**1) FM is advantageous over AM as**

**a.** The amplitude of FM is constant. So transmitter power remains unchanged in FM but it changes in AM

**b.** The depth of modulation in FM can be changed to any value by changing the frequency deviation. So the signal is not distorted

**c.** There is less possibility of adjacent channel interference due to presence of guard bands

**d.** All of the above

**ANSWER: All of the above**

**2) VCO is used to generate**

**a.** Direct FM

**b.** Indirect FM

**c.** SSB-SC

**d.** DSB-SC

**ANSWER: (a) Direct FM**

**3) Change in instantaneous phase of the carrier with change in amplitude of the modulating signal generates**

**a.** Direct FM

**b.** Indirect FM

**c.** SSB-SC

**d.** DSB-SC

**ANSWER: (b) Indirect FM**

**4) Phase-locked loop can be used as**

**a.** FM demodulator

**b.** AM demodulator

**c.** FM receiver

**d.** AM receiver

**ANSWER: (a) FM demodulator**

**5) The increase or decrease in the frequency around the carrier frequency is termed as**

**a.** Figure factor

**b.** Frequency deviation

**c.** Modulation index

**d.** Frequency pectrum

**ANSWER: (b) Frequency deviation**

**6) Carson’s rule is used to calculate**

**a.** Bandwidth of FM signal

**b.** Signal to noise ratio

**c.** Modulation index

**d.** Noise figure

**ANSWER: (a) Bandwidth of FM signal**

**7) The ratio of maximum peak frequency deviation and the maximum modulating signal frequency is termed as**

**a.** Frequency deviation

**b.** Deviation ratio

**c.** Signal to noise ratio

**d.** Frequency spectrum

**ANSWER: (b) Deviation ratio**

**8) The equation v(t) = A cos [ω _{c}t + k_{p} Φ(t)] represents the signal as**

**a.** Phase modulated signal

**b.** SSBSC signal

**c.** DSB SC signal

**d.** None of the above

**ANSWER: (a) Phase modulated signal**

**9) Calculate the maximum frequency deviation for the FM signal
v(t) = 10 cos (6000t+ 5sin2200t)**

**a.** 2200 Hz

**b.** 6000 Hz

**c.** 1750 Hz

**d.** 11000 Hz

**ANSWER: (c) 1750 Hz**

**Explanation: **

A standard FM signal is represented by

v(t) = A_{c} cos(2πf_{c}t + k_{f}sin2πf_{m}t)

A_{c} = carrier amplitude

f_{c} = carrier frequency

k_{f} = modulation index

f_{m }= modulating frequency = 2200/2π = 350 Hz

k_{f} = frequency deviation/modulating frequency

5 = freq deviation/ 350

Therefore, deviation = 5 * 350

= 1750Hz

**10) Calculate the dissipation in power across 20Ω resistor for the FM signal
v(t)= 20 cos(6600t+ 10sin2100t)**

**a.** 5W

**b.** 20W

**c.** 10W

**d.** 400W

**ANSWER: (a) 5W**

**Explanation: **

A standard FM signal is represented by

v(t) = A_{c }cos(2πf_{c}t + k_{f}sin2πf_{m}t)

A_{c} = carrier amplitude

f_{c} = carrier frequency

k_{f} = modulation index

f_{m }= modulating frequency

k_{f }= frequency deviation/modulating frequency

the power dissipated across 20Ω resistor is given by

V_{rms}^{2}/R

=(20/√2)^{2}/R

= 5W

**11) What is the value of carrier frequency in the following equation for the FM signal?
v(t)= 5 cos(6600t+ 12sin2500t)**

**a.** 1150 Hz

**b.** 6600 Hz

**c.** 2500 Hz

**d.** 1050 Hz

**ANSWER: (d) 1050 Hz**

**Explanation: **

A standard FM signal is represented by

v(t) = A_{c} cos(2πf_{c}t + k_{f}sin2πf_{m}t)

A_{c} = carrier amplitude

f_{c} = carrier frequency

k_{f} = modulation index

f_{m }= modulating frequency

k_{f} = frequency deviation/modulating frequency

therefore, f_{c }= 6600/2π

= 1050Hz

**12) Calculate the modulation index in an FM signal when f _{m} (modulating frequency) is 250Hz and Δf (frequency deviation) is 5KHz.**

**a.** 20

**b.** 35

**c.** 50

**d.** 75

**ANSWER: (a) 20**

**Explanation: **

Modulation index is the measure of how much the modulation parameter changes from its un modulated value. The modulation index of FM is given by

