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Solution3

Solution3 - EE 200 – Fall 2008(Weber Homework 3 Solutions...

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Unformatted text preview: EE 200 – Fall 2008 (Weber) Homework 3 Solutions 1. a. s(0) = s(1) = 1 0 1 0 1 1 1 0 = 1 0 The state does not change when the input is zero so for n ∈ Integers, s(n) = b. s(0) = s(1) = s(2) = s(3) = 0 1 1 0 1 0 1 0 1 1 1 1 1 1 0 1 1 1 2 1 = = = 1 1 2 1 3 1 1 0 From this we can conclude that for n ∈ Integers, s(n) = c. s(0) = s(1) = s(2) = s(3) = 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1 1 2 1 3 1 = = = 2 1 3 1 4 1 n 1 From this we can conclude that for n ∈ Integers, s(n) = n+1 1 1 2. a. A= 0 0 1 0 ,b = x(n − 2) x(n − 1) 0 1 + ,c = 0 1 1/3 1/3 x(n) = , d = 1/3 x(n − 1) x(n) 01 00 1/3 1/3 b. A= 1/2 1/2 −1/2 −1/2 1/3 0 3. For the case of A0 A0 = If the claim is true for n = k then Ak+1 Ak A = x(n − 2) x(n − 1) + 1/3x(n) = 1/3(x(n − 2) + x(n − 1) + x(n)) ,b = + 1 1 1 1 ,c = 1/3 0 , d = 1/3 = x(n) + x(n − 1) x(n) − x(n − 1) 1/2 1/2 −1/2 −1/2 x(n − 1) + x(n − 2) x(n − 1) − x(n − 2) x(n) = x(n − 1) + x(n − 2) x(n − 1) − x(n − 2) x(n − 1) + x(n) −x(n − 1) + x(n) + 1/3x(n) = 1/3(x(n − 1) + x(n − 2) + x(n)) = 1 0 0 1 cos 0 − sin 0 sin 0 cos 0 = cos(kω ) − sin(kω ) sin(kω ) cos(kω ) cos(ω ) − sin(ω ) sin(ω ) cos(ω ) = cos(kω ) cos(ω ) − sin(kω ) sin(ω ) − cos(kω ) sin(ω ) − sin(kω ) cos(ω ) sin(kω ) cos(ω ) + cos(kω ) sin(ω ) − sin(kω ) sin(ω ) + cos(kω ) cos(ω ) cos((k + 1)ω ) − sin((k + 1)ω ) sin((k + 1)ω ) cos((k + 1)ω ) = This shows the claim is true for all n = 0, 1, 2, . . .. 4. For the function to be periodic √ x1 (t) = x1 (t + p) = sin(2π (t + k1 p1 ) + sin( 2π (t + k2 p2 )) √ where p1 = 1 is the period of the ﬁrst sine function and p2 = 2 is the period of the second one and there exist integer values k1 and k2 such that p = k1 p1 = k2 p2 . This says that a period p is equal to k1 periods of the ﬁrst sine function and also equal to k2 periods of the second one. √ k1 p2 2 k1 p1 = k2 p2 ⇒ = = k2 p1 1 √ Since 2 is an irrational number and can’t be represented by a fraction of two integers, there are no values of k1 and k2 that work. The function x1 (t) is therefore not periodic. Doing the same for the second function √ √ x2 (t) = sin(2 2πt) + sin( 2πt) √ 1 P1 = √2 is the period of the ﬁrst sine function and p2 = 2 is the period of the second one. √ p2 k1 2 = = 1 =2 √ k2 p1 2 √ 1 The period of x2 (t) is then k1 p1 = 2 √2 = 2 2 5. a. The fundamental frequency is the highest frequency for which the frequencies ω1 and ω2 are integer multiples of it. ω1 ω2 = = 76π = (2 × 2 × 19)π 190π = (5 × 2 × 19)π The fundamental frequency (ω0 ) is therefore 38π radians/second. b. If ω0 = 38π , then f0 = 19 Hz and the period is 1/19 seconds. c. x(t) = = = 3 cos(ω1 t) + 7 sin(ω2 t) 3 cos(76πt) + 7 cos(190πt − π/2) 3 cos(2ω0 t) + 7 cos(5ω0 t − π/2) A2 = 3, A5 = 7, all other Ak = 0. φ5 = −π/2 and all other φk = 0. 3 ...
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