Solution3

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online