Bandwidth Utilization Quiz

Assume that a voice channel occupies a bandwidth of 4 kHz. We need to combine three voice channels into a link with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the configuration, using the frequency domain. Assume there are no guard bands.

Answer explanation

To combine three 4 kHz voice channels into a 12 kHz link, allocate frequencies from 20 to 32 kHz: Channel 1: 20-24 kHz, Channel 2: 24-28 kHz, Channel 3: 28-32 kHz. This configuration uses the entire bandwidth without guard bands.

Five channels, each with a 100-kHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 kHz between the channels to prevent interference?

Answer explanation

To calculate the minimum bandwidth, add the bandwidths of the channels and the guard bands. For 5 channels: 5 x 100 kHz = 500 kHz. There are 4 guard bands of 10 kHz: 4 x 10 kHz = 40 kHz. Total = 500 kHz + 40 kHz = 540 kHz.

Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration, using FDM.

Answer explanation

Use FDM to allocate 4 channels within the 1 MHz bandwidth. Each channel can occupy 250 kHz, allowing simultaneous transmission of 1 Mbps per channel, effectively utilizing the satellite channel.

The Advanced Mobile Phone System (AMPS) uses two bands. The first band of 824 to 849 MHz is used for sending, and 869 to 894 MHz is used for receiving. Each user has a bandwidth of 30 kHz in each direction. How many people can use their cellular phones simultaneously?

Answer explanation

The total bandwidth for sending is 25 MHz (824-849 MHz) and for receiving is 25 MHz (869-894 MHz). Each user requires 30 kHz. Thus, 25,000 kHz / 30 kHz = 833.33. Therefore, 833 users can use their phones simultaneously.

In synchronous TDM, the data rate of the link is n times faster, and the unit duration is n times shorter. In Figure 6.13, the data rate for each input connection is 3 kbps. If 1 bit at a time is multiplexed (a unit is 1 bit), what is the duration of (a) each input slot, (b) each output slot, and (c) each frame?

Answer explanation

Each input slot duration is 1/3 ms (since 3 kbps = 1/3 ms per bit). Each output slot is also 1/3 ms. The frame duration, with n inputs (n=3), is 3*(1/3 ms) = 1 ms.

Find (a) the input bit duration, (b) the output bit duration, (c) the output bit rate, and (d) the output frame rate.

Answer explanation

(a) Input bit duration is the time taken for one bit to be transmitted. (b) Output bit duration is the time for one output bit. (c) Output bit rate is the number of bits transmitted per second. (d) Output frame rate is the number of frames per second.

Four 1-kbps connections are multiplexed together. A unit is 1 bit. Find (a) the duration of 1 bit before multiplexing, (b) the transmission rate of the link, (c) the duration of a timeslot, and (d) the duration of a frame.

Answer explanation

(a) Duration of 1 bit = 1 ms; (b) Transmission rate = 4 kbps; (c) Duration of a timeslot = 1 ms; (d) Duration of a frame = 4 ms. Each connection transmits 1 bit in 1 ms, and all 4 bits are sent in 4 ms.

If each channel sends 100 bytes/s and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the duration of a frame, the frame rate, and the bitrate for the link.

Answer explanation

Each channel sends 100 bytes/s, multiplexing 1 byte/channel means 100 channels. Frame size = 100 bytes. Duration = 1s, Frame rate = 100 frames/s, Bitrate = 10000 bits/s (100 bytes/s * 8 bits/byte).

A multiplexer combines four 100-kbps channels using a time slot of 2 bits. Show the output with four arbitrary inputs. What is the frame rate? What is the frame duration? What is the bit rate? What is the bit duration?

Answer explanation

The multiplexer outputs 8 bits per frame (2 bits x 4 channels). Frame rate = 100 kbps / 8 b = 12.5 kframes/s. Frame duration = 1/12.5 k = 80 µs. Bit rate = 100 kbps. Bit duration = 1/100 kbps = 10 µs.

We have four sources, each creating 250 characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (a) the data rate of each source, (b) the duration of each character in each source, (c) the frame rate, (d) the duration of each frame, (e) the number of bits in each frame, and (f) the data rate of the link.

Answer explanation

(a) Data rate = 250 chars/s. (b) Duration = 1/250 s. (c) Frame rate = 250 Hz. (d) Frame duration = 1/250 s. (e) Frame size = 250 chars + 1 bit = 2001 bits. (f) Link data rate = 1000 bits/s.