📝 New protocols for real time multimedia (25 MCQs)

📖 From Data Communication and Networks • 28. Multimedia • 25 questions available

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What is New protocols for real time multimedia?

New protocols for real time multimedia include modern standards like WebRTC, QUIC-based transports, and SRT that address limitations of legacy systems by integrating encryption, congestion control, NAT traversal, and low-latency delivery natively, enabling secure, high-quality interactive media directly in browsers and across challenging network paths without proprietary plugins.

Easy

Medium

Hard

📝 All New protocols for real time multimedia MCQs

Q1. In the DCT matrix for N=4, how does the magnitude of the first element compare to the other elements in its row?

A.It is larger than the others

B.It is equal to the others ✅

C.It is smaller than the others

D.It varies depending on the input

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Because the DCT matrix is orthogonal and its first row contains four identical values, each entry has the same magnitude. The equality ensures that the average (DC) component is captured uniformly, which is why the correct answer is that the first element is equal to the others.

Q2. If the DCT matrix is orthogonal, what can be deduced about the dot product of any two distinct rows?

A.It is zero ✅

B.It is one

C.It is negative one

D.It is undefined

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Orthogonal matrices have rows (and columns) that are mutually perpendicular, meaning the dot product of any two different rows equals zero. This property guarantees that the rows are linearly independent and that the matrix inverse equals its transpose, confirming the deduction.

Q3. Apply the DCT to the sequence (100,101,102,103). Which transformed sequence best matches the example in the text?

A.(203, ‑2.22, 0.00, ‑0.16)

B.(200, 0, 0, 0)

C.(400, ‑4.44, 0.00, ‑0.32) ✅

D.(203, 2.22, 0.00, 0.16)

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The example explicitly shows that applying the one‑dimensional DCT to (100,101,102,103) yields (203, ‑2.22, 0.00, ‑0.16). Recognizing this exact pattern demonstrates understanding of how the transform redistributes energy, making option C the correct match.

Q4. When source data items are very similar, what effect does the DCT have on rows with alternating signs?

A.Values close to zero ✅

B.Values become large

C.No change

D.Values become negative

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: If the input values differ only slightly, the alternating‑sign rows of the DCT matrix cause positive and negative contributions to cancel each other out, producing results near zero. This cancellation is the intended effect, indicating redundancy in the original data.

Q5. Given the orthogonal nature of the DCT matrix, what is the sign pattern of the second row for N=4?

A.+ ‑ + ‑ ✅

B.‑ + ‑ +

C.+ + ‑ ‑

D.‑ ‑ + +

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The second row of the 4×4 DCT matrix alternates between positive and negative entries: + ‑ + ‑. This pattern ensures that high‑frequency components are captured with opposite signs, leading to near‑zero results when the data are smooth.

Q6. If the first row of the DCT matrix sums to 1, what does this imply about the average value of the transformed sequence?

A.It equals the average of the original sequence

B.It is zero ✅

C.It is double the average

D.It equals the maximum value

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When the first row sums to 1, the corresponding coefficient represents the average (DC) component of the original data. Therefore the transformed sequence’s first value equals the original average, not zero or any other value.

Q7. Suppose the source data is (100,100,100,100). Predict the DCT coefficients.

A.(400, 0, 0, 0)

B.(200, ‑2, 2, ‑2)

C.(100, 100, 100, 100)

D.(0, 0, 0, 0) ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: A constant input yields only a non‑zero DC term; all alternating‑sign rows produce exact cancellation, resulting in coefficients (400, 0, 0, 0). The first coefficient captures the total energy, while the remaining coefficients are zero because no variation exists.

Q8. Compare the DCT transformation matrix to the identity matrix. Which statement is true?

A.Both are orthogonal

B.DCT is orthogonal while identity is not

C.Identity is orthogonal while DCT is not ✅

D.Neither is orthogonal

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Both matrices satisfy $M^T M = I$ ; the identity matrix trivially does, and the DCT matrix does by construction. Therefore the correct statement is that both are orthogonal.

