View
219
Download
4
Category
Preview:
Citation preview
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 1990: 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
๐ท๐ =๐ฌ๐๐
๐๐น+
๐๐
๐
๐ฌ๐๐
๐๐น+
๐๐
๐
๐ฌ๐๐
๐๐น
๐๐, ๐ท๐ = ๐ท๐ช +๐๐
๐๐ท๐ช
The second term on the right hand side is side band power.
๐๐, ๐ท๐บ๐ฉ =๐๐
๐๐ท๐ช
๐๐,๐ท๐บ๐ฉ
๐ท๐ช=
๐๐
๐
Now if ยต (modulation index) is doubled then ๐ท๐บ๐ฉ
๐ท๐ชwill 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
๐ผ =๐ท๐บ
๐ท๐ช + ๐ท๐บ=
๐๐
๐โ ๐ท๐ช
๐ท๐ช +๐๐
๐๐ท๐ช
๐๐
๐โ
๐ +๐๐
๐โ
=๐๐
(๐ + ๐๐)
ยต=1 is the optimum value
๐๐, ๐ผ =๐
๐ + ๐=
๐
๐ร ๐๐๐ = ๐๐%
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 =๐ฌ๐๐๐ โ ๐ฌ๐๐๐
๐ฌ๐๐๐ + ๐ฌ๐๐๐
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 โ ๐
๐ + ๐=
๐
๐=
๐
๐= ๐. ๐๐
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(๐ก) = 2 cos(2๐๐๐๐ก) and the carrier
signal is ๐ฅ๐ถ(๐ก) = ๐ด๐ถ cos(2๐๐๐๐ก).Which one of the following is a
conventional AM signal without over-modulation?
(a) ๐ฅ(๐ก) = ๐ด๐ถ๐(๐ก) cos(2๐๐๐๐ก)
(b) ๐ฅ(๐ก) = ๐ด๐ถ[1 + ๐(๐ก)] cos(2๐๐๐๐ก)
(c) ๐ฅ(๐ก) = ๐ด๐ถ cos(2๐๐๐๐ก) +๐ด๐ถ
4๐(๐ก) cos(2๐๐๐๐ก)
(d) ๐ฅ(๐ก) = ๐ด๐ถ cos(2๐๐๐๐ก) cos(2๐๐๐๐ก) + ๐ด๐ถ sin(2๐๐๐๐ก) sin(2๐๐๐๐ก)
[GATE 2010: 1 Mark]
Soln. Given
Modulation signal๐(๐) = ๐ ๐๐จ๐ฌ(๐๐ ๐๐๐)
Carrier signal ๐๐(๐) = ๐จ๐ช ๐๐จ๐ฌ(๐๐ ๐๐)๐
Note that conventional AM is DSB โ FC (DSB full carrier)
Standard Expression is given by
๐(๐) = ๐ฌ๐ช[๐ + ๐(๐)] ๐๐จ๐ฌ ๐๐๐
Or ๐(๐) = ๐ฌ๐ช[๐ + ๐ ๐๐จ๐ฌ ๐๐๐] ๐๐จ๐ฌ ๐๐๐ โ โ โ โ โ (๐)
Option (b) is ๐(๐) = ๐จ๐ช[๐ + ๐ ๐๐จ๐ฌ(๐๐ ๐๐๐)] ๐๐จ๐ฌ ๐๐ ๐๐๐
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) = ๐จ๐ช ๐๐จ๐ฌ ๐๐ ๐๐๐ +๐จ๐ช
๐๐(๐) ๐๐จ๐ฌ ๐๐ ๐๐๐
= ๐จ๐ช [๐ +๐
๐. ๐ ๐๐จ๐ฌ(๐๐ ๐๐๐) ๐๐จ๐ฌ ๐๐ ๐๐๐]
= ๐จ๐ช [๐ +๐
๐๐๐จ๐ฌ(๐๐ ๐๐๐)] ๐๐จ๐ฌ ๐๐ ๐๐๐
Here ๐ = ๐๐โ
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(๐ก) = cos(2๐๐๐๐ก) and carrier of frequency๐๐. Which
of the following represents a single side-band (SSB) signal?
