Conventional Amplitude Modulation MCQ Quiz in বাংলা - Objective Question with Answer for Conventional Amplitude Modulation - বিনামূল্যে ডাউনলোড করুন [PDF]

Advantage of AM:

• The advantages of AM radio are that it is relatively easy to detect with simple equipment, even if the signal is not very strong. 
• It has a narrower bandwidth than FM, and wider coverage compared with FM radio.
• Its use avoids receiver complexity, thus reducing the cost at the receiver end. This makes them suitable for broadcasting purposes.

Disadvantages of AM:

• In AM most of the transmitted power is in a carrier that contains no information.
• It is affected by noise.
• In AM if there are two more signals received at the same frequency, then both will be demodulated which can lead to interference.

Important differences between AM and FM are elaborated in the table below:

Concept:

Amplitude Modulation:

The amplitude of the carrier is varied according to the amplitude of the message signal.

The modulation index is a measure of the extent of modulation done on a carrier signal. In Amplitude modulation, it is defined as the ratio of the amplitude of the modulating signal to that of the carrier signal.

 \(\mu = \frac{{{{\rm{A}}_{\rm{m}}}}}{{{{\rm{A}}_{\rm{c}}}}}\)

Calculation:

Given,

Message Signal frequency, f_m = 1 kHz

Carrier frequency, f_c = 240 kHz

Peak Voltage of the message signal, A_m = 3 V

Peak Voltage of the carrier wave, A_c = 10 V

The modulation index is given by,

\(\mu = \frac{{{{\rm{A}}_{\rm{m}}}}}{{{{\rm{A}}_{\rm{c}}}}} = \frac{{3}}{{10}} = 0.3\)

∴ The correct option is 2.

Concept:

The total transmitted power for an AM system is given by:

\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

Pc = Carrier Power

μ = Modulation Index

Calculation:

Given μ = 80% = 0.8

Pc = 100 W

\(P_t= {100}\left( {1 + \frac{{{0.8^2}}}{2}} \right)\)

\(P_t= 100\times1.32 ~W\)

P_t = 132 W

Concept:

For envelope detector, the time constant must satisfy the following relation:

\(RC \le \frac{1}{{{\omega _m}}}\left[ {\frac{{\sqrt {1 - {μ^2}} }}{μ}\;} \right]\)  ---(1)

For the good performance of the envelope detector:

\(\frac{1}{\omega_c}<

Application:

Given:

f_m = 10 kHz

modulation index (μ) = 0.6

From (1), we get

\(RC \le \frac{1}{{{2\pi \times 10}}}\left[ {\frac{{\sqrt {1 - {0.6^2}} }}{0.6}\;} \right]\)

= 0.02 msec

Derivation of RC time constant formula for the envelope detector: )

For the output of the envelope detector to closely follow the modulating signal, the slope of vo is always greater than that of the envelope of AM signal input.

The output vo is the voltage across the capacitor which is exponentially decaying

\(\begin{array}{l} {v_0}\left( t \right) = {v_i}{e^{ - \frac{t}{{RC}}}}\\ = {V_i}\left[ {1 - \frac{t}{{RC}} + \frac{1}{{2!{{\left( {\frac{t}{{RC}}} \right)}^2}}} + \ldots } \right] \end{array}\)

Neglecting higher power terms as RC>>t

\(= {V_i}\left[ {1 - \frac{t}{{RC}}} \right]\)

For successful detection, the rate of discharge of the capacitor i.e. magnitude of the slope of v0(t) must be more than that of the envelope

The envelope of the AM signal is \({V_i}\left( t \right) = {A_c}\left[ {1 + {A_m}\cos {\omega _m}t} \right]\)

The slope of the message signal

\( \frac{{d{V_i}\left( t \right)}}{{dt}} = - {A_c}{A_m}\sin {\omega _m}t\)

The slope of output voltage

\(\frac{{d{v_0}}}{{dt}} = \frac{{{-V_i}}}{{Rc}}\)

For DETECTION slope of capacitor voltage should be always more than the envelope of AM signal

\(\begin{array}{l} \frac{{{V_i}}}{{RC}} \ge - {A_C}{A_m}\sin {\omega _m}t\\ \frac{{{A_c}}}{{RC}}\left[ {1 + {A_m}\cos {\omega _m}t} \right] \le {A_c}{A_m}\sin {\omega _m}t\\ RC \le \frac{{1 + {A_m}\cos {\omega _m}t}}{{{A_m}{\omega _m}\sin {\omega _m}t}} \end{array}\)

 RC must be less than the minimum of R.H.S.

\(RC \le \frac{1}{{{\omega _m}}}\left[ {\frac{{\sqrt {1 - {μ^2}} }}{μ}\;} \right]\)

Frequency modulation is a technique or a process of encoding information on a particular signal (analogue or digital) by varying the carrier wave frequency in accordance with the frequency of the modulating signal.

