T324 activities block1

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T324 activities block1

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T324 activities block1

  1. 1. T324 Activities of Block 1 Page 1 of 10 T324 Activities of Block 1 Activity 4, B1, p26 If each UHF TV channel occupies the same bandwidth in the 470-854 MHz frequency range, this frequency range is shared out equally into 48 channels, what is the bandwidth available to each channel? Which frequencies are occupied by channel 21? Which frequencies are occupied by channel 68? Answer: ℎ= 2− 1 854 − 470 = 384 The bandwidth available for each channel 384 =8 48 The bandwidth available for channel 21 As this channel is the lowest frequency, so The lowest frequency (470 MHz) up to (8 MHz) plus 470 + 8 = 478 So the frequency will be 470 – 478 MHz The bandwidth available for channel 68 As this channel is the highest frequency, so The lowest frequency (854 MHz) up to (8 MHz) minus 854 – 8 = 846 So the frequency will be 846-854 MHz Activity 6, B1, p32 If an antenna in free space receives 16 µW of power at distance of 2 km from an isotropic transmitter, how much will it receive at 4 km? How much at 8 km? Answer: At distance (4 km) Disclaimer: this paper(s) is prepared by a AOU student of KSA, (Yaseen Alhashim). Moreover, I am not responsible of any mistake (if any). So, kindly consider it as un-official paper & refer to the book for more information. You might notify me for any mistake by e- mail: yalhashim@gmail.com , http://alhashem.blogSpot.com | May, 2010
  2. 2. T324 Activities of Block 1 Page 2 of 10 1 2 = 2 1 16 4 = 2 2 16 4 = 2 1 4 2 = 16 16 2= 4 2=4µ At distance (8 km) 1 2 = 2 1 16 8 = 2 2 16 16 = 2 1 16 2 = 16 16 2= 16 2=1µ Activity 7 A 1.8 GHz radio wave propagates 16 km through the atmosphere, and then through two brick walls, each of 100 mm thickness. Rain leads to an atmospheric loss of 1.5 dB/km, and brick attenuates at 40 dB/m at this frequency. Calculate the total power loss in dB over this path that is due to attenuation by the atmosphere and be the wall. (Ignore the inverse square law). Answer Given information: (1) At atmosphere; 16 km (2) Atmosphere loss; 1.5 dB/km (3) 2 brick wall; 100 mm (100 x 10-3 m) (4) Brick attenuation; 40 dB/m Requests: (1) Power loss in dB by (a) atmosphere, (b) by the wall =? (2) Total power loss (a+b) in dB = ? Disclaimer: this paper(s) is prepared by a AOU student of KSA, (Yaseen Alhashim). Moreover, I am not responsible of any mistake (if any). So, kindly consider it as un-official paper & refer to the book for more information. You might notify me for any mistake by e- mail: yalhashim@gmail.com , http://alhashem.blogSpot.com | May, 2010
  3. 3. T324 Activities of Block 1 Page 3 of 10 NOTE: to find the total loss of a path in dB, you simply multiply the attenuation coefficient in dB/km by the path length in km or path in dB/m in path length in m (a) 16 1.5 = 24 Refers to 2 brick wall (b) 2 100 10 40 =8 To convert (mm) to (m) So, the total power loss = 24 + 8 = 32 dB Activity 11, B1, p50 Estimate the physical length of a physical length of a λ/4 rod to be used for a 100 MHz FM radio station. How long would it be for a 2.4 GHz Wi-Fi link? (Hint: in each case, assue a free space wavelength). Answer: In a free space wavelength c = 3 x 108 3 10 λ= = =3 100 10 λ 3 . : . = = 0.75 4 4 For 2.4 GHz Wi-Fi 3 10 λ= = = 0.125 = 0.125 10 = 125 2.4 10 λ 125 .:. = = 31.25 4 4 Activity 3, B1, p69 What is the rate of transmission if the signaling rate is 4500 baud, and each symbol represents 4 bits? Answer 4 x 4500 = 18000 bit/s or 18 kbit/s Disclaimer: this paper(s) is prepared by a AOU student of KSA, (Yaseen Alhashim). Moreover, I am not responsible of any mistake (if any). So, kindly consider it as un-official paper & refer to the book for more information. You might notify me for any mistake by e- mail: yalhashim@gmail.com , http://alhashem.blogSpot.com | May, 2010
