ECE6374_Homework_05_Fall2023_Solution

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Georgia Institute of Technology School of Electrical and Computer Engineering ECE6374 Cyber-Physical Security in Electric Energy Systems Fall 2023 Homework Assignment #5 SOLUTION Problem P1 (10 points): Demetrius creates an Elgamal cryptosystem as follows: Public key: (293,21,4) His private key is: 15. He sends the public key to Helen. Helen has a 9-element alphanumeric password that wants to send to Demetrius. She converts each alphanumeric character into a number, using the ASCII code shown in the table, i.e. an “a” becomes “97”, an “A” becomes “65”, etc. Helen uses the public key she received from Demetrius and encrypts each number. The generated cipher text is send to Demetrius, shown below: (31,33), (31,9), (31,204), (31,23), (31,67), (31,77), (31,160), (31,155), (31,155) Help Demetrius to decrypt the message and uncover the 9-element alphanumeric password. Hint: decrypt each one of the numbers and then use the ASCII code to convert the number into an alphanumeric. From the private and public keys: x = 15, p = 293, g = 21, y = g x mod ( p ) = 21 15 mod ( 293 ) = 4 For any encrypted information received, (a,b), the decrypted message is calculated using the following equation: m = b ( a x ) − 1 mod ( p ) First, find the expression: c = ( a x ) − 1 mod ( p ) = ( 31 15 ) − 1 mod ( 293 )= 233 Then,  (31,33): m = 33 ( 31 15 ) − 1 mod ( 293 ) = 33 ∗ 233 mod 293 = 71 →' G '  (31,9): m = 9 ( 31 15 ) − 1 mod ( 293 )= 9 ∗ 233 mod 293 = 46 →' .'  (31,204): m = 204 ( 31 15 ) − 1 mod ( 293 )= 204 ∗ 233 mod 293 = 66 →' B'  (31,23): m = 23 ( 31 15 ) − 1 mod ( 293 )= 23 ∗ 233 mod 293 = 85 →' U '  (31,67): m = 67 ( 31 15 ) − 1 mod ( 293 )= 67 ∗ 233 mod 293 = 82 →' R'

 (31,77): m = 77 ( 31 15 ) − 1 mod ( 293 )= 77 ∗ 233 mod 293 = 68 →' D'  (31,160): m = 160 ( 31 15 ) − 1 mod ( 293 )= 160 ∗ 233 mod 293 = 69 →' E'  (31,155): m = 155 ( 31 15 ) − 1 mod ( 293 )= 155 ∗ 233 mod 293 = 76 →' L'  (31,155): m = 155 ( 31 15 ) − 1 mod ( 293 )= 155 ∗ 233 mod 293 = 76 →' L' Final Answer: G.BURDELL

Problem P2 (10 points): One method to guard against transmission errors and provide additional security of packets, is the use of CRC (Cyclic Redundancy Check). 1. Describe in your own words how CRC works and the algorithm for computing the CRC- n. Hint: search the web (or Wikipedia) for text describing the CRC. Study the text and then describe how CRC works in your own words. Be Concise. 2. Consider the following 32 bit message: 1001 0101 0110 0101 0101 0011 0101 1101 Compute the CRC-5 using the following generator polynomial 111010. A description of CRC from Wikipedia is provided as a separate document. 1. Visit https://en.wikipedia.org/wiki/Cyclic_redundancy_check 2. 10010101011001010101001101011101 00000 (padding with 5 0's) 111010 01111101011001010101001101011101 00000 111010 0001001011001010101001101011101 00000 111010 0111111001010101001101011101 00000 111010 000101001010101001101011101 00000 111010 010011010101001101011101 00000 111010 01110010101001101011101 00000 111010 0000110101001101011101 00000 111010 001111001101011101 00000 111010 0001101101011101 00000 111010 0011001011101 00000 111010 00100011101 00000 111010 011001101 00000 111010 00100101 00000 11 1010 01 10100 1 11010 0 01110 CRC-5 = 01110

