T305 T325 B BLOCK 3 4 PART III T325 Summary Session 11 Block III Part 3 Access & Modulation [Type Dr. Saatchi, your address] Seyed Mohsen [Type your phone number] [Type your e-mail address] Prepared by: Dr. Saatchi, Seyed Mohsen T325
Block III Part 3 Access & Modulation Q1: TDMA stands for Time Division Multiple Access Q2: CDMA stands for Code Division Multiple Access Q3: The key Access technology in GSM system is the Time division multiple access (TDMA) Q4: Operator s band of frequencies for uplink and downlink is divided into a set of channels for use within a cell Q5: Operator s band of frequencies for uplink and downlink is divided into a set of channels and each channel is shared by a few users simultaneously using timedivision multiplexing Q6: The sharing of frequency channels by several users in GSM means that TDMA is combined with frequency division multiple access (FDMA) Prepared by: Dr. Saatchi, Seyed Mohsen 2
Q7: In the uplink and downlink directions, communication on each of the frequency channels is divided (in GSM) into frames of 4.615 ms duration. Q8: in previous figure, The frames are further subdivided into slots, with eight time slots per frame. Prepared by: Dr. Saatchi, Seyed Mohsen 3
Q9: One frame duration 4.615 ms Q10: each time slot allocated to user within frequency channels, and it is 200Khz Q11: Slots are allocated to individual users Q12: In TDMA, users DO NOT literally have simultaneous access Q13: Why in TDMA, users DO NOT literally have simultaneous access? Because the users rapidly take turns to have exclusive use of the particular frequency channel they have been allocated to in a cell. These turns come round every 4.615ms (the duration of the frame), and the turn lasts for approximately half a millisecond (the duration of the slot). Q14: 1 in 13 frames is reserved entirely for control data. Q15: Time slot & GSM channel, Should not to be confused with a frequency channel Q16: Each frequency channel carries eight GSM channels Q17: A frame lasts 4.615ms. How many frames are there per second? Number of frames per second is = 1 / 4.615 = 216.7 Q18: A slot in a frame carries 114bits Prepared by: Dr. Saatchi, Seyed Mohsen 4
Q19: A frame lasts 4.615ms. How many frames are there per second? A. A slot in a frame carries 114bits. For a single user, what is the data rate in bits per second of their GSM channel if no frames are robbed for control data? B. 1 in 13 frames is reserved for control data. Hence, what is the actual data rate for a GSM channel? Number of frames per second is = 1 / 4.615 = 216.7 A) Hence, per GSM channel the rate is: 216.7 x 114 bit/s = 24 700 bit/s in one direction, the same rate will be for other direction = 24 700 bit/s B) One frame in 13 is reserved for control data; the previous answer needs to be reduced by a factor of 12/13 to arrive at the actual data rate per GSM channel: (24 700 x 12) / 13 = 22.8 Kbit/s Q20: In GPRS, higher data rates are achieved by allocating a user more than one GSM channel. Q21: A single GSM channel has a capacity of 22.8 kbit/s Q22: There are often constraints that limit the number of GSM channels that can be allocated to one user, such as using EDGE (2.75 G technology) increases the data rate further by using multilevel signaling Q23: Multilevel signaling in EDGE enables 3 bits of data to be transmitted in the time it takes to transmit 1 bit in GSM and GPRS. Prepared by: Dr. Saatchi, Seyed Mohsen 5
Q24: If the signaling medium could adopt four different states, each state could represent a combination of 2 bits of data. (2 2 = 4) Q25: If the medium could adopt 16 states, then each one could represent a combination of 4 bits of data (2 4 = 16) Q26: Each state in a signaling medium is referred to as a symbol, and the duration of a symbol is the symbol period. Q27: The number of symbols per second is the symbol rate. Symbol rate = 1/ symbol period Q28: Using multilevel signalling does not increase the bandwidth requirement of the channel carrying the signal, but increasing the data rate in EDGE. Q29: Using many levels of singal generally increases vulnerability to noise compared with fewer states Prepared by: Dr. Saatchi, Seyed Mohsen 6
