mcq cs analog comm system i

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Amplitude Modulated Systems (a) Modulation 1. In commercial TV transmission in India, picture and speech signals are modulated respectively (Picture) (Speech) (a) VSB and VSB (b) VSB and SSB (c) VSB and FM (d) FM and VSB [GATE: 2 Marks] Soln. Note that VSB modulation is the clever compromise between SSB and DSB. Since TV bandwidth is large so VSB is used for picture transmission. Also, FM is the best option for speech because of better noise immunity Option (c) 2. In a double side-band (DSB) full carrier AM transmission system, if the modulation index is doubled, then the ratio of total sideband power to the carrier power increases by a factor of ____________ . [GATE 2014: 1 Mark] Soln. The AM system is Double side band (DSB) with full carrier. The expression for total power in such modulation signal is = + +

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Amplitude Modulated Systems

(a) Modulation

1. In commercial TV transmission in India, picture and speech signals are

modulated respectively

(Picture) (Speech)

(a) VSB and VSB

(b) VSB and SSB

(c) VSB and FM

(d) FM and VSB

[GATE: 2 Marks]

Soln. Note that VSB modulation is the clever compromise between SSB and

DSB. Since TV bandwidth is large so VSB is used for picture

transmission. Also, FM is the best option for speech because of better

noise immunity

Option (c)

2. In a double side-band (DSB) full carrier AM transmission system, if the

modulation index is doubled, then the ratio of total sideband power to the

carrier power increases by a factor of ____________.

[GATE 2014: 1 Mark]

Soln. The AM system is Double side band (DSB) with full carrier. The

expression for total power in such modulation signal is

๐‘ท๐’• =๐‘ฌ๐’„๐Ÿ

๐Ÿ๐‘น+

๐๐Ÿ

๐Ÿ’

๐‘ฌ๐’„๐Ÿ

๐Ÿ๐‘น+

๐๐Ÿ

๐Ÿ’

๐‘ฌ๐’„๐Ÿ

๐Ÿ๐‘น

๐’๐’“, ๐‘ท๐’• = ๐‘ท๐‘ช +๐๐Ÿ

๐Ÿ๐‘ท๐‘ช

The second term on the right hand side is side band power.

๐’”๐’, ๐‘ท๐‘บ๐‘ฉ =๐๐Ÿ

๐Ÿ๐‘ท๐‘ช

๐’๐’“,๐‘ท๐‘บ๐‘ฉ

๐‘ท๐‘ช=

๐๐Ÿ

๐Ÿ

Now if ยต (modulation index) is doubled then ๐‘ท๐‘บ๐‘ฉ

๐‘ท๐‘ชwill be 4 times

So, it is factor of 4

Ans. Factor of 4

3. The maximum power efficiency of an AM modulator is

(a) 25%

(b) 50%

(c) 33%

(d) 100%

[GATE 1992: 2 Marks]

Soln. Efficiency of modulation can be given as

๐œผ =๐‘ท๐‘บ

๐‘ท๐‘ช + ๐‘ท๐‘บ=

๐๐Ÿ

๐Ÿโ„ ๐‘ท๐‘ช

๐‘ท๐‘ช +๐๐Ÿ

๐Ÿ๐‘ท๐‘ช

๐๐Ÿ

๐Ÿโ„

๐Ÿ +๐๐Ÿ

๐Ÿโ„

=๐๐Ÿ

(๐Ÿ + ๐๐Ÿ)

ยต=1 is the optimum value

๐’”๐’, ๐œผ =๐Ÿ

๐Ÿ + ๐Ÿ=

๐Ÿ

๐Ÿ‘ร— ๐Ÿ๐ŸŽ๐ŸŽ = ๐Ÿ‘๐Ÿ‘%

Option (c)

4. Consider sinusoidal modulation in an AM systems. Assuming no over

modulation , the modulation index (ยต) when the maximum and minimum

values of the envelope, respectively, are 3V and 1V is ____________

[GATE 2014: 1 Mark]

Soln. As given is the problem the modulation is sinusoidal this is also called

tone modulation.