Q9. Evaluate the energy‑compaction property of the DCT using the example coefficients. Which observation is correct?

A.The first coefficient is much larger than the others ✅

B.All coefficients have similar magnitude

C.The last coefficient is the largest

D.The second coefficient dominates

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: In the example, the DC term (203) far exceeds the remaining values (‑2.22, 0.00, ‑0.16). This demonstrates the DCT’s ability to concentrate most of the signal’s energy into the first few coefficients.

Q10. Differentiate between forward and inverse DCT formulas regarding the scaling factor $C(m)$ .

A.Both use the same

C(m)

B.Forward uses

C(m)

, inverse uses

1/C(m)

✅

C.Inverse omits

C(m)

entirely

D.Forward omits

C(m)

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The forward DCT includes the scaling factor $C(m)$ to normalize each basis function, while the inverse DCT also uses $C(m)$ to maintain orthogonality. However, the inverse formula effectively divides by the same factor, making the relationship a reciprocal scaling.

Q11. Given the DCT formula $T(m,n)=C(m)\cos\!\left[\frac{\pi n(2m+1)}{2N}\right]$ , analyze how increasing N from 4 to 8 affects frequency resolution.

A.Resolution improves

B.Resolution degrades

C.Resolution unchanged ✅

D.Resolution becomes infinite

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Increasing N doubles the number of sampling points, which narrows the spacing between frequency bins. The cosine argument’s denominator $2N$ grows, allowing finer discrimination of higher frequencies, thereby improving frequency resolution.

Q12. Apply the orthogonality property to explain why the DCT is reversible. Which reason is correct?

A.The transpose equals the inverse

B.The matrix is diagonal

C.The determinant is zero

D.The matrix is sparse ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Because an orthogonal matrix satisfies $M^{-1}=M^{T}$ , the inverse transformation can be performed by simply transposing the matrix. This eliminates the need for a separate inverse calculation and guarantees lossless reconstruction.

Q13. Synthesize how the DCT identifies redundant data in a sequence of similar values. Which description best fits?

A.High‑frequency coefficients near zero indicate redundancy

B.All coefficients become large ✅

C.Only the DC term is zero

D.Redundancy cannot be detected by DCT

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When the input values are nearly identical, the alternating‑sign rows generate coefficients that cancel out, leaving small high‑frequency values. This pattern signals that much of the original data is redundant, allowing compression.

Q14. Explain the role of the scaling factor $C(m)$ in ensuring orthogonality of the DCT matrix. Which statement is accurate?

A.It normalizes each basis vector ✅

B.It adds a constant offset

C.It rotates the matrix

D.It zeros out the diagonal

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: $C(m)$ is defined as $\frac{1}{\sqrt{2}}$ for $m>0$ and $\frac{1}{\sqrt{N}}$ for $m=0$ . This choice guarantees that each row (or column) has unit length, making the matrix orthogonal.

Q15. If a signal contains a sudden change, how would its DCT coefficients differ from those of a smooth signal?

A.Higher high‑frequency coefficients

B.Lower high‑frequency coefficients

C.All coefficients become zero

D.Only the DC term changes ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: A sudden change introduces high‑frequency content, which the DCT captures in the higher‑order coefficients. Consequently, those coefficients increase in magnitude relative to a smooth signal where they remain near zero.

Q16. Design a protocol that uses DCT to compress telemetry data. Which step guarantees lossless reconstruction?

A.Using the orthogonal DCT matrix

B.Quantizing coefficients to integers

C.Discarding high‑frequency coefficients ✅

D.Applying a non‑linear transform

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Lossless reconstruction requires that the inverse operation exactly undo the forward transform. This is ensured only when the original orthogonal DCT matrix is retained without discarding any coefficients; quantization or truncation would introduce irreversible errors.

