Technologies for Digital Media Business Transmission

Question: Discuss about the Technologies for Digital Media of Business Transmission. Answer: Calculate the total number of bits in this GOP. I1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P13 P14 P15 57648 11488 5208 3760 2608 1984 1320 840 1144 648 448 800 932 856 880 Total number of bits = 90564 bits Knowing that the 15 frames in Table 1 should be transmitted in 0.5s, what is the required bit rate (in bits/s) to successfully transmit the GOP of Table 1 within the required time limit? Required bit rate = What can you conclude concerning the transmission of this sequence over the link between BS1 and the mobile user in Fig. 1 (be reminded that the data rate over that link is 100 kbits/s)? The required data rate is 176.883kbps which is greater than the available data rate of 100kbps. Hence data can not be transmitted within the time frame of 0.5 seconds over the link between the BS1 and the mobile user. Description 1 I1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P13 P14 P15 28704 8024 3400 2352 1720 1344 720 592 640 352 128 392 608 480 424 Description 2 I1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P13 P14 P15 28944 3464 1808 1408 888 640 600 248 504 296 320 408 324 376 456 Calculate the total number of bits in the GOP of D1 (Table 2). Total number of bits in the GOP of D1 = 49880 bits Knowing that the 15 frames in Table 2 should be transmitted in 0.5s, what is the required bit rate (in bits/s) to successfully transmit the GOP of Table 2 within the required time limit? Required bit rate to transmit GOP of Table 2 = Calculate the total number of bits in the GOP of D2 (Table 3). Total number of bits in the GOP of D2 = 40684 bits Knowing that the 15 frames in Table 3 should be transmitted in 0.5s, what is the required bit rate (in bits/s) to successfully transmit the GOP of Table 3 within the required time limit? Required bit rate to transmit GOP of Table 2 = What can you conclude concerning the transmission of the video sequence, after splitting it into two descriptions, knowing that the achievable data rate over the link between BS1 and the mobile user is 100 kbits/s, and the achievable data rate over the link between BS2 and the mobile user is 95 kbits/s? After splitting the video sequence into two descriptions, the required bit rate of 97.4kbps and 79.kbps for D1 and D2 are less than the available data rate of 100Kbps for the link between BS1 and mobile, and 95kbps for the link between BS2 and mobile respectively. Hence the data can be transmitted successfully within the time frame of 0.5 seconds over the available links. Assuming a simple form of encryption that consists of performing a XOR operation with the Key: Perform ECB encryption for plaintext Blocks 1, 2, and 3. Key 1 0 1 0 1 0 1 0 Plaintext Block1 1 1 1 0 0 1 1 0 Plaintext Block2 1 0 0 0 1 0 0 1 Plaintext Block3 1 1 1 0 0 1 1 0 XOR operation is performed between the plaintext and the key to encrypt the data and produce the ciphertext. Ciphertext Block1 0 1 0 0 1 1 0 0 Ciphertext Block2 0 0 1 0 0 0 1 1 Ciphertext Block3 0 1 0 0 1 1 0 0 What can you remark, especially concerning Blocks 1 and 3? Same data is encrypted as same ciphertext with the key. Hence the Electronic Code Book encryption is deterministic. Hence, with a small number of trial and errors, the plain text can be predicted from the ciphertext. Perform CBC encryption for plaintext Blocks 1, 2, and 3. XOR operation is performed between the Initialization Vector (IV), the key and the Plaintext Block1. This will result in the Ciphertext Block1. Then, XOR operation is performed between the Ciphertext Block1, Plaintext Block2 and the key. This will result in the Ciphertext Block2. Similarly, the Ciphertex Block2, Plaintext Block3 and the key are XORed to generate the Ciphertext Block3. The advantage of Cipher Block Chaining (CBC) encryption is that changing the initialization vector results in different ciphertext for identical messages. Initialization Vector 1 0 0 0 0 0 0 1 Key 1 0 1 0 1 0 1 0 Plaintext Block1 1 1 1 0 0 1 1 0 Plaintext Block2 1 0 0 0 1 0 0 1 Plaintext Block3 1 1 1 0 0 1 1 0 Ciphertext Block1 1 1 0 0 1 1 0 1 Ciphertext Block2 1 1 1 0 1 1 1 0 Ciphertext Block3 1 0 1 0 0 0 1 0 What can you conclude, especially in comparison with the ECB results? Block1 and Block3 are identical data for which the Electronic Code Book (ECB) encryption produces identical Ciphertexts. But for the identical blocks, the Cipher Block Chaining (CBC) encryption produces different Ciphertexts depending upon the Initialization Vector. Hence, it is difficult for the attacker to recover the Plaintext. The disadvantage of CBC is that the error, if any too gets propagated to few further blocks due to chaining effect.