查看“︁訊號與系統/傅立葉轉換的範例”︁的源代码

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== 範例5.1 ==
試求y(t)=<math> e^{-\alpha t}u(t)</math> ，<math>\alpha>0</math>之傅立葉轉換，並繪出y(t)的頻譜。
    

                                          © G. E. Carlson, Signal and Linear System Analysis, 2nd ed., John Wiley & Sons, 1998.


 【解】 <math>\int_{-\infty}^{\infty}y(t) e^{-j2\pi ft}\, dt</math>
        =<math>\int_{-\infty}^{\infty}y(t) e^{-\alpha t} u(t)  e^{-j2\pi ft}\, dt</math>
       =<math>\int_{-\infty}^{\infty}y(t)e^{-(\alpha+j2\pi f)t} \, dt</math>
       =<math>\frac{-1}{\alpha+j2\pi f} e^{-(\alpha+j2\pi f)t}\, dt\mid_0^\infty</math>

       =<math>\frac{1}{\alpha+j2\pi f} </math>

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== 範例5.1(續) ==
<math>\mid Y(f)\mid=\mid\frac{1}{\alpha+j2\pi f}\mid=\frac{1}{\sqrt{\alpha^2+4\pi^2 f^2}}</math>

<math>\angle Y(f)=\angle(\frac{1}{\alpha+j2\pi f})=\angle1-\angle(\alpha+j2\pi f)</math>

<math>=0-\tan^{-1}(\frac{2\pi f}{\alpha})</math>

<math>=-\tan^-{1}(\frac{2\pi f}{\alpha})</math>


                                  © G. E. Carlson, Signal and Linear System Analysis, 2nd ed., John Wiley & Sons, 1998.
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== 範例5.2 ==
試求<math>\chi_2(t) =e^{-a\mid t\mid}</math>，a>0之傅立葉轉換

【解】 <math>\chi_2(f)</math>=<math>\Im</math>{<math>\chi_2(t)</math>} =<math>\int_{-\infty}^{\infty}y(t)\chi_2(t) e^{-j2\pi ft}\, dt</math>
  = <math>\int_{-\infty}^{0}e^{a t} e^{-j2\pi ft}\, dt</math>+<math>\int_{0}^{\infty}e^{-a t}  e^{-j2\pi ft}\, dt</math>
  = <math>\frac{1}{a-j2\pi f} </math> +<math>\frac{1}{a+j2\pi f} </math>
  =<math>\frac{2a}{a^2+(2\pi f)^2} </math>

  假設a=2 


                                          ©余兆棠、李志鵬，信號與系統， 滄海書局，2007。
     
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== 範例5.3 ==
試求<math>\chi(t)=rect\frac{t}{\tau}</math>的傅立葉轉換。

【解】


                                 © B. P. Lathi, Signal Processing and Linear Systems, Berkeley-Cambridge Press, 1998.


            (1)用頻率f：<math>\Chi(f)</math>=<math>\Im</math>{<math>\chi_2(t)</math>} =<math>\int_{-\infty}^{\infty}\chi(t) e^{-j2\pi ft}\, dt</math>
=<math>\int_{-\infty}^{\infty}rect\frac{t}{\tau} e^{-j2\pi ft}\, dt</math>
=<math>\int_{\frac{-\tau}{2}}^{\frac{\tau}{2}}1 e^{-j2\pi ft}\, dt</math>
=<math>\frac{1}{-j2\pi f}e^{-j2\pi ft}\mid_\frac{\tau}{2}^\frac{\tau}{2}</math>

=<math>\frac{1}{-j2\pi f}e^{-j2\pi ft} </math>(<math>e^{j\pi f\tau}- e^{j\pi f\tau} </math>) 

=<math>\tau\frac{\sin(\pi f\tau)}{\pi f\tau}</math>=<math>\tau sinc(\tau f)</math>
             sinc(x)=<math>\frac{\sin(\pi\chi)}{\pi\chi}</math>
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== 範例5.3(續) ==
(2)用角頻率<math>\omega</math>：
<math>\Chi_\omega(\omega) =\int_{-\infty}^{infty} e^{-j\omega t}\, dt</math>

<math>=\int_{-\infty}^{infty}rect(\frac{t}{\tau}) e^{-j\omega t}\, dt</math>

=<math>\int_{-\frac{t}{\tau}}^{\frac{t}{\tau}}1 e^{-j\omega t}\, dt</math>

=<math>\frac{1}{-j\omega}</math>(<math>e^{-j\omega\frac{t}{\tau}}-e^{-j\omega\frac{t}{\tau}}</math>)

=<math>\tau\frac{\sin\frac{\omega\tau}{2}}{\frac{\omega\tau}{2}}</math>

=<math>\tau sa(\frac{\omega\tau}{2})</math>
            <math>sa(y)=\frac{\sin(y)}{y}</math>
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== 範例5.3(續) ==
定義：<math>sinc(x)=\frac{\sin(\pi x)}{\pi x}</math>
     <math>Sa(y)=\frac{\sin(y)}{y}</math>
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== 範例5.3(續) ==



                             © B. P. Lathi, Signal Processing and Linear Systems, Berkeley-Cambridge Press, 1998.