μ = frequency deviation/ modulating frequency

= Δf/ f_{m}

Where Δf is the peak frequency deviation i.e. the deviation in the instantaneous value of the frequency with modulating signal.

f_{m} is the value of modulating frequency

μ = 5000/250

= 20

**13) After passing the FM signal through mixer, what is the change in the frequency deviation Δ when the modulating frequency is doubled?**

**a.** Becomes 2 Δ

**b.** Becomes Δ /2

**c.** Becomes Δ^{2}

**d.** Remains unchanged

**ANSWER: (d) Remains unchanged**

**14) In frequency modulation,**

**a.** Armstrong method is used for generation

**b.** Multiple side bands are generated

**c.** The FM signal has infinite bandwidth

**d.** All of the above

**ANSWER: (d) All of the above**

**15) Maximum frequency deviation and the maximum bandwidth allowed for commercial FM broadcast is**

**a.** 80KHz, 160Khz

**b.** 75KHz, 200Khz

**c.** 60KHz, 170Khz

**d.** 75KHz, 250Khz

**ANSWER: 7 (b) 5KHz, 200Khz**

**16) Guard bands are provided in FM signal to**

**a.** Prevent interference from adjacent channels

**b.** To increase the noise

**c.** To increase bandwidth

**d.** None of the above

**ANSWER: (a) Prevent interference from adjacent channels**

**17) For a FM signal v(t) = 15 cos ( 10 * 10 ^{8}t + 10 sin 1220t), calculate
1. Carrier frequency
2. Modulating frequency**

**a.** 159.1MHz, 194.1Hz

**b.** 185.5MHz, 200.15Hz

**c.** 350.1MHz, 200.1Hz

**d.** 159.1Hz, 194.1Hz

**ANSWER: 1 (a) 59.1MHz, 194.1Hz**

**18) For a FM signal v(t) = 25 cos (15 * 10 ^{8}t + 10 sin 1550t), calculate:**

**Modulation index****Maximum frequency deviation**

**a.** 10, 3000.1Hz

**b.** 20, 1550.9Hz

**c.** 10, 2465.9Hz

**d.** 10, 2000.0Hz

**ANSWER: (c) 10, 2465.9Hz**

**Explanation: **

Standard expression for FM signal is given by

v(t) = A cos ( ω_{c}t + mf sin ω_{m}t)

Comparing with the given equation,

Modulation index m_{f} = 10

Maximum frequency deviation is given by

m_{f} = Δf/f_{m}

Δf = m_{f} * f_{m}

Here f_{m} = 1550/2Π = 246.59 Hz

Δf = 10 * 246.59

= 2465.9Hz

**19) For a FM signal v(t) = 20 cos ( 10 * 10 ^{8}t + 30 sin 3000t), calculate the power dissipated by the FM wave in a 20Ω resistor.**

**a.** 100 Watts

**b.** 10 Watts

**c.** 200 Watts

**d.** 20 Watts

**ANSWER: (b) 10 Watts**

**Explanation: **

Standard expression for FM signal is given by

v(t) = A cos ( Ω_{c}t + m_{f} sin Ω_{m}t)

Comparing with the given equation,

A = 20

The dissipated power is given by P = V^{2}_{rms}/R

= (20/√2)^{2}/ 20

= 10Watts

**20) A 100MHz carrier is frequency modulated by 5 KHz wave. For a frequency deviation of 100 KHz, calculate the carrier swing of the FM signal.**

**a.** 2000 KHz

**b.** 100 KHz

**c.** 105 KHz

**d.** 200 KHz

**ANSWER: (d) 200 KHz**

**Explanation: **

Carrier frequency f_{c} = 100MHz

Modulating frequency f_{m} = 5 KHz

Frequency deviation Δf = 100 KHz

Carrier swing of the FM signal = 2 * Δf

= 2 * 100

= 200 KHz

**21) A 100MHz carrier is frequency modulated by 10 KHz wave. For a frequency deviation of 50 KHz, calculate the modulation index of the FM signal.**

**a.** 100

**b.** 50

**c.** 70

**d.** 90

**ANSWER: (b) 50**

**Explanation: **

Carrier frequency f_{c} = 100MHz

Modulating frequency f_{m} = 10 KHz

Frequency deviation Δf = 500 KHz

Modulation index of FM signal is given by

m_{f} = Δf/f_{m}

= 500 * 10^{3}/ 10 * 10^{3}

= 50

**22) Narrow band FM has the characteristics:**

**a.** The frequency sensitivity k_{f} is small

**b.** Bandwidth is narrow

**c.** Both a and b

**d.** None of the above

**ANSWER: (c) Both a and b**

**23) Wide band FM has the characteristics:**

**a.** The frequency sensitivity k_{f} is large

**b.** Bandwidth is wide

**c.** Both a and b

**d.** None of the above

**ANSWER: (c) Both a and b**

**24) Determine the Bandwidth of a FM wave when the maximum deviation allowed is 75KHz and the modulating signal has a frequency of 10KHz.**

