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Q. Explain Source encoding or data compression?
The sequence of binary digits from the source encoder, known as the information sequence, is passed on to the channel encoder. The purpose of the channel encoder is to introduce some redundancy in a controlled manner in the binary information sequence, so that the redundancy can be used at the receiver to overcome the effects of noise and interference encountered in the transmission of the signal through the channel. Thus, redundancy in the information sequence helps the receiver in decoding the desired information sequence, thereby increasing the reliability of the received data and improving the fidelity of the received signal.
The binary sequence at the output of the channel encoder is passed on to the digitalmodulator, which functions as the interface to the communication channel. The primary purpose of the digital modulator is to map the binary information sequence into signal waveforms, since nearly all the communication channels used in practice are capable of transmitting electric signals (waveforms). Because themessage has only two amplitudes in a binary system, themodulation process is known as keying. In amplitude-shift keying (ASK), a carrier's amplitude is shifted or keyed between two levels. Phase-shift keying (PSK) involves keying between two phase angles of the carrier,whereas frequency-shift keying (FSK) consists of shifting a carrier's frequency between two values.Many other forms of modulation are also possible.
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The focal length of a concave spherical mirror is equal to 1 meter. What is the radius of curvature of this mirror? Ans: The radius of curvature of this mirror is 2 meter.
Q. For the networks shown in Figure, determine the transfer function G(s) = V o (s)/V i (s).
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what are the advantages and disadvantages of Thevenin theorm over Norton theorem
Q. Consider the synchronous counter shown in Figure of the text. (a) Draw its timing diagram. (b) Show the implementation of the same synchronous counter using D flip-flops.
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Schematic Symbols The junction gate field-effect transistor or JFET gate is sometimes drawn in the middle of the channel (in place of at the drain or source electrode as in t
Q. In a CD amplifier,given Rs =4k?, µ=50 and rd= 35k μ. Find the voltage gain Av. The voltage gain, A v = Vo / V i = µR s / (µ+1) R s + rd = 50*4* (103 )/ (50+1)*4*103
Draw the hysteresis loop for a soft magnetic material and compare it with the hysteresis loop of hard magnetic material. Give two examples of each. Soft and hard magnetic mate
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