訊號與系統/傅立葉轉換的範例

範例5.1

試求y(t)=$e^{-\alpha t}u(t)$ ，$\alpha >0$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】 $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}y(t)e^{-j2\pi ft}dt$
       =$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}y(t)e^{-\alpha t}u(t)e^{-j2\pi ft}dt$
      =$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}y(t)e^{-(\alpha +j2\pi f)t}dt$
      =$\frac{-1}{\alpha +j2\pi f}{e}^{-(\alpha +j2\pi f)t}dt\underset{0}{\overset{\mathrm{\infty }}{\mid }}$

      =$\frac{1}{\alpha +j2\pi f}$

---

範例5.1(續)

$\mid Y(f)\mid =\mid \frac{1}{\alpha +j2\pi f}\mid =\frac{1}{\sqrt{{\alpha }^2+4{\pi }^2{f}^2}}$

$\mathrm{\angle }Y(f)=\mathrm{\angle }(\frac{1}{\alpha +j2\pi f})=\mathrm{\angle }1-\mathrm{\angle }(\alpha +j2\pi f)$

$=0-{\tan }^{-1}(\frac{2\pi f}{\alpha })$

$=-{\tan }^{-}1(\frac{2\pi f}{\alpha })$

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

---

範例5.2

試求${\chi }_2(t)=e^{-a\mid t\mid }$，a>0之傅立葉轉換

【解】 ${\chi }_2(f)$=$\mathrm{\Im }${${\chi }_2(t)$} =$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}y(t){\chi }_2(t)e^{-j2\pi ft}dt$

 = $\underset{-\mathrm{\infty }}{\overset{0}{\int }}{e}^{at}{e}^{-j2\pi ft}dt$+$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-at}{e}^{-j2\pi ft}dt$
 = $\frac{1}{a-j2\pi f}$ +$\frac{1}{a+j2\pi f}$
 =$\frac{2a}{a^2+(2\pi f{)}^2}$

 假設a=2 

                                         ©余兆棠、李志鵬，信號與系統， 滄海書局，2007。
    

---

範例5.3

試求$\chi (t)=rect\frac{t}{\tau }$的傅立葉轉換。

【解】

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

           (1)用頻率f：$\mathrm{X}(f)$=$\mathrm{\Im }${${\chi }_2(t)$} =$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\chi (t)e^{-j2\pi ft}dt$

=$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}rect\frac{t}{\tau }{e}^{-j2\pi ft}dt$ =$\underset{\frac{-\tau }{2}}{\overset{\frac{\tau }{2}}{\int }}1e^{-j2\pi ft}dt$ =$\frac{1}{-j2\pi f}{e}^{-j2\pi ft}\underset{\frac{\tau }{2}}{\overset{\frac{\tau }{2}}{\mid }}$

=$\frac{1}{-j2\pi f}{e}^{-j2\pi ft}$($e^{j\pi f\tau }-e^{j\pi f\tau }$)

=$\tau \frac{\sin (\pi f\tau )}{\pi f\tau }$=$\tau sinc(\tau f)$

            sinc(x)=$\frac{\sin (\pi \chi )}{\pi \chi }$

---

範例5.3(續)

(2)用角頻率$\omega$： ${\mathrm{X}}_{\omega }(\omega )=\underset{-\mathrm{\infty }}{\overset{infty}{\int }}{e}^{-j\omega t}dt$

$=\underset{-\mathrm{\infty }}{\overset{infty}{\int }}rect(\frac{t}{\tau })e^{-j\omega t}dt$

=$\underset{-\frac{t}{\tau }}{\overset{\frac{t}{\tau }}{\int }}1e^{-j\omega t}dt$

=$\frac{1}{-j\omega }$($e^{-j\omega \frac{t}{\tau }}-e^{-j\omega \frac{t}{\tau }}$)

=$\tau \frac{\sin \frac{\omega \tau }{2}}{\frac{\omega \tau }{2}}$

=$\tau sa(\frac{\omega \tau }{2})$

           $sa(y)=\frac{\sin (y)}{y}$

---

範例5.3(續)

定義：$sinc(x)=\frac{\sin (\pi x)}{\pi x}$

    $Sa(y)=\frac{\sin (y)}{y}$

---

範例5.3(續)

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

---

範例5.4

試求$x_b(t)=rect(t+\frac{1}{2})-rect(t-\frac{1}{2})$的傅立葉轉換

【解】$x_b(f)=\underset{-1}{\overset{0}{\int }}{e}^{-j2\pi ft}dt-\underset{0}{\overset{1}{\int }}{e}^{-j2\pi ft}dt$

=$-\frac{1}{j2\pi f}$[$e^{-j2\pi ft}{\mid }_{-}{1}^0-e^{-j2\pi ft}\underset{0}{\overset{1}{\mid }}$]

=$-\frac{1}{j2\pi f}$($1-e^{j2\pi ft}-e^{-j2\pi ft}+1$)