There is no over modulation means that modulation index is less than or

equal to 1.

In such case the formula for modulation index is given by

๐ =๐‘ฌ๐’Ž๐’‚๐’™ โˆ’ ๐‘ฌ๐’Ž๐’Š๐’

๐‘ฌ๐’Ž๐’‚๐’™ + ๐‘ฌ๐’Ž๐’Š๐’

Where Emax is the maximum value of the envelope

Emin is the minimum value of the envelope.

This method is popular when the modulated waveform is observed is

CRO

๐ =๐Ÿ‘ โˆ’ ๐Ÿ

๐Ÿ‘ + ๐Ÿ=

๐Ÿ

๐Ÿ’=

๐Ÿ

๐Ÿ= ๐ŸŽ. ๐Ÿ“๐ŸŽ

Modulation index is 0.50

5. Which of the following analog modulation scheme requires the minimum

transmitted power and minimum channel band-width?

(a) VSB

(b) DSB-SC

(c) SSB

(d) AM

[GATE: 2005 1 Mark]

Soln. Modulation type BW Power

Conventional AM 2 fm Maximum power

DSB SC 2 fm (Less power)

VSB fm + vestige

SSB fm Less & power

So, SSB least power & bandwidth

Option (c)

6. Suppose that the modulating signal is ๐‘š(๐‘ก) = 2 cos(2๐œ‹๐‘“๐‘š๐‘ก) and the carrier

signal is ๐‘ฅ๐ถ(๐‘ก) = ๐ด๐ถ cos(2๐œ‹๐‘“๐‘๐‘ก).Which one of the following is a

conventional AM signal without over-modulation?

(a) ๐‘ฅ(๐‘ก) = ๐ด๐ถ๐‘š(๐‘ก) cos(2๐œ‹๐‘“๐‘๐‘ก)

(b) ๐‘ฅ(๐‘ก) = ๐ด๐ถ[1 + ๐‘š(๐‘ก)] cos(2๐œ‹๐‘“๐‘๐‘ก)

(c) ๐‘ฅ(๐‘ก) = ๐ด๐ถ cos(2๐œ‹๐‘“๐‘๐‘ก) +๐ด๐ถ

4๐‘š(๐‘ก) cos(2๐œ‹๐‘“๐‘๐‘ก)

(d) ๐‘ฅ(๐‘ก) = ๐ด๐ถ cos(2๐œ‹๐‘“๐‘š๐‘ก) cos(2๐œ‹๐‘“๐‘๐‘ก) + ๐ด๐ถ sin(2๐œ‹๐‘“๐‘š๐‘ก) sin(2๐œ‹๐‘“๐‘๐‘ก)

[GATE 2010: 1 Mark]

Soln. Given

Modulation signal๐’Ž(๐’•) = ๐Ÿ ๐œ๐จ๐ฌ(๐Ÿ๐…๐’‡๐’Ž๐’•)

Carrier signal ๐’™๐’„(๐’•) = ๐‘จ๐‘ช ๐œ๐จ๐ฌ(๐Ÿ๐…๐’‡๐’„)๐’•

Note that conventional AM is DSB โ€“ FC (DSB full carrier)

Standard Expression is given by

๐’†(๐’•) = ๐‘ฌ๐‘ช[๐Ÿ + ๐’Ž(๐’•)] ๐œ๐จ๐ฌ ๐Ž๐’„๐’•

Or ๐’†(๐’•) = ๐‘ฌ๐‘ช[๐Ÿ + ๐ ๐œ๐จ๐ฌ ๐Ž๐’Ž๐’•] ๐œ๐จ๐ฌ ๐Ž๐’„๐’• โˆ’ โˆ’ โˆ’ โˆ’ โˆ’ (๐Ÿ)

Option (b) is ๐’™(๐’•) = ๐‘จ๐‘ช[๐Ÿ + ๐Ÿ ๐œ๐จ๐ฌ(๐Ÿ๐…๐’‡๐’Ž๐’•)] ๐œ๐จ๐ฌ ๐Ÿ๐…๐’‡๐’„๐’•

