- Find the modulation index if the amplitude of message signal is Two thirds of the amplitude of carrier signal ———————.
- 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 ω is carrier frequency both in rad/sec) ( )
a. RC<1/W b. RC>1/W c. None of the above d. RC>1/ω.
- Given AM signal is XAM(t)= 10 (1+0.25 sin 2πfmt) cos 2πfct , then The average side band power given for the above AM signal is ( )
a. 25W b. 12.5W c.1.5625W d.3.125W.
- Given AM signal is XAM(t) = 100 (1+0.85 cos 2πfmt) ,then The total power required for the above AM signal is ( )
a. 25W b. 12.5W c.6.806KW d. None of the above.
- Consider the AM signal Ac cos Wct + 2 cos Wct cos Wmt for the demodulation of the signal using envelope detector the minimum value of Ac should be ( ).
a. 2 b. 1 c. 0.5 d. 0.
- Given AM signal is SAM(t) = 100 (1+0.3 cos 2πfmt +0.4 sin 2πfmt) cos 2πfct i. i.The total power required for the above AM signal is ( )
a. 5.625W b. 5.625KW c.6.806KW d. None of the above.
ii.Modulation index is ( )
a. 0.53 b. 0.5 c. 0.2 d. None of the above.
iii.The carrier power is ( )
a. 5KW b.6KW c. 7KW d.100KW.
iv. Total current flowing through the transmitter if carrier current is 5A ( ).
a. 5mA b.5.303A c. none of these d. 25A.
- If the band width of message signal is 5KHz and the carrier frequency is 200KHz then upper sideband frequencies are( )
a. 205KHz,190KHz b.205KHz,-205KHz
c. 205 KHz,-195 KHz d. None of the above.
- If the highest frequency of message signal is 5KHz and the carrier frequency is 200KHz then lower sideband frequencies are( )
a. 205KHz,190KHz b.205KHz,-205KHz
c. 195 KHz,-195 KHz d. None of the above.
- If the bandwidth of message signal is 500Hz then the bandwidth of Amplitude Modulated signal is ( )
a. 500Hz b. 1000Hz c.2KHz d. None of the above.
- If the message m(t)= 10cos2000πt and carrier signal is C(t) = 25 cos 200000 πt then draw the amplitude spectrum of AM signal————— .
UNIT 2- QUIZ 2
- Angle modulation is a technique in which the ————— of the is varied with respect to instantaneous values of ————————— by keeping as constant. 2M.
- Write the expression for Angle Modulated signal –. 1M.
- An Angle Modulated signal is given as 𝑥(𝑡) = 100 cos(2𝜋𝑓𝑐 𝑡 + 4 sin(1000𝜋𝑡)) where 𝑓𝑐 = 10 𝑀𝐻𝑧. 6M.
i. The Peak frequency deviation is ( )
a. 2K b. 4000 c. 4π d. 8π.
ii. The Peak- phase deviation is ( ).
a. 4 b. 6 c. 0 d. None of the above.
iii. The power of the Modulated signal is ( ).
a. 10KW. b. 5 W c. 5 KW d. 50W.
- The amount of change in carrier frequency produced by modulating signal is known as ( ). 1M.
a. phase deviation b. amplitude deviation
c. Frequency deviation d. none of the above.
- The total Transmitted power in FM is equal to the power of ( ) 1M.
a. An AM signal. b. an unmodulated carrier
c. Message signal d. all of the above.
- A 20 MHz carrier is frequency modulated by a sinusoidal signal with frequency 1KHz such that peak frequency deviation is 100KHz what will be the modulation index ( ) 2M.
a. 100 b.101 c. 99 d.200.
- The bandwidth for above FM system will be ( ) 2M.
a. 101 KHz b. 202 KHz c. 99 KHz d. 100 KHz.
- Which one of the following is an indirect method of generating FM ( ) 1M.
a. Armstrong method b. Varactor diode modulator
c. Reactance BJT modulator. d. Reactance FET Modulator.
- In which of the Modulation system when the modulating frequency is doubled the modulation index reduces to half while modulating voltage remains constant ( ) 2M. a. Phase b. Amplitude c. Frequency d. None of the above.
- In FM, the frequency deviation is ( ) 2M. a. Proportional to modulating frequency. b. Proportional to amplitude of modulating signal. c. Constant. d. Zero.
- In indirect method of FM generation FM is obtained from ( ) 1M. a. AM b. PM c. DSB d. FM
- Write Carson’s rule –. 1M.
- The Bandwidth of NBFM is given as –. 1M.
- A 25 MHz carrier is modulated by a 400Hz audio sine wave. The carrier voltage is 4V and the maximum deviation is 10 KHz. The modulation index will be( ) 2M. a. 2.5 b. 5 c. 15 d. 25
- For the above problem write the expression of FM wave will be———————————————————————————————–.1M.
- For the problem in 14 write the expression of PM wave———————————————————————————————————.1M.
- Standard FM broadcast stations uses a maximum bandwidth of ( ) 1M. a. 150 KHz b.75KHz. c. 200KHz d. 15KHz.
- Which type of oscillators are preferred for carrier generation because of their good frequency stability ( ) 1M. a. LR b.LC c. Crystal d. RC.