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Problem P3 (10 points) : Two common line protection functions are: (a) line percentage differential protection and (b) alpha plane line differential. A brief description of these two protection functions are provided at the end of the problem. A transmission line is protected with these two protection functions, one acts as the primary protection and the other as the backup protection. Primary/backup protection is normally used to improve reliability of the protection system. The settings for each one of the two protection functions for the transmission line of the figure are as follows: Line Percentage Differential: CT ratio 1200:5, Minimum pickup time 0.7 Amperes, percentage: 25%. This means: 0, 5 /1200, 0.7 , 0.25 set k i A p    Alpha Plane Line Differential: Inner radius: 1/3, Outer radius: 3.0, Total angular extent: 210 degrees. Relay trips when ratio is outside the shaded area (relay characteristic). Assume that the GPS receiver/clock in one terminal of the line, as shown in Figure P3.1, has been spoofed and the intruder simply inserts a time delay in the time signal in the GPS receiver. Determine the minimum time delay that will cause the misoperation of both relay functions (primary and backup) for a line external fault. The operating conditions during this fault are shown in Figure P3.2. In other words, it is given that the data shown in Figure P3.2 represent an external fault; under normal circumstances, the two protective functions will not trip the line. Figure P3.1

Figure P3.2 Appendix A: Line Differential Protection Differential protection for transmission lines presents challenges because of the distance between the two terminals of the line. Direct application of differential protection will require wires from one end of the line to the other. These wires will parallel the power line and during faults may experience induced voltages. Differential protection schemes do provide more certainty in identifying whether the fault is within the zone of protection. However, because of the geographic extent of transmission lines, differential relaying cannot be applied easily, except for relatively short lines. This can be easily understood since for a differential scheme the currents should be measured at the two ends of the line simultaneously and be brought to the location of the differential relay. The wires that are required by the traditional differential protection will present additional technical, logistical and cost issues. During faults the induced voltages on these wires may be excessive and should be mitigated by appropriate designs. Without going into more discussion of the issues, it should be understood that conventional differential protection for transmission lines is not practical or economical. Differential protection schemes have attractive advantages. For this reason, approaches and systems have been developed over the years to enable some of these advantages for the protection of transmission lines. Since the basic obstacle in providing differential protection for lines is the communication of the voltages and currents from the terminals of the line to the relay, the developments were along the lines of how much information can be communicated from one end of the line to another, with what communication means and how this information will be utilized. We refer to all these developments (that will be introduced later) as pilot relaying. Recently, the introduction of GPS synchronized measurements and fiber optic communications provided the enabling technology for true differential protection of transmission lines. Specifically, GPS synchronized measurements enable the simultaneous measurement of voltages

and currents at remote terminals of a line. These measurements are time tagged and can be communicated via fiberoptic to the location of the relay. The relay can time align the measurements and perform a true differential protection function. An illustration of this approach is given in Figure A-1. The fiberoptic communication can be provided in the shield wires of an overhead line, or by fiberoptic lines embedded in the cable for UG transmission. This approach provides a secure protection method, as long as the GPS signal is available and the fiberoptic communication is operational. The reliability of these systems can be very high, for example for GPS availability, one can provide low cost equipment that will maintain GPS time for substantial periods in case of lost satellite signals and one can use redundant fiberoptic lines. (a) (b) Figure A-1: Illustration of Line Differential Protection (a) Line Differential with Fast Communications (b) GPS Synchronized Measurements Line Percentage Differential Protection Differential relays detect internal faults in devices (or more general in a protection zone) such as transformers, generators, motors, reactors, capacitors etc., by monitoring the electric currents on all terminals of a device and forming the sum of all currents. The sum of all these currents should be identical to zero. Thus, differential relays are so designed as to “see” zero current under normal operating conditions or external faults. In case of an internal fault the relay will “see” a substantial current and it will trip the device. Differential relays can detect small internal device fault currents and thus are useful in isolating a faulted device from the power system.

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Percentage differential relays restrict the operation of the relay unless the relay operating current is above a specified percentage of the phase currents in the protection zone. This approach makes the relay more secure. In the case of line percentage differential relaying, the relay will trip if for at least one phase the following holds:     0 1 2 0 0, 0 1 2 0.5 R R set R R I k I I I i I pk I I             Where k is the CT ratio, 0, set i is the operating current setting, p is the percentage setting in pu, i indicates the phase (a, b, and c), and 1 and 2 indicate the two ends of the line. The indicated line currents flow into the line. Alpha Plane Line Differential Protection This method is based on transmitting the measured current phasor from one end of the line to the other and taking the ratio of the current phasors at the two ends of the line. For an ideal line and neglecting the capacitive current of the line, this ratio will be exactly -1.0. If there is an internal fault in the line the ratio of the current phasor will be different than -1.0. For an external fault the ratio will be close to -1.0. For a three phase line, the method uses three-phase currents at both sides of the line.

