无线射频2016期末试题

I、Fill in the blank(Each blank 1 mark, in total 4 marks)

1.The work principle of an inductive coupled RF module is similar to the work principle of a

transformer .

2.When a RFID middleware works in the interactive mode, readers see each command from

host as a single task, execute by order and return result immediately.

3.The EIRP limit of RFID readers is 36dBm according to the FCC in USA.

4.In even parity-check, there is only one check bit, not matter how many bits in the source code. The

check bit make the number of “1” in the coded message to be even .

II、True (T) or False (F) Questions（1 mark for each question, in total 4 marks）

1.The work distance of the HF reader can be larger than 1m. ( F )

2.The limitation effect of link budget for a RFID system is the backward link energy. ( F )

3.The throughput of the S-ALOHA algorithm is always the twice of the pure ALOHA algorithm. ( F )

4.OOK can always perform very well for digital modulation on RFID reader signal. ( F )

III、Multiple-choice, only ONE answer.（Each blank 1 mark, in total 4 marks）

1.( B ) is NOT the feature of RFID.

A、It can permanently store a certain amount of data

B、It works on the contact condition

C、It has simple logical processing ability

D、It has fast reading speed

2.Among the following options, ( D ) is NOT the function of RF tags.

A、Store goods-related information.

B、Absorb energy from electromagnetic field generated by the reader.

C、Collision Concessions.

D、Connecting peripherals.

3. In CLC (Cyclic Redundancy Check)，the polynomial function corresponding to 11001 is ( C )

A、x5+x4+ x

B、x6+x4+x2+ 1

C、x4+x3 + 1

D、x5+x3+ 1

4. ( C ) is NOT a direct impact factor on the effective communication range of a RFID system

A. Frequency band

B. transmission power

C. Upper application

D. Coupling way of the antenna

IV、Short Answer Questions（in total 13 marks）

Question 1. (4 marks)

Please analyse the energy consumption processes of the backward link of RFID systems based on the following figure.

Energy consumption process of the forward/backward link

Answer: For the backward link, tag’s receive energy is -10dBm, but actual reflection energy is about

1/3, i.e., -15dBm (with 5dB attenuation). We can assume that the required energy threshold for a

reader to recognize tags is -75dBm. Based on the transmit distance of 3m i n the forward link, reader’s receiver can obtains energy -55dBm, which is 20dB more than the required energy.

Question 2 (5 marks)

What is the maximum channel utilization ratio of S-ALOHA, please give the brief deduction process of S-ALOHA’s channel u tilization formula.

Answer:

The maximum utilization ratio is 1/e = 0.368

Deduction process:

1) Suppose frames come following Poisson distribution, and the mean of the distribution is G. Then the probability of k frames is Pr[k]=(G^k e^(-G))/k!

2) Therefore, the probability of there being zero transmission-attempts between t and t+T is

P_0=e^(-G).

3) The throughput can be calculated as the rate of transmission-attempts multiplied by the probability of success, and it can be concluded that the throughput is S=Ge^(-G).

4) The maximum throughput is 1/e frames per frame-time (reached when G=1), which is approximately 0.368 frames per frame-time. This means that, in S-ALOHA, about 36.8% of the time is used for successful transmissions.

Question 3 (4 marks)

Let there be n=4 transponders in the interrogation zone of the reader. The transponders possess unique serial numbers in the range 00H~ffH. Please complete the following procedures of binary search algorithm (use the