Comparing this expression with the standard one given equation (I)

We get ยต = 2 i.e. conventional AM with over modulation

Option (c)

๐’™(๐’•) = ๐‘จ๐‘ช ๐œ๐จ๐ฌ ๐Ÿ๐… ๐’‡๐’„๐’• +๐‘จ๐‘ช

๐Ÿ’๐’Ž(๐’•) ๐œ๐จ๐ฌ ๐Ÿ๐… ๐’‡๐’„๐’•

= ๐‘จ๐‘ช [๐Ÿ +๐Ÿ

๐Ÿ’. ๐Ÿ ๐œ๐จ๐ฌ(๐Ÿ๐…๐’‡๐’Ž๐’•) ๐œ๐จ๐ฌ ๐Ÿ๐…๐’‡๐’„๐’•]

= ๐‘จ๐‘ช [๐Ÿ +๐Ÿ

๐Ÿ๐œ๐จ๐ฌ(๐Ÿ๐…๐’‡๐’Ž๐’•)] ๐œ๐จ๐ฌ ๐Ÿ๐…๐’‡๐’„๐’•

Here ๐ = ๐Ÿ๐Ÿโ„

So, this represents conventional AM without over modulation.

Option (d) is non standard expression

So, correct option is option (c)

7. For a message signal ๐‘š(๐‘ก) = cos(2๐œ‹๐‘“๐‘๐‘ก) and carrier of frequency๐‘“๐‘. Which

of the following represents a single side-band (SSB) signal?

(a) cos(2๐œ‹๐‘“๐‘š๐‘ก) cos(2๐œ‹๐‘“๐‘๐‘ก)

(b) cos(2๐œ‹๐‘“๐‘๐‘ก)

(c) cos[2๐œ‹(๐‘“๐‘ + ๐‘“๐‘š)๐‘ก]

(d) [1 + cos(2๐œ‹๐‘“๐‘š๐‘ก)]. cos(2๐œ‹๐‘“๐‘๐‘ก)

[GATE 2009: 1 Mark]

Soln. Option (a) in the problem represents AM signal DSB-SC. If will have

both side bands

option (b) represents only the carrier frequency

Option (c), ๐œ๐จ๐ฌ[๐Ÿ๐…(๐’‡๐’„ + ๐’‡๐’Ž)๐’•] represents upper side band (SSB-SC).

It represent SSB signal

Option (d) represents the conventional AM signal

Ans. Option (c)

8. A DSB-SC signal is generated using the carrier cos(๐œ”๐ถ๐‘ก + ๐œƒ) and

modulating signal x(t). The envelop of the DSB-SC signal is

(a) ๐‘ฅ(๐‘ก)

(b) |๐‘ฅ(๐‘ก)|

(c) Only positive portion of x(t)

(d) ๐‘ฅ(๐‘ก) cos ๐œƒ

[GATE 1998: 1 Mark]

Soln. Given

Carrier ๐’„(๐’•) = ๐œ๐จ๐ฌ(๐Ž๐’„๐’• + ๐œฝ)

Modulating signal ๐’Ž(๐’•) = ๐’™(๐’•)

DSB SC modulated signal is given by ๐’„(๐’•). ๐’Ž(๐’•) = ๐’”(๐’•)

= ๐’™(๐’•) ๐œ๐จ๐ฌ(๐Ž๐’„๐’• + ๐œฝ)

