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ELEC 221 Signals & Systems EXAM SOLUTIONS

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1. “Short” Answers: a) (6pts) For 𝑥𝑥(𝑡𝑡) = 3 − 2cos �83𝜋𝜋 𝑡𝑡� + cos �35𝜋𝜋 𝑡𝑡 − 𝜋𝜋4�, determine the fundamental frequency 𝜔𝜔𝑥𝑥 and compute the complex exponential Fourier coefficients {Xk} as well as the power, 𝑃𝑃𝑥𝑥. 𝜔𝜔1 = 8𝜋𝜋 3 ; 𝜔𝜔2 = 3𝜋𝜋 5 ⇒ 𝜔𝜔 𝑥𝑥 = 𝐺𝐺𝐺𝐺𝐺𝐺(𝜔𝜔1, 𝜔𝜔2) = 𝜋𝜋 15 ⇒ 𝜔𝜔1 = 40𝜔𝜔𝑥𝑥; 𝜔𝜔2 = 9𝜔𝜔𝑥𝑥 𝑥𝑥(𝑡𝑡) = 3 − �𝑒𝑒 𝑗𝑗8𝜋𝜋𝑡𝑡 3 + 𝑒𝑒−𝑗𝑗8𝜋𝜋𝑡𝑡 3 � + 𝑒𝑒𝑗𝑗�35𝜋𝜋𝑡𝑡−𝜋𝜋4�+𝑒𝑒−𝑗𝑗�35𝜋𝜋𝑡𝑡−𝜋𝜋4� 2 ; 𝑃𝑃𝑥𝑥 = 9 + 22 2 + 12 2 = 11 1 2 ⇒ 𝜔𝜔 𝑥𝑥 = 𝜋𝜋 15 ; 𝑋𝑋𝑘𝑘 = ⎧⎪⎨⎪⎩ 3 if 𝑘𝑘 = 0 𝑒𝑒 ∓𝑗𝑗𝑗𝑗 4 2 if 𝑘𝑘 = ±9 −1 if 𝑘𝑘 = ±40 0 otherwise⎭⎪ ⎬ ⎪ ⎫ = ⎩ ⎨ ⎧ 3 if 𝑘𝑘 = 0 1∓𝑗𝑗 2√2 if 𝑘𝑘 = ±9 −1 if 𝑘𝑘 = ±40 0 otherwise⎭⎬ ⎫; 𝑃𝑃𝑥𝑥 = 11 1 2 b) (10pts) For 𝑦𝑦(𝑡𝑡) = 𝑥𝑥2(𝑡𝑡) using the signal from (a) above, determine the fundamental frequency 𝜔𝜔𝑦𝑦 and compute the complex exponential Fourier coefficients {Yk} as well as the power, 𝑃𝑃𝑦𝑦. [Aside: I apologise as the calculations were a little more tedious than expected. The following simplifying identities will help: • (𝑎𝑎 + 𝑏𝑏 + 𝑐𝑐)2 = 𝑎𝑎2 + 𝑏𝑏2 + 𝑐𝑐2 + 2(𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑐𝑐) • cos(𝑎𝑎) cos(𝑏𝑏) = cos(𝑎𝑎+𝑏𝑏)+cos (𝑎𝑎−𝑏𝑏) 2 �⇒ cos2 𝑎𝑎 = cos 2𝑎𝑎+1 2 � • −cos(𝑎𝑎) = cos (𝑎𝑎 ± 𝜋𝜋) ] ⇒ 𝑦𝑦(𝑡𝑡) = 9 + 4cos2(40𝜔𝜔𝑥𝑥𝑡𝑡) + cos2 �9𝜔𝜔𝑥𝑥𝑡𝑡 − 𝜋𝜋4� + 2 �− 6cos(40𝜔𝜔𝑥𝑥𝑡𝑡) + 3 cos �9𝜔𝜔𝑥𝑥𝑡𝑡 − 𝜋𝜋4� − 2cos(40𝜔𝜔𝑥𝑥𝑡𝑡) cos �9𝜔𝜔𝑥𝑥𝑡𝑡 − 𝜋𝜋4�� = 9 + 4 �cos 80𝜔𝜔𝑥𝑥𝑡𝑡+1 2 � + �cos�18𝜔𝜔𝑥𝑥𝑡𝑡−𝜋𝜋2�+1 2 � − 12cos(40𝜔𝜔𝑥𝑥𝑡𝑡) + 6 cos �9𝜔𝜔𝑥𝑥𝑡𝑡 − 𝜋𝜋4� − 2 �cos �49𝜔𝜔𝑥𝑥𝑡𝑡 − 𝜋𝜋4� + cos �31𝜔𝜔𝑥𝑥𝑡𝑡 + 𝜋𝜋4�� = 11 1 2 + 2 cos 80𝜔𝜔 𝑥𝑥𝑡𝑡 + sin(18𝜔𝜔𝑥𝑥𝑡𝑡) 2 − 12cos(40𝜔𝜔𝑥𝑥𝑡𝑡) + 6 cos �9𝜔𝜔𝑥𝑥𝑡𝑡 − 𝜋𝜋4� + 2 cos �49𝜔𝜔𝑥𝑥𝑡𝑡 + 34𝜋𝜋� + 2 cos �31𝜔𝜔𝑥𝑥𝑡𝑡 − 34𝜋𝜋� 𝑃𝑃𝑦𝑦 = �11 1 2�2 + 1 2 �4 + 1 4 + 144 + 36 + 4 + 4� = �121 + 1 4 + 11� + 1 2 �192 1 4� = 228 3 8 ⇒ 𝜔𝜔 𝑦𝑦 = 𝜔𝜔𝑥𝑥 = 𝜋𝜋 15 ; 𝑌𝑌𝑘𝑘 = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ 11 1 2 if 𝑘𝑘 = 0 3𝑒𝑒∓𝑗𝑗4𝑗𝑗 if 𝑘𝑘 = ±9 ∓ 𝑗𝑗 4 if 𝑘𝑘 = ±18 𝑒𝑒∓𝑗𝑗34𝜋𝜋 if 𝑘𝑘 = ±31 −6 if 𝑘𝑘 = ±40 𝑒𝑒±𝑗𝑗34𝜋𝜋 if 𝑘𝑘 = ±49 1 if 𝑘𝑘 = ±80 0 otherwise ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎫⎭; 𝑃𝑃𝑥𝑥

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