newell_lec11-15_100713

It can also be written in polar form as rei where x r

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: v ￿ e−p/2 t + p2 /4 ve−p/2 t . Substitute into d 2 y2 dy2 +p + p2 /4 y2 = 0 dt2 dt and find p2 −p/2 t ve 4 p2 +p(v ￿ e−p/2 t − p/2ve−p/2 t + q ve−p/2 t = 0, 4 e−p/2 t v ￿￿ − pv ￿ e−p/2 t + or, cleaning up, v ￿￿ = 0 or v = At + B. Then y2 (t) = (At + B )e−p/2 t is a second solution. In particular, the solution with A = 1, B = 0, namely y2 (t) = te−p/2 t is “essentially different”. Please do examples #1,3,5 on NSS 165. We now return to Case 3, p2 < 4q . Case 3. p2 < 4q , complex roots. See Notes on Complex Arithmetic. ￿ ￿ √ 1￿ r1 ,2 = −p ± i 4q − p2 , i = −1. 2 Solution: √ √ 1 1 2 2 y (t) = Ae 2 (−p+i 4q−p )t + Be 2 (−p−i 4q−p )t There are questions. To discuss things we must know a little complex arithmetic. A complex number z is an ordered pair (x, y ) of real numbers and can be written as x + iy . It can also be written in polar form as reiθ where x = r cos θ and y = r sin θ. Here we have used the identity eiθ = cos θ + i sin θ. 10 y✻ ✻ y ❄ ✚ ✛ r✚ ✚ ✚θ ✚ x ✚ ✚ ✚ θ ￿z = x + iy = re ✲ x ✲ Also prove cos θ = (eiθ + e−iθ )/2 and sin θ = (eiθ − e−iθ )/2i. The complex conjugate of a complex number z is denoted z ∗ or z and is defined as ¯ z ∗ = z = x − iy . Show z ∗ = re−iθ . ¯ Given z , to find z ∗ , simply change i to −i in the expression for z , e.g. (sin(1+ i))∗ = sin(1 − i). The sum of z and its complex conjugate z ∗ is real and equals 2x = 2 real z . √ √ 1 1 2) 2 If y (t) is real, and y (t) = Ae 2 (−p+i 4q−p√ t + Be 2 (−p−i 4q−p )t , then, because √ i i 2 2 e− 2 4q−p t is the complex conjugate of e 2 4q−p t , B must be the complex conju∗ gate A of A. Therefore, for p2 < 4q , the general solution of (6.7) is √ √ 1 1 2 2 y (t) = Ae 2 (−p+i 4q−p )t + A∗ e 2 (−p−i 4q−p )t (6.14) where A is in general complex. It is often simpler to write (6.14) in another form. Let A = C −iD . Then A∗ = C +iD , 2 2 when C and D are real. Then ￿ ￿ p t t y (t) = e− 2 t (C cos 4q − p2 + D sin 4q − p2 ). 2 2 There is also another convenient way to write this expression which gives one a better picture of what the solution does. A little trigonometry: C cos θ + D sin θ can be written ￿ ￿ ￿ C D C 2 + D2 √ cos θ + √ sin θ . C 2 + D2 C 2 + D2 Now define the angle φ as cos φ = √ C , + D2 sin φ = √ C2 11 D + D2 C2 and then C cos θ + D sin θ = ￿ C 2 + D2 cos(θ − φ). Therefore we can rewrite (6.14) as ￿￿ ￿ ￿ p t 2 + D 2 e− 2 t cos 2 −φ y ( t) = C 4q − p 2 where cos φ = = −2 Im A. √ C C 2 +D 2 D and sin φ √C 2 +D2 , C = A + A∗ = 2 Rl A, D = i(A − A∗ ) We see that y (t) is the product of an ￿ ￿ amplitude or envelope ￿ p C 2 + D2 exp(− t) 2 which decays, and an oscillatory sinusoidal part ￿ ￿...
View Full Document

Ask a homework question - tutors are online