Because of the capacitive current of the line, the ratio of the currents will deviate from -1.0. The longer the line is the higher the deviation will be. In addition, the ratio will vary depending on the loading of the line. In lighter loaded line the ratio of the currents at the two ends may deviate much more than the -1.0 value. Another issue that affect the accuracy of the ratio is the time synchronization of the measurements. Any time errors will cause the ratio to deviate from the -1.0 value. For all of these reasons, the relay is set to operate only when the ratio has a value that practically guarantees that an internal fault is causing the deviation. Based on many studies, the recommended relay characteristic is shown in Figure A-2. Note the relay will trip when the ratio is outside the shaded area. Figure A-2: Line Differential Protection Characteristic in alpha-plane

Answer: A- Normal Operation (No Spoofing Attack) i- Line Differential Protection Get ~ I 0 : ~ I k ∗( ¿¿ 1 A + ~ I 2 A )= 0.003 + j 0.038 A ~ I 0 A = ¿ ~ I k ∗( ¿¿ 1 B + ~ I 2 B )= 0.017 − j 0.015 A ~ I 0 B = ¿ ~ I k ∗( ¿¿ 1 C + ~ I 2 C )= 0.027 + j 0.022 A ~ I 0 C = ¿ Check conditions #1: ~ ¿ I 0 A ∨ ¿ 0.038 > 0.7 →False →No trigger ~ ¿ I 0 B ∨ ¿ 0.022 > 0.7 →False →Notrigger ~ ¿ I 0 A ∨ ¿ 0.035 > 0.7 →False →Notrigger Check conditions #2: ~ ¿ I ¿ p ∗ k ∗ ¿ ~ ¿ I 0 A ∨ ¿ 0.038 > ¿ ~ ¿ I ¿ p ∗ k ∗ ¿ ~ ¿ I 0 B ∨ ¿ 0.022 > ¿ ~ ¿ I ¿ p ∗ k ∗ ¿ ~ ¿ I 0 C ∨ ¿ 0.035 > ¿ We conclude that LDP relay will not trigger during normal operation. ii- Alpha Plane Differential Protection

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Get α : α A = ~ I 1 A / ~ I 2 A = 0.9796 @ − 177 D→ RestrainRegion α B = ~ I 1 B / ~ I 2 B = 0.9988 @ − 179.94 D →RestrainRegion α C = ~ I 1 C / ~ I 2 C = 0.9964 @ − 178.28 D →RestrainRegion No trip for the APDP relay. B- With Spoofing Attack By trial and error, insert 4762 us (102.8592D) at Bus 2. Update the values at Bus 2. ~ I 2 A = 133.5 + 169.7 j ~ I 2 B =− 746.6 − 3357 j ~ I 2 C = 229.8 + 153.35 j i- Line Differential Protection Get ~ I 0 : ~ I k ∗( ¿¿ 1 A + ~ I 2 A )= 1.3669 + 0.3607 j ~ I 0 A = ¿ ~ I k ∗( ¿¿ 1 B + ~ I 2 B )=− 17.4232 − 14.0823 j ~ I 0 B = ¿ ~ I k ∗( ¿¿ 1 C + ~ I 2 C )= 1.7667 − 0.1743 j ~ I 0 C = ¿ Check conditions #1: ~ ¿ I 0 A ∨ ¿ 1.4137 > 0.7 →True ~ ¿ I 0 B ∨ ¿ 22.4026 > 0.7 →True ~ ¿ I 0 A ∨ ¿ 1.7752 > 0.7 →True

Check conditions #2: ~ ¿ I ¿ p ∗ k ∗ ¿ ~ ¿ I 0 A ∨ ¿ 1.4137 > ¿ ~ ¿ I ¿ p ∗ k ∗ ¿ ~ ¿ I 0 B ∨ ¿ 22.4026 > ¿ ~ ¿ I ¿ p ∗ k ∗ ¿ ~ ¿ I 0 C ∨ ¿ 1.7752 > ¿ We conclude that LDP relay will trip at all of the three phases. ii- Alpha Plane Differential Protection Get α : α A = ~ I 1 A / ~ I 2 A = 0.9796 @ − 74.9808 D →TripRegion α B = ~ I 1 B / ~ I 2 B = 0.9988 @ − 77.0808 →TripRegion α C = ~ I 1 C / ~ I 2 C = 0.9964 @ − 78.8608 D →TripRegion We conclude that APDP relay will trip at all of the three phases. Both relays have tripped given a spoofing attack of 4726 us.