- The oscillator whose frequency is varied by an input voltage is called as ————————————————. 1M.
- Maximum deviation results at what point on modulating signal if the system is FM( ) 1M. a. Zero crossing of m(t) b. Peak negative amplitude and peak positive amplitude of m (t). c. None of the above. d. Both a and b.
(Radio receivers and Transmitters)
- Radio receivers are classified into how many types ( ). 2M.
a. Three b. two c. four d. none of the above.
- The ability of a radio receiver to amplify weak signals is called as ( ). 2M.
a. Fidelity b. Selectivity c. sensitivity d. all of the above.
- The phenomenon of Picking up of same short wave station at two nearby points on the receiver dial is known as ( ). 2M.
a. Fidelity b. sensitivity c. Double spotting d. selectivity.
- The ability of a receiver to reject unwanted signals is called ( ). 2M.
a. Selectivity b. Fidelity c. sensitivity d. Double spotting
- Standard broadcast AM receivers tuned in the frequency range of 540 KHz to 1640 KHz has an intermediate frequency of ( ). 2M.
a. 455 KHz b.1MHz c. 20Hz d. 200Hz.
- Standard broadcast FM receivers tuned in the frequency range of 88MHz -108 MHz has an intermediate frequency of ( ). 2M.
a. 455 KHz b.1MHz c. 20Hz d. 10.7MHz.
- Television receivers in the VHF band(54MHz-223MHz) and in the UHF band(470MHz-940MHz) use an IF between 26MHz and 46MHz with the two most popular values ( ). 3M.
a. 36 MHz and 46 MHz b. 455 KHz and 46 kHz.
c. 36 KHz and 46 KHz. d. none of the above.
- In a broadcast FM receiver if the local oscillator is tuned to 98.7 MHz then the image frequency is ( ). 3M.
a. 88MHz b. 109.4MHz c. 96 MHz d. none of the above.
- In a broadcast AM receiver if the signal is tuned to 530 KHz then Intermediate frequency, local oscillator frequency and image frequency are( ). 3M.
a. 200 kHz, 730 KHz and 1000 kHz.
b. 10.7MHz, 15.37MHz and 1000 KHz.
c. 455 KHz, 985 KHz and 1440 KHz.
d. None of the above.
- In communications, Audio frequency range is —————-. 2M.
- In communications, Radio frequency range is—————–. 2M.
- Draw the radio frequency spectrum with detailed values starting from Very Low Frequencies (VLF) to Extreme High Frequencies (EHF). 5M.
- Draw the block diagram of TRF receiver (only diagram). 5M.
Pulse Modulation Techniques
- In Pulse Position Modulation, the drawbacks are ( ) 2M. a.Synchronization is required between transmitter and receiver b. Large bandwidth is required as compared to PAM c. None of the above d. Both a and b.
- In PWM signal reception, the Schmitt trigger circuit is used ( ) 2M. a. To remove noise b. To produce ramp signal c. For synchronization d. None of the above.
- In pulse width modulation, ( ) 2M. a. Synchronization is not required between transmitter and receiver b. Amplitude of the carrier pulse is varied c. Instantaneous power at the transmitter is constant d. None of the above.
- In different types of Pulse Width Modulation, ( ) 2M. a. Leading edge of the pulse is kept constant b. Trailing edge of the pulse is kept constant c. Centre of the pulse is kept constant d. All of the above.
- In Pulse time modulation (PTM), ( ) 2M. a. Amplitude of the carrier is constant b. Position or width of the carrier varies with modulating signal c. Pulse width modulation and pulse position modulation are the types of PTM d. All of the above.
- Drawback of using PAM method is ( ) 2M. a. Bandwidth is very large as compared to modulating signal b. Varying amplitude of carrier varies the peak power required for transmission c. Due to varying amplitude of carrier, it is difficult to remove noise at receiver d. All of the above.
- Pulse time modulation (PTM) includes ( ) 2M a. Pulse width modulation b. Pulse position modulation c. Pulse amplitude modulation d. Both a and b.
- Calculate the Nyquist rate for sampling when a continuous time signal is given by x(t) = 5 cos 100πt +10 cos 200πt – 15 cos 300πt ( ) 3M. a. 300Hz b. 600Hz c. 150Hz d. 200Hz.
- Calculate the minimum sampling rate to avoid aliasing when a continuous time signal is given by x(t) = 5 cos 400πt ( ) 3M. a. 100 Hz b. 200 Hz c. 400 Hz d. 250 Hz.
- A distorted signal of frequency fm is recovered from a sampled signal if the sampling frequency fs is ( ) 2M. a. fs > 2fm b. fs < 2fm c. fs = 2fm d. fs ≥ 2fm.
- The desired signal of maximum frequency wm centered at frequency w=0 may be recovered if ( ) 2M. a. The sampled signal is passed through low pass filter b. Filter has the cut off frequency wm c. Both a and b d. None of the above.
- The frequency spectrum of x(t) is X(f) is given as follows 6M.
Draw the frequency spectrum of sampled signal by assuming suitable values for sampling frequency under the following conditions
i. Over sampling ii. Under sampling iii. fs = 2fm.