                                          $e^{j2\pi ft}+e^{-j2\pi ft}$

                                                =$2\cos 2\pi f$

=$-\frac{1}{j2\pi f}$[$2-2\cos (2\pi f)$]

=$\frac{j}{\pi f}(1-\cos 2\pi f)$

                                        $1-\cos 2\pi f$

                                         =$2{\sin }^2(\pi f)$

=$j2\frac{{\sin }^2\pi f}{\pi f}$

=$j2\pi fsin{c}^2(f)$

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

---

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：${\mathrm{X}}_b(f)=j2\pi fsin{c}^2(f)$

    $\Rightarrow {\mathrm{X}}_b\omega (\omega )=j\omega (\frac{\sin \pi f}{\pi f}{)}^2$
                                =$j\omega$[$\frac{\sin (\frac{\omega }{2})}{\frac{\omega }{2}}$]$2^{}$                                 
                                =$j\omega s{a}^2\frac{\omega }{2}$

---

範例5.5

試求直流訊號$\chi (t)=1$,$-\mathrm{\infty }<t<\mathrm{\infty }$的傅立葉轉換

【解】$\mathrm{X}(f)$=$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}1e^{-j2\pi ft}dt$

    =$\underset{T\to \mathrm{\infty }}{\lim }\underset{-\frac{t}{2}}{\overset{\frac{t}{2}}{\int }}1e^{-j2\pi ft}dt$

    =$\underset{T\to \mathrm{\infty }}{\lim }\frac{1}{-j2\pi f}{e}^{-j2\pi ft}dt{\mid }_{-}{\frac{t}{2}}^{\frac{t}{2}}$

   =$\underset{T\to \mathrm{\infty }}{\lim }\frac{1}{-j2\pi f}$[$e^{-j2\pi ft}-e^{j2\pi ft}$]

=$\frac{1}{\pi f}\underset{T\to \mathrm{\infty }}{\lim }\sin (\pi fT)$

因為$\underset{T\to \mathrm{\infty }}{\lim }\sin (\pi fT)$不存在，故$\chi (t)=1$ 的傅立葉轉換不收斂。 由此範例可知，直流訊號$\chi (t)=1$的傅立葉轉換不存在。很多常用的訊號 模型 如$\cos ,\sin ,\delta (t),u(t)$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

---

廣義的傅立葉轉換

為了讓一些常用的訊號模型如$\sin ,\cos$複指數訊號$(e^{-j2\pi ft}),\delta (t),u(t)$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)$\mathrm{\Im }$ {$\cos 2\pi f_{0}t$} =$\underset{\alpha \to 0}{\lim }\mathrm{\Im }$  {$exp(-\alpha \mid t\mid )\cos 2\pi f_{0}t$} ,$\alpha >0$

     =$\frac{1}{2}$ {$\delta (f-f_0)+\delta (f+f_0)$}

    (2)$\mathrm{\Im }$ {$\delta (t)$}= $\underset{ϵ\to 0}{\lim }$$\mathrm{\Im }$ {${\delta }_{ϵ}(t)=rect(t/2ϵ)/(2ϵ)$} =1
 
    (3)$\mathrm{\Im }$ {A} =$\underset{T\to \mathrm{\infty }}{\lim }\mathrm{\Im }$ {$Arect(t/T)$}=A$\delta (f)$

---

範例5.6

試求函數$\delta (t)$的傅立葉轉換。

【解】$\mathrm{\Im }$ {$\delta (t)$ }= $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (t)e^{-j2\pi ft}dt$

                                 $\chi (t)\delta (t)=\chi (0)\delta (t)$

   = $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (t)e^{-j2\pi f0}dt$
   =$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (t),dt$=1

所以：$\delta (t)\leftrightarrow 1$

                                        ©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$\delta (t)$包含所有頻率且不同頻率之振幅均相等。

---

範例5.7

試求$\delta (f)$的傅立葉逆轉換

【解】${\mathrm{\Im }}^{-1}$ {$\delta (f)$ }= $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (f)e^{j2\pi ft}df$=$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (f)e^{j2\pi 0t}df$=$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (f)df=1$

所以：$1\leftrightarrow \delta (f)$ 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。$Im$ {$1$}=$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j\pi ft}dt=\delta (f)$ 積分公式：$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j\pi ft}dt=\delta (f)$

---

範例5.8

試求$2\pi \delta (\omega )$的傅立葉逆轉換。

【解】${\mathrm{\Im }}^{-1}$ {$2\pi \delta (\omega )$}=$\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}2\pi \delta (\omega )e^{j\omega t}d\omega =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (\omega )e^{j0t}d\omega =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (\omega )d\omega =1$ 所以：$1\leftrightarrow 2\pi \delta (\omega )$

 注意 ：根據上述關係可知$Im$ {1} =$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}1e^{-j\omega t}dt=2\pi \delta (\omega )$

