📝 RTP Real time Transport Protocol explained (22 MCQs)

📖 From Data Communication and Networks • 28. Multimedia • 22 questions available

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What is RTP Real time Transport Protocol explained?

RTP Real time Transport Protocol explained is a network protocol that delivers audio and video streams over IP by adding sequence numbers, timestamps, and payload type identifiers to each packet, enabling receivers to reconstruct timing, detect losses, and synchronize multiple media streams without guaranteeing delivery or ordering, which are handled separately by underlying transport layers.

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📝 All RTP Real time Transport Protocol explained MCQs

Q1. If the scaling factor $C(m)$ equals 1 for $m=0$ and $\sqrt{2/N}$ for $m>0$ , what happens to the DC coefficient when applying the inverse DCT compared to the forward DCT?

A.It is multiplied by

\sqrt{2/N}

B.It is halved.

C.It is unchanged. ✅

D.It becomes zero.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The DC coefficient uses the scaling factor $C(0)=1$ in both forward and inverse transforms. Because no additional scaling is applied for the zero frequency, the value remains exactly the same after the inverse DCT, preserving the average intensity of the block.

Q2. Suppose all high‑frequency coefficients (where $m>0$ and $n>0$ ) are set to zero before the inverse DCT. What is the most likely effect on the reconstructed block?

A.Increased sharpness.

B.Smoother appearance with loss of detail. ✅

C.Color inversion.

D.No change.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Zeroing high‑frequency components removes fine‑grained variations, leaving only low‑frequency content that represents smooth gradients. Consequently, the reconstructed image appears blurred or smoother, and details such as edges are diminished, which is a typical result of aggressive frequency truncation in compression.

Q3. Given that the DCT basis functions are symmetric, if a block is symmetric about its vertical axis, which DCT coefficients are guaranteed to be zero?

A.All coefficients with odd

m

B.All coefficients with odd

n

. ✅

C.All coefficients with even

m

and

n

D.No coefficients are guaranteed zero.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Vertical symmetry forces the horizontal odd‑indexed basis functions to have zero inner product with the block, because their sign changes across the axis. Therefore, any coefficient whose horizontal index $n$ is odd cancels out, ensuring those coefficients are exactly zero.

Q4. If the scaling factor $C(m)$ is mistakenly applied as $\sqrt{N/2}$ for all $m$ , how does this affect the energy of the reconstructed block after inverse DCT?

A.Energy is unchanged.

B.Energy is doubled.

C.Energy is multiplied by

N

. ✅

D.Energy becomes zero.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Applying $\sqrt{N/2}$ uniformly amplifies every basis function by the same factor. Since the DCT is orthogonal, the total energy scales with the square of the amplification, i.e., $(\sqrt{N/2})^{2}=N/2$ . Combined with the forward scaling, the net effect is an overall multiplication of the block’s energy by roughly $N$ .

Q5. Because the DCT basis functions are orthogonal, what is the inner product of two distinct basis functions over the block?

A.Zero ✅

B.One

N

D.Depends on the indices

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Orthogonality means that the sum of the product of two different basis functions across all pixel positions equals zero. This property is essential for separating frequency components without interference, allowing each coefficient to represent an independent contribution to the signal.

Q6. When performing the forward 2‑D DCT on integer pixel values, rounding the cosine terms to the nearest integer introduces which type of error in the frequency domain?

A.No error

B.Quantization error causing spectral leakage ✅

C.Systematic bias that only affects the DC term

D.Only affects high frequencies

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Rounding the cosine values replaces exact real numbers with approximate integers, creating quantization noise. This noise does not stay confined to a single coefficient; it leaks into neighboring frequencies, distorting the spectral representation and potentially reducing compression efficiency.

Q7. If the block size $N$ is 8, how many DCT coefficients are produced by the 2‑D DCT?

