# 问题的提出

## 求解分段函数的第1部分对应的h1(t)

### 利用傅里叶变换的对称性：

{ f ( t )   ⇔ F ( ω ) F ( t )   ⇔ 2 π f ( − ω ) \begin{cases} f(t)\, \Leftrightarrow F(\omega) \\ F(t)\, \Leftrightarrow 2\pi f(-\omega) \end{cases}

f ( t ) = { E   , ∣ t ∣ ≤ τ 2 0   , 其 他 ⇔ F ( ω ) = 2 E sin ⁡ ω τ 2 ω f(t)=\begin{cases} E\,,|t| \leq {\large \tau \over 2} \\ 0\,,其他 \end{cases} \Leftrightarrow F(\omega)=2E{\sin{\large \omega \tau \over 2} \over \omega}

### 推出：(1)

H 1 ( ω ) = { T b   , 0 ≤ ∣ ω ∣ ≤ π ( 1 − α ) T b 0   , 其 他 ⇔ h 1 ( t ) = 2 T b sin ⁡ π ( 1 − α ) t T b 2 π t = T b sin ⁡ π ( 1 − α ) t T b π t ⇒ H 1 ( f ) = { T b   , 0 ≤ ∣ f ∣ ≤ ( 1 − α ) 2 T b 0   , 其 他 ⇔ h 1 ( t ) = 2 T b sin ⁡ π ( 1 − α ) t T b 2 π t = T b sin ⁡ π ( 1 − α ) t T b π t \begin{aligned} H_1(\omega) &= \begin{cases} T_b\,,0\leq |\omega| \leq \large{ \pi (1-\alpha) \over T_b} \\ 0\,,其他 \end{cases} \Leftrightarrow \large h_1(t)=2T_b{\sin \large{ \pi (1-\alpha)t \over T_b } \over 2\pi t} = T_b{\sin \large{ \pi (1-\alpha)t \over T_b } \over \pi t} \\ \Rightarrow H_1(f) &= \begin{cases} T_b\,,0\leq |f| \leq \large{ (1-\alpha) \over 2T_b} \\ 0\,,其他 \end{cases} \Leftrightarrow \large h_1(t)=2T_b{\sin \large{ \pi (1-\alpha)t \over T_b } \over 2\pi t} = T_b{\sin \large{ \pi (1-\alpha)t \over T_b } \over \pi t} \end{aligned}

## 求解分段函数的第2部分对应的h2(t)

H 2 ( f ) = T b 2 { 1 + cos ⁡ [ π T b α ( ∣ f ∣ − 1 − α 2 T b ) ] }   , 1 − α 2 T b ≤ ∣ f ∣ ≤ 1 + α 2 T b ⇒ H 2 ( ω ) = T b 2 { 1 + cos ⁡ [ T b 2 α ( ∣ ω ∣ − π ( 1 − α ) T b ) ] }   , π ( 1 − α ) T b ≤ ∣ ω ∣ ≤ π ( 1 + α ) T b \large \begin{aligned} H_2(f) &= {T_b \over 2}\{1+\cos[{\pi T_b \over \alpha}(|f|-{1-\alpha \over 2T_b})]\} \,,{1-\alpha \over 2T_b} \leq |f| \leq {1+\alpha \over 2T_b} \\ \Rightarrow H_2(\omega) &= {T_b \over 2}\{1+\cos[{ T_b \over 2\alpha}(|\omega|-{\pi(1-\alpha) \over T_b})]\} \,,{\pi(1-\alpha) \over T_b} \leq |\omega| \leq {\pi(1+\alpha) \over T_b} \end{aligned}

### 借助以下公式：

∫ cos ⁡ ( a ω ± b ) e j ω t d ω = cos ⁡ ( a ω ± b ) e j ω t j t + ∫ a s i n ( a ω ± b ) j t e j ω t d ω = cos ⁡ ( a ω ± b ) e j ω t j t − a s i n ( a ω ± b ) t 2 e j ω t + ∫ a 2 c o s ( a ω ± b ) t 2 e j ω t d ω \begin{aligned} \int\cos(a\omega \pm b)e^{j\omega t}d\omega &= {\cos(a\omega \pm b)e^{j\omega t} \over jt}+\int{asin(a\omega \pm b) \over jt}e^{j\omega t}d\omega \\ &= {\cos(a\omega \pm b)e^{j\omega t} \over jt} - {asin(a\omega \pm b) \over t^2}e^{j\omega t}+\int{a^2cos(a\omega \pm b) \over t^2}e^{j \omega t}d\omega \end{aligned}