= ๐’™(๐’•){๐œ๐จ๐ฌ ๐œฝ . ๐œ๐จ๐ฌ ๐Ž๐’„๐’• โˆ’ ๐ฌ๐ข๐ง ๐œฝ ๐ฌ๐ข๐ง ๐Ž๐’„๐’•}

= ๐’™(๐’•) ๐œ๐จ๐ฌ ๐œฝ . ๐œ๐จ๐ฌ ๐Ž๐’„๐’• โˆ’ ๐’™(๐’•). ๐ฌ๐ข๐ง ๐œฝ ๐ฌ๐ข๐ง ๐Ž๐’„๐’•

Envelope of ๐’”(๐’•) = โˆš[๐’™(๐’•) ๐œ๐จ๐ฌ ๐œฝ]๐Ÿ + [๐’™(๐’•) ๐ฌ๐ข๐ง ๐œฝ]๐Ÿ

= โˆš๐’™๐Ÿ(๐’•)(๐’„๐’๐’”๐Ÿ๐œฝ + ๐’”๐’Š๐’๐Ÿ๐œฝ)

= ๐’™(๐’•)

Option (b) |๐’™(๐’•)|

9. A 1 MHz sinusoidal carrier is amplitude modulated by a symmetrical square

wave of period 100 ยตsec. Which of the following frequencies will not be

present in the modulated signal?

(a) 990 kHz

(b) 1010 kHz

(c) 1020 kHz

(d) 1030 kHz

[GATE 2002: 1 Mark]

Soln. Frequency of carrier signal is ๐Ÿ๐‘ด๐‘ฏ๐’› = ๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ ๐‘ฒ๐‘ฏ๐’›

Modulation signal is square wave of period 100 ยตS.

Frequency =๐Ÿ

๐Ÿ๐ŸŽ๐ŸŽร—๐Ÿ๐ŸŽโˆ’๐Ÿ”= ๐Ÿ๐ŸŽ ๐‘ฒ๐‘ฏ๐’›

Since modulation signal is symmetrical square wave it will contain only

odd harmonics i.e. 10 KHz, 30 KHz, 50 KHz -----etc.

Thus the modulated signal has

๐’‡๐’„ ยฑ ๐’‡๐’Ž = (๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ ยฑ ๐Ÿ๐ŸŽ๐‘ฒ๐‘ฏ๐’›) = ๐Ÿ๐ŸŽ๐Ÿ๐ŸŽ๐‘ฒ๐‘ฏ๐’› & ๐Ÿ—๐Ÿ—๐ŸŽ ๐‘ฒ๐‘ฏ๐’›

๐’‡๐’„ ยฑ ๐Ÿ‘๐’‡๐’Ž = (๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ ยฑ ๐Ÿ‘๐ŸŽ๐‘ฒ๐‘ฏ๐’›) = ๐Ÿ๐ŸŽ๐Ÿ‘๐ŸŽ๐‘ฒ๐‘ฏ๐’› & ๐Ÿ—๐Ÿ•๐ŸŽ ๐‘ฒ๐‘ฏ๐’›

So, 1020 KHz will not be present in modulated signal

Option (c)

10. A message signal given by ๐‘š(๐‘ก) = (1

2) cos ๐œ”1๐‘ก โˆ’ (

1

2) sin ๐œ”2๐‘ก is amplitude

modulated with a carrier of frequency ฯ‰c to generate

๐‘ (๐‘ก) = [1 + ๐‘š(๐‘ก)] cos ๐œ”๐‘ ๐‘ก

What is the power efficiency achieved by this modulation scheme?

(a) 8.33%

(b) 11.11%

(c) 20%

(d) 25%

[GATE 2009: 2 Marks]

Soln. Given

๐’Ž(๐’•) =๐Ÿ

๐Ÿ๐œ๐จ๐ฌ ๐Ž๐Ÿ๐’• โˆ’

๐Ÿ

๐Ÿ๐ฌ๐ข๐ง ๐Ž๐Ÿ๐’•

๐’”(๐’•) = [๐Ÿ + ๐’Ž(๐’•)] ๐œ๐จ๐ฌ ๐Ž๐’„๐’•

Note that the modulation frequency are ๐Ž๐Ÿ๐’‚๐’๐’… ๐Ž๐Ÿ i.e. multitone

modulation

Net modulation index is ๐ = โˆš๐๐Ÿ๐Ÿ + ๐๐Ÿ

๐Ÿ + โˆ’ โˆ’ ๐๐’๐Ÿ

Here, ๐ = โˆš(๐Ÿ

๐Ÿ)

