1. A synchronous TDM combines 10 digital sources where each has data rate of 500-kbps. Each output slot carries 4 bits from each digital source, but one extra bit is added to each frame for synchronization. Determine the frame size and output data rate of this TDM. (10 points) 2. Assume synchronous time division multiplexer with three inputs – channel 1, channel 2, and channel 3. The output of this multiplexer produces synchronous 18-bit frames that are provisioned for stuffing as shown in the figure. The stuffing process can occur with each frame. The output link of this multiplexer operates at 36 Mbps. Determine the upper and lower bound of data rate at each input channel.  (10 points) PLEASE SEE ATTACHED FOR THE IMAGE/DIAGRAM 3. Find the random numbers generating by x3 + x2 + x + 1 using Viterbi algorithm. Is it primitive? (5 points) 4. Assume that the BER is 10-6. (a) Determine the probability that the 1000-bit frame would be received without an error. (5 points)

1. In a synchronous Time Division Multiplexing (TDM) system, multiple digital sources are combined to create a single stream of data for transmission. Each digital source in this particular scenario has a data rate of 500 kbps. The TDM system combines 10 of these digital sources.

To determine the frame size and output data rate of this TDM system, we need to consider the number of digital sources and the amount of data carried in each output slot. In this case, each output slot carries 4 bits from each digital source, with an additional bit added for synchronization.

Given that there are 10 digital sources, each carrying 4 bits per slot, the total number of bits carried in each frame can be calculated as follows:
10 (digital sources) x 4 (bits per slot) = 40 bits per frame.

Adding the extra synchronization bit, the frame size for this TDM system is 41 bits.

To determine the output data rate, we need to consider the time taken to transmit each frame. Since the frame size is 41 bits, and the data rate of each digital source is 500 kbps, the time taken to transmit one frame can be calculated as follows:
41 (bits per frame) / 500,000 (bps) = 0.000082 seconds.

To calculate the output data rate, we can divide the frame size by the time taken to transmit one frame. Thus, the output data rate of this TDM system is:
41 (bits per frame) / 0.000082 (seconds) = 500,000 bps, or 500 kbps.

Therefore, in this TDM system, the frame size is 41 bits and the output data rate is 500 kbps.

2. In this scenario, we have a synchronous Time Division Multiplexer (TDM) with three input channels – channel 1, channel 2, and channel 3. The output of this multiplexer produces synchronous 18-bit frames, with provision for stuffing as shown in the attached figure.

To determine the upper and lower bounds of data rate at each input channel, we need to consider the output link’s operating rate, which is given as 36 Mbps.

From the figure, we can observe that each frame consists of 18 bits, with some bits reserved for stuffing. The exact number of bits reserved for stuffing can vary from frame to frame, depending on the data being transmitted.

In a TDM system, the data rate at each input channel is dependent on the number of frames allotted to that particular channel. The upper bound of data rate at each channel occurs when the maximum number of frames are allotted to that channel, and the lower bound occurs when the minimum number of frames are allotted.

To calculate the upper and lower bounds of data rate at each channel, we need to consider the number of frames transmitted per second. The output link operates at 36 Mbps, which means it can transmit 36 million bits per second.

Let’s assume that the stuffing process occurs with each frame. This means that each frame transmitted on the output link consists of 18 bits, plus some additional bits for stuffing. The exact number of stuffing bits can vary, depending on the data being transmitted.

To find out the number of frames transmitted per second, we need to divide the output link’s data rate by the number of bits in each frame, including the stuffing bits.

Let’s denote the number of stuffing bits per frame as ‘S’. Therefore, the total number of bits in each frame, including the stuffing bits, can be given by 18 + S.

Now, we can calculate the number of frames transmitted per second as:
36 million (bits per second) / (18 + S) (bits per frame) = number of frames per second.

To determine the upper and lower bounds of data rate at each input channel, we need to consider the number of frames allotted to each channel. Let’s denote the number of frames allotted to channel 1, channel 2, and channel 3 as ‘N1’, ‘N2’, and ‘N3’, respectively.

The upper bound of data rate at each channel occurs when the maximum number of frames are allotted to that channel. Therefore, the upper bound of data rate at each channel can be calculated as:
(N1 x (18 + S)) (bits per second) for channel 1
(N2 x (18 + S)) (bits per second) for channel 2
(N3 x (18 + S)) (bits per second) for channel 3.

The lower bound of data rate at each channel occurs when the minimum number of frames are allotted to that channel. Therefore, the lower bound of data rate at each channel can be calculated as:
(N1 x (18 + S)) (bits per second) for channel 1
(N2 x (18 + S)) (bits per second) for channel 2
(N3 x (18 + S)) (bits per second) for channel 3.

Thus, the upper and lower bounds of data rate at each input channel depend on the number of frames allotted to each channel, the number of stuffing bits per frame, and the output link’s data rate.

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