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== 範例5.4 ==
試求<math>x_b(t)=rect(t+\frac{1}{2}) -rect(t-\frac{1}{2})</math>的傅立葉轉換


【解】<math>x_b(f)=\int_{-1}^{0} e^{-j2\pi ft}\, dt-\int_{0}^{1} e^{-j2\pi ft}\, dt</math>

=<math>-\frac{1}{j2\pi f}</math>[<math>e^{-j2\pi ft}\mid_-1^0-e^{-j2\pi ft}\mid_0^1</math>]

=<math> -\frac{1}{j2\pi f} </math>(<math>1-e^{j2\pi ft}-e^{-j2\pi ft}+1</math>)
                                           <math>e^{j2\pi ft}+e^{-j2\pi ft}</math>

                                                 =<math>2\cos2\pi f</math>
=<math> -\frac{1}{j2\pi f}</math>[<math>2-2\cos(2\pi f)</math>]

=<math>\frac{j}{\pi f}(1-\cos2\pi f)</math>
                                         <math>1-\cos2\pi f</math>

                                          =<math>2\sin^2(\pi f)</math>
=<math>j2\frac{\sin^2\pi f}{\pi f}</math>

=<math>j2\pi fsinc^2(f)</math>



                                © Rodger E. Ziemer, William H. Tranter, D. Ronald Fannin, Signals & Systems:  Continuous and Discrete, 4th ed., Prentice Hall International, 1998.

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== 範例5.4(續) ==

                                      © Rodger E. Ziemer, William H. Tranter, D. Ronald Fannin, Signals & Systems:  Continuous and Discrete, 4th ed., Prentice Hall International, 1998.


說明：<math>\Chi_b(f)=j2\pi fsinc^2(f)</math>
     <math>\Rightarrow        \Chi_b\omega(\omega)=j\omega(\frac{\sin\pi f}{\pi f})^2</math>
                                 =<math>j\omega</math>[<math>\frac{\sin(\frac{\omega}{2})} {\frac{\omega}{2}}</math>]<math>^2</math>                                 
                                 =<math>j\omega sa^2\frac{\omega}{2}</math>
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== 範例5.5 ==
試求直流訊號<math>\chi(t)=1</math>,<math>-\infty<t<\infty</math>的傅立葉轉換

【解】<math>\Chi(f)</math>=<math>\int_{-\infty}^{\infty} 1e^{-j2\pi ft}\, dt</math>

     =<math>\lim_{T\to \infty}\int_{-\frac{t}{2}}^{\frac{t}{2}} 1e^{-j2\pi ft}\, dt</math>

     =<math>\lim_{T\to \infty}\frac{1}{-j2\pi f}e^{-j2\pi ft}\, dt \mid_-\frac{t}{2}^\frac{t}{2}</math>

    =<math>\lim_{T\to \infty}\frac{1}{-j2\pi f} </math>[<math>e^{-j2\pi ft} - e^{j2\pi ft}</math>]

=<math>\frac{1}{\pi f}\lim_{T\to \infty}\sin(\pi fT) </math>

因為<math>\lim_{T\to \infty}\sin(\pi fT)</math>不存在，故<math>\chi(t)=1</math> 的傅立葉轉換不收斂。
由此範例可知，直流訊號<math>\chi(t)=1</math>的傅立葉轉換不存在。很多常用的訊號
模型 如<math>\cos,\sin,\delta(t),u(t)</math>等其傅立葉轉換均不存在!
下節將介紹廣義的傅立葉轉換(generalized Fourier transform)來解決此一問題

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== 廣義的傅立葉轉換 ==
為了讓一些常用的訊號模型如<math>\sin,\cos</math>複指數訊號<math>(e^{-j2\pi ft}),\delta(t),u(t)</math>可作傅立葉轉換，將其擴展為包含極限的定義，也就是廣義的傅立葉轉換。

例如：
      (1)<math>\Im</math> {<math>\cos2\pi f_0 t</math>} =<math>\lim_{\alpha\to 0}\Im</math>  {<math>exp(-\alpha\mid t\mid)\cos2\pi f_0 t</math>} ,<math>\alpha>0</math>