Q30: Mention some refinement (enhuncement) of EDGE: Availability of extra coding schemes compared with GSM and GPRS, with different coding rates Coding rate can be more carefully matched to the radio-channel conditions Q31: CDMA was an alternative air interface to TDMA in 2G Q32: In 3G, CDMA is used exclusively Q33: List one of the major advantages & disadvantages it offers by CDMA: Advantages: Is a less definite limit on capacity. CDMA enables adjacent cells to operate on the same frequency band which offers many advantages, including soft handover from cell to cell, whereby an existing connection does not need to be broken before a new one is established with another cell. disadvantages: CDMA is complex and requires an amount of processing power o This is one reason why TDMA was adopted for GSM Q34: List two main sorts of code in CDMA: Channelisation codes Scrambling codes Q35: Suppose the base station transmits simultaneously to two users, A and B. If the data is sent unmodified and simultaneously, As it is shown in figure, what will be the result in ordinary result? Prepared by: Dr. Saatchi, Seyed Mohsen 7
Q36: Decoding the superposed signal (provided there are two sets of data): If result = 2 A = 1 and B = 1 If result = 0 A = 0 and B = 0 BUT if result = 1 (A = 1 and B = 0) or (A = 0 and B = 1) Q37: The idea with CDMA is to encode the 1s and 0s in such a way that the receiver can always undo the superposition of coded data and get back to the original data streams Q38: Coding creates separate channels of communication It makes the data streams Orthogonal Q39: What does it mean by term Orthogonality? When multiple signals are conveyed through a shared medium, the signals should not interfere with each other in a way that prevents data recovery. Q40: TDMA, CDMA and OFDMA enable signals to remain orthogonal (even when they are sharing the same transmission medium) Prepared by: Dr. Saatchi, Seyed Mohsen 8
Q41: The CDMA encoding consists of replacing the 1s and 0s of a user s binary data stream by patterns of shorter chips prior to transmission. Q42: The chips, like binary data itself, can exist in either of two states: represented as 1 and -1, rather than 1 and 0. Q43: Try to channelize the binary number 11000 to bipolar form and code chip for A and Bipolar. Data for A in binary form (a) Bipolar form (b) Code chips for A Prepared by: Dr. Saatchi, Seyed Mohsen 9
Q44: Try to channelize the binary number shown in figure: Prepared by: Dr. Saatchi, Seyed Mohsen 10
Q45: The transmission of A s and B s data is accomplished by transmitting a superposition of the two encoded versions Q46: How to reconstruct A s and B s data from the superposed version of their encoded data which arrives at A s and B s receivers? By carrying out a mathematical operation called correlation A s receiver knows that the code used to encode the data intended for A was 1, -1,1, -1. Prepared by: Dr. Saatchi, Seyed Mohsen 11
The superposed set of chips received is 2, -2, 0, 0 To extract A s data, we multiply, in order, each chip in the received signal by the corresponding chip in A s code. Example: the first chip of the signal is 2. The first chip of A s code is 1. We multiply them and store the result; that is, 2 x 1= 2. Prepared by: Dr. Saatchi, Seyed Mohsen 12
The result of correlation will be: You can see that the result of the correlation process has been to recover a bipolar form of A s original data, scaled by a factor of 4. Q47: Replacing bits by chips Spreading the frequency spectrum of the transmitted signal over a wider band of frequencies. Q48: This gives enhanced immunity to certain types of noise, which is known as processing gain. Q49: Processing gain accounts for the robustness of CDMA in the presence of noise. Q50: In downlink in CDMA using Channelization and in Uplink using Scrambling. Q51: In the uplink, the scrambling code distinguishes users Channelisation codes are nevertheless still needed in the uplink Prepared by: Dr. Saatchi, Seyed Mohsen 13
Q52: Suppose we have a user C who has been assigned the code: 1,1,-1, -1 There is no data for C in the encoded stream What happens when the correlation process is carried out using this code on the stream in which there is only data for A and B? In the first period, A s and B s superposed data consisted of the chip sequence 2, -2, 0, 0. Carrying out correlation using C s chip sequence on this interval gives (2 x1) + (-2 x1) + (0 x -1) + (0 x -1) = 0 C has no data in the transmitted chips, and correlation yields no data for C. All the channelization codes in CDMA must be mutually orthogonal. Two codes are said to be orthogonal their correlation result should be ZERO Walsh codes: set of mutually orthogonal codes Q53: In practice, the number of chips used in W-CDMA varies from 4 to 256 (and up to 512 for downlink only). Q54: For the uplink there is a maximum of 256 parallel codes available per user and for the downlink there is a maximum of 512 parallel codes available per cell. Prepared by: Dr. Saatchi, Seyed Mohsen 14