A.8

B.16

C.64 ✅

D.128

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: A two‑dimensional DCT processes an $N \times N$ block, generating one coefficient for each spatial position. With $N=8$ , the total number of coefficients equals $8 \times 8 = 64$ , covering all possible frequency pairs in the horizontal and vertical directions.

Q8. Which of the following best describes the main difference between the forward and inverse 2‑D DCT formulas?

A.Inverse uses cosine of negative angles.

B.Inverse includes scaling factor

C(m)C(n)

. ✅

C.Forward uses sine functions.

D.Both are identical.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The forward DCT (type‑II) omits the normalization factors, while the inverse DCT (type‑III) multiplies each term by $C(m)C(n)$ to undo the scaling applied during the forward transform, ensuring perfect reconstruction when no quantization is performed.

Q9. How does increasing the block size $N$ from 8 to 16 affect the frequency resolution of the DCT coefficients?

A.It becomes coarser.

B.No change.

C.It becomes finer with more coefficients. ✅

D.It reduces total energy.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Larger blocks contain more sample points, allowing the DCT to represent a denser set of spatial frequencies. Doubling $N$ quadruples the number of coefficients (from 64 to 256), providing finer granularity in the frequency domain and enabling more precise modeling of image detail.

Q10. The naive implementation of a 2‑D DCT has computational complexity $O(N^{4})$ . Which algorithm reduces this to $O(N^{2}\log N)$ ?

A.Direct matrix multiplication.

B.Fast DCT algorithm using separability and FFT‑like tricks. ✅

C.Gaussian elimination.

D.Monte Carlo integration.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: By exploiting the separable nature of the 2‑D DCT—applying a 1‑D DCT to rows then columns—and using fast cosine transform techniques analogous to the FFT, the computation drops dramatically to $O(N^{2}\log N)$ , making it practical for real‑time video processing.

Q11. The forward 2‑D DCT given in the text corresponds to which type of DCT, and what is its inverse counterpart?

A.DCT‑I and DCT‑IV.

B.DCT‑II and DCT‑III. ✅

C.DCT‑III and DCT‑II.

D.DCT‑IV and DCT‑I.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The presented formulas match the standard DCT‑II (forward) and DCT‑III (inverse) pair. DCT‑II is widely used for compression because of its energy‑compaction properties, while DCT‑III reverses the process, restoring the original block when the scaling factors are correctly applied.

Q12. If a block of size 8×8 is zero‑padded to 16×16 before applying the 2‑D DCT, what is the effect on the resulting frequency representation?

A.All original coefficients are scaled by 2.

B.New low‑frequency coefficients appear, original ones stay the same. ✅

C.Original coefficients are shifted to higher frequencies.

D.No effect.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Zero‑padding adds surrounding zeros, which introduces additional low‑frequency components due to the larger spatial extent. The original 8×8 coefficients remain unchanged within the central region, but the spectrum now contains extra coefficients that represent the padded zeros, subtly altering the overall frequency layout.

Q13. If the cosine term in the DCT formula is normalized by dividing by $\sqrt{N}$ instead of the standard factor, how does this change the energy compaction property?

A.Improves compaction.

B.Reduces compaction, spreading energy more evenly. ✅

C.No effect.

D.Causes energy loss.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Changing the normalization weakens the concentration of signal energy into the low‑frequency coefficients. The altered scaling distributes energy more uniformly across all frequencies, diminishing the DCT’s ability to concentrate most of the signal’s power into a few coefficients, which is detrimental for compression efficiency.

Q14. In matrix notation, the forward DCT can be written as $X = T\,P\,T^{\top}$ . What is the corresponding inverse operation?

P = T\,X\,T^{\top}

P = T^{\top}\,X\,T

✅

P = T^{-1}\,X\,T

P = X

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Because the transform matrix $T$ is orthogonal ( $T^{\top}=T^{-1}$ ), the inverse operation simply swaps the order of multiplication: $P = T^{\top} X T$ . This restores the original pixel block from the transformed coefficients without additional scaling.

