Re: [理工] [工數] O.D.E. (和訊號與系統??)

[BLUEBL00D (藍血魂)]

※ 引述《BLUEBL00D (藍血魂)》之銘言：
: Let H{。}　be a linear time invarient system such
: that given an input function x(t) : (D = d/dt)
: H{x(t)}= (D^4 + D^3 + D^2 + D + 1)x(t)
: (a)Does the system H{。} has an impulse response?
:    If yes , find the impulse response ; otherwise
:    give your reasons in detail .
       ┌───────────┐
 input │    L.T.I. system     │ output
------>│                      │-------->
 X(jω)│        H(jω)        │ Y(jω)
       └───────────┘
Y(jω) = H(jω) X(jω)  , by Convolution Thm.

H(jw)=(jw)^4 + (jw)^3 + (jw)^2 + (jw)^1 + 1

這個地方我不太肯定是不是這樣 訊號從大2修完就沒碰了

令 F{。} 表 Fourier Transform and F{δ(t)} = 1

   F^-1{。} 表 inverse Fourier Transform

Y(jω) = H(jw)*1 = H(jw)

       = (jw)^4 + (jw)^3 + (jw)^2 + (jw)^1 + 1

impulse response :  y(t)   = F^-1{Y(jω)}

還有我也不知道是不是每個系統都有impulse response??

: (b)Find all the possible real function x(t) such
:    that H{x(t)} = sin(t) .

這邊我不知道怎麼由訊號與系統角度寫這題 我直接當成O.D.E.解

(D^4 + D^3 + D^2 + D + 1)x(t)=sin(t)

<1>求xh: let x=exp(mt) 帶入 O.D.E.

         => m^4 + m^3 + m^2 + m + 1 = 0

         又m^5 - 1 = (m-1)(m^4 + m^3 + m^2 + m + 1)

         由m^5 - 1 = 0 得5個複根(高中複數n次方根)

         會在m = (R,θ) = (1,2kπ/5) , k=0,1,2,3,4,...

         然後再去掉m=1 ( k=0 ) 的解

1.k=1,4 => x1=exp[cos(2π/5)t]{c1cos[sin(2π/5)]t + c2sin[sin(2π/5)]t}

2.k=2,3 => x2=exp[cos(4π/5)t]{c3cos[sin(4π/5)]t + c4sin[sin(4π/5)]t}

<2>求xp:反微算子=>xp=sin(t)

<3> x=xh+xp

:                 ∞
: (c)Let y(t)  =  Σ  δ(t-k) , where δ(t) is the
:              k=-∞
:    Dirac delta function.
:                        ∞
:    Prove that y(t)  =  Σ exp(i2kπt) in the
:                     k=-∞
:    sense of distributions.

y(t)展成Fourier-series , 又y(t)的fundamental period T=1

ωk=2kπ/T=2kπ

           ∞
let y(t) = Σ ck exp(i2kπt)
        k=-∞

                0.5
其中ck =  (1/T)  ∫ δ(t) exp(-i2kπt) dt  = 1  (T=1,k為整數)
               -0.5
              ∞               ∞
故知 y(t)  =  Σ  δ(t-k)   =  Σ exp(i2kπt)
           k=-∞            k=-∞

: (d)Use the above result to find all the possible
:    real input function x(t) such that H{x(t)}= y(t).
  1. 由(b)小題已知xh

  2. let L(D) = (D^4 + D^3 + D^2 + D + 1)

     xp = [1/L(D)]y(t)

           ∞
        =  Σ {  [(i2kπ) - 1] / [(i2kπ)^5 - 1]  } exp(i2kπt)
        k=-∞

  3. x=xh+xp

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◆ From: 220.138.18.5
推 shinyhaung:原來要變成m^5-1=0才能算= =" 卡很久...
→ doom8199:要寫 h(t) , H(jw) 會稱做 h(t)的 frequency response
※ 編輯: BLUEBL00D       來自: 61.224.230.27        (03/15 02:13)
※ 編輯: BLUEBL00D       來自: 61.224.230.27        (03/15 02:15)
※ 編輯: BLUEBL00D       來自: 61.224.230.27        (03/15 02:21)
※ 編輯: BLUEBL00D       來自: 61.224.230.27        (03/15 02:29)