Q55: Explain why shorter codes are associated with higher data rates? They use fewer chips to represent a bit of data, and The chip rate is constant for all codes Q56: Walsh codes can lose their orthogonality if they become unsynchronized Q57: If the codes for B and C were put out of synchronism by an amount equal to the duration of 1 chip. The orthogonality of B and C depends on the synchronization of the codes. Q58: SCH stands for Using synchronisation channels Q59: The figure showing the long code synchronized, two shorter codes and four even shorter codes. Q60: Replacing each bit by several chips increases the bandwidth of the signal carrying the data. Prepared by: Dr. Saatchi, Seyed Mohsen 15
Q61: In UMTS, the chip rate is fixed at 3.84 Mchip/s, In the commonest variant of UMTS, the downlink occupies 5MHz and the uplink occupies 5MHz, As in GSM, data is chunked into frames at the physical level.in UMTS, the frames are 10ms long. there are 100 frames per second, what is the number of chips per frame: number of chips per frame is (3.84x10 6 )/100 = 38 400 chips Q62: List some spreading factors of UMTS Q63: Spreading factor is sometimes expressed in terms of chip rate and data rate as follows: Q64: Channelization codes are also named : orthogonal variable spreading factor (OVSF) codes Prepared by: Dr. Saatchi, Seyed Mohsen 16
Q65: Mention the reason of using scrambling codes in downlink and uplink? Downlink: Adjacent cells use the same frequency in UMTS users can receive signals from more than one Node B. Uplink: channelisation codes are not unique to a given source, the Node Bs cannot reliably determine which user a signal arrives from. Scrambling codes uniquely identify sources of signals in either direction User equipment can distinguish one Node B from another through the Node B s different scrambling codes A Node B can distinguish one piece of user equipment from another through the different scrambling codes allocated to user equipment. Prepared by: Dr. Saatchi, Seyed Mohsen 17
66: Scrambling codes have a pseudo-random character that makes them look like noise known as pseudo-noise codes or pseudo-random codes. Prepared by: Dr. Saatchi, Seyed Mohsen 18
Q67: The figure showing the encoding of A s and B s data, and recovery of A s data, Synchronized streams with walsh codes used Q67: The figure showing the encoding of A s and B s data, and recovery of A s data: Non- synchronized streams with walsh codes used Q68: The figure showing the Gold codes: a set of scrambling codes named after the person who devised them. Prepared by: Dr. Saatchi, Seyed Mohsen 19
Q67:List the difference between scrambling code (Gold Code) and channelization code (Walsh code): Q68: Synchronisation achieves two main outcomes, list those two achiements: Enables receiving devices to work out where the frame boundary is; that is, when frames start Enables a receiving device to work out the scrambling code of the Node B it is receiving a signal from. Q69: Synchronisation takes place using two special synchronisation channels transmitted from the Node B: list those two special synchronization channels: Primary synchronisation channel (P-SCH) Secondary synchronisation channel (S-SCH) Note: Each of these transmits fixed data, using a channelisation code but no scrambling code Prepared by: Dr. Saatchi, Seyed Mohsen 20
Note: The primary synchronisation channel transmits a fixed data word repeatedly once per slot Q70: How the Synchronisation of user equipment to Node B works with primary and secondary synchronization? 1. When the user equipment is switched on, it scans the radio channels in the 3G band, based on a stored list of preferred frequencies. Priority is given to the user equipment s home network; that is, the one with which the user is registered as a customer. 2. Once the user equipment has found a frequency on which transmissions are taking place, it looks for the primary synchronisation channel and uses the fact that the fixed data word is transmitted once per slot to align itself to the slot boundary. 3. The UE then looks for the secondary synchronisation channel. The secondary synchronisation channel transmits a sequence which occupies a whole frame. Unlike the case with the primary synchronisation code, the sequence in the secondary synchronisation channel varies from cell to cell. However, there are only 64 possible sequences, so the user equipment is able to work out which sequence is in use. Q72: List two Achieving high data rates in UMTS: HSDPA(high-speed downlink packet access) HSUPA (high-speed uplink packet access) Q73: OFDMA is based on orthogonal FDM (OFDM). Q74: The essence of OFDM is the simultaneous use of many wireless carrier signals, known as subcarriers, each with a different frequency, and with the frequencies chosen in a way that gives particular properties. Prepared by: Dr. Saatchi, Seyed Mohsen 21
