56851. In digital TV digital imaging standard comprises
56852. The 2 out of 5 code has
56853. Consider the following statements as regards BPSK and QPSK: In BPSK we deal individually with each bit of duration TbIn QPSK we lump two bits to form a symbolIn both BPSK and QPSK signal changes occur at bit rateIn QPSK the symbol can have any one of four possible values Which of the above statements are correct?
56854. In equally tampered musical scale, an octave is divided into
56855. Entropy gives
56856. Quantization noise occurs in
56857. AC hum is picked up from power cables
56858. PIL tube means
56859. In an SSB transmitter one is most likely to find
56860. For a given data rate, the bandwidth βp of a BPSK signal and the bandwidth β0 of the OOK signal are related as
56861. A direct radiating cone type loudspeaker attenuates :
56862. The advantage of pulse modulation is
56863. Sparking affects amplitude modulation.
56864. For low impedance SAW filter the most suitable material for substrate of SAW filter is
56865. A telephone channel requires a bandwidth of about
56866. For a given carrier wave, maximum undistorted power is transmitted when value of modulation is
56867. As per Shannon-Hartley theorem, a noise less Gaussian channel has
56868. DVD uses
56869. Fourier analysis indicate that a square wave can be represented as
56870. FM transmitting and receiving equipment as compared to AM equipment is
56871. An audio signal (say from 50 Hz to 10000 Hz) is frequency translated by a carrier having a frequency of 106 Hz. The values of initial (without frequency translation) and final (after frequency translation) fractional change in frequency from one band edge to the other are
56872. If in a broadcasting studio, a 1000 kHz carrier is modulated by an audio signal of frequency range 100-5000 kHz, the width of channel is __________ kHz.
56873. Which one of the following is analog?
56874. The disadvantage of FM over AM is that
56875. Which of the following is a digital modulation technique?
56876. Which of the following is used to generate PDM?
56877. Which is the most effective strategy for prevention of neonatal sepsis?
56878. A zero mean white Gaussian noise is passed through an ideal low pass filter of bandwidth 10 kHz. The output of the samples so obtained would be
56879. What is the purpose of peak clipper circuits in radio transmitters?
56880. In case of low level amplitude modulation system, the amplifiers following the modulated stage must be
56881. Under ordinary circumstances, impulse noise can be reduced in
56882. In case of frequency modulation, modulating voltage remains constant if the modulating frequency is lowered, then
56883. If sampling is done at the rate of 10 kHz. The bandwidth required is
56884. It is found that a ship to ship communication suffers from fading. This can be avoided by using
56885. Skip distance depends on time of day and angle of incidence.
56886. In practical commercial FM system, channel bandwidth is
56887. In EM waves, polarization
56888. The maximum power output of a standard A earth station over the total band allocated to satellite communication is about
56889. What is the commonest cause for neonatal mortality in India?
56890. The characteristic impedance of a twin wire feeder used for TV signals is about
56891. In a FM receiver, amplitude limiter
56892. A buffer amplifier is
56893. Leak type bias is used in plate modulated class C amplifier to
56894. The direction of rotation of a CD is
56895. What is the age at which “object permanence”is normally attained?
56896. Full duplex operation-permits transmission in both directions at the same time.
56897. In a TV receiver antenna the length of reflector rod
56898. The modulation index of an FM is changed from 0 to 1. How does the transmitted power change?
56899. For AM receivers the standard IF frequency is
56900. For a plate-modulated class C amplifier the plate supply voltage is E. The maximum plate cathode voltage could be almost high as