__UNIT-1__

__QUIZ__

- 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—————————–

__UNIT-2__

__QUIZ 2__

- 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 ) [**Hint: Shannon’s bound**C_{infinity}= S/N_{o}**] 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
- 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.
- Given a Binary Symmetric Channel, the expression for Entropy is (
**p**is conditional probability of error ——————- 3M.

__Digital Communications Unit-3 Quiz with Solutions__

__UNIT3__

__QUIZ3[CO3]__

- The number of Parity check bits in an (n, k) Linear Block codes are (
)*b* - 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
) a.Generator matrix & Generator polynomial. b.Generator matrix & Generator matrix. c.Generator polynomial & Generator matrix. d.None of the above.*——————-.( c* - The rate of a Block code is the ratio of(
) a.Message length to Block length. b.Block length to message length. c.Message weight to Block length. d.None of the mentioned.*c* - The syndrome in LBC is calculated using , where Y represents received code word (
)*a***2M**a. S= Y H^{T}b. S = YH c. S= Y^{T}H d. S= Y^{T}H^{T} - 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 (
), where E represents error vector. a. X= E + Y b. X = X.Y c. X= E.Y d. X= X/Y*a* - 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 [110] and [111]——————-. Ans: [110110], [111000].
- For the Q.8. Find the syndrome value when the received code word is 001111—-
**4M**. Ans: syndrome value [0 0 1] - 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.
__UNIT-4____QUIZ-4__- 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 E
_{b}, the energy per bit of a binary digital signal, is 10^{-5}watt-sec and the one-sided power spectral density of the white noise, N_{o}= 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 N
_{o}/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 P
_{e}= (½) erfc(x) thenthe same P_{e}interms of Q-function is ( b ) a. Q ( 3^{1/2}* x) b. Q (2^{1/2}* x) c. Q(x) d. none of these.

- The modulation technique that provides minimum probability of error is ( b ). a. ASK