๐Ÿ+ (

๐Ÿ

๐Ÿ)

๐Ÿ=

๐Ÿ

โˆš๐Ÿ

๐œผ =๐๐Ÿ

๐๐Ÿ + ๐Ÿร— ๐Ÿ๐ŸŽ๐ŸŽ%

=(๐Ÿ/โˆš๐Ÿ)

๐Ÿ

(๐Ÿ/โˆš๐Ÿ)๐Ÿ

+ร— ๐Ÿ๐ŸŽ๐ŸŽ% = ๐Ÿ๐ŸŽ%

Option (c)

11. A 4 GHz carrier is DSB-SC modulated by a low-pass message signal with

maximum frequency of 2 MHz. The resultant signal is to be ideally sampled.

The minimum frequency of the sampling impulse train should be

(a) 4 MHz

(b) 8 MHz

(c) 8 GHz

(d) 8.004 GHz

[GATE: 1990 2 Mark]

Soln. Given

๐’‡๐’„ = ๐Ÿ’ ๐‘ฎ๐‘ฏ๐’› = ๐Ÿ’๐ŸŽ๐ŸŽ๐ŸŽ ๐‘ด๐‘ฏ๐’›

๐’‡๐’Ž = ๐Ÿ ๐‘ด๐‘ฏ๐’› (๐’๐’๐’˜ ๐’‘๐’‚๐’”๐’” ๐’Ž๐’†๐’”๐’”๐’‚๐’ˆ๐’† ๐’”๐’Š๐’ˆ๐’๐’‚๐’)

Such a signal is amplitude modulated (DSB-SC) i.e. two side bands

(๐’‡๐’„ + ๐’‡๐’Ž)&(๐’‡๐’„ โˆ’ ๐’‡๐’Ž)

i.e. 4002 & 3998 or 4 MHz = BW

so, min. sampling frequency should be (Nyquist Rate)

option (b) ๐’‡๐’”(๐’Ž๐’Š๐’) = ๐Ÿ ร— ๐Ÿ’ = ๐Ÿ– ๐‘ด๐‘ฏ๐’›

(b) Demodulation

12. Consider the amplitude modulated (AM) signal๐ด๐ถ cos ๐œ”๐‘๐‘ก +

2 cos ๐œ”๐‘š ๐‘ก cos ๐œ”๐‘๐‘ก. For demodulating the signal using envelope detector, the

minimum value of AC should be

(a) 2

(b) 1

(c) 0.5

(d) 0

[GATE 2008: 1 Mark]

Soln. Modulated signal is given as

๐‹๐‘จ๐‘ด(๐’•) = ๐‘จ๐’„ ๐œ๐จ๐ฌ ๐Ž๐’„๐’• + ๐Ÿ ๐œ๐จ๐ฌ ๐›š๐ฆ๐ญ. ๐œ๐จ๐ฌ ๐Ž๐’„๐’•

๐‹๐‘จ๐‘ด(๐’•) = [๐‘จ๐’„ + ๐Ÿ ๐œ๐จ๐ฌ ๐Ž๐’„๐’•] ๐œ๐จ๐ฌ ๐Ž๐’Ž๐’•

Note that for envelope detection the modulation should not go beyond

full modulation i.e. ๐ = ๐Ÿ, so amplitude of baseband signal has to be

less than the carrier amplitude (Ac)

|๐’‡(๐’•)|๐’Ž๐’‚๐’™ โ‰ค ๐‘จ๐’„

i.e. |๐Ÿ ๐œ๐จ๐ฌ ๐Ž๐’Ž๐’•|๐’Ž๐’‚๐’™ = ๐Ÿ โ‰ค ๐‘จ๐’„

or๐‘จ๐’„ โ‰ฅ ๐Ÿ

option (a)

13. Which of the following demodulator (s) can be used for demodulating the

signal

๐‘ฅ(๐‘ก) = 5(1 + 2 cos 200 ๐œ‹๐‘ก)๐‘๐‘œ๐‘ 20000๐œ‹๐‘ก

(a) Envelope demodulator

(b) Square-law demodulator

(c) Synchronous demodulator

(d) None of the above

[GATE 1993: 2 Marks]