Q17. What is the value of the scaling factor $C(0)$ for N=4?

\frac{1}{\sqrt{4}}

\frac{1}{2}

\sqrt{2}

✅

D.1

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For the DC term, $C(0)=\frac{1}{\sqrt{N}}$ . With $N=4$ , this evaluates to $\frac{1}{\sqrt{4}} = \frac{1}{2}=0.5$ . The answer that matches this expression is the third option.

Q18. If two rows of the DCT matrix multiply the same source vector, why can their results differ in sign?

A.Because the rows have alternating signs ✅

B.Because the matrix is not orthogonal

C.Because of different scaling factors

D.Because of rounding errors

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Each row of the DCT matrix represents a distinct basis function with its own sign pattern. When the same vector is projected onto these bases, the alternating signs cause some components to be added and others subtracted, leading to results that may have opposite signs.

Q19. Evaluate the effect of rounding transformed coefficients to two decimal places on DCT reversibility. Which outcome is correct?

A.It loses some information, making perfect reversal impossible

B.It remains perfectly reversible ✅

C.It improves reconstruction accuracy

D.It eliminates all high‑frequency content

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Rounding introduces quantization error; the inverse DCT will reconstruct a signal that approximates the original but is not identical. Therefore, perfect lossless reversal is compromised, and the statement about losing information is accurate.

Q20. Explain how the DCT can be used to detect edges in image processing. Which mechanism is appropriate?

A.High‑frequency coefficients highlight abrupt intensity changes

B.Low‑frequency coefficients indicate edges

C.The DC term alone reveals edges ✅

D.DCT cannot detect edges

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Edges correspond to rapid spatial variations, which manifest as high‑frequency components in the DCT domain. By examining coefficients beyond the DC term, one can locate regions where these values are significant, indicating edge presence.

Q21. What does DCT stand for?

A.Discrete Cosine Transform ✅

B.Digital Compression Technique

C.Direct Correlation Theory

D.Data Conversion Tool

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The abbreviation DCT refers to the Discrete Cosine Transform, a widely used method for converting spatial data into frequency components.

Q22. If the source values increase linearly, what pattern do the DCT coefficients exhibit?

A.A dominant low‑frequency coefficient with small higher‑order terms

B.All coefficients are equal

C.Only high‑frequency coefficients are non‑zero

D.Coefficients alternate between positive and negative large values ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: A linear ramp introduces a strong DC component and a predictable pattern in the first few frequencies, but the alternating‑sign rows produce coefficients that follow a sinusoidal pattern, often alternating in sign while decreasing in magnitude.

Q23. Contrast the effect of using the forward DCT matrix versus its transpose on a data vector. Which statement is correct?

A.Both produce the same result because the matrix is symmetric

B.The forward matrix spreads energy, the transpose concentrates it

C.The transpose yields the inverse transformation ✅

D.The forward matrix is orthogonal while the transpose is not

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Applying the transpose of an orthogonal DCT matrix is equivalent to applying its inverse. Therefore, using the transpose reconstructs the original data from the transformed coefficients, whereas the forward matrix computes those coefficients.

Q24. Propose a method to select DCT coefficients to retain for lossy compression while preserving perceptual quality. Which approach is most effective?

A.Keep the largest‑magnitude coefficients and discard the rest ✅

B.Retain only the DC term

C.Randomly select coefficients

D.Discard all high‑frequency coefficients

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Human perception is less sensitive to high‑frequency details. By sorting coefficients by magnitude and preserving the largest ones, the most visually important information is kept, while the smaller, less noticeable components can be discarded with minimal quality loss.

Q25. In the example, what is the second transformed value?

A.-2.22 ✅

B.0

C.-0.16

D.203

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The example explicitly lists the transformed sequence as (203, ‑2.22, 0.00, ‑0.16). Hence the second value is \-2.22, matching option A.

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