CS 898T Mobile and Wireless Networks - Homework 2 - Name:_________________ Assigned: Monday, February 23 Due: Wednesday, March 3 (in class) Please show all work on a separate sheet attached to this sheet. For each question write key points. (50 points - 10 points for each question) 1. What is direct sequence spread spectrum? List its advantages and disadvantages. What is frequency hopping spread spectrum? List its advantages and disadvantages. Give an example system which uses DSSS or FHSS respectively. Ans: 1) Direct Sequence Spread Spectrum (DSSS) is a transmission technology used in wireless transmissions where a data signal at the sending station is combined with a higher data rate bit sequence, or chipping sequence (code), that divides the user data according to a spreading factor (ratio). The chipping code is a redundant bit pattern for each bit that is transmitted, which increases the signal's resistance to interference. If one or more bits in the pattern are damaged during transmission, the original data can be recovered due to the redundancy of the transmission. 2) Advantages of DSSS: o More restance to fading and multi-path effects o More efficient use of channel bandwidth o Short latency time o Constant processing gain - a better signal to noise ratio o Quick Lock-In as radio synchronize o No recynchronization with other radio necessary o Long outdoor range o Greater overall data throuput Disadvantage of DSSS: Complex design 3) Frequency Hopping Spread Spectrum (FHSS) is a transmission technology used in wireless transmissions where the data signal is modulated with a narrowband carrier signal that "hops" in a random but predictable sequence from frequency to frequency as a function of time over a wide band of frequencies. The signal energy is spread in time domain rather than chopping each bit into small pieces in the frequency domain. This technique reduces interference because a signal from a narrowband system will only affect the spread spectrum signal if both are transmitting at the same frequency at the same time. If synchronized properly, a single logical channel is maintained. 4) Advantage of FHSS: o Fundamentally much simpler to implement o Better range, due to lower receiver sensitivity o Good rejection of in band interference o Good performance in multipath environments o No "near/far" problems Disadvanges Of FHSS: o Long latency time o Slow Lock-In, must search a channel o No processing gain o Must recynchronization with other after every hop o short outdoor range o Lower overall data throuput 5) Example for DSSS: CDMA, Wireless LAN Example for FHSS: GSM, Bluetooth 2. What is the cellular system for mobile communications? List its advantages and disadvantages. Describe two types of the frequency assignment. Ans: 1) Cellular systems for mobile communications implement SDM. Each transmitter, typically called a base station, covers a certain area, a cell. 2) Advantages of cellular systems: o Higher capacity o Less transmission power o Local interference only o Robustness Disadvantages of cellular systems: o Infrastructure needed o Handover needed o Frequency planning 3) Two types of frequency assignment are fixed channel allocation (FCA) and dynamic channel allocation (DCA). 3. Describe the problems when CSMA/CD is applied to wireless networks. Describe how MACA solve the above problems. Ans: 1) Two problems occur: o Hidden and exposed terminals - Carrier sensing may fail to detect another terminal or dectect a terminal outside the interference range. o Near and far terminals - The local signal might drown out the remote transmission. 2) When a station is ready for transmission, it sends a request to send (RTS) frame to the receiver and waits to receive a clear to send (CTS) frame from the receiver. As a result, all stations within the range will refrain from transmitting a data frame. Once CTS is received, the sender can send packets. In this way, the CTS frame can be heard by the hidden terminals and the medium for future use by other sending terminal is reserved. The exposed terminal won't react to RTS and doesn't receive CTS because the exposed terminal is not the receiver. The near and far terminals could be solved in the similar way. 4. What is a good code for CDMA? Redo the CDMA example of section 3.5, but now add random 'noise' to the transmitted signal (-2, 0, 0, -2, +2, 0). Add, for example, (1, -1, 0, 1, 0, -1). In this case, what can the receiver detect for sender A and B repectively? Now include the near/far problem. How does this complicate the situation? What would be possible countermeasures? Ans: 1) A good code for CDMA should have a good autocorrelation and should be orthogonal to other codes. 2) The transmitted signal in this simplified example is (-2,0,0,-2,+2,0) + (1,-1,0,1,0,-1) = (-1,-1,0,-1,+2,-1). The receiver will calculate for A: (-1,-1,0,-1,+2,-1) * (-1,+1,-1,-1,+1,+1) = 1-1+0+1+2-1 = 2. For B the result is: (-1,-1,0,-1,+2,-1) * (+1,+1,-1,+1,-1,+1) = -1-1+0-1-2-1 = -6. The receiver can decide more easily for the binary 0 in case of B compared to the binary 1 in case of A. Noise can obviously affect the signal. But still the receiver can distinguish between the two signals - our simple example uses perfectly synchronised signals (the spread symbols are in phase). Adding the near/far problem to our simplified example does not change much: still the receiver can detect the signal - unless noise becomes too strong compared to the signal. Simply multiply the noise and B's signal by, let us say, 20. The transmitted signal is then: As+20*Bs+20*noise = (-1,+1,-1,-1,+1,+1) + (-20,-20,+20,-20,+20,-20) + (+20,-20,0,+20,0,-20) = (-1,-39,19,-1,+21,-39). The receiver then receives for A: (-1,-39,19,-1,+21,-39) * (-1,+1,-1,-1,+1,+1) = 1-39-19+1+21-39 = -74, and for B: (-1,-39,19,-1,+21,-39) * (+1,+1,-1,+1,-1,+1) = -1-39-19-1-21-39 = -120. Both results are negative, the receiver can not reconstruct the original data of A, but that of B. If A is a near terminal, the reciver can reconstruct A but not that of B. This example should just give a rough feeling what the problems are. For our simple problem here we don't see all the effects: the spreading codes are much too short, everything is synchronised. 5. Name the main elements of the GSM system architecture and describe their functions. What are the advantages of specifying not only the radio interface but also all internal interfaces of the GSM system? Ans: See figure 4.4. Specifying all (or at least many) internal interfaces allows for a larger variety of vendors. As long as vendors stay with the standardised interfaces equipment of different vendors can be combined and network operators are not completely dependent from one manufacturer. However, reality often looks different and network operators often use only equipment from one or two vendor(s).