Soln. The modulated signal given is ๐’™(๐’•) = ๐Ÿ“(๐Ÿ + ๐Ÿ ๐œ๐จ๐ฌ ๐Ÿ๐ŸŽ๐ŸŽ๐…๐’•). ๐œ๐จ๐ฌ ๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ๐…๐’•

The standard equation for AM is

๐‘ฟ๐‘จ๐‘ด(๐’•) = ๐‘จ๐’„(๐Ÿ + ๐ ๐œ๐จ๐ฌ ๐Ž๐’Ž๐’•) ๐œ๐จ๐ฌ ๐Ž๐’„๐’•

If we compare the two equation we find ๐ = ๐Ÿ.

The modulation index is more than 1 here, so it is the case of over

modulation.

When modulation index is more than 1 (over modulation) then detection

is possible only with, Synchronous modulation, such signal can not be

detected with envelope detector.

Option (c)

14. The amplitude modulated wave form ๐‘ (๐‘ก) = ๐ด๐ถ[1 + ๐พ๐‘Ž๐‘š(๐‘ก)] cos ๐œ”๐ถ๐‘ก is

fed to an ideal envelope detector. The maximum magnitude of ๐พ๐‘Ž๐‘š(๐‘ก) is

greater than 1. Which of the following could be the detector output ?

(a) ๐ด๐‘๐‘š(๐‘ก)

(b) ๐ด๐‘2[1 + ๐พ๐‘Ž๐‘š(๐‘ก)]2

(c) |๐ด๐ถ[1 + ๐พ๐‘Ž๐‘š(๐‘ก)]|

(d) ๐ด๐ถ|1 + ๐พ๐‘Ž๐‘š(๐‘ก)|2

[GATE 2000: 1 Mark]

Soln. Given

|๐‘ฒ๐’‚๐’Ž(๐’•)|>|

For the above condition the AM signal is over modulated. Envelope

detector will not be able to detect over modulated signal correctly.

Non of the above options

15. The diagonal clipping in Amplitude Demodulation (using envelope

detector) can be avoided if RC time-constant of the envelope detector

satisfies the following condition, (here W is message bandwidth and ฯ‰ is

carrier frequency both in rad/sec)

(a) ๐‘…๐ถ <1

๐‘Š

(b) ๐‘…๐ถ >1

๐‘Š

(c) ๐‘…๐ถ <1

๐œ”

(d) ๐‘…๐ถ >1

๐œ”

[GATE 2006: 2 Marks]

Soln. It is seen that to avoid negative peak clipping also said diagonal clipping

the RC time constant of detector should be

Or ๐‰ <๐Ÿ

๐’‡๐’Ž

Note fm is maximum modulating frequency i.e. the bandwidth w

So, ๐‘น๐‘ช <๐Ÿ

๐’˜

16. An AM signal is detected using an envelope detector. The carrier frequency

and modulation signal frequency are 1 MHz and 2 KHz respectively. An

appropriate value for the time constant of the envelope detector is

(a) 500 ยตsec

(b) 20 ยตsec

(c) 0.2 ยตsec

(d) 1 ยตsec

[GATE 2004: 1 Mark]

Soln. Note that the time constant RC should satisfy the following condition

๐Ÿ

๐’‡๐’„< ๐‘น๐‘ช <

๐Ÿ

๐’‡๐’Ž

๐Ÿ

๐Ÿร—๐Ÿ๐ŸŽ๐Ÿ”< ๐‘น๐‘ช <

๐Ÿ

๐Ÿร—๐Ÿ๐ŸŽ๐Ÿ‘

Or ๐Ÿ ๐๐’” < ๐‘น๐‘ช < ๐ŸŽ. ๐Ÿ“๐’Ž๐’”

Option (b)

17. A DSB-SC signal is to be generated with a carrier frequency ๐‘“๐‘ = 1๐‘€๐ป๐‘ง

using a non-linear device with the input-output characteristic

๐‘‰0 = ๐‘Ž0๐‘ฃ๐‘– + ๐‘Ž1๐‘ฃ๐‘–3

Where a0 and a1 are constants. The output of the non-linear device can be

filtered by an appropriate band-pass filter.