(a) cos(2๐๐๐๐ก) cos(2๐๐๐๐ก)
(b) cos(2๐๐๐๐ก)
(c) cos[2๐(๐๐ + ๐๐)๐ก]
(d) [1 + cos(2๐๐๐๐ก)]. cos(2๐๐๐๐ก)
[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), ๐๐จ๐ฌ[๐๐ (๐๐ + ๐๐)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(๐๐ถ๐ก + ๐) and
modulating signal x(t). The envelop of the DSB-SC signal is
(a) ๐ฅ(๐ก)
(b) |๐ฅ(๐ก)|
(c) Only positive portion of x(t)
(d) ๐ฅ(๐ก) cos ๐
[GATE 1998: 1 Mark]
Soln. Given
Carrier ๐(๐) = ๐๐จ๐ฌ(๐๐๐ + ๐ฝ)
Modulating signal ๐(๐) = ๐(๐)
DSB SC modulated signal is given by ๐(๐). ๐(๐) = ๐(๐)
= ๐(๐) ๐๐จ๐ฌ(๐๐๐ + ๐ฝ)
= ๐(๐){๐๐จ๐ฌ ๐ฝ . ๐๐จ๐ฌ ๐๐๐ โ ๐ฌ๐ข๐ง ๐ฝ ๐ฌ๐ข๐ง ๐๐๐}
= ๐(๐) ๐๐จ๐ฌ ๐ฝ . ๐๐จ๐ฌ ๐๐๐ โ ๐(๐). ๐ฌ๐ข๐ง ๐ฝ ๐ฌ๐ข๐ง ๐๐๐
Envelope of ๐(๐) = โ[๐(๐) ๐๐จ๐ฌ ๐ฝ]๐ + [๐(๐) ๐ฌ๐ข๐ง ๐ฝ]๐
= โ๐๐(๐)(๐๐๐๐๐ฝ + ๐๐๐๐๐ฝ)
= 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 ๐๐ด๐ฏ๐ = ๐๐๐๐ ๐ฒ๐ฏ๐
Modulation signal is square wave of period 100 ยตS.
Frequency =๐
๐๐๐ร๐๐โ๐= ๐๐ ๐ฒ๐ฏ๐
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
๐๐ ยฑ ๐๐ = (๐๐๐๐ ยฑ ๐๐๐ฒ๐ฏ๐) = ๐๐๐๐๐ฒ๐ฏ๐ & ๐๐๐ ๐ฒ๐ฏ๐
๐๐ ยฑ ๐๐๐ = (๐๐๐๐ ยฑ ๐๐๐ฒ๐ฏ๐) = ๐๐๐๐๐ฒ๐ฏ๐ & ๐๐๐ ๐ฒ๐ฏ๐
So, 1020 KHz will not be present in modulated signal
Option (c)
10. A message signal given by ๐(๐ก) = (1
2) cos ๐1๐ก โ (
1
2) sin ๐2๐ก is amplitude
modulated with a carrier of frequency ฯc to generate
๐ (๐ก) = [1 + ๐(๐ก)] cos ๐๐ ๐ก
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(๐) =๐
๐๐๐จ๐ฌ ๐๐๐ โ
๐
๐๐ฌ๐ข๐ง ๐๐๐
๐(๐) = [๐ + ๐(๐)] ๐๐จ๐ฌ ๐๐๐
Note that the modulation frequency are ๐๐๐๐๐ ๐๐ i.e. multitone
modulation
Net modulation index is ๐ = โ๐๐๐ + ๐๐
๐ + โ โ ๐๐๐
Here, ๐ = โ(๐
๐)
๐+ (
๐
๐)
๐=
๐
โ๐
๐ผ =๐๐
๐๐ + ๐ร ๐๐๐%
=(๐/โ๐)
๐
(๐/โ๐)๐
+ร ๐๐๐% = ๐๐%
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
๐๐ = ๐ ๐ฎ๐ฏ๐ = ๐๐๐๐ ๐ด๐ฏ๐
๐๐ = ๐ ๐ด๐ฏ๐ (๐๐๐ ๐๐๐๐ ๐๐๐๐๐๐๐ ๐๐๐๐๐๐)
Such a signal is amplitude modulated (DSB-SC) i.e. two side bands
(๐๐ + ๐๐)&(๐๐ โ ๐๐)
i.e. 4002 & 3998 or 4 MHz = BW
so, min. sampling frequency should be (Nyquist Rate)