      =<math>\frac{1}{2}</math> {<math>\delta(f-f_0)+\delta(f+f_0)</math>}

     (2)<math>\Im</math> {<math>\delta(t)</math>}= <math>\lim_{\epsilon\to 0}</math><math>\Im </math> {<math>\delta_\epsilon(t)=rect(t/2\epsilon)/(2\epsilon)</math>} =1
  
     (3)<math>\Im</math> {A} =<math>\lim_{T\to \infty}\Im</math> {<math>A rect(t/T)</math>}=A<math>\delta(f)</math>

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== 範例5.6 ==
試求函數<math>\delta(t)</math>的傅立葉轉換。

【解】<math>\Im</math> {<math>\delta(t)</math> }= <math>\int_{-\infty}^{\infty}\delta(t) e^{-j2\pi ft}\, dt</math>
                                  <math>\chi(t)\delta(t)=\chi(0)\delta(t)</math>

    = <math>\int_{-\infty}^{\infty}\delta(t) e^{-j2\pi f0}\, dt</math>
    =<math>\int_{-\infty}^{\infty}\delta(t), dt</math>=1

所以：<math>\delta(t)\leftrightarrow 1</math>

                                         ©余兆棠、李志鵬，信號與系統， 滄海書局，2007。
由上圖知，單位脈衝訊號<math>\delta(t)</math>包含所有頻率且不同頻率之振幅均相等。

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== 範例5.7 ==
試求<math>\delta(f)</math>的傅立葉逆轉換

 【解】<math>\Im^{-1}</math> {<math>\delta(f)</math> }= <math>\int_{-\infty}^{\infty}\delta(f) e^{j2\pi ft}\, df</math>=<math>\int_{-\infty}^{\infty}\delta(f) e^{j2\pi 0t}\, df</math>=<math>\int_{-\infty}^{\infty}\delta(f)\, df=1</math>

所以：<math>1\leftrightarrow \delta(f)</math>
物理意義：大小為1的常數訊號為一直流訊號，故其頻譜只在f=0處存在
                     一單位脈衝函數
                            
                                             ©余兆棠、李志鵬，信號與系統， 滄海書局，2007。


注意：根據上述關係可知。<math>Im</math> {<math>1</math>}=<math>\int_{-\infty}^{\infty} e^{-j\pi ft}\, dt=\delta(f)</math>
積分公式：<math>\int_{-\infty}^{\infty} e^{-j\pi ft}\, dt=\delta(f)</math>

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== 範例5.8 ==
試求<math>2\pi\delta(\omega)</math>的傅立葉逆轉換。

【解】<math>\Im^{-1}</math> {<math>2\pi\delta(\omega)</math>}=<math>\frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\delta(\omega) e^{j\omega t}\, d\omega=\int_{-\infty}^{\infty}\delta(\omega) e^{j0 t}\, d\omega=\int_{-\infty}^{\infty}\delta(\omega)\, d\omega=1</math>
所以：<math>1\leftrightarrow2\pi\delta(\omega)</math>
  注意 ：根據上述關係可知<math>Im</math> {1} =<math>\int_{-\infty}^{\infty}1e^{-j\omega t}\, dt=2\pi\delta(\omega)</math>
積分公式：<math>\int_{-\infty}^{\infty} e^{-j\omega t}\, dt=\delta(\omega)</math>

                                      © B. P. Lathi, Signal Processing and Linear Systems, Berkeley-Cambridge Press, 1998.

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== <math>\delta(k\chi)v.s.\delta(\chi)</math> ==
        <math>\delta(k\chi)=\frac{1}{\left| k\right|}\delta(\chi)</math>    <math>k\ne 0</math>

【證明】
(1)假設k>0       <math>\int_{-\infty}^{\infty} \delta(k\chi)\, dx</math>
                                                  令y=<math>k\chi</math>
                                                   dy=<math>k d\chi</math>

                           =<math>\int_{-\infty}^{\infty} \delta(y)\, (\frac{y}{k})</math>  

                           =<math>\frac{1}{k}\int_{-\infty}^{\infty} \delta(y)\, dy</math>

                           =<math>\frac{1}{k}\int_{-\infty}^{\infty} \delta(x)\, dx</math>

                           =<math>\int_{-\infty}^{\infty} \frac{1}{k}\delta(x)\, dx</math>

                         故<math>\delta(k\chi)=\frac{1}{k}\delta(\chi)</math>   k>0
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== <math>\delta(k\chi)v.s.\delta(\chi)</math>(續) ==
 (2)假設k<0         <math>\int_{-\infty}^{\infty} \delta(k\chi)\, dx</math>
                                            令y=<math>k\chi</math>
                                             dy=<math>kd\chi</math>
                    
