ECE_313_SP2023_HW8sol

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University of Illinois, Urbana Champaign *

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Electrical Engineering

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Jan 9, 2024

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University of Illinois Spring 2023 ECE 313: Problem Set 8: Problems and Solutions Due: Thursday, March 23 at 5:00:00 p.m. Reading: ECE 313 Course Notes, Sections 3.6.2 and 3.6.3 Note on reading: For most sections of the course notes there are short answer questions at the end of the chapter. We recommend that after reading each section you try answering the short answer questions. Do not hand in; answers to the short answer questions are provided in the appendix of the notes. Note on turning in homework: Homework is assigned on a weekly basis on Thursdays, and is due by 5:00 p.m. on the following Thursday. You must upload handwritten homework to Gradescope. No typeset homework will be accepted. No late homework will be accepted. Please write on the top right corner of the first page: NAME AS IT APPEARS ON Canvas NETID SECTION PROBLEM SET # Page numbers are encouraged but not required. Five points will be deducted for improper headings. 1. [Gaussian Distribution] Suppose X is a N ( − 2 , 4) random variable. Compute the following quantities. (a) P { X ≥ − 2 } . Solution: We are given that X has a Gaussian distribution with mean − 2 and variance 4. Then X +2 2 is a standard normal distribution. P { X ≥ − 2 } = P X − ( − 2) 2 ≥ − 2 − ( − 2) 2 = P X + 2 2 ≥ 0 = 1 2 (b) P ( X ≥ − 2 | X ≥ − 4).

Solution: P ( X ≥ − 2 | X ≥ − 4) = P ( X ≥ − 2 ∩ X ≥ − 4) P ( X ≥ − 4) = P ( X ≥ − 2 ∩ X ≥ − 4) P ( X ≥ − 4) = P ( X ≥ − 2) P ( X ≥ − 4) = P ( X +2 2 ≥ 0) P ( X +2 2 ≥ − 1) = 1 / 2 Q ( − 1) = 1 2Φ(1) (c) P { X 2 < X + 2 } . Solution: P { X 2 < X + 2 } = P { X 2 − X − 2 < 0 } = P { ( X − 2)( X + 1) < 0 } = P {− 1 < X < 2 } = P { 0 . 5 < X + 2 2 < 2 } = Φ(2) − Φ(0 . 5) (d) E [( X + 2) 2 ]. Solution: Recall that Var( X ) = E [( X − E [ X ]) 2 ] . Hence, E [( X + 2) 2 ] = E [( X − ( − 2)) 2 ] = Var( X ) = 4 . Alternate solution: E [( X + 2) 2 ] = Z ∞ −∞ ( x + 2) 2 f X ( x ) dx = Z ∞ −∞ ( x + 2) 2 1 √ 2 πσ 2 exp − ( x − µ ) 2 2 σ 2 dx = Z ∞ −∞ ( x + 2) 2 1 √ 8 π exp − ( x + 2) 2 8 dx 2

Using change of variable u = x + 2 and doing integration by parts, E [( X + 2) 2 ] = Z ∞ −∞ u 2 1 √ 8 π exp − u 2 8 du = − 4 √ 8 π Z ∞ −∞ u − u 4 exp − u 2 8 du = − 4 √ 8 π u exp − u 2 8 ∞ −∞ − Z ∞ −∞ exp − u 2 8 du ! = 0 + 4 Z ∞ −∞ 1 √ 2 π 4 exp − u 2 2 · 4 du = 4 2. [Communication in Gaussian Noise] A wireless communication system consists of a transmitter and a receiver. The transmitter sends a signal x , and the receiver observes Y = x + Z, where Z is a noise term, modeled as a Gaussian random variable with mean µ Z = 0 and variance σ 2 Z = 1. (a) Suppose the transmitted signal is x = 1. What is the pdf of the received signal Y ? Solution: The received signal Y follows a Gaussian distribution with µ Y = 1 and σ 2 Y = 1 . The pdf is P ( Y = y ) = 1 √ 2 π exp − ( y − 1) 2 2 (b) Now suppose the transmitted signal can be either x = − 1 or x = 1. The receiver uses the following decoding rule: if Y > 0, it declares that x = 1; if Y ≤ 0, it declares that x = − 1. Assuming that the transmitter sends − 1 or +1 with probability 1 / 2 each, what is the receiver’s error probability? Solution: Let Y 1 and Y − 1 be the received signal when x = 1 and x = − 1 are transmitted respectively. Their distributions are Y 1 ∼ N (1 , 1) Y − 1 ∼ N ( − 1 , 1) The individual probabilities of error for the two symbols transmitted can be written as P ( Y = − 1 | x = 1) = P ( Y 1 < 0) = P ( Y 1 − 1 < − 1) = Φ( − 1) = Q (1) 3

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