積分公式：$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j\omega t}dt=\delta (\omega )$

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

---

$\delta (k\chi )v.s.\delta (\chi )$

       $\delta (k\chi )=\frac{1}{\left|k\right|}\delta (\chi )$    $k\ne 0$

【證明】 (1)假設k>0 $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (k\chi )dx$

                                                 令y=$k\chi$
                                                  dy=$kd\chi$

                          =$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (y)(\frac{y}{k})$  

                          =$\frac{1}{k}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (y)dy$

                          =$\frac{1}{k}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (x)dx$

                          =$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{k}\delta (x)dx$

                        故$\delta (k\chi )=\frac{1}{k}\delta (\chi )$   k>0

---

$\delta (k\chi )v.s.\delta (\chi )$(續)

(2)假設k<0         $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (k\chi )dx$
                                           令y=$k\chi$
                                            dy=$kd\chi$
                   
                 因為k<0
                     
                          =$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (y)\frac{dy}{k}$  

                          =-$\frac{1}{k}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (y)dy$

                          =-$\frac{1}{k}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta (x)dx$

                          =$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}-\frac{1}{k}\delta (x)dx$

                        故$\delta (k\chi )=-\frac{1}{k}\delta (\chi )$   k<0

結論： (1)$\delta (k\chi )=\frac{1}{\left|k\right|}\delta (\chi )$ $k\ne 0$

       (2)k=-1

$\delta (-\chi )=\delta (\chi )$即$\delta (\chi )$為一偶函數。

---

範例5.9

試求$\chi (t)=e^{j2\pi f_{0}t}$之傅立葉轉換，其中$f_0$為一固定頻率。

【解】$\mathrm{X}(f)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\chi (t)e^{-j2\pi ft}dt$

                 =$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\chi (t)e^{j2\pi f_{0}t}{e}^{-j2\pi ft}dt$
   
                 =$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\chi (t)e^{-j2\pi (f-f_0)t}dt$
                                           根據積分公式$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\chi (t)e^{-j2\pi ft}dt=\delta (f)$

                 =$\delta (f-f_0)$ 

${\mathrm{X}}_{\omega }(\omega )=\mathrm{X}(\frac{\omega }{2\pi })=\delta (\frac{\omega -{\omega }_0}{2\pi })$

     $\omega =2\pi f$
      ${\omega }_0=2\pi f_0$

                    =$2\pi \delta (\omega -{\omega }_0)$

由$\mathrm{X}(f)$(或${\mathrm{X}}_{\omega }(\omega )$ )可知，$\chi (t)=e^{j2\pi f_{0}t}$為僅具有單一頻率$f_0$的訊號

---

範例5.10

試求符號函數(sign function)y(t)=$\mathrm{sgn}(t)$的傅立葉轉換。

【解】 符號函數

                   $sgn(t)=$1  t>0 ;  -1  t<0

$\Rightarrow \mathrm{sgn}(t)=$ [$\underset{a\to 0}{\lim }{e}^{-at}u(t)-e^{-a(-t)}u(-t)$] a>0

Y(f)=$\mathrm{\Im }$ {$sgn(t)$}= [$\underset{a\to 0}{\lim }{e}^{-at}u(t)-e^{-a(-t)}u(-t)$]

                                                 © Charls L. Phillips, John M. Parr, Eve A. Riskin, Signals, Systems, and Transforms, 3rd ed., Pearson Education, 2003.

---

範例5.10(續)

$\mathrm{\Im }$ [e^{-at}u(t)-e^{-a(-t)}u(-t)]

=$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}$ { $e^{-at}u(t)-e^{-a(-t)}u(t)$ } $e^{-j2\pi ft}dt$

=$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-at}u(t)e^{-j2\pi ft}dt$-$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-a(-t)}u(-t)e^{-j2\pi ft}dt$

=$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-at}{e}^{-j2\pi ft}dt$-$\underset{-\mathrm{\infty }}{\overset{0}{\int }}{e}^{-a(-t)}u(t)e^{-j2\pi ft}dt$

=$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-(a+j2\pi f)t}dt$-$\underset{-\mathrm{\infty }}{\overset{0}{\int }}{e}^{-(-a+j2\pi f)t}dt$

=$\frac{1}{a+j2\pi f}+\frac{1}{-a+j2\pi f}=\frac{j4\pi f}{-[a^2+4{\pi }^2{f}^2]}$

---

範例5.10(續)

Y(f)=$\mathrm{\Im }$ { sgn(t) }=$\underset{a\to 0}{\lim }\frac{j4\pi f}{-(a^2+4{\pi }^2{f}^2)}$=$\frac{j4\pi f}{-4{\pi }^2{f}^2}$=$\frac{1}{j\pi f}$

所以sgn(t)$\leftrightarrow \frac{1}{j\pi f}(=\frac{2}{j\omega },\omega =2\pi f)$

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

---