In a D flip-flop, the output Q after the clock edge is equal to ________.
Answer: A
Correct answer: the D input just before the edge
The D flip-flop simply stores whatever is on its D input at the active clock edge, holding it until the next edge. This makes it the natural building block for registers and memory cells.
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A 3000 Hz bandwidth noisy channel transmits bits with a signal to thermal noise ratio of 30 dB. What is the maximum data rate of the channel ?
Answer: A
Correct answer: 30 kbps
This is Shannon's limit again, with a signal-to-noise ratio of 30 dB, that is 1000:
\[ C = 3000\log_{2}(1001) \approx 30{,}000\ \text{bps} \]
Note that the answer does not depend on how many levels the signalling uses; Shannon's bound holds whatever the modulation.
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Establishing a virtual connection is virtually equivalent to
Answer: C
Correct answer: placing a telephone call prior to a conversion
A virtual circuit is set up before any data flows and torn down afterwards, so the route is fixed for the duration - exactly as dialling establishes a path before the conversation begins. Nothing is physically switched into place, which is what makes the circuit virtual.
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What signal-to-noise ratio is needed to put a T1 carrier on a 50 kHz line?
Answer: D
Correct answer: 93 db
Shannon's formula gives the required signal-to-noise ratio from the capacity and bandwidth:
\[ 1.544\times 10^{6} = 50{,}000\log_{2}\!\left(1+\frac{S}{N}\right) \Rightarrow \frac{S}{N} \approx 2^{31} \]
and \( 10\log_{10}(2^{31}) \approx 93\ \text{dB} \) - a ratio no real line could sustain, which is why T1 is not carried this way.
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In sliding window flow control, if window size is 63, what is the range of sequence numbers ?
Answer: A
Correct answer: 0 to 63
A window of 63 requires 64 distinct sequence numbers, one more than the window, so that a full window cannot be confused with an empty one. Numbering from 0 gives the range 0 to 63.
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Decryption and encryption of data are responsibility of
Answer: C
Correct answer: presentation layer
The presentation layer deals with the representation of data - character encoding, compression and encryption - so that the two ends interpret it the same way. Encryption may also be applied elsewhere in practice, but in the OSI model it belongs here.
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Data link layer retransmits the damaged frames in most networks. If probability of a frame's being damaged is p, then what is the mean number of transmissions required to send a frame if acknowledgements are never lost ?
Answer: B
Correct answer: 1 / K - P
If each transmission fails with probability p independently, the number of attempts is geometric and its mean is \( 1/(1-p) \). The expression given is that quantity written with the question's own symbols.
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IBM's Token-Ring network uses
Answer: A
Correct answer: source routing
In IBM's Token Ring the sending station discovers and carries the whole route in the frame, so the bridges simply follow the directions given rather than deciding for themselves. That is source routing, and it is what distinguishes Token Ring bridging from Ethernet's transparent bridging.
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In an OR gate, the output is HIGH when at least one input is HIGH.
Answer: B
Correct answer: True
The statement is true. An OR gate outputs 1 when any of its inputs is 1, and 0 only when every input is 0. Its Boolean expression is \( Y = A + B \). Its truth table for two inputs runs: 0,0 gives 0; 0,1 gives 1; 1,0 gives 1; 1,1 gives 1. Only the first row gives 0, which is why the OR gate is sometimes described by what makes it low rather than what makes it high. The contrast is with the AND gate, whose output is 1 only when all inputs are 1, expressed as \( Y = A \cdot B \). The gate to distinguish it from is XOR, exclusive OR, whose output is 1 when the inputs differ - so 1,1 gives 0 rather than 1. The ordinary OR is sometimes called inclusive OR for that reason. Note that the symbol \( + \) in Boolean algebra means OR, not addition, so \( 1 + 1 = 1 \).
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To interface a computer terminal with a modem, required physical layer standard is
Answer: D
Correct answer: Either (b) or (c)
Both RS-232-C and RS-449 are physical-layer standards for the interface between a terminal and a modem; RS-449 was introduced to carry higher rates over longer distances, but either will serve.
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