**a.** 170 KHz

**b.** 200 KHz

**c.** 100 KHz

**d.** 1000 KHz

**ANSWER: (a) 170 KHz**

**Explanation: **

Modulating frequency f_{m} = 10 KHz

Frequency deviation Δf = 75 KHz

According to Carson s rule, BW = 2(Δf + f_{m})

= 2 (75 + 10)

= 170 KHz

**25) FM signal is better than AM signal because**

**a.** Less immune to noise

**b.** Less adjacent channel interference

**c.** Amplitude limiters are used to avoid amplitude variations

**d.** All of the above

**ANSWER: (d) All of the above**

**26) FM is disadvantageous over AM signal because**

**a.** much wider channel bandwidth is required

**b.** FM systems are more complex and costlier

**c.** Adjacent channel interference is more

**d.** Both a and b

**ANSWER: (d) Both a and b**

**27) For a three stage cascade amplifier, calculate the overall noise figure when each stage has a gain of 12 DB and noise figure of 8dB.**

**a.** 12

**b.** 24

**c.** 13.55

**d.** 8

**ANSWER: (c) 13.55**

**Explanation: **

As the signal passes through various stages of an amplifier, the output has the original signal and some noise that gets amplified at different stages of amplifiers. So the final noise figure of the cascaded amplifier is obtained by

F_{N} = F_{1} + (F_{2} – 1)/ G_{1} + (F_{3} – 1)/ G_{1}G_{2}+……+(F_{N}– 1)/ G_{1}G_{2}G_{3} G_{N}

F_{1}, F_{2}, F_{3} .. F_{N}, G_{1},G_{2}, G_{3}…. G_{N} are the noise figures and the gains respectively of the amplifiers at different stages.

F_{1} = 12, F_{2} = 12, F_{3} = 12

G_{1} = 8, G_{2} = 8, G_{3} = 8

F_{N }= 12 + (12- 1)/ 8+ (12- 1)/ 8 * 8

= 12 + 11/8 + 11/64

= 13.55

**28) The Hilbert transform of the signal sinω1t + sinω _{2}t is**

**a.** sinω_{1}t + sinω_{2}t

**b.** cosω_{1}t + cosω_{2}t

**c.** sinω_{2}t + cosω_{2}t

**d.** sinω_{1}t + sinω_{1}t

**ANSWER: (b) cosω _{1}t + cosω_{2}t**

**Explanation: **

In Hilbert transform, the signal gets shifted by 900.

So the signal sinω_{1}t+ sinω_{2}t gets shifted by 900

sinω_{1}(t+900)+ sinω_{2}(t+900)

= cosω_{1}t+ cosω_{2}t

**29) The noise due to random behaviour of charge carriers is**

**a.** Shot noise

**b.** Partition noise

**c.** Industrial noise

**d.** Flicker noise

**ANSWER: (a) Shot noise**

**30) Transit time noise is**

**a.** Low frequency noise

**b.** High frequency noise

**c.** Due to random behavior of carrier charges

**d.** Due to increase in reverse current in the device

**ANSWER: (b) High frequency noise**

**31) Figure of merit γ is**

**a.** Ratio of output signal to noise ratio to input signal to noise ratio

**b.** Ratio of input signal to noise ratio to output signal to noise ratio

**c.** Ratio of output signal to input signal to a system

**d.** Ratio of input signal to output signal to a system

**ANSWER: (a) Ratio of output signal to noise ratio to input signal to noise ratio**

**32) Signum function sgn(f), for f>0, f=0 and f<0, has the values:**

**a.** -1 to +1

**b.** +1, 0, -1 respectively

**c.** -∞ to + ∞

**d.** 0 always

**ANSWER: (b) +1, 0, -1 respectively**

**Explanation: **

The sgn(f) is a signum function that is defined in the frequency domain as

sgn(f) = 1, f> 0

= 0, f = 0

= -1, f< 0

Mathematically, the sign function or signum function is an odd mathematical function which extracts the sign of a real number and is often represented as sgn

