Engin112_HW5_solution_2023

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ENGIN 112 University of Massachusetts Amherst ENGIN 112: Homework 5 Solution Due date: 10/20/23 5:00 p.m. Please submit your answers via Gradescope. You can write your answers electronically or by hand and submit a scan or photo. Question 1 Assume you make a phone call between two telephones that are connected to the telephone network via the Internet using a VoIP adapter for each phone. Each phone uses a regular, U.S. phone number. (a) Show an illustration that shows what that connection looks like. You may show the Internet and the phone network as a “cloud” (i.e., no internal details required). (b) The voice information for this phone call can be carried in three forms: analog, digital in the phone network, and digital in the Internet. State the order in which form the voice information for this call is transferred. The phone call is in form of analog at phone side, and then it converted to digital in VoIP adapter. The voice information transmits in the internet in form of digital and it converted back to analog in VoIP adapter in the receiver side. It is analog in the receiver’s telephone. – analog ( 1 ⃝ in figure) – digital in the Internet ( 2 ⃝ and 3 ⃝ in figure) – analog ( 4 ⃝ in figure) Question 2 Answer the following questions about frequency allocations. Homework 5 ©2015–23 Tilman Wolf, Rui Ma 1

ENGIN 112 University of Massachusetts Amherst (a) Download the U.S. Frequency Allocation Chart: https://www.ntia.doc.gov/files/ntia/publications/january_2016_spectrum_wall_chart.pdf Identify a frequency range used for cellular telephony in the chart (e.g., 809MHz–849MHz). What is the label for this frequency range in the chart? Fixed Land Mobile (b) Consider the frequency allocation for the United Kingdom: http://www.onlineconversion.com/downloads/uk_frequency_allocations_chart.pdf What would you perceive as a challenge when building cell phone infrastructure that crosses national boundaries? (2–3 sentences) Comparing the allocation chart of U.S. and U.K., we can see that frequency uses are quite different between these two countries. The challenge to build a cell phone infrastructure across national boundaries is to identify suitable frequencies that are available in each country and that can be sent and received by the mobile phone. Ideally, the same frequencies can be uses in many countries to simplify the handset implementation. Question 3 Assume you have two oscillators that generate a carrier signal at 850 MHz ± 80 ppm. Assume that at time t=0 both oscillators are synchronized. Also assume the worst case difference in frequency between the oscillators (i.e., one oscillating as slow as possible within specification and the other oscillating as fast as possible within specification). (a) At what time are the two oscillators exactly opposite to each other (i.e., 180 degrees out of phase)? You can calculate this time by assuming that the slower oscillator will have completed x cycles at that time and the faster oscillator will have completed x + 1 2 cycles. Define the frequency of the faster signal as f f and the slower signal as f s . The cycle of the slower signal is denoted by x . When the two oscillators are exactly opposite to each other for the first time, the faster signal has been transmitted x + 1 2 cycles. The time T that the two oscillators are opposite to each other equals to: T = x f s = x + 1 2 f f (1) = ⇒ x × f f = ( x + 1 2 ) × f s = ⇒ x = f s 2( f f − f s ) (2) From (1) and (2): T = x f s = 1 2( f f − f s ) For the worst case difference in frequency: f s = 850 MHz − 80 ppm = 850 × 10 6 Hz − 850 × 80 Hz = 849932000 Hz f f = 850 MHz + 80 ppm = 850 × 10 6 Hz + 850 × 80 Hz = 850068000 Hz Therefore, plug f s and f f in: T = x f s = 1 2( f f − f s ) = 3 . 6765 × 10 − 6 s = 3 . 6765 µs (b) Assume that you transmit 640 kilobits of binary data per second on this carrier signal. Also assume that the data can be received correctly by the receiver when the phase difference between the sender and the receiver is between 0 degrees and 90 degrees. How many bits can be transmitted correctly before the sender and receiver are out of sync by more than 90 degrees? Homework 5 ©2015–23 Tilman Wolf, Rui Ma 2

ENGIN 112 University of Massachusetts Amherst From part (a), we know that T = 3 . 6765 µs to reach 180 degrees phase difference. Since the oscillators are stable, it will cost T 2 = 1 . 8382 µs to reach the 90 degrees phase difference. During this time period, the total bits can be transmitted correctly are: 640 kbps × 1 . 8382 µs = 640 × 10 3 bits/s × 1 . 8382 × 10 − 6 s = 1 . 1765 bits, which is less than 2 bits. Therefore, only 1 bit will be correctly transmitted before the sender and receiver are out of sync by more than 90 degrees. Question 4 Please answer the following questions regarding modulation of bits on a signal transmission. (a) Assuming frequency-shift keying (higher frequency indicating a ‘1’), what bit sequence is encoded in the following transmission pattern? You may assume that 3 oscillations of the lower frequency are transmitted per bit or 6 oscillations of the higher frequency per bit. The transmission pattern is representing the bit sequence: 01001111 (b) How is the bit sequence ‘1101110110’ encoded with on-off keying? (c) What problem do you foresee if you were to transmit 100,000 consecutive bits of value 0 (zero) using on-off keying? Would this problem go away if you were to use amplitude-shift keying or frequency-shift keying? Homework 5 ©2015–23 Tilman Wolf, Rui Ma 3

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ENGIN 112 University of Massachusetts Amherst Difficult to tell if its just zero being transmitted or the system is non-responsive. Since AM or FM uses carrier frequency, or amplitude, it will resolve this issue. The signal will not be zero, even if it is transmitting zero. Question 5 Consider the Hamming (7,4) code discussed in the lecture: (a) What is the 7-bit encoding of 0000 when using this code? The 7-bit encoding of 0000 is 0000000. (b) What is the 7-bit encoding of 0001 when using this code? The 7-bit encoding of 0000 is 0001111. (c) What is the Hamming distance between 7-bit code words from (a) and (b)? The Hamming distance between 7-bit code words from (a) and (b) is 4 (4 different bits). (d) Noting that 0000 and 0001 differ by only 1 bit and considering your answer to (c), explain why a 1-bit transmission error can be corrected with a Hamming (7,4) code. The Hamming distance between 7-bit code words from (a) and (b) is 4, which is greater than 3. The minimal Hamming distance between any two correct code words is 3. Therefore, a single-bit flip can be corrected by identifying the code word that is closest (Slides Page 56). Question 6 Assume an 800 MHz electromagnetic signal arrives at a receiver via two paths that differ in length by a distance of ∆ d = 10 m . Assume a propagation speed that corresponds to that of light in vacuum. (a) What is the wavelength of this signal? The propagation speed is 3 × 10 8 m/s (the speed of light in vacuum). Therefore, the wavelength of this signal is: λ = c f = 3 × 10 8 800 × 10 6 = 0 . 375 m Homework 5 ©2015–23 Tilman Wolf, Rui Ma 4

ENGIN 112 University of Massachusetts Amherst (b) How much later does the signal along the longer path arrive at the receiver? Based on the distance difference is ∆ d = 10 m and the propagation speed 3 × 10 8 m/s , the time delay is: t = ∆ d c = 10 3 × 10 8 = 3 . 33 × 10 − 8 s. (c) What is the phase difference between the signal that arrives first and the signal that arrives later? The phase difference is: ∆ d λ = 10 0 . 375 = 26 . 67 wavelengths. Homework 5 ©2015–23 Tilman Wolf, Rui Ma 5