Let ๐‘‰๐‘– = ๐ด๐‘๐‘– cos(2๐œ‹๐‘“๐‘

๐‘–๐‘ก) + ๐‘š(๐‘ก) where m(t) is the message signal. Then the

value of ๐‘“๐‘๐‘– (in MHz) is

(a) 1.0

(b) 0.333

(c) 0.5

(d) 3.0

[GATE 2003: 2 Marks]

Soln. ๐‘ฝ๐ŸŽ = ๐‹๐ŸŽ[๐‘จ๐’„๐’Š ๐œ๐จ๐ฌ(๐Ÿ๐…๐’‡๐’„

๐’Š ๐’•) + ๐’Ž(๐’•)]

+๐‹๐Ÿ[๐‘จ๐’„๐’Š ๐‹๐‘จ๐‘ด(๐’•) = ๐‘จ๐’„ ๐œ๐จ๐ฌ ๐Ž๐’„๐’• ๐Ÿ ๐œ๐จ๐ฌ ๐Ž๐’„๐’•]

= ๐‹๐ŸŽ[๐‘จ๐’„๐’Š ๐œ๐จ๐ฌ(๐Ÿ๐…๐’‡๐’„

๐’Š ๐’•) + ๐’Ž(๐’•)]

+๐‹๐Ÿ [(๐‘จ๐’„๐’Š )

๐Ÿ‘๐’„๐’๐’”๐Ÿ‘(๐Ÿ๐…๐’‡๐’„

๐’Š ๐’•) + ๐’Ž๐Ÿ‘(๐’•)]

+๐Ÿ‘. ๐‘จ๐’„๐’Š ๐œ๐จ๐ฌ(๐Ÿ๐…๐’‡๐’„

๐’Š ๐’•). ๐’Ž๐Ÿ(๐’•) + ๐Ÿ‘. (๐‘จ๐’„๐’Š )

๐Ÿ๐’„๐’๐’”๐Ÿ(๐Ÿ๐…๐’‡๐’„

๐’Š ๐’•). ๐’Ž(๐’•)

AM โ€“ DSB โ€“ SC signal lies is

๐‹๐Ÿ. ๐Ÿ‘(๐‘จ๐’„๐’Š )

๐Ÿ๐’Ž(๐’•)๐’„๐’๐’”๐Ÿ(๐…๐’‡๐’„

๐’Š ๐’•)

For DSB โ€“ SC the last term is important

๐Ÿ‘๐‹๐Ÿ(๐‘จ๐’„๐’Š )

๐Ÿ๐’„๐’๐’”๐Ÿ๐Ÿ๐…๐’‡๐’„

๐’Š ๐’•. ๐’Ž(๐’•)

๐Ÿ‘ ๐‹๐Ÿ(๐‘จ๐’„๐’Š )

๐Ÿ. ๐’Ž(๐’•). [๐Ÿ + ๐œ๐จ๐ฌ ๐Ÿ๐…(๐Ÿ๐’‡๐’„

๐’Š )๐’•]

Note ๐’Ž(๐’•) ๐œ๐จ๐ฌ ๐Ž๐’„๐’• โ†’ ๐’‡๐’„ (๐’„๐’‚๐’“๐’“๐’Š๐’†๐’“) = ๐Ÿ๐‘ด๐‘ฏ๐’›

For ๐’„๐’๐’”๐Ÿ term as expended the term is having ๐Ÿ๐’‡๐’„๐’Š

๐Ÿ๐’‡๐’„๐’Š = ๐Ÿ๐‘ด๐‘ฏ๐’› ๐’”๐’, ๐’‡๐’„

๐’Š = ๐ŸŽ. ๐Ÿ“ ๐‘ด๐‘ฏ๐’›

Option (c)