The standard representation of FM signal is,

 FM = E _c cos(ω_C t + m_f sin(ω_m t)

 m_f is the modulation of  FM signal

 Carrier frequency, f_c = ω_C / 2π

 Maximum frequency deviation can be calculated as, m_f × fm

Given equation,

v(t) = 20 cos(25 × 108 t + 10sin 1450t)

On comparing with the above equations, we get

modulation index m_f = 10.

Message frequency (\(f_m\))= 1450/2π = 230.77466 Hz

Maximum frequency deviation =  mf × fm

 = 10 × 230.77466 Hz

 = 2307.746 Hz

Conventional AM signals are easily demodulated by an envelope detector.

It consists of a diode and an RC circuit, which is a simple low-pass filter as shown:

• During the positive half-cycle of the input signal, the diode conducts and the capacitor charges up to the peak value of the input signal.
• When the input falls below the voltage on the capacitor, the diode becomes reverse-biased and the input disconnects form the output.
• During this period, the capacitor discharges slowly through resistor R.
• On the next cycle of the carrier, the diode again conducts when the input signal exceeds the voltage across the capacitor.

The time constant given by RC must be selected to follow the variations in the envelope of the modulated signal.

• If RC is too large, then the discharge of the capacitor is too slow and the output will not be able to follow the envelope.
• If RC is too small, then the output of the filter falls very rapidly after each peak and will not follow the envelope closely.

This is explained by the given figure:

So, for the good performance of the envelope detector:

\(\frac{1}{\omega_c}<

Concept:

For a single-tone sinusoidal signal, the expression for amplitude modulated wave is given by:

\({x_{AM}}\left( t \right) = {A_c}\left( {1 + {m_a}\cos {\omega _m}t} \right)\cos {\omega _c}t\)

\({A_c}\cos {\omega _c}t + \frac{{{A_c}{m_a}}}{2}\cos \left( {{\omega _c} + {\omega _m}} \right)t + \frac{{{A_c}{m_a}}}{2}\cos \left( {{\omega _c} - {\omega _m}} \right)t\)

Calculation:

Comparing the given expression with the standard expression, the given AM signal can be written as:

\({x_{AM}}\left( t \right) = 4\left( {1 + \frac{1}{2}\cos {\omega _m}t} \right)\cos {\omega _c}t\)

\(So,\;{m_a} = \frac{1}{2}\;and\;{A_c} = 4\)

For a sinusoidal signal ratio of sideband power to carrier power is given:

=\(\frac{P_{SB}}{P_C}\)

=  \(\frac{P_Cm_a^2}{2}/P_c\)

=\(\frac{1}{8}\)

= 0.125

Special Note: It is an Official GATE Question. The question was challenged to GATE regarding the answer key. If we consider the sideband power as the message signal power, we'll get 0.125 as the ratio, but if we solve it by comparing the given expression with the standard expression, we'll 0.25 as explained in the solution.

Concept:

The total transmitted power for an AM system is given by:

\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

Pc = Carrier Power

μ = Modulation Index

Analysis:

When μ = 0, the transmitted power will be:

\({P_t} = {P_c}\left( {1 + \frac{{{0^2}}}{2}} \right)=P_c\)

When μ = 1, the transmitted power will be:

\({P_t} = {P_c}\left( {1 + \frac{{{1^2}}}{2}} \right)=\frac{3}{2}P_c\)

The % increase in the modulated signal power is given by:

\(=\rm \frac{{\frac{3}{2}{P_C} - {P_C}}}{{{P_C}}} \times 100\%\)

\(\rm = \frac{{\frac{1}{2}{P_C}}}{{{P_C}}} \times 100 = 50\%\)

Concept:

The equation of the multitone AM modulated wave is given by 

y(t) = A_c(1+ μ₁cosω_m1t +  μ₂cosωm2 t) cosω_ct 

\({μ _1} = \frac{{{A_m}_1}}{{{A_c}}}\)

\({μ _2} = \frac{{{A_m}_2}}{{{A_c}}}\)

The amplitude of sidebands is given by \(\frac{A_c\mu}{2}\)

Calculations:

Comparing with the standard equation of AM 

The amplitude of carrier A_c = 25

The modulation indices are 0.7 and 0.3

The amplitude of the signal is given by (A_c μ)

\(\therefore\)The amplitude of the signals are (0.7 × 25) V and (0.3 × 25) V

The amplitude of sidebands is given by \(\frac{A_c\mu}{2}\)

∴ The amplitude of sidebands are \(\left( {{{0.7 \times 25} \over 2}} \right)\) V and \(\left( {{{0.3 \times 25} \over 2}} \right)\) V

= 8.75 V and 3.75 V

Concept :

Power transmitted by an Antenna is given as :

P_t = \(\frac{I_o^{2}}{2}R_{rad}\)

R_rad : Radiation Resistance

I_o : Peak Current through antenna

P_t  \(\propto\) (I_o)²

Calculation :

Pt  \(\propto\) (Io)2