Angle Modulation MCQ || Angle Modulation Questions and Answers

Answer.2. Halved

Explanation

β_FM = Δf/f_m = K_fA_m/f_m

β_PM = Δφ = K_PA_m

Given that,

β_FM = β_PM

if the modulation frequency is doubled then

β’_FM = β_FM /2 and β’_PM = β_PM

β_FM/β_PM = 1/2

Answer.1. Integrating

Explanation

Frequency modulated signal is regarded as the phase-modulated signal in which the modulating wave is integrated before modulation. This means that an FM signal can be generated by first integrating the message signal and then using the result as an input to a phase modulator.

Answer.3. Depends on both A_m and f_m

Explanation

If m(t) is the message single & c(t) = A_c cos(ω_ct) is the Carrier signal, then the General expression for a Phase modulated signal is:

S_PM(t) = A_c cos[ω_ct + k_pm(t)]

Where k_p = Phase Sensitivity (rad/Volts)

Analysis:

S_PM(t) = A_ccos[ω_ct + k_pm(t)] = A_c cos(θ_i)

Instantaneous Phase θ_i(t) = ω_c(t) + k_pm(t)

Instantaneous frequency ω_i(t) = $\frac{{d{\theta _i}}}{{dt}}$

ω_i(t) = ω_c + k_p $\frac{{d[\left( {m\left( t \right)} \right]}}{{dt}}$

• Phase deviation Δϕ = k_p m(t)
• Frequency deviation Δω = ${k_p}\frac{{d\left( {m\left( t \right)} \right)}}{{dt}}$
• Deviation ratio β = $\frac{{{{\left| {{\rm{\Delta }}\omega } \right|}_{max}}}}{{{\omega _m}}} = \;\frac{{\left( {{k_p}{\omega _m}{A_m}} \right)}}{{{\omega _m}}}$ {If m(t) = A_m cos (ω_mt)}

According to Carson formula, the Bandwidth of a phase-modulated signal is:

BW = 2(β + 1) ω_m = 2(k_pA_m + 1)ω_m

Hence the Bandwidth depends on both A_m & ω_m

Answer.3. 15 kHz

Explanation

The Bandwidth of an FM signal is approximated by Carlson’s rule and is given by

B.W. = 2(Δf + f_m).

Where Δf = maximum frequency deviation from the carrier frequency and f_m = message signal frequency.

Calculation:

Given, S_FM(t) = 10 cos (2π × 10⁸ t + 0.5 sin (10⁴πt))

The instantaneous frequency is given by:

${f_i} = \frac{1}{{2\pi }}\left( {\frac{{d{\phi _i}}}{{dt}}} \right) = \frac{1}{{2\pi }}\frac{d}{{dt}}\left[ {2\pi \times {{10}^8}t + 0.5\sin \left( {{{10}^4}\pi t} \right)} \right]$

$= \frac{1}{{2\pi }}\left[ {2\pi \times {{10}^8} + 0.5 \times {{10}^4}\pi .\cos {{10}^4}\pi t} \right]$

⇒ Maximum instantaneous frequency ⇒ ${f_{i\left( {{\rm{max}}} \right)}} = \frac{1}{{2\pi }}\left[ {2\pi \times {{10}^8} + 0.5 \times {{10}^4}\pi } \right]$

Maximum frequency deviation

${\rm{\Delta }}f = \frac{1}{{2\pi }}\left[ {0.5 \times {{10}^4}\pi } \right]$

$= \frac{1}{{2\pi }} \times 5000\pi$

= 2.5 kHz

Given, ${f_m} = \frac{{{{10}^4}\pi }}{{2\pi }} = 5\;kHz$

So, Approximate bandwidth according to Carson’s rule

⇒ 2(Δf + f_m) = 2(2.5 + 5) = 2 (7.5 K) = 15 K

Answer.2. Amplitude of message signal

Explanation

Frequency Modulation (FM) is the encoding of information in a carrier wave by changing the instantaneous frequency of the wave. FM technology is widely used in the fields of computing, telecommunications, and signal processing. The frequency modulation index defines the relationship between the amplitude of the message signal and the bandwidth of the transmitted signal.

If the modulating signal is a low pass signal, the maximum bandwidth of the modulating signal is equal to the highest frequency component present in the modulating signal.

Answer.4. To maintain the same volume of the output when stations of different strengths are received

Explanation

• Automatic gain control (AGC) works in FM radio transmitter/receiver that maintains Automatic controlling of weak and strong signals which are received by the radio receiver.
• The automatic frequency control voltage of the FM transmitter VCO is DC voltage.
• AGC maintains a constant level of the output signal based on the received signal nature, i.e. it maintains the same volume of the output when stations of different strengths are received.
• AGC adjusts the gain of RF and IF amplifiers according to need.
• AGC can handle problems like overloading and fading in the receiver.

Answer.2. 83%

Explanation

Modulation Index (β) in FM is given by:

β = Frequency Deviation/Message frequency

β = Δf/fm

The maximum frequency band permitted by FCC is 75 kHz.

Calculation:

The carrier swing given is 125 kHz, i.e.

fmax – fmin = 125 kHz

2 × Δf = 125 kHz

Δf = 62.5 kHz

The modulation index will now be:

β = 62.5/75

β = 0.833 or 83%

Answer.1. Carson’s

Explanation

FM bandwidth is approximated using Carson’s rule. Carson’s rule states that the bandwidth required to transmit an angle modulated wave is twice the sum of the peak frequency deviation and highest modulating signal frequency.

Answer.2. 10

Explanation

The instantaneous phase for the given frequency modulated wave is:

θ_i = 2π × 105t + 5 sin (2π × 1500t) + 7.5 sin (2π × 1000t)

The instantaneous frequency will be:

ωi = dθi/dt

ωi = 2π × 10⁵ + 15000πCos(2π × 1500t) + 15000πCos(2π × 1000t)

Maximum frequency deviation will be:

Δω = 15000π + 15000π

Δω = 30000π

Δf = 15000

Now, the baseband message signal bandwidth is f_m = 1500 Hz.

∴ The modulation index will be:

β = Δf/fm = 15000/1500 = 10

Answer.3. 10 kHz

Explanation

According to Carson’s rule, if β>1

BW is given by

BW = 2(β + 1)f_m

Where,

f_m = highest modulating frequency

Δ­­_f = β.f_m

Where,

Δ_f  = peak deviation and

β = Modulation index of FM

if β<1 then it is called NBFM and for NBFM bandwidth is 2fm.

Calculation:

Given:

Δ_f  = 3 kHz, f_m = 5 kHz

therefore β = 0.6 which is less than one. hence the given signal is NBFM

so  B.W = 2f_m which is equal to 10K Hz.