For a given metal and frequency of incident radiation, the number of photoelectrons ejected per second is directly proportional to the intensity of the incident light.
2. For a given metal, a certain minimum frequency of the incident radiation below which no emission of photoelectrons take place. The is the threshold frequency.
3. Above the threshold frequency, the maximum kinetic energy of the emitted photoelectron is independent of the intensity of the incident light and is dependent only upon the frequency or wavelength of the incident light.
Charge (q) is the same but the mass of both particles is different (m1 > m2).
The De-broglie wavelength given by,
The slope of the graph of λ versus =
The slope of the smaller mass is larger; therefore, plat A in the above graph represents mass m2.
A moving charge produces both electric and magnetic fields, and an oscillating charge produces oscillating magnetic and electric fields. These oscillating electric and magnetic fields with respect to space and time produce electromagnetic waves.
The propagation of electromagnetic waves can be shown as:
Maxwell's generalization of ampere's circuital law given by,
Consider that a parallel capacitor C is charging in a circuit.
The magnitude of electric field between the two plates will be,
, is perpendicular to the surface of the plate.
Electric flux through the surface will be,
i) The function of three segments of a transistor are:
Emitter: Emits the majority charge carriers
Collector: Collects the majority charge carriers
Base: Base provides the interaction between the collector and the base
ii) An n-p-n transistor is as shown below:
Input characteristics: The variation between the base current and the base-emitter voltage is obtained. Input characteristics is used to find input dynamic resistance of the transistor as it is represented by the slope.
Output characteristics: A graph representing the variation of the collector current and emitter voltage is obtained, keeping base current fixed. The slope fo the output characteristic graph gives us the output dynamic resistance.
Radius of curvature, R = 20 cm
So, focal length, f = R/2 = - 10 cm
Since the image obtained is real, therefore magnification of the image, m = -2
Now, using the formula,
Therefore, the distance of the object is 15 cm in front of the mirror and the position of the image is 30 cm, formed in front of the mirror.
For a convex mirror,
Focal length, f > 0
Position of the object, u < 0
Using mirror formula, we have
That is, the image formed by a convex lens is always behind the mirror and hence is virtual.