Analog Transmission Example 1 What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)? Solution The sampling rate must be twice the highest frequency in the signal: Sampling rate = 2 x (11,000) = 22,000 samples/s

Sample Rate = 2 X Highest Frequency Used Sampling Rate According to Nyquist theorem sampling rate must be at least 2 times the highest frequency in the signal. Reason Perfect reconstruction is possible if the signal is sampled at this rate. Sample Rate = 2 X Highest Frequency Used

Analog Transmission Example 2 A signal is sampled. Each sample requires at least 11 levels of precision (-5, -4, …, -1, 0, +1, …, +5) How many bits should be sent for each sample? Solution We need 4 bits. A 3-bit value can represent 23 = 8 levels (000 to 111), which is not enough. A 4-bit value can carry up to 24 = 16 values. Note: A signal with L levels actually can carry log2L bits per level

Analog Transmission Example 3 We want to digitize the human voice [range: up to 4000 Hz]. What is the bit rate, assuming 8 bits per sample? Solution The human voice normally contains frequencies from 0 to 4000 Hz. Sampling rate = 4000 x 2 = 8000 samples/s Bit rate = sampling rate x number of bits per sample = 8000 x 8 = 64,000 bps = 64 Kbps

Analog Transmission Example 4 An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate Solution Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 = 4000 bps

Analog Transmission Example 5 The bit rate of a signal is 3000 bits per second. If each signal unit carries 6 bits, what is the baud rate? Solution Baud rate = 3000 / 6 = 500 baud/s

Amplitude Shift Keying (ASK) Bandwidth is proportional to the signal rate (baud rate). ASK is normally implemented using only 2 levels. When each symbol is binary it carries just one bit, so baud and bit rate are equal.

Analog Transmission Example 6 Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half-duplex (i.e. in one direction at a time). Solution In ASK the baud rate and bit rate are the same. The baud rate is therefore 2000 per second. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz.

Analog Transmission Example 7 Given a bandwidth of 5000 Hz for an ASK signal with 2 signal levels, what are the baud rate and bit rate? Solution In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000 per second. But because the baud rate and the bit rate are also the same for ASK with 2 signal levels, the bit rate is 5000 bps.

Analog Transmission Example 7 Find the minimum bandwidth for an FSK signal using 2 signal levels transmitting at 2000 bps. Transmission is in half-duplex mode, and the carriers are separated by 3000 Hz. Solution For FSK BW = baud rate + fc1 - fc0 BW = bit rate + fc1 - fc0 = 2000 + 3000 = 5000 Hz fc1 = higher frequency signal level fc0 = lower frequency signal level The difference is the bandwidth (3000 Hz)

Analog Transmission Example 8 Find the maximum bit rates for an FSK signal with 2 signal levels if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode (i.e. simultaneous, two-way). Solution Because the transmission is full duplex, only 6000 Hz is allocated for each direction. BW = baud rate + fc1 - fc0 Baud rate = BW - (fc1 - fc0 ) = 6000 - 2000 = 4000 But because the baud rate is the same as the bit rate, the bit rate is 4000 bps.

Analog Transmission Example 9 Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode. Solution For 4-PSK the baud rate is one half of the bit rate. Therefore the baud rate is 1000 per second. A PSK signal requires a bandwidth equal to its baud rate. Therefore the bandwidth is 1000 Hz.

Analog Transmission Example 9 :4PSK method

Analog Transmission Example 10 A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate? Solution The constellation indicates 8-PSK with the points 45 degrees apart. Since 23 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is 4800 / 3 = 1600 baud per second

Analog Transmission Example 10: 8 PSK Method

Analog Transmission Example 11 Compute the bit rate for a 1000-baud 16-QAM signal. Solution A 16-QAM signal has 4 bits per signal unit since log216 = 4. Thus, (1000)(4) = 4000 bps

Multiplexing Example 1 Solution 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? Solution For five channels, we need at least four guard bands. This means that the required bandwidth is at least 5 x 100 + 4 x 10 = 540 KHz,

Multiplexing Example 1 Five channels of 100 KHz Guard Band of10 KHz

Multiplexing Example 2 Solution The Advanced Mobile Phone System (AMPS) uses two bands. The first band, 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. The 3-KHz voice is modulated using FM, creating 30 KHz of modulated signal. How many people can use their cellular phones simultaneously? Solution Each band is 25 MHz. If we divide 25 MHz into 30 KHz, we get 833.33. In reality, the band is divided into 832 channels.

Multiplexing Example 3 Solution Four 1-Kbps connections are multiplexed together. A unit is 1 bit. Find (1) the duration of 1 bit before multiplexing, (2) the transmission rate of the link, (3) the duration of a time slot, and (4) the duration of a frame? Solution The duration of 1 bit is 1/1 Kbps, or 0.001 s (1 ms). 2. The rate of the link is 4 Kbps. 3. The duration of each time slot 1/4 ms or 250 ms. 4. The duration of a frame 1 ms.

Multiplexing Example 3 – some points Bit duration is inverted bit rate = 1/1kbps = 1ms Each frame carries 4 time slots so the duration of each output time slot is 1/4th of the input time slot = 1/4 ms or 250 ms. Each frame carries 4 time slots so duration is 4 X 1/4 ms or 1ms . The duration of the frame is same as the duration of the input unit. Frame

Multiplexing Example 4 Solution Four channels are multiplexed using TDM. 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 bit rate for the link. Solution

Multiplexing Example 5 Solution 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? Solution

Multiplexing Example 6 Solution We have four sources, each creating 250 8-bit characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (1) the data rate of each source, (2) the duration of each character in each source, (3) the frame rate, (4) the duration of each frame, (5) the number of bits in each frame, and (6) the data rate of the link. Solution See next slide.

Solution (continued) 1. The data rate of each source is 2000 bps = 2 Kbps. 2. The duration of a character is 1/250 s, or 4 ms. 3. The link needs to send 250 frames per second. 4. The duration of each frame is 1/250 s, or 4 ms. 5. Each frame is 4 x 8 + 1 = 33 bits. 6. The data rate of the link is 250 x 33, or 8250 bps.

Multiplexing Example 7 Solution Two channels, one with a bit rate of 100 Kbps and another with a bit rate of 200 Kbps, are to be multiplexed. How this can be achieved? What is the frame rate? What is the frame duration? What is the bit rate of the link? Solution We can allocate one slot to the first channel and two slots to the second channel. Each frame carries 3 bits. The frame rate is 100,000 frames per second because it carries 1 bit from the first channel. The frame duration is 1/100,000 s, or 10 ms. The bit rate is 100,000 frames/s x 3 bits/frame, or 300 Kbps.