  4. 4. T324 Activities of Block 1 Page 4 of 10 Activity 4, B1, p69 A carrier is modulated at signaling rate of 6000 baud. It is used with a modulation system known as 16- QAM, in which there are 16 symbols. a) How many bits does each symbol represent? b) What is the data rate of this system in bits per second (or kilobits per second)? Answer: Given information: (1) 6000 baud (bit/s) (2) 16 symbols Requests: (1) bit rate = [ log10 number of symbols ] / 0.301 (2) data rate = symbols per second x bit per symbol (bit rate) a) 16 = 0.301 =4 b) So, = 6000 4 = 24000 / 24 / Activity 11, B1, p82 Suppose that a version of MSK operates at 10 kbaud and one symbol uses a segment of a 10 MHz sinusoid. If the other symbol uses a segment of a higher frequency, what frequency should that higher frequency be? Answer: The frequency spacing is half the signaling rate, so is 5 kHz (half of 10 kbaud), which is 0.005 MHz. Hence the other symbol should be a segment of a sinusoid with frequency 10 MHz + 0.005 MHz = 10.005 MHz. The frequency spacing is half the signaling rate 10 kbaud = 10 kHz So, the half is 5 kHz As we have the segments in MHz, so we will convert the 5 kHz to MHz 5000 Hz x 10-6 = 0.005 MHz .:. 10 MHz + 0.005 MHz = 10.005 MHz Disclaimer: this paper(s) is prepared by a AOU student of KSA, (Yaseen Alhashim). Moreover, I am not responsible of any mistake (if any). So, kindly consider it as un-official paper & refer to the book for more information. You might notify me for any mistake by e- mail: yalhashim@gmail.com , http://alhashem.blogSpot.com | May, 2010
  5. 5. T324 Activities of Block 1 Page 5 of 10 Activity 12, B1, p89 A transmission at 2 GHz has a bandwidth of 4 MHz. Which of the receiver responses is the best suited for a receiver of this transmission mentioned figures (a, b and c)? Figure a: f1=1.999, f2=2.001 GHz, Figure b: f1=1.998, f2=2.002 GHz, Figure c: f1=1.6, f2=2.4 GHz Answer: We are requested to find the bandwidth of each figure. The bandwidth which is close to given bandwidth in the question (4 MHz) is the best suited for receiver. The a) 2 − 1 = 2.001 − 1.999 = 0.002 = 0.002 10 = 2 b) 2 − 1 = 2.002 − 1.998 = 0.002 = 0.004 10 = 4 c) 2 − 1 = 2.4 − 1.6 = 0.8 = 0.8 10 = 800 So, the beast one suited to transmission with bandwidth of 4 MHz is (b) Activity 17, B1, p97 (a) In a 64-QAM system, how many bits are there per symbol? (b) If the symbol rate is 10000 baud, what is the data rate? Answer: (a) Number of symbol = 2 2 = 64 .:. there is 6 bits per symbol You might find it by [ log10 64 ] / 0.301 = 6 (b) data rate = symbol per second * bit per symbol = 10000 x 6 = 60000 bit/s Activity 19, B1, p98 What is the spectral efficiency if (f2 = 701.02 MHz, f1 = 701 MHz) and data rate is 800 kbit/s? Answer: Spectral efficiency = = 2− 1 = 701.02 − 701 = 0.02 = 0.02 10 = 20 / = = 40 / / = 40 bit/s/Hz Disclaimer: this paper(s) is prepared by a AOU student of KSA, (Yaseen Alhashim). Moreover, I am not responsible of any mistake (if any). So, kindly consider it as un-official paper & refer to the book for more information. You might notify me for any mistake by e- mail: yalhashim@gmail.com , http://alhashem.blogSpot.com | May, 2010
