Tutorial 9 - UMTS and GSM PDF

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UMTS and GSM...

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¨nchen Technische Universit ¨at Mu ¨ r Nachrichtentechnik Lehrstuhl fu Prof. Dr. sc. techn. Gerhard Kramer July 11th, 2017

Mobile Communications: Tutorial 9 Exam 08/09

UMTS and GSM

In the following we want to compare UMTS and GSM using a transmission over a Rayleigh fading channel with two taps. For simplification we assume the UMTS bandwidth WU = 3.84 MHz and the GSM bandwidth WG = 0.24 MHz. (a) Calculate the noise power in dBm for both cases (noise power density N = kT = −173.8 dBm/Hz). Due to shadowing and path loss the power at the receiver PRX = −114 dBm and for this task we assume a spreading factor SF = 256. (b) Calculate first the CINR of the given UMTS system (neglecting the interference in comparison to the thermal noise) and then calculate the SNR. Furthermore, give also the value for the SNR of the GSM system. (c) Considering the given numbers and assuming UMTS downlink (QPSK) with the given spreading factor what is the raw data rate (this is the rate after channel coding and rate matching in the transmitter) in kbps which can be transmitted? What is the raw data rate in this GSM system (WG = 0.24 MHz and GMSK with 1 bit/symbol). (d) The given values are typically used for voice transmission. However, looking on the result of (c), GSM has a much higher data rate. What access scheme is applied in GSM to obtain a similar data rate as in UMTS. How many users have to share in our scheme one channel to obtain a similar raw data rate? Now we look onto the receiver of the GSM and the UMTS system. We have a channel where many (almost infinite!) smaller paths contribute leading finally to two discrete taps having a delay of τ0 = 0 µs and τ1 = 0.26 µs, and an average amplitude of h0 = 0.8 and h1 = 0.6. (e) First we look onto the GSM system. Can you resolve the two taps? What kind of channel model would you assume now? Please calculate the BER in case of the calculated SNR (if you did not solve task (b) take SNR = 6 dB)? (f) Now we look on the UMTS system. For simplification we use SF = 4 with the spreading code c = 1 −1 −1 1 . Give another code which you would use in case of a single tap fading channel in the downlink? Give a reason for your choice.

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(g) How many taps can you resolve now in UMTS with the given channel? What problem occurs in case of the found second sequence? (h) In the following we consider one user who uses the spreading code given in (f) and we further   simplify it by using BPSK. We transmit the 3 symbols u = u0 u1 u2 = 1 1 −1 . What is the sequence x after spreading? Calculate the correlation function CF in the range of n = −2 to n=2(i.e., CF (−2)....CF (2)) between the middle symbol u1 spreaded with c and the whole sequence x (no transmission over a channel). Think about how to calculate CF(0) in this case! What is your conclusion from the result? (i) Using the given channel and the CINR of 0 dB we receive the following sequence:   y = 0.5 1.0 −1.3 0.4 0.6 0.6 −0.6 0.2 0.0 0.3 1.3 0.2 −1.0 If you apply MRC assuming ideal channel estimation (you know h0 = 0.8 and h1 = 0.6), what is your estimate for the middle symbol u1 after despreading. Now we calculate the error probability for this channel. It is done in two tasks (j) and (k), where first the estimate u˜n is calculated and then the error probabilities based on the estimate. (j) Show the MRC solution for one sample y˜m before despreading in our case with the two channel taps in the form y˜m = y˜0,m + y˜1,m . Now take only the first part (also called Rake finger) of this MRC and extend it the result of the first  by showing  Rake finger as a function of the channel h = h0 h1 , any arbitrary transmitted sequence x and the noise w. Then, extend it by the spreading and despreading operation to obtain u˜1,n . Use this first result to obtain the overall MRC solution u˜n dependent on the transmitted symbol sequence u, the channel h, and the noise sequence w. (k) Calculate the error probability of this Rake receiver assuming the fixed channel coefficients h and the given CNR of 0 dB. If you did not solve task (j), take the following result: u˜n = 4un + 0.48(un−1 − un ) + 0.48(un+1 − un ) + 2w ˜n With σw2˜m = σ 2wm = N0 /2. Use the table for the Q functions given in the appendix. (l) Looking at the calculations and compare the two systems with the used setup. Please show one disadvantage and one advantage of CDMA in this concrete setup.

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