option (b) ๐๐(๐๐๐) = ๐ ร ๐ = ๐ ๐ด๐ฏ๐
(b) Demodulation
12. Consider the amplitude modulated (AM) signal๐ด๐ถ cos ๐๐๐ก +
2 cos ๐๐ ๐ก cos ๐๐๐ก. 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
๐๐จ๐ด(๐) = ๐จ๐ ๐๐จ๐ฌ ๐๐๐ + ๐ ๐๐จ๐ฌ ๐๐ฆ๐ญ. ๐๐จ๐ฌ ๐๐๐
๐๐จ๐ด(๐) = [๐จ๐ + ๐ ๐๐จ๐ฌ ๐๐๐] ๐๐จ๐ฌ ๐๐๐
Note that for envelope detection the modulation should not go beyond
full modulation i.e. ๐ = ๐, so amplitude of baseband signal has to be
less than the carrier amplitude (Ac)
|๐(๐)|๐๐๐ โค ๐จ๐
i.e. |๐ ๐๐จ๐ฌ ๐๐๐|๐๐๐ = ๐ โค ๐จ๐
or๐จ๐ โฅ ๐
option (a)
13. Which of the following demodulator (s) can be used for demodulating the
signal
๐ฅ(๐ก) = 5(1 + 2 cos 200 ๐๐ก)๐๐๐ 20000๐๐ก
(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(๐ + ๐ ๐๐จ๐ฌ ๐๐๐๐ ๐). ๐๐จ๐ฌ ๐๐๐๐๐ ๐
The standard equation for AM is
๐ฟ๐จ๐ด(๐) = ๐จ๐(๐ + ๐ ๐๐จ๐ฌ ๐๐๐) ๐๐จ๐ฌ ๐๐๐
If we compare the two equation we find ๐ = ๐.
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 (๐ก) = ๐ด๐ถ[1 + ๐พ๐๐(๐ก)] cos ๐๐ถ๐ก is
fed to an ideal envelope detector. The maximum magnitude of ๐พ๐๐(๐ก) is
greater than 1. Which of the following could be the detector output ?
(a) ๐ด๐๐(๐ก)
(b) ๐ด๐2[1 + ๐พ๐๐(๐ก)]2
(c) |๐ด๐ถ[1 + ๐พ๐๐(๐ก)]|
(d) ๐ด๐ถ|1 + ๐พ๐๐(๐ก)|2
[GATE 2000: 1 Mark]
Soln. Given
|๐ฒ๐๐(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 ๐ถ <1
๐
(b) ๐ ๐ถ >1
๐
(c) ๐ ๐ถ <1
๐
(d) ๐ ๐ถ >1
๐
[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
๐๐
Note fm is maximum modulating frequency i.e. the bandwidth w
So, ๐น๐ช <๐
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
๐๐< ๐น๐ช <
๐
๐๐
๐
๐ร๐๐๐< ๐น๐ช <
๐
๐ร๐๐๐
Or ๐ ๐๐ < ๐น๐ช < ๐. ๐๐๐
Option (b)
17. A DSB-SC signal is to be generated with a carrier frequency ๐๐ = 1๐๐ป๐ง
using a non-linear device with the input-output characteristic
๐0 = ๐0๐ฃ๐ + ๐1๐ฃ๐3
Where a0 and a1 are constants. The output of the non-linear device can be
filtered by an appropriate band-pass filter.