Q75: the way the frequencies are chosen in OFDM differentiates it from ordinary FDM. Q76: Sometimes in OFDM all the subcarriers are allocated to a single data stream. Prepared by: Dr. Saatchi, Seyed Mohsen 22
Q77: Modulation: is the process of modifying a radio wave to enable it to carry data. Prepared by: Dr. Saatchi, Seyed Mohsen 23
Q78: There are benefits in using multilevel signalling, as more data can be conveyed per symbol. Q79: Binary phase-shift keying is sometimes regarded as a special type of amplitude modulation (amplitude changes by a factor of -1) Q80: This way of thinking about amplitude change is useful in connection with quadrature amplitude modulation (QAM) techniques. Prepared by: Dr. Saatchi, Seyed Mohsen 24
Q81: Two sine waves different in phase by 90º are orthogonal, where as sine waves different in phase by 180º are not orthogonal. Prepared by: Dr. Saatchi, Seyed Mohsen 25
Q82: The I and Q axes represent the orthogonal carrier waves, and graduations on these axes represent available amplitudes. 4-QAM constellation diagram Q83: The previous figure showing the modulation method is known as quadrature phase-shift keying (QPSK). Q84: In practice, the four blobs in QPSK are usually rotated relative to the axes Q85: List one of the drawbacks of 16-QAM (and 64-QAM): is that the relative closeness of the constellation points increases the likelihood of one point being mistaken for another when the symbols are subject to noise interference (as they usually are in radio transmission). Q86: How many additional bits of data are represented by each blob in 16-QAM compared with (a) QPSK? (b) BPSK? Prepared by: Dr. Saatchi, Seyed Mohsen 26
(a) In 16-QAM there are 16 blobs, or states. Hence, each blob can represent 4bits of data(2 4 = 16). In QPSK there are four blobs; hence, each blob represents 2 bits of data (2 2 = 4). Thus, each blob in 16 QAM carries two more bits of data than does each blob in QPSK. (b) In BPSK there are only two blobs, so each blob represents 1 bit of data. Hence, each blob in 16-QAM represents three more bits of data than does each blob in BPSK. Q87: Can we use the following two codes for a CDMA protocol to avoid collision between two stations? Justify your answer. (Hint: CDMA uses orthogonal codes to avoid collision) c1 = (1;-1;-1;-1; 1; 1;-1;-1) c2 = (1; 1;-1; 1; 1;-1; 1; -1) The codes must be orthogonal: (1 * 1) + (-1 * 1) + (-1 * -1) + (-1 * 1) + (1 * 1) + (1 * -1) + (-1 * 1) + (-1 * -1) = 1-1 + 1-1 + 1 1-1 +1 = 0 => orthogonal b. Calculate the spreading factor in the above coding Each bit is represented using 8 chips => spreading factor = 8 c. In UMTS, the chip rate is fixed at 3.84 Mchip/s, and the frame rate is 100 frames per second. Using these numbers and the spreading factor obtained from b, calculate: i. The number of chips per frame Number of chips per frame = 3.84 * 10^6 / 100 = 38400 chips per frame ii. the number of bits per frame Number of bits per frame = 38400 / 8 = 4800 bits per frame iii. the data rate Data rate = 4800 * 100 = 480000 bits/s Prepared by: Dr. Saatchi, Seyed Mohsen 27
Q88: Explain the principle of ON-OFF keying (OOK) and Phase-Shift Keying (PSK). Use diagrams to support your answer. OOK: transmit a burst of a signal ( on ) to represent a data 1 and send nothing ( off ) to represent a data 0. (PSK) uses segments of sinusoids that have the same frequency and amplitude but differ in phase Q89: Consider the two codes: (1,-1, 1,-1, 1, -1, -1, 1) and (1, -1,-1, 1, 1, -1, 1, -1). i. Show that these codes are orthogonal. (1*1) + (-1*-1)+ (1*-1)+ (-1*1)+ (1*1)+ (-1*-1)+ (1*-1)+ (-1*1) = 0 => the two codes are orthogonal ii. ii. Suppose that these codes are used in CDMA. What is the spreading factor associated to this coding. The spreading factor is 8 because each bit is spread among 8 chips iii. Suppose that the data rate is 2 Mbps. What will be the chip rate using this coding scheme? Spreading factor = chip rate / data rate => chip rate = data rate * spreading factor = 16 Mbps Good Luck,,, Prepared by: Dr. Saatchi, Seyed Mohsen 28