  6. 6. T324 Activities of Block 1 Page 6 of 10 Activity 21, B1, p105 Calculate the signal-to-noise ratio for devices A, B and C, and hence stat which of A, B and C has the best signal-to-noise ratio. • Bandwidth of A device = 10 MHz, • Bandwidth of B device = 20 MHz, • Bandwidth of C device = 30 MHz, • Signal = 150 pW/Hs, • Noise = 50 pW/Hz Answer: Single-to-noise ratio = Signal power = ℎ Noise power = ℎ For (A) device: Signal power = 150 x 10 = 1500 W = 1500 x 10-6 = 1.5 x 10-3 µW = 1.5 mW Noise power = 50 x 10 = 500 W = 500 x 10-6 = 5 x 10-4 µW . Single-to-noise ratio = . = 30 = 30 10 = 3 For (B) device: Signal power = 150 x 20 = 3000 W = 3000 x 10-6 = 3 x 10-3 µW = 3 mW Noise power = 50 x 20 = 1000 W = 1000 x 10-6 = 1 x 10-3 µW = 1 mW Single-to-noise ratio = =3 For (C) device: For device C, the passband is 30 MHz wide, but the signal spectrum occupies only 20 MHz of this. Hence the signal power is exactly as for device B, 3 mW Signal power = 150 x 20 = 3000 W = 3000 x 10-6 = 3 x 10-3 µW = 3 mW Noise power = 50 x 30 = 1500 W = 1500 x 10-6 = 1.5 x 10-3 µW = 1.5 mW Single-to-noise ratio = =2 . Devices A and B equally have the best signal-to-noise ration of 3:1 Disclaimer: this paper(s) is prepared by a AOU student of KSA, (Yaseen Alhashim). Moreover, I am not responsible of any mistake (if any). So, kindly consider it as un-official paper & refer to the book for more information. You might notify me for any mistake by e- mail: yalhashim@gmail.com , http://alhashem.blogSpot.com | May, 2010
  7. 7. T324 Activities of Block 1 Page 7 of 10 Activity 24 If S/N has a value 10, and W is 5 MHz, what is the theoretical maximum data rate? = 2 1+ =5 2 1 + 10 =5 2 11 11 =5 2 = 5 3.45 = 17.29 / = 17.29 10 / Activity 33, B1, p127 If 16-QAM is used, what is the signaling rate for each subcarrier? Use the approximate data rate per subcarrier of 110 bit/s. Answer: Each signaling represents 4 bits (24 = 16). The signaling rate = 110/4 = 27.5 baud (bit/s) Activity 9, B1, p152 A signal has a flat power spectrum extending from 3 GHz to 8 GHz. The power density is -20 dBm/MHz. what is the total power in the signal? Answer Note: 0 dBm is equal to 1 mW, and each additional 10 dBm represents an increase of 10 times the power. Subtracting is similar. So, 1 W is 30 dBm, and 1 µW is -30 dBm. • dBm = log10 (mW)*10 • mW =10^(dBm/10) -20 dBm = 10(-20/10) = 10(-2) = 0.01 mW (milliwatt) The bandwidth = 8-3 = 5 GHz = 5000 MHz The total power = 0.01 x 5000 = 50 mW Disclaimer: this paper(s) is prepared by a AOU student of KSA, (Yaseen Alhashim). Moreover, I am not responsible of any mistake (if any). So, kindly consider it as un-official paper & refer to the book for more information. You might notify me for any mistake by e- mail: yalhashim@gmail.com , http://alhashem.blogSpot.com | May, 2010
