Analog Communication Test Questions - Set 2

ANSWER: 79.36 W

Explanation:
Modulation Index = 0.8
P_c = 124W
Power in sidebands may be calculated as = m² P_c/4
= (0.8)² * 124/4
= 79.36 W

ANSWER: 0.8544

Explanation:
Here, m₁ = 0.8
m₂ = 0.3

total modulation index m_t = √( m₁² + m₂² )
= √( 0.8² + 0.3² )
= √ 0.73
= 0.8544 = 85.44%

ANSWER: 2000.35, 1999.65 and 2000.6, 1999.4

Explanation:
The frequencies obtained in the spectrum after the amplitude modulation are
f_c + f_m and f_c + f_m

therefore,

the available frequencies after modulation by 0.350 KHz are
2000KHz + 0.350 KHz = 2000.35 and 2000KHz - 0.350 KHz = 1999.65

the available frequencies after modulation by 0.6 KHz are
2000KHz + 0 .6 KHz = 2000.6 and 2000KHz - 0.6 KHz = 1999.4

ANSWER: 1.6 MHz and 8 KHz, 0.2, 16 KHz

Explanation:
The amplitude modulated wave equation is
v = ( 10 + 2 Sin( 2Π * 8 * 10³ t) ) * Sin (2Π * 1.6 * 10⁶ t) volts

Instantaneous value of AM signal is represented by the equation
v = {V_c + V_m Sin ( ω_m t )} * Sin (ω_c t )
comparing it with the given equation,

V_c = 10 V
V_m = 2V
ω_c (= 2Π f_c) = 2Π * 1.6 * 10⁶
ω_m (= 2Π f_m) = 2Π * 8 * 10³

(i) The carrier frequency f_c is = 1.6 * 10⁶ = 1.6 MHz
The modulating frequency f_m is = 8* 10³ = 8 kHz

(ii) The modulation index m = V_m/V_c = 2/10 = 0.2

(iii) The bandwidth BW = 2 f_m = 16 kHz

ANSWER: 43.59 W, 2.205 W

Explanation:
Given that P_t = 48 W

Modulation index m = 0.45

The total power in an AM is given by
P_t = P_c ( 1 + m²/2)
= P_c ( 1 +0.45₂/2)
48 = P_c * 1.10125

Therefore, P_c = 48/ 1.10125
= 43.59 W
The total power in two sidebands is 4843.59 = 4.41 W
So the power in each sideband is 4.41/2 = 2.205 W

ANSWER: 90%, 95%

Explanation:
i) The total power in an AM is given by
P_t = P_c ( 1 + m²/2)

Given: m = 0.45
Therefore P_t = P_c ( 1 + 0.45²/2 )
P_t= P_c *1.10125
P_c/ P_t = 1/1.10125
= 0.908
= 90%

This shows that the carrier occupies 90% of total power. So 90% of total power may be saved if carrier is suppressed in the AM signal.

(ii) If one of the sidebands is also suppressed, half of the remaining power will be saved
i.e., 10/2 = 5 %. So a total of 95% (90% + 5% ) will be saved when carrier and a side band are suppressed.

ANSWER: 825 Hz

Explanation:
Upper frequency = 850Hz

Bandwidth = 50Hz

Therefore lower Frequency = 850 - 50= 800 Hz

Carrier Frequency = (850-800)/2
= 825 Hz

ANSWER: 1

Explanation:
A figure of merit used to describe the performance of a system. The figure of merit γ is the ratio of output signal to noise ratio to input signal to noise ratio of a receiver system. Figure of merit for SSB modulation is always 1.

ANSWER: Class A

Explanation:
When the modulation takes place prior to the output element of the final stage of the amplifier, it is low level modulation. Class A amplifiers are used for this purpose.

ANSWER: 28.5%

Explanation:
I_t = I_c √(1+ m²/2)
10.4= 10 √(1+ m²/2)
√ (1+ m²/2) = 1.04

Therefore m = 0.285
= 28.5%

ANSWER: 50%

Explanation:
P_t = P_c ( 1 + m²/2)

P_t= P_c ( 1 + 0²/2) = P_c ..(1)

New total power P_t1= P_c ( 1 + 1²/2)
= P_c *3/2 ..(2)
(2) / (1),

We get , P_t1/ P_t= 3/2= 1.5
P_t1= 1.5 P_t
i.e. there is increase in total power by 50%

ANSWER: Multiplication of two signals

Explanation:
RF mixer translates the frequencies of the two incoming signals from one band to another by multiplying them. When the two incoming signals are mixed, the new frequencies obtained are the sum and difference of incoming signal frequencies .i.e., if the two input signal frequencies are f₁ and f₂, then the output of the mixer will have signals at frequencies (f₁ + f₂) and (f₁ - f₂).

ANSWER: Poor selectivity

Explanation:
Frequency selectivity means to separate out closely lying components of frequency. If the receiver has poor selectivity, it has poor capacity of blocking adjacent channel interference.

ANSWER: All of the above

Explanation:
The main advantage of high frequency signals is that the signal may be transmitted over very long distances and thus dissipates very less power. The antenna height required for transmission is reduced at higher frequencies. Thus, the audio signals must be sent with the high frequency carrier signals for communication purpose.

ANSWER: Both a and b

Explanation:
In AM, the signal appears in the form of components at frequencies slightly higher and lower than that of the carrier frequency. These sidebands carry the information.

Both the sidebands are similar to each other, so one of the sidebands may be rejected. Vestigial Side Band transmission is similar to single-sideband (SSB) transmission, in which the second sideband is not completely removed, but is filtered to remove the range of frequencies that are not required.

ANSWER: All of the above

Explanation:
Modulation is required as with modulation, the signal is transmitted at higher frequency that reduces the probability of addition of noise while transmission and reduces the height of antenna at the transmitter. With modulation, there is reduction in the complexity of circuitry and more signals may be transmitted using multiplexing technique.

ANSWER: f_m

Explanation:
In an amplitude modulated wave, total bandwidth required is from f_c + f_m to f_c - f_m

i.e. BW = 2f_m where f_c is carrier frequency.

In SSB-SC transmission, as the carrier and one of the sidebands are suppressed, the bandwidth remains as f_m.

ANSWER: m > 1

Explanation:
In AM, the modulation index m is defined as the ratio of the amplitude modulation signal to the amplitude of carrier signal. When m > 1, it is called over-modulation and it generates modulation with distortions in the envelope of the modulated signal, from which it is difficult to recover the information.

For Under modulation m < 1

For Critical modulation m = 1

For Over modulation m > 1

ANSWER: Both a and b

Explanation:
Demodulation is the process of recovering the original information from a modulated carrier wave. Systems are designed to be used as demodulators that detect the information signal from the carrier. The envelope detector and product detector are few of the AM detectors.

ANSWER: 6.25 W

Explanation:
The side band power is given by
P_c m²/2
= 50 * (0.5) ²/2
= 6.25W