- The band width needed to transmit Television video plus audio signals of bandwidth 4.2 MHz using Binary PCM quantization level of 512 is ( c ) a. 2 MHz b. 25.6MHz c.37.8MHz d.75.6MHz
- A signal m(t)has a bandwidth of 1.0 KHz and exhibits a maximum rate of change of 2.0 volts/sec. The signal is sampled at a sampling frequency of 20 KHz and quantized using delta modulator. The minimum step size to avoid slope overload is ( b ) a. 1.0mV b. 0.1mV c.10.0mV d.0.01mV
- For a 10 bit PCM system, the signal to Quantization noise ratio is 62 dB. If the number of bits are increased by 2, then the signal to Quantization noise ratio will be ( c ) a. Increased by 6 dB b. Decrease by 6 dB c. Increased by 12 dB d. Decrease by 12 dB
- Write the condition to eliminate slope overload error in a Delta modulation system—————————
- Write the condition to eliminate slope overload error in a Delta modulation system if the input is a single-tone signal—————————
- For which value of A , A-law has linear transfer characteristics——————–.
- For which value of µ , µ-law has linear transfer characteristics——————–.
- In a PCM system if the step size is 5V , then the Quantization noise in dB is ( d ) a. -5 dB. b. -3.18dB c. -10dB d. 3.18dB.
- Draw the characteristics of Mid-rise and Mid-tread type Quantizers ————–
- The maximum slope of the signal ———————-
- The sampling rate of the signal ————————–
- The step size in a 8- bit pcm system if the input signal to PCM is oscillating between [+4V,-4V] is—————————–
- Entropy is a measure of ——————————– 3M.
- The capacity of a band – limited AWGN channel in terms of kbps if the average received signal power to noise power spectral density is 1000 and the bandwidth is approximately infinite is ( a ) a. 1.44 b. 1.08 c. 0.72 d. 0.36
- If Y= g(X) where g denotes a deterministic function, then the conditional entropy H(Y/X) is ( b ) 3M. a. ≠ 1 b. = 0 c. = 1 d. ≠ 0
- A Source generates three symbols with probability of 0.25, 0.25, and 0.50 at a rate of 3000 symbols per second. Assuming the symbols are generated independently from the source, the most efficient source encoder would have average bit rate of ( b ) 4M. a. 6000 bits/sec b. 4500 bits/sec c. 3000 bits/sec d. 1500 bit/sec.
- A source generates four equi-probable symbols. If the source coding is adopted, the average length of the code for 100% efficiency is ( c ) a. 6 bits / symbol b. 3 bits / symbol c. 2 bits / symbol d. 4 bits / symbol
- Draw the channel diagram of Binary Erasure channel and write its channel matrix ………………………..
- For a Binary Symmetric Channel the entropies of X & Y are if the conditional probability p=0.5, where X & Y are input & output random variables of BSC ( d )3M a. 1,0 b. 0,0 c. 0,1 d. 1,1
- Given a Binary Symmetric Channel, the expression for Entropy is (pis conditional probability of error ——————- 3M.
- Match the following ( d ) a. Lossless channel 1. only one non-zero element in each row b. Deterministic channel 2. Only one non-zero element in each row and each column. c. Noiseless channel 3. Only one Non-zero element in each column. d. for a noiseless channel 4. H(Y/X)=0 and H(X/Y)= 0.
a. none of these b. b-3,c-4,a-1,d-2 c. c-4,d-3,a-1,b-2 d. b-1,c-2,a-3,d-4.
- The number of Parity check bits in an (n, k) Linear Block codes are ( b ) a.n b. (n-k) c. (n+k) d. k
- The Hamming Weight of the following code words 10011101 & 00111100 is ( c ) 2M. a. None of these b. 4, 5 c.5, 4 d.3,4
- A cyclic code can be generated using———————— and A block can be generated using——————-.( c ) a.Generator matrix & Generator polynomial. b.Generator matrix & Generator matrix. c.Generator polynomial & Generator matrix. d.None of the above.
- The rate of a Block code is the ratio of( c ) a.Message length to Block length. b.Block length to message length. c.Message weight to Block length. d.None of the mentioned.
- The syndrome in LBC is calculated using , where Y represents received code word ( a ) 2M a. S= Y HT b. S = YH c. S= YT H d. S= YT HT
- A non-Zero value of Syndrome in a Block code represents ( b ) 2M. a.No error during transmission. b.An error occurred during transmission. c.Both a and b d.None of the above.
- The transmitted code word(X) in an LBC can be obtained from received code word( Y) by using the equation ( a ), where E represents error vector. a. X= E + Y b. X = X.Y c. X= E.Y d. X= X/Y
- The parity check matrix of a (6,3) block code if Generator matrix is G 3M. ——————-.
- In the above question find the Code words corresponding to message vectors and ——————-. Ans: , .
- For the Q.8. Find the syndrome value when the received code word is 001111—-4M. Ans: syndrome value
- For a (7,4) cyclic code , the generator polynomial is given as find the codeword for the data 1100
- Non systematic codeword is—————- .Ans: 1011100
- Systematic codeword is——————–.Ans: 1011100.
- The modulation technique that provides minimum probability of error is ( b ). a. ASK b. PSK c. DPSK d.FSK
- At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by ( c ) a. 6 dB b. 2 dB c. 3 dB d. 0 dB e. None of these.
- If Eb, the energy per bit of a binary digital signal, is 10-5watt-sec and the one-sided power spectral density of the white noise, No= 10-6 W/Hz, then the output SNR of the matched filter is ( d ) a. 26 dB b. 20 dB c. 10 dB d. 13 dB e. None of these
- In which system, bit stream is portioned into even and odd stream ( c ) a. BPSK b. MSK c. QPSK d. FSK
- Optimum filter can be called as———when the input noise is white noise ( a ) 2M. a. Matched filter b. High pass filter c. Low pass filter d. None of these
- The probability of error of ASK is ————————————.
- The probability of error of FSK is ————————————.
- Write the expression for QPSK modulation —————————————-2M.
- Draw the block diagram of DPSK system————————————.
- A pulse g(t) = A cos(πt/2T) for 0 ≤ t ≤ T is transmitted over an AWGN channel with two sided noise power spectral density No/2 Watts/Hz. The impulse response of the matched filter is ( a ) a. A sin (πt/2T) b. A rec (πt/2T) c. A rec (π(T-t)/2T) d. A sin (π(T-t)/2T)
- If the probability of error function of a modulation scheme is Pe = (½) erfc(x) thenthe same Pe interms of Q-function is ( b ) a. Q ( 31/2 * x) b. Q (21/2 * x) c. Q(x) d. none of these.