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πŸ“ IPSec security services (12 MCQs)

πŸ“– From Data Communication and Networks β€’ 32. Internet Security β€’ 12 questions available

What is IPSec security services?

IPSec delivers essential security services including access control, connectionless integrity, data origin authentication, protection against replay attacks, confidentiality via encryption, and limited traffic flow confidentiality.

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Easy
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Medium
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Hard

πŸ“ All IPSec security services MCQs

Q1. In the delta modulation encoder, what does the parameter D represent?

A.Step size used for updating y_n βœ…
B.Initial decoder output y_0
C.Quantization threshold for e_n
D.Sampling frequency of the input signal
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: The parameter D is the step size that determines how much the reconstructed value y_n changes each iteration. It multiplies the quantized direction q_n in the update rule yn=ynβˆ’1+DΓ—qny_n = y_{n-1} + D \times q_n, directly influencing the slope of the reconstructed signal.

Q2. If the error signal e_n is positive at iteration n, what is the resulting value of q_n?

A.1 βœ…
B.-1
C.0
D.It depends on C_n
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: When en>0e_n > 0 the encoder decides to increase the reconstruction, so it sets the quantized direction qnq_n to +1. This choice makes the decoder add DD in the next step, moving y_n upward toward the original signal.

Q3. For a slowly increasing input x_n, why does quantizing xnβˆ’ynβˆ’1x_n - y_{n-1} produce a self‑correcting reconstruction?

A.Because the error e_n becomes zero quickly
B.Because the step size D adapts automatically
C.Because the difference xnβˆ’ynβˆ’1x_n - y_{n-1} directly reflects the accumulated error, causing q_n to push y_n toward x_n each cycle βœ…
D.Because quantizing xnβˆ’xnβˆ’1x_n - x_{n-1} already corrects errors
πŸ’‘ Difficulty: medium | βœ… Correct: C

πŸ“– Explanation: The term xnβˆ’ynβˆ’1x_n - y_{n-1} measures the current mismatch between the true signal and its reconstruction. When this difference is quantized, the resulting q_n forces the decoder to add or subtract D, reducing the error in the next iteration and thus self‑correcting the trajectory.

Q4. Which statement best compares quantizing xnβˆ’xnβˆ’1x_n - x_{n-1} with quantizing xnβˆ’ynβˆ’1x_n - y_{n-1} for a slow‑varying signal?

A.Both methods generate identical staircases
B.Quantizing xnβˆ’xnβˆ’1x_n - x_{n-1} accumulates error over time
C.Quantizing xnβˆ’ynβˆ’1x_n - y_{n-1} reduces error by referencing the previous reconstruction βœ…
D.Neither method can track the signal accurately
πŸ’‘ Difficulty: medium | βœ… Correct: C

πŸ“– Explanation: Quantizing xnβˆ’ynβˆ’1x_n - y_{n-1} uses the previous reconstructed value y_{n-1} as a reference, so each update corrects the accumulated deviation. In contrast, quantizing the raw difference xnβˆ’xnβˆ’1x_n - x_{n-1} does not account for past reconstruction errors, allowing those errors to accumulate.

Q5. How does selecting a large step size D affect stability when the input signal is a fast‑falling function?

A.It improves tracking speed without side effects
B.It may cause the decoder to overshoot and become unstable βœ…
C.It has no impact on stability
D.It forces the error e_n to remain positive
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: A large D causes the decoder to change y_n by a big amount each iteration. For a rapidly decreasing x_n, this can make y_n overshoot the true value, producing alternating large errors that grow rather than diminish, ultimately leading to instability in the reconstruction.

Q6. If the control flag C_n equals 0, what is the update rule for y_n?

A.yn=ynβˆ’1+Dy_n = y_{n-1} + D
B.yn=ynβˆ’1βˆ’Dy_n = y_{n-1} - D βœ…
C.yn=ynβˆ’1y_n = y_{n-1}
D.yn=ynβˆ’1+2Dy_n = y_{n-1} + 2D
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: When C_n = 0 the encoder indicates a negative direction, so the decoder subtracts the step size. The update becomes yn=ynβˆ’1βˆ’Dy_n = y_{n-1} - D, moving the reconstructed value downward to follow a decreasing portion of the original signal.

Q7. Given an alternating sequence of q_n = +1, -1, +1, -1,… and D = 0.5, what is the long‑term behavior of y_n?

A.It diverges to infinity
B.It converges to a constant value
C.It oscillates within a bounded range βœ…
D.It settles at zero
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: Each iteration adds or subtracts 0.5, so y_n moves up half a unit then down half a unit, repeatedly. The net change over two steps is zero, keeping y_n confined to a narrow band. Thus the reconstruction oscillates but never grows without bound.

Q8. With y_{n-1}=3, q_n=+1, D=2, and C_n=1, what is the next reconstructed value y_n?

A.5 βœ…
B.1
C.3
D.7
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: Using the update rule yn=ynβˆ’1+DΓ—qny_n = y_{n-1} + D \times q_n (since C_n=1 indicates a positive direction), we compute yn=3+2Γ—(+1)=5y_n = 3 + 2 \times (+1) = 5. This moves the reconstruction upward toward the original signal.

Q9. When x_n is a linear ramp, how does the staircase obtained by quantizing xnβˆ’xnβˆ’1x_n - x_{n-1} compare with the original function?

A.It exactly matches the ramp
B.It lags behind the ramp by one step βœ…
C.It overshoots the ramp at each interval
D.It produces a smoother curve than the ramp
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Quantizing the difference between successive samples produces a step of fixed height D each time the ramp exceeds the previous step. Consequently the staircase trails the ideal linear ramp, creating a lag of roughly one quantization interval.

Q10. Why must the encoder know the initial value y_0 and the step size D before encoding begins?

A.To compute the error e_n correctly βœ…
B.To set the sampling frequency
C.To determine the bandwidth of the channel
D.To choose the modulation scheme
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: The error signal is defined as en=xnβˆ’ynβˆ’1e_n = x_n - y_{n-1}. Without an initial reconstruction y_0, the first error cannot be calculated. Similarly, D determines how much y_n changes for each quantized direction, so both parameters are essential for accurate encoding.

Q11. If at iteration n the error e_n is negative and C_n = 0, what are the values of q_n and the subsequent change in y_n?

A.q_n = +1, y_n increases by D
B.q_n = -1, y_n decreases by D
C.q_n = 0, y_n unchanged βœ…
D.q_n = -1, y_n unchanged
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: A negative error indicates the reconstruction is above the true signal. With C_n = 0 the encoder signals a negative direction, so q_n is set to 0 (representing a negative step). The decoder then updates yn=ynβˆ’1βˆ’Dy_n = y_{n-1} - D, decreasing the reconstruction.

Q12. In delta modulation, what does the term β€œself‑correcting” imply about the relationship between successive errors e_n and e_{n+1}?

A.Errors increase exponentially over time
B.Errors tend to diminish as the system iterates βœ…
C.Errors remain constant after each step
D.Errors are unrelated to previous values
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: β€œSelf‑correcting” means that each quantized update reduces the magnitude of the error. When ene_n is positive, q_n pushes y_n upward; when negative, it pushes downward. This feedback drives the subsequent error en+1e_{n+1} toward zero, gradually improving the reconstruction.

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