                  因為k<0
                      
                           =<math>\int_{-\infty}^{\infty} \delta(y)\, \frac{dy}{k}</math>  

                           =-<math>\frac{1}{k}\int_{-\infty}^{\infty} \delta(y)\, dy</math>

                           =-<math>\frac{1}{k}\int_{-\infty}^{\infty} \delta(x)\, dx</math>

                           =<math>\int_{-\infty}^{\infty} -\frac{1}{k}\delta(x)\, dx</math>

                         故<math>\delta(k\chi)=-\frac{1}{k}\delta(\chi)</math>   k<0

結論：   (1)<math>\delta(k\chi)=\frac{1}{\left|  k \right|}\delta(\chi)</math>    <math>k\ne 0</math>
        
        (2)k=-1

<math>\delta(-\chi)=\delta(\chi)</math>即<math>\delta(\chi)</math>為一偶函數。
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== 範例5.9 ==
試求<math>\chi(t)=e^{j2\pi f_0t}</math>之傅立葉轉換，其中<math>f_0</math>為一固定頻率。

【解】<math>\Chi(f)=\int_{-\infty}^{\infty}\chi(t) e^{-j2\pi ft}\, dt</math>

                  =<math>\int_{-\infty}^{\infty}\chi(t)e^{j2\pi f_0t} e^{-j2\pi ft}\, dt</math>
    
                  =<math>\int_{-\infty}^{\infty}\chi(t) e^{-j2\pi (f-f_0)t}\, dt</math>
                                            根據積分公式<math>\int_{-\infty}^{\infty}\chi(t) e^{-j2\pi ft}\, dt=\delta(f)</math>

                  =<math>\delta(f-f_0)</math> 

<math>\Chi_\omega(\omega)=\Chi(\frac{\omega}{2\pi})=\delta(\frac{\omega-\omega_0}{2\pi})</math>
      <math>\omega=2\pi f</math>
       <math>\omega_0=2\pi f_0</math>

                     =<math>2\pi\delta(\omega-\omega_0)</math>

由<math>\Chi(f)</math>(或<math>\Chi_\omega(\omega)</math> )可知，<math>\chi(t)=e^{j2\pi f_0t}</math>為僅具有單一頻率<math>f_0</math>的訊號
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== 範例5.10 ==
試求符號函數(sign function)y(t)=<math>\sgn(t)</math>的傅立葉轉換。
 
【解】     符號函數
                    <math>sgn(t)=</math>1  t>0 ;  -1  t<0

<math>\Rightarrow\sgn(t)=</math> [<math>\lim_{a \to 0}e^{-at}u(t)-e^{-a(-t)}u(-t)</math>] a>0

Y(f)=<math>\Im</math> {<math>sgn(t)</math>}= [<math>\lim_{a \to 0}e^{-at}u(t)-e^{-a(-t)}u(-t)</math>]
       
                                                  © Charls L. Phillips, John M. Parr, Eve A. Riskin, Signals, Systems, and Transforms, 3rd ed., Pearson Education, 2003.

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== 範例5.10(續) ==
<math>\Im</math> [e^{-at}u(t)-e^{-a(-t)}u(-t)]

=<math>\int_{-\infty}^{\infty}</math> { <math> e^{-at}u(t)-e^{-a(-t)}u(t)</math> }
<math>e^{-j2\pi ft}\, dt</math> 

=<math>\int_{-\infty}^{\infty}e^{-at}u(t) e^{-j2\pi ft}\, dt</math>-<math>\int_{-\infty}^{\infty}e^{-a(-t)}u(-t) e^{-j2\pi ft}\, dt</math>

=<math>\int_{0}^{\infty}e^{-at} e^{-j2\pi ft}\, dt</math>-<math>\int_{-\infty}^{0}e^{-a(-t)}u(t) e^{-j2\pi ft}\, dt</math>

=<math>\int_{0}^{\infty} e^{-(a+j2\pi f)t}\, dt</math>-<math>\int_{-\infty}^{0} e^{-(-a+j2\pi f)t}\, dt</math>

=<math>\frac{1}{a+j2\pi f}+\frac{1}{-a+j2\pi f}=\frac{j4\pi f}{-[a^2+4\pi^2 f^2]}</math>
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== 範例5.10(續) ==

Y(f)=<math>\Im</math> { sgn(t) }=<math>\lim_{a \to 0}\frac{j4\pi f}{-(a^2+4\pi^2 f^2)}</math>=<math>\frac{j4\pi f}{-4\pi^2 f^2}</math>=<math>\frac{1}{j\pi f}</math>

所以sgn(t)<math>\leftrightarrow\frac{1}{j\pi f} (=\frac{2}{j\omega },\omega=2\pi f)</math>

                                             © G. E. Carlson, Signal and Linear System Analysis, 2nd ed., John Wiley & Sons, 1998.

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