Question

A transmitter uses 64-QAM symbols with a data rate of 960,000 bits/second. The transmitter sends 5 images, wherein each image is of size 1000 X 1000 pixels and is represented using 16 bits/pixel. The image is split into multiple frames with each frame a maximum of 840 bytes, of which 40 bytes is dedicated to header and trailers at various levels of the protocol stack. Assume that the remaining 800 bytes of each frame contain the image pixel data payload.

What is the symbol rate of the transmitter?
How long does it take to transmit a frame at the data link layer?
If the data link layer employs a stop-and-wait flow control, how long does it take to transmit all 5 images? The acknowledgment frames are 50 bytes long. The propagation time is 0.1ms each way. ignore processing time. assume there are no errors in the transmission.

Answer 1

The symbol rate of the transmitter is 160,000 symbols/second, and it takes approximately 7 milliseconds to transmit a frame at the data link layer. The time to transmit all 5 images with stop-and-wait flow control includes frame transmission, acknowledgments, and propagation delays.

Step-by-step explanation:

The symbol rate of the transmitter, also known as the baud rate, can be calculated by dividing the data rate by the number of bits per symbol. In the case of 64-QAM, each symbol represents 6 bits (because 26 = 64). Therefore, the symbol rate is 960,000 bits/second divided by 6 bits/symbol, resulting in 160,000 symbols/second.

To calculate how long it takes to transmit a frame at the data link layer, we take the frame size (including header and trailer) and divide it by the data rate. At 840 bytes per frame and 8 bits per byte, we have 6,720 bits per frame. Dividing this by the data rate of 960,000 bits/second gives us approximately 0.007 seconds or 7 milliseconds per frame.

To transmit all 5 images using stop-and-wait flow control, we need to consider the time needed for each frame (including acknowledgments and propagation time). Each image is 1000x1000 pixels at 16 bits/pixel, which is 16,000,000 bits per image or 2,000,000 bytes. With 800 bytes of image data per frame, that's 2,500 frames per image. The total time includes the transmission time for 2,500 frames, the time for transmitting 2,500 acknowledgment frames (50 bytes or 400 bits long), and the round-trip propagation time for each frame and acknowledgment. With the given data rate, propagation time, and neglecting processing time and errors, the total time would be substantial.