1. The bandwidth is 4 kHz., If
8 differentThe physical status indicates data, and the signal-to-noise ratio is 30db. therefore, the maximum data transmission rate is calculated based on the neeness criterion and Shannon theorem. ① c = 2 F log2n = 2*4 K * log28 = 24 kbps ② dB calculation is: 10lgs/N, that is, this article entitled: 10lgs/n = 30 then: s/N = 103c = f log2 (1 + s/n) = 4 K * log21001 = 40 kbps
2. For a 6 MHz channel, if four different States are used to represent data, what is the maximum data transmission rate of the channel without considering hot noise?
A:
The formula C = 2hlog2n = 2*6 m * log24 = 24 Mbps indicates that the maximum data transmission rate of the channel is 24 Mbps.3. A
The modem uses both the amplitude shift key and phaseShift key, using 0, Wu/2, Wu and 3/2 Wu four phases, each phase has two different amplitude, q: What is the data rate when the baud rate is 1200? A: log28 * 1200 = 3600b/s
4. If the channel bandwidth is 3 kHz and the signal-to-noise ratio is 30 dB, how many BITs can be sent per second?
A:
Shannon formula with hot noise:
C = hlog2 (1 + s/n) = 3 K * log2 (1 + 1030/10) <3 K * log2210 = 30 kbps,
Therefore, the number of BITs that can be sent per second cannot exceed 30 kbps.5.
8 phasesAnd the first phase has two ways of PAM modulation.
1200baud Signal TransmissionData that can be achieved at a specific rate
Transmission RateWhat is it? My answer is: S = B · log2n = Export xlog2 8 = 3600bps
6. The Pam modulation method with two amplitude values for each phase is used to transmit digital signals over an 8 kHz non-noise channel. To achieve a data rate of 64 Kbps, how many different phases are required?
A:
Obtained from the nequest formula of the non-noise channel: c = 2hlog2n:
N = 2C/2 h = 264 K/(2*8 K) = 24 = 16, number of BITs = 16/2 = 8
That is, at least eight different phases are required.
7. The data rate is ipvbps, using non-validation, one-bit stop bit asynchronous transmission (that is, the Group Synchronization P22-23), ask how many Chinese characters can be transmitted within 1 minute (double byte )? [If it is changed to an ascii or ebcdic code?]
A:
The maximum number of Chinese characters that can be transmitted within 1 minute is:
1200*60/[2*(1 + 8 + 1)] = 7200/20 = 3600 (s)
8. Draw a 1011001 waveform with standard Manchester encoding and differential Manchester encoding respectively.
9. What is the baud rate of a 10 Mbps LAN encoded in Manchester?
A:
The Manchester-encoded baud rate is twice the original code baud rate, that is, 20 Mbps.
10 Brief description of characters in asynchronous transmission modeAnd describes the functions of each part. A: One start bit, one-to-two stop bits, 5-8 data bits, and one-bit parity check bits (Optional). Group Synchronization relies on the Start bit and stop bits to synchronize characters.
11. The source transmits data in bytes (8 bits). If the data rate is B (BPS), the effective data transmission rate is calculated for the following two cases:
(1) asynchronous serial transmission, with no checkpoint and one stop bit;
(2) Synchronous Serial Transmission, each frame contains 48-bit control bit and 4096-Bit Data bit.
Answer: (1) valid data transmission rate: 8/(1 + 8 + 1) = 8/10
(2) valid data transmission rate: 4096/(48 + 4096) = 4096/414412. At the same data rate,
Transmission of large volumes of data using asynchronous and synchronous protocolsData: What is the transmission efficiency ratio of the two? Answer: The effective data transmission efficiency of the asynchronous protocol is 8-Bit Data bit/10-bit total data bit = 8/10 The effective data transmission efficiency of the synchronization protocol is effective data bit (up to several thousand bits)/(valid data bit + end of the frame header and other related control bits, the value is very small) = 1/1 (approx)Percentage of effective data transmission efficiency of asynchronous and synchronous protocols = (8/10)/1 * 100% = 80%
13. For voice signals with a bandwidth of 4 kHz, the PCM method with a quantization level of 128 is used for encoding. How much transmission rate can be used to transmit the binary bit produced? P23-24)
Answer: Because H = 4 kHz, n = 128
So S = 2hlog2n = 2*4klog2128 = 56 K (BPS)14. Briefly describe the implementation principles of OFDM and TDM. Answer: Principle of OFDM multiplexing: when the available bandwidth of a physical channel exceeds the bandwidth required by a single original signal, the total bandwidth of the physical channel can be divided into several sub-channels with the same (or slightly the same) bandwidth as the transmission of a single signal. Each sub-channel transmits one channel of signal. TDM time division multiplexing principle: If the bit transmission rate that the media can reach exceeds the data transmission rate required for data transmission, TDM can be used for time division multiplexing, it also divides a physical channel into several time slices and distributes them to multiple signals in turn. Each time slice is occupied by a multiplexing signal.15. Calculate the encoding efficiency and overhead rate of the T1 and E1 carriers respectively. T1 carrier coding efficiency = 7*24/(8*24 + 1) = 168/193 Overhead rate = (1*24 + 1)/193 = 25/193E1 carrier encoding efficiency = 8*30/(8 + 8 + 8*30) = 240/256 Overhead rate = (8 + 8)/256 = 16/256
16. 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?
Answer: 1) B = s/log2n = 50 K/log22 = 50 K (baud)
2) H = B/2 = 50 K/2 = 25 K (HZ)
3) C = hlog2 (1 +S/N), C = 1.544 Mbps = 1544 kbps
S/N= 2C/H-1 = 21544 K/25k-1 = 261.76-1
10lg (S/N) = 10lg (261.76-1) = 10*18.6 = 186 (db) (too noisy !)
19. Compare the working principles of the ARQ and FEC methods to illustrate their differences. A: How ARQ automatically resends a request: when the receiving end detects an error, it tries to notify the sender to resend the request until the correct code word is received. Working principle of FEC forward correction: the receiving end can not only detect errors, but also determine the location where the binary code element has an error to correct it.
20. It is known that the generated polynomial is X4 + X3 + X2 + 1, and the CRC code of information bit 1010101 is obtained.
Solution: 1) 10101010000/11101 = 1110101 ...... 1001
2) CRC code: 1010101100121. If
1000 kilometers apartOf
The two locations must be 3 KData can be transmitted by cable at a data rate of 48 KB/s or by satellite channels at a data rate of 50 kb/s, which method does it take a short time to send data from the sender to the receiver to receive all the data? The time from which all data is sent to the receiver = propagation time + transmission time through the ground cable: t = 1000/5000000 + 3/4. 8 = 0.6252s via satellite: t = 0.27 + 3/50 = 0.33s. Therefore, the satellite transmission time is short. 22. A simple telephone system consists of two terminal bureaus and one long-distance Bureau,
Each terminal uses a 1 MHz full-duplex trunk cable.Connect to the long-distance Bureau. The average number of calls per phone is 4 in 8 hours, and each call lasts for 6 minutes on average. among them, 10% of the calls are long-distance calls (that is, after the long-distance Bureau). If the bandwidth of each voice is 4 kHz, ask how many telephone phones can be supported by each terminal Bureau: multiplexing by frequency. Each terminal can support a maximum of 1 MHz/4 kHz = 250 calls at the same time, and each terminal board can support a maximum of 8 × 60 calls) /(4 × 6 × 10%) = 200 telephony. Therefore, each terminal can support a maximum of 250 × 200 = 50000 telephony.