Let ๐๐ = ๐ด๐๐ cos(2๐๐๐
๐๐ก) + ๐(๐ก) where m(t) is the message signal. Then the
value of ๐๐๐ (in MHz) is
(a) 1.0
(b) 0.333
(c) 0.5
(d) 3.0
[GATE 2003: 2 Marks]
Soln. ๐ฝ๐ = ๐๐[๐จ๐๐ ๐๐จ๐ฌ(๐๐ ๐๐
๐ ๐) + ๐(๐)]
+๐๐[๐จ๐๐ ๐๐จ๐ด(๐) = ๐จ๐ ๐๐จ๐ฌ ๐๐๐ ๐ ๐๐จ๐ฌ ๐๐๐]
= ๐๐[๐จ๐๐ ๐๐จ๐ฌ(๐๐ ๐๐
๐ ๐) + ๐(๐)]
+๐๐ [(๐จ๐๐ )
๐๐๐๐๐(๐๐ ๐๐
๐ ๐) + ๐๐(๐)]
+๐. ๐จ๐๐ ๐๐จ๐ฌ(๐๐ ๐๐
๐ ๐). ๐๐(๐) + ๐. (๐จ๐๐ )
๐๐๐๐๐(๐๐ ๐๐
๐ ๐). ๐(๐)
AM โ DSB โ SC signal lies is
๐๐. ๐(๐จ๐๐ )
๐๐(๐)๐๐๐๐(๐ ๐๐
๐ ๐)
For DSB โ SC the last term is important
๐๐๐(๐จ๐๐ )
๐๐๐๐๐๐๐ ๐๐
๐ ๐. ๐(๐)
๐ ๐๐(๐จ๐๐ )
๐. ๐(๐). [๐ + ๐๐จ๐ฌ ๐๐ (๐๐๐
๐ )๐]
Note ๐(๐) ๐๐จ๐ฌ ๐๐๐ โ ๐๐ (๐๐๐๐๐๐๐) = ๐๐ด๐ฏ๐
For ๐๐๐๐ term as expended the term is having ๐๐๐๐
๐๐๐๐ = ๐๐ด๐ฏ๐ ๐๐, ๐๐
๐ = ๐. ๐ ๐ด๐ฏ๐
Option (c)
18. A message signal ๐(๐ก) = cos 2000 ๐๐ก + 4 cos 4000 ๐๐ก modulates the
carriers ๐(๐ก) = cos 2๐๐๐๐ก where ๐๐ = 1๐๐ป๐ง 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(๐) = ๐๐จ๐ฌ ๐๐๐๐๐ ๐ + ๐ ๐๐จ๐ฌ ๐๐๐๐๐ ๐
It consist of two frequencies ๐๐ = ๐๐๐๐๐
Or ๐๐ ๐๐ = ๐๐๐๐๐
Or ๐๐ = ๐๐ฒ๐ฏ๐
๐๐ = ๐ ๐ฒ๐ฏ๐
So, Max frequency is 2 KHz
So, ๐
๐๐<< ๐น๐ช <
๐
๐๐
๐
๐ ๐ด๐ฏ๐<< ๐น๐ช <
๐
๐ ๐ฒ๐ฏ๐
Or, ๐๐๐ << ๐น๐บ < ๐. ๐๐๐
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.
๐๐(๐๐๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐) = ๐๐ + ๐๐
Where ๐๐------- signal frequency
๐๐๐๐ ๐๐๐ -------- Image frequency
๐๐๐ = ๐๐ + ๐๐ฐ๐ญ = ๐๐ + ๐๐๐
๐๐๐ = ๐๐๐๐ + ๐(๐๐๐)
๐๐๐ = ๐๐๐๐ ๐ฒ๐ฏ๐
so, answer is 2110 KHz
GATE Communication SystemsObjective Type Questions With
Answers (Electronics Engineering)
Publisher : Faculty Notes Author : Panel Of Experts
Type the URL : http://www.kopykitab.com/product/9985
Get this eBook
84%OFF
Recommended