Computer Network Principle calculation questions

1. Set the channel bandwidth to 3400Hz and adopt PCM encoding. The sampling period is 125b/S. If the sample size is 128, the data rate of the channel is ()?

Resolution:
The sampling period is 125b/s, so it is 8000Hz, that is, F = 1/T = 1/0. 000125 = 8000Hz, 128 quantization levels, requires 7-bit encoding (that is, the 7th Power of 2 ).

R (data transmission rate) = 1/T * log2n = 8000*7 = 56kb/s ---> B (modulation rate) = 1/T baud

PS. Because the sampling theorem is defined, the sampling frequency is twice the bandwidth to complete sampling. This bandwidth is actually the bandwidth of the voice frequency range, 300Hz-3400Hz. However, the sampling frequency is 8 KB. Therefore, this bandwidth is not required.

2. When a channel with a bandwidth of 3 kHz without noise transmits binary signals, the maximum data transmission rate is ___. A channel with a bandwidth of 3 kHz and a signal-to-noise ratio of 30 dB can reach the maximum data transmission rate of ___. The above results indicate _____. in order to ensure the transmission quality, the bandwidth required to reach the data transmission rate of 3khbps is ____, And the binary signal is transmitted on an infinite bandwidth-free channel, when the signal bandwidth is 3 kHz, the maximum data rate is _ kbps.

Resolution:
(1) According to the nequest first theorem, the data transmission rate that can be achieved when binary signals are transmitted through an ideal low-pass channel is 2B (bandwidth ).
(2) A channel with a bandwidth of 3 kHz and a signal-to-noise ratio of 30 dB can achieve a maximum data transmission rate of 3 kHz * log2 (1 + 1000) = 29.9 kbit/s ≈ 30 kbps
(3) Shannon formula is used for noisy channels.
(4) According to the nequest first theorem, if the digital signal data rate is W and the transmission system bandwidth is 2 w, satisfactory services can be provided.
(5) Transmit binary signals over an infinite bandwidth-free channel. When the signal bandwidth is 3 kHz, the maximum data transmission rate is 6 kbps.

PS. The data rate of the signal has a direct relationship with the bandwidth. The higher the data rate of the signal, the wider the effective bandwidth. That is to say, the wider the bandwidth provided by the transmission system, the higher the signal data rate that the system can transmit.

If the data is set to W, the bandwidth of the transmission system is usually 2 W, and satisfactory communication service is provided. the bandwidth required for 3 kbps data transmission is 2*3 = 6Hz. The binary signal is a discrete pulse. Each pulse can represent a binary bit. The time width is the same. The time width T = 1/f, the reciprocal of the time is the data transmission rate (1/T). According to the nequest theorem, when the signal bandwidth is 3 kHz, the maximum data transmission rate C = 2 (1/T) can be reached) = 6 kbps

3. Three analog signals are provided with a bandwidth of 2 kHz. 4 kHz, 2 kHz, 8-way digital signal, data rate is BPS, when the TDM method is used to reuse it on a communication line, assuming that the multiplexing is digital transmission, if the PCM method is used to quantify the level-16 of the analog signal, what is the minimum communication capability required for the multiplexing line?

Resolution:
The PCM mode is used to convert three analog signals into digital signals. The sampling frequency is 4 kHz, 8 kHz, and 4 kHz respectively. The PCM method is used to quantify the level of 16 analog signals, the required data rates are 16 kbps and 32 Kbps and 16 kbps respectively.
For 8-channel digital signals, 8x7200 = 57.6 kbps, the answer is 128 Kbps.

4. If you want to transmit a T1 carrier of 1.544mbps over a 50 Kbps channel using two physical states, how much signal-to-noise ratio of the channel is required?

Resolution:
The Shannon formula is used to calculate the signal-to-noise ratio:
C = H * log2 (1 + S/N)
Relationship between modulation rate and data transmission rate:

R = B * log2n (BPS) ---> B = r/log2n (baud)
1) B = r/log2n = 50 K/log22 = 50 K (baud)
2) B = 2 * h ---> H = B/2 = 50 K/2 = 25 K (HZ)
3) C = H * log2 (1 + S/N), c = 1.544 Mbps = 1544 kbps
S/N = 2C/H-1 = 21544 K/25k-1 = 261.76-1 = 2.3058430092137*1018
Then convert the above S/N to the DB format:

Conversion formula:

10lg (S/N) = 10lg (2.3058430092137*1018)

= 10x18.363

= 183.63 DB
10lg (S/N) = 10lg (261.76-1) = 10*18.6 = 186 (db) (too noisy !)

5. A modem uses both amplitude shift keying and phase shift keying, and uses four phases: 0, Wu/2, Wu and 3/2 Wu. Each phase has two different amplitude values, what is the data rate when the baud rate is 1200?

R = B * log2n

= 1200 * log28

= 3600bps

6. 8 phases and each phase has two amplitude Pam modulation methods. How much data transmission rate can be achieved at the signal transmission rate of 1200baud?

R = B * log2n

= 1200 * log28

= 3600bps

7. transmit digital signals over an 8 kHz non-noise channel with two different bandwidths for each phase. To achieve a data rate of 64 Kbps, how many different phases are required for the PAM modulation method?
Known: c = 64 Kbps H = 8 kbps

C = 2 * H * log2n
64 = 2*8 * log2n
Log2n = 64/16 = 4
N = 4*2 = 8
8. If the channel bandwidth is 3 kHz and the signal-to-noise ratio is 30 dB, the number of BITs that can be sent per second will not exceed the number of BPS?
C = H * log2 (1 + S/N)
C = 3 K * log2 (1 + 1030/10)
C = 30 kbps
9. The bandwidth is 4 kHz. If eight different physical states represent data and the signal-to-noise ratio is 30db, the maximum data transmission rate is calculated based on the neeness criterion and Shannon theorem.
Native principle: c = 2hlog2n
C = 2*4 * log28
C = 24 kbps
Shannon theorem: c = H * log2 (1 + S/N)
C = 4 * log2 (1 + 1030/10)
C = 40 kbps

10. Use a 12 MHz sampling frequency to sample the signal. If the quantization level is 4, calculate the data transmission rate and the required channel bandwidth in the non-noise channel. (Write the computing process)

12 MHz is the sampling frequency

Data transmission rate = sampling frequency * log2 (4) = 2 * sampling frequency = 24 Mbps

The bandwidth mentioned here is the bandwidth of the analog channel, that is, the bandwidth of the sampled signal.

According to the sampling law: the bandwidth of the sampled signal = sampling frequency/2 = 6 MHz

To accommodate this signal, the channel bandwidth is 6 MHz.

11. The transmission rate of the modem is 4800bps, and the time required to transmit 2400 Chinese characters is calculated based on the asynchronous transmission mode of one start bit, one stop bit, and one parity bit. (Write the computing process)

Because a Chinese character occupies two bytes, you must:

2400*[(1 + 1 + 8 + 1) * 2]/4800bps

= 11 S

12. Station A and Station B are located at both ends of the 2Km long baseband bus LAN, Station C is located between Station A and Station B, and the data transmission rate is 10 Mbps, the signal transmission speed is 200 Mb/s. After Station B receives a frame of data from Station A, it takes 80 μs to obtain the data frame length; if both sites A and C send a frame of data to each other at the same time, the two sites will find a conflict after 4 μs, and the distance between the two sites A and C will be obtained. (Write the computing process)

Total time between sending and receiving of two sites = data transmission latency + signal transmission latency

Signal Propagation Rate: 2000/200 Mb/s = 10 Mb/s

Transmission Time: X/10 Mbps + 10 m/μs = 80 μs