**33) In Hilbert transform of a signal, the phase angles of all components of a given signal are shifted by**

**a.** +/- π

**b.** +/- π/4

**c.** +/- π/2

**d.** Any angle from 00 to 3600

**ANSWER: (c) +/- π/2**

**34) The noise voltage (V _{n}) and the signal bandwidth (B) are related as**

**a.** V_{n} is directly proportional to bandwidth

**b.** V_{n} is directly proportional to √bandwidth

**c.** V_{n} is inversely proportional to absolute temperature

**d.** V_{n} is inversely proportional to bandwidth

**ANSWER: (b) V _{n} is directly proportional to √bandwidth**

**35) Noise factor for a system is defined as the ratio of**

**a.** Input noise power (P_{ni}) to output noise power (P_{no})

**b.** Output noise power (P_{no}) to input noise power (P_{ni})

**c.** Output noise power (P_{no}) to input signal power (P_{si})

**d.** Output signal power (P_{so}) to input noise power (P_{ni})

**ANSWER: (b) Output noise power (P _{no}) to input noise power (P_{ni})**

**36) Noise Factor(F) and Noise Figure(NF) are related as**

**a.** NF = 10 log_{10}(F)

**b.** F = 10 log_{10}(NF)

**c.** NF = 10 (F)

**d.** F = 10 (NF)

**ANSWER: (a) NF = 10 log _{10}(F)**

**37) The Noise Factor for cascaded amplifiers (F _{N}) is given by (F_{1}, F_{2}, F_{3} .. F_{N}, G_{1}, G_{2}, G_{3}….G_{N}) are the noise factors and the gains of the amplifiers at different stages:**

**a.** F_{N} = F_{1} + F_{2}/ G_{1} + F_{3}/ G_{1}G_{2}+ ..+ F_{N}/ G_{1}G_{2}G_{3}G_{N}

**b.** F_{N} = F_{1} + (F_{2} – 1)/ G_{1} + (F_{3} – 1)/ (G1+G2)+ ..+(F_{N} – 1)/ (G1+G2+G3+…+G_{N})

**c.** F_{N} = F_{1} + F_{2}/ G_{1} + F_{3}/ (G1+G2) +…+ F_{N}/ (G1+G2+G3+…+G_{N})

**d.** F_{N} = F_{1} + (F_{2} – 1)/ G_{1} + (F_{3} – 1)/ G_{1}G_{2}+…+(F_{N} – 1)/ G_{1}G_{2}G_{3}G_{N}

**ANSWER: (d) F _{N} = F_{1} + (F_{2} – 1)/ G_{1} + (F_{3} – 1)/ G_{1}G_{2}+…+(F_{N} – 1)/ G_{1}G_{2}G_{3}G_{N}**

**38) For a two stage amplifier, first amplifier has Voltage gain = 20, Input Resistance R _{in1}=700Ω, equivalent Resistance R_{eq1}=1800Ω and Output Resistor R_{o1} = 30KΩ. The corresponding values of second amplifier are : 25, 80 KΩ, 12 KΩ, 1.2 MΩ respectively. What is the value of equivalent input noise resistance of the given two stage amplifier?**

**a.** 2609.1Ω

**b.** 2607.1Ω

**c.** 107.1Ω

**d.** 2107.1Ω

**ANSWER: (b) 2607.1Ω**

**Explanation: **

R_{1} = R_{in1} + R_{eq1} = 700 + 1800 = 2500Ω

R_{2} = (R_{o1} R_{in2})/ (R_{o1} + R_{in2}) + R_{eq2} = 30 * 80/(30 + 80) + 12 = 40.92KΩ

R_{3} = R_{o2} = 1.2MΩ

Equivalent input noise resistance of a two stage amplifier is given by

R_{eq} = R_{1} + R_{2}/ A^{2}_{1} + R_{3}/ (A^{2}_{1} A^{2}_{2})

= 2500 + 40.92 * 10^{3}/(20)^{2} + 1.2 * 10^{6}/(20)^{2}(25)^{2}

= 2607.1Ω

**39) The noise temperature at a resistor depends upon**

**a.** Resistance value

**b.** Noise power

**c.** Both a and b

**d.** None of the above

**ANSWER: (b) Noise power**

**40) Noise voltage V _{n} and absolute temperature T are related as**

**a.** V_{n} = 1/ √(4RKTB)

**b.** V_{n} = √(4RK)/ (TB)

**c.** V_{n} = √(4RKTB)

**d.** V_{n} = √(4KTB)/R

**ANSWER: (c) V _{n} = √(4RKTB)**