Kerala Engineering Entrance 2007 Questions on Communication Systems

“I have no special talents. I am only passionately curious.”

– Albert Einstein

Questions on communication systems at the level expected from you are generally simple and interesting. Today we will discuss three questions (MCQ) on communication systems which appeared in Kerala Engineering Entrance 2007 question paper.

(1) The time variations of signals are given as in (A), (B) and (C).

Point out true statement from the following

(a) A, B and C are analog signals

(b) A and B are analog, but C is digital signal

(c) A and C are digital, but B is analog signal

(d) A and C are analog, but B is digital signal

(e) A, B and C are digital signals (e)

Digital signals will have two discrete levels only, corresponding to the zero level and one level as shown in graph (B). Analog signals have various instantaneous values, as shown in graphs (A) and (C). The correct option therefore is (d).

(2) A photo detector used to detect the wave length of 1700 nm, has energy gap of about

(a) 0.073 eV

(b) 1.2 eV

(c) 7.3 eV

(d) 1.16 eV

(e) 0.73 eV

The energy gap of a photo detector should be equal to or less than the energy of the photon which it is intended to detect. The product of the energy of a photon in electron volt and the wave length in Angstrom is 12400. Therefore, the energy of the 1700 nm photon is 12400/ 17000 electron volt = 0.73 eV. So, the correct option is (e). Option (a) is not suitable since a semiconductor with a gap as low as 0.073 eV (if at all available) will be unreliable due to the breaking of bonds by thermal excitation.

[You can use the expression hc/λ for calculating the energy of the photon in joule and then convert it into electron volts by dividing it by the electronic charge of 1.6×10^–19 joule. But it will be more difficult and time consuming].

(3) The optical fibres have an inner core of refractive index n₁ and a cladding of refractive index n₂ such that

(a) n₁ = n₂ (b) n₁ ≤ n₂ (c) n₁ < n₂ (d) n₁ > n₂ (e) n₁ ≥ n₂

The cladding has to be rarer than the core so as to totally reflect the beam of light entering the optical fibre. So, the correct option is (d).

Kerala Medical Entrance 2008 (KEAM 2008) Questions on Communication Systems

The following multiple choice questions on communication systems appeared in the Kerala Medical Entrance 2008 (KEAM-2008) examination question paper:

(1) A 1000 kHz carrier wave is modulated by an audio signal of frequency range 100-5000 Hz. Then the width of the channel in kHz is

(a)10

(b) 20

(c) 30

(d) 40

(e) 50

Let us assume that the system uses amplitude modulation of the usual double sideband type. (It should have been mentioned in the question)

The channel width is twice the highest modulating signal frequency and is therefore equal to 2×5000 Hz = 10000 Hz = 10 kHz.

[Remember that in the standard AM sound broadcast systems the channel band width allotted to a station is 10 kHz].

(2) If the critical frequency for sky wave propagation is 12 MHz, then the maximum electron density in the ionosphere is

(a) 1.78×10¹²/m³

(b) 0.178×10¹⁰/m³

(c) 1.12×10¹²/m³

(d) 0.56×10¹²/m³

(e) 0.148×10¹²/m³

The critical frequency f_c for reflection by the ionosphere is given by

f_c = 9 N^1/2 where N is the maximum electron number density.

Therefore, N = f_c²/81 = (12×10⁶)² /81 = 1.78×10¹²/m³

Two Questions (MCQ) on Photo Diodes

“He who joyfully marches in rank and file has already earned my contempt. He has been given a large brain by mistake, since for him the spinal cord would suffice.”

– Albert Einstein

As you know, photo diodes are special purpose p-n junction diodes. They are very popular photo detectors used for detecting optical signals in communication systems.

Today we will discuss two multiple choice questions on photo diodes:

(1) In using a photo diode as a photo detector, it is invariably reverse biased. Why?

(a) The power consumption is much reduced compared to reverse biased condition

(b) Electron hole pairs can be produced by the incident photons only if the photo diode is reverse biased

(c) Light variations can be converted into current variations only if the photo diode is reverse biased

(d) When photons are incident on the diode, the fractional change in the reverse current is much greater than the fractional change in the forward current

(e) The photo diode will be spoilt if it is operated under forward biased condition

Whether the photo diode is reverse biased or forward biased, the number of electron hole pairs produced by the incident photons is the same. In other words, the change in the diode current is the same in both cases. But in the reverse biased condition the current drawn by the diode in the absence of the photons is extremely small, of the order of nanoamperes or microamperes where as in the forward biased condition this current is significant, of the order of tens of milliamperes. The fractional change in the current because of the incident photons is therefore large and easily measurable if the photo diode is reverse biased. The correct option is (d).

(2) The maximum wave length of photons that can be detected by a photo diode made of a semiconductor of band gap 2 eV is about

(a) 620 nm

(b) 700 nm

(c) 740 nm

(d) 860 nm

(e)1240 nm

The wave length λ (in Angstrom unit) of a photon of energy E (in electron volt) is given by

λE = 12400, very nearly.

Therefore, λ = 12400/E

[The above expression can be easily obtained by remembering that a photon of energy 1 eV has wave length 12400 Ǻ and the energy is inversely proportional to the wave length].

Since E = 2 eV we have λ = 12400/2 = 6200 Ǻ = 620 nanometre.

Photons with wave length greater than 640 nm will have energy less than 2 eV so that they will be unable to produce electron hole pairs in the semiconductor of band gap 2 eV. So the correct option is (a).