Q15. Why is the 2‑D DCT particularly suitable for JPEG image compression compared to the DFT?

A.DCT uses complex numbers, making it more efficient.

B.DCT concentrates energy in a few real coefficients, enabling quantization. ✅

C.DFT has better energy compaction.

D.DCT requires larger blocks, reducing compression.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The DCT yields real‑valued coefficients and packs most of the image’s energy into the low‑frequency components, which can be heavily quantized with minimal visual impact. This property aligns with JPEG’s block‑based approach, whereas the DFT’s complex coefficients and less efficient energy distribution make it less practical for standard compression pipelines.

Q16. The term $\cos\!\left[\dfrac{m\pi(2n+1)}{2N}\right]$ in the DCT formula defines which characteristic of the basis function?

A.Amplitude scaling.

B.Spatial frequency (horizontal) of the basis. ✅

C.Phase shift only.

D.Temporal frequency.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This cosine argument determines how rapidly the basis function oscillates along the horizontal axis for a given vertical index $m$ . Larger values of $n$ increase the argument, producing higher spatial frequencies, which correspond to finer detail in the horizontal direction of the image block.

Q17. If the scaling factor $C(m)$ is altered to be 1 for all $m$ , what property of the DCT basis set is lost?

A.Completeness

B.Orthogonality ✅

C.Symmetry

D.Periodicity

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Uniform scaling removes the normalization that makes each basis vector unit‑length. Consequently, the set is no longer orthonormal; inner products between distinct basis functions remain zero (orthogonal) but their lengths differ, breaking orthogonality in the strict sense required for perfect energy preservation.

Q18. In RTP streaming of video, how does applying a 2‑D DCT before packetization affect bandwidth usage?

A.Increases bandwidth due to extra computation.

B.Reduces bandwidth by enabling compression of DCT coefficients. ✅

C.Has no effect on bandwidth.

D.Causes packet loss.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Transforming each video block with a DCT allows the encoder to quantize and discard high‑frequency coefficients, which often contain less perceptually important information. This reduces the amount of data that must be transmitted, thereby lowering the required bandwidth while preserving visual quality.

Q19. In the 2‑D DCT formulas, what does the variable $N$ represent?

A.Number of color channels.

B.Block dimension. ✅

C.Sampling frequency.

D.Frame rate.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: $N$ denotes the size of the square block (e.g., 8 for an 8×8 block). Both the forward and inverse transforms operate on an $N \times N$ pixel region, and the cosine arguments are scaled by $2N$ to ensure orthogonality across the block.

Q20. Why does the DCT align well with the human visual system's sensitivity to spatial frequencies?

A.Humans are equally sensitive to all frequencies.

B.DCT spreads energy uniformly, matching perception.

C.DCT concentrates energy in low frequencies, matching sensitivity. ✅

D.DCT ignores low frequencies.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The human eye is most sensitive to low‑frequency variations and less to high‑frequency details. The DCT naturally concentrates most of an image’s energy into a few low‑frequency coefficients, allowing quantization to remove high‑frequency components with little perceived loss, which mirrors visual sensitivity.

Q21. Which trigonometric function is used in the forward 2‑D DCT formula?

A.Sine

B.Cosine ✅

C.Tangent

D.Cotangent

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The DCT employs the cosine function because it yields real‑valued, even‑symmetry basis functions that are well‑suited for representing image data. Sine functions would introduce odd symmetry and complex values, which are unnecessary for typical image compression tasks.

Q22. In the forward DCT equation, what does $p(k,l)$ represent?

A.Frequency coefficient

B.Pixel intensity ✅

C.Scaling factor

D.Quantization step

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: $p(k,l)$ denotes the original pixel value at spatial coordinates $(k,l)$ within the block. The DCT computes a weighted sum of these pixel intensities with cosine basis functions to produce the frequency‑domain coefficients.

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