  8. 8. T324 Activities of Block 1 Page 8 of 10 Activity 12, B1, p189 Given that the slot time is 20 µs and range of multipliers is 0 to 31, calculate the range of values for backoff time. Answer Minimum time is 0, and maximum time is give by the product of the contention window and the slot time = . = . = . Activity 13, B1, p190 Calculate the transmission time for a frame consisting of 2364 byte over a wireless link operating at a data rate of 2 Mbit/s. Repeat the calculation for a frame consisting of 14 bytes. Answer: = = Note that it is important to compatible units, so we will convert the byte to bit. 2364 8 = 2 10 = 9.384 10 = 9.384 For 14 byte 14 8 = 2 10 = 5.6 10 = 5.6 10 10 = 56 µ Activity 14, B1, p190 Calculate the propagation time for signals over a wireless link of 20 m. you should assume that the speed of propagation is 3 x 108 m/s. = = Disclaimer: this paper(s) is prepared by a AOU student of KSA, (Yaseen Alhashim). Moreover, I am not responsible of any mistake (if any). So, kindly consider it as un-official paper & refer to the book for more information. You might notify me for any mistake by e- mail: yalhashim@gmail.com , http://alhashem.blogSpot.com | May, 2010
  9. 9. T324 Activities of Block 1 Page 9 of 10 20 = 3 10 = 6.667 10 = 6.667 10 10 = 66.67 Activity 15, B1, p191 Given a data rate of 2 Mbit/s, calculate how many bits are transmitted within 66.67 ns. Repeat the calculation for a data rate of 34 Mbit/s. Answer: = = 2 10 66.67 10 = 0.1333 For data rate 34 Mbit/s = 34 10 66.67 10 = 2.267 For data rate 54 Mbit/s = 54 10 66.67 10 = 3.6 Activity 16, B1, p192 Calculate the minimum amount of time that is wasted if a 2364 byte frame is involved in a collision. You should assume that the data rate is 2 Mbit/s and the period of time a station waits for an acknowledgment is 100 µs. repeat the calculation for a 14 byte. Answer: Note that it is important to compatible units, so we will convert the byte to bit. = 2364 8 = 2 10 = 9.384 10 = 9.384 Transmission time for 2364 byte frame at 2 Mbit/s is 9.384 ms The minimum wasted time is 9384 + 100 = 9484 µ Disclaimer: this paper(s) is prepared by a AOU student of KSA, (Yaseen Alhashim). Moreover, I am not responsible of any mistake (if any). So, kindly consider it as un-official paper & refer to the book for more information. You might notify me for any mistake by e- mail: yalhashim@gmail.com , http://alhashem.blogSpot.com | May, 2010
  10. 10. T324 Activities of Block 1 Page 10 of 10 Activity 25, B1, p211 Let P=0111 0100 0110 1000 and K=0011 0001 0011 0100. Calculate C= P + K and confirm P = C + K Answer: C=P+K 0 1 1 1 0 1 0 0 0 1 1 0 1 0 0 0 0 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 1 0 1 0 1 1 1 0 0 P=C+K 0 1 0 0 0 1 0 1 0 1 0 1 1 1 0 0 0 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 1 0 1 0 0 0 1 1 0 1 0 0 0 Activity 26, B1, p211 The initialization vector is 24 bits long. Assuming that the initialization vector is incremented by 1 for each frame transmitted by an access point operating at 11 Mbit/s and frames are, on average, 2000 bytes long, estimate how long it will take for an initialization vector value to be repeated. Answer: With 24 bits, the number of different initialization vector value is 2 = 16777216 ℎ ℎ ℎ The maximum rate of transmitting frame is 11 10 = = 687.5 / 2000 8 The time taken to repeat a value 16777216 24403 = 24403 = = 6.77 ℎ 687.5 60 60 Disclaimer: this paper(s) is prepared by a AOU student of KSA, (Yaseen Alhashim). Moreover, I am not responsible of any mistake (if any). So, kindly consider it as un-official paper & refer to the book for more information. You might notify me for any mistake by e- mail: yalhashim@gmail.com , http://alhashem.blogSpot.com | May, 2010

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