18. A message signal ๐‘š(๐‘ก) = cos 2000 ๐œ‹๐‘ก + 4 cos 4000 ๐œ‹๐‘ก modulates the

carriers ๐‘(๐‘ก) = cos 2๐œ‹๐‘“๐‘๐‘ก where ๐‘“๐‘ = 1๐‘€๐ป๐‘ง to produce an AM signal. For

demodulating the generated AM signal using an envelope detector, the time

constant RC of the detector circuit should satisfy

(a) 0.5 ms< RC < 1 ms

(b) 1 ยตs << RC < 0.5 ms

(c) RC<< 1 ยตs

(d) RC >> 0.5 ms

[GATE 2011: 2 Marks]

Soln. Message signal is

๐’Ž(๐’•) = ๐œ๐จ๐ฌ ๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ๐…๐’• + ๐Ÿ’ ๐œ๐จ๐ฌ ๐Ÿ’๐ŸŽ๐ŸŽ๐ŸŽ๐…๐’•

It consist of two frequencies ๐Ž๐Ÿ = ๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ๐…

Or ๐Ÿ๐…๐’‡๐Ÿ = ๐Ÿ๐ŸŽ๐ŸŽ๐ŸŽ๐…

Or ๐’‡๐Ÿ = ๐Ÿ๐‘ฒ๐‘ฏ๐’›

๐’‡๐Ÿ = ๐Ÿ ๐‘ฒ๐‘ฏ๐’›

So, Max frequency is 2 KHz

So, ๐Ÿ

๐’‡๐’„<< ๐‘น๐‘ช <

๐Ÿ

๐’‡๐’Ž

๐Ÿ

๐Ÿ ๐‘ด๐‘ฏ๐’›<< ๐‘น๐‘ช <

๐Ÿ

๐Ÿ ๐‘ฒ๐‘ฏ๐’›

Or, ๐Ÿ๐๐’” << ๐‘น๐‘บ < ๐ŸŽ. ๐Ÿ“๐’Ž๐’”

Option (b)

(c) Receivers

19. A super heterodyne radio receiver with an intermediate frequency of 455

KHz is tuned to a station operating at 1200 KHz. The associated image

frequency is -----------KHz

[GATE 1993: 2 Marks]

Soln. In most receivers the local oscillator frequency is higher than incoming

signal i.e.

๐’‡๐ŸŽ(๐’‡๐’“๐’†๐’’๐’–๐’†๐’๐’„๐’š ๐’๐’‡ ๐’๐’๐’„๐’‚๐’ ๐’๐’”๐’„๐’Š๐’๐’๐’‚๐’•๐’๐’“) = ๐’‡๐’” + ๐’‡๐’Š

Where ๐’‡๐’”------- signal frequency

๐’‡๐’Š๐’๐’“ ๐’‡๐’”๐’Š -------- Image frequency

๐’‡๐’”๐’Š = ๐’‡๐’” + ๐Ÿ๐‘ฐ๐‘ญ = ๐’‡๐’” + ๐Ÿ๐’‡๐’Š

๐’‡๐’”๐’Š = ๐Ÿ๐Ÿ๐ŸŽ๐ŸŽ + ๐Ÿ(๐Ÿ’๐Ÿ“๐Ÿ“)

๐’‡๐’”๐’Š = ๐Ÿ๐Ÿ๐Ÿ๐ŸŽ ๐‘ฒ๐‘ฏ๐’›

so, answer is 2110 KHz

20. In a super heterodyne AM receiver the image channel selectivity is

determined by

(a) The pre-selector, RF and IF stages

(b) The pre-selector, RF and IF stages

(c) The IF stages

(d) All the stages

[GATE 1987: 2 Marks]

Soln. Image rejection depends on front end selectivity of receiver and must be

achieved before If stage. So image channel selectivity depends upon pre

selector & RF amplifier only.

NOTE:- Similar problems have appeared in GATE 1995, 1996 and 1987.

Only problem statement is different otherwise they are same problems.