∫ cos ⁡ ( a ω ± b ) e j ω t d ω = cos ⁡ ( a ω ± b ) j t e j ω t − a sin ⁡ ( a ω ± b ) t 2 e j ω t 1 − a 2 t 2 \int\cos(a\omega \pm b)e^{j\omega t}d\omega ={\large{ {\cos(a\omega \pm b) \over jt}e^{j\omega t} - {a \sin(a\omega \pm b) \over t^2}e^{j\omega t} \over 1-{a^2 \over t^2}}}

### 推出：(3)

∫ − 1 + α T b π − 1 − α T b π cos ⁡ [ T b 2 α ( − ω − π ( 1 − α ) T b ) ] e j ω t d ω = ∫ − 1 + α T b π − 1 − α T b π cos ⁡ [ − T b 2 α ω − π ( 1 − α ) 2 α ] e j ω t d ω = cos ⁡ [ − T b 2 α ω − π ( 1 − α ) 2 α ] j t e j ω t − − T b 2 α sin ⁡ [ − T b 2 α ω − π ( 1 − α ) 2 α ] t 2 e j ω t 1 − ( − T b 2 α ) 2 t 2 ∣ − 1 + α T b π − 1 − α T b π = 1 1 − T b 2 4 α 2 t 2 [ e j ( − 1 − α T b π ) t j t − ( − e j ( − 1 + α T b π ) t j t ) ] = 1 1 − T b 2 4 α 2 t 2 [ e j ( − 1 − α T b π ) t j t + e j ( − 1 + α T b π ) t j t ] \begin{aligned} & \int_{\large -{1+\alpha \over T_b}\pi}^{\large -{1-\alpha \over T_b}\pi}\cos[{ T_b \over 2\alpha}(-\omega-{\pi(1-\alpha) \over T_b})]e^{j\omega t}d\omega \\ &= \int_{\large -{1+\alpha \over T_b}\pi}^{\large -{1-\alpha \over T_b}\pi}\cos[-{ T_b \over 2\alpha}\omega-{\pi(1-\alpha) \over 2\alpha}]e^{j\omega t}d\omega \\ &= {\Large{ {\cos[-{ T_b \over 2\alpha}\omega-{\pi(1-\alpha) \over 2\alpha}] \over jt}e^{j\omega t} - { -{ T_b \over 2\alpha} \sin[-{ T_b \over 2\alpha}\omega-{\pi(1-\alpha) \over 2\alpha}] \over t^2}e^{j\omega t} \over 1-{(-{ T_b \over 2\alpha})^2 \over t^2}}} {\Big |}_{\large -{1+\alpha \over T_b}\pi}^{\large -{1-\alpha \over T_b}\pi} \\ &= { 1 \over {\large 1 - {T_b^2 \over 4\alpha^2 t^2}}}[{e^{j ({\large -{1-\alpha \over T_b}\pi}) t} \over jt} -(-{e^{j ({\large -{1+\alpha \over T_b}\pi}) t} \over jt})] \\ &= { 1 \over {\large 1 - {T_b^2 \over 4\alpha^2 t^2}}}[{e^{j ({\large -{1-\alpha \over T_b}\pi}) t} \over jt} + {e^{j ({\large -{1+\alpha \over T_b}\pi}) t} \over jt}] \end{aligned}

### 推出：(4)

∫ 1 − α T b π 1 + α T b π cos ⁡ [ T b 2 α ( ω − π ( 1 − α ) T b ) ] e j ω t d ω = ∫ − 1 + α T b π − 1 − α T b π cos ⁡ [ T b 2 α ω − π ( 1 − α ) 2 α ] e j ω t d ω = cos ⁡ [ T b 2 α ω − π ( 1 − α ) 2 α ] j t e j ω t − T b 2 α sin ⁡ [ T b 2 α ω − π ( 1 − α ) 2 α ] t 2 e j ω t 1 − ( T b 2 α ) 2 t 2 ∣ 1 − α T b π 1 + α T b π = 1 1 − T b 2 4 α 2 t 2 [ − e j ( 1 + α T b π ) t j t − e j ( 1 − α T b π ) t j t ] \begin{aligned} & \int_{\large {1-\alpha \over T_b}\pi}^{\large {1+\alpha \over T_b}\pi}\cos[{ T_b \over 2\alpha}(\omega-{\pi(1-\alpha) \over T_b})]e^{j\omega t}d\omega \\ &= \int_{\large -{1+\alpha \over T_b}\pi}^{\large -{1-\alpha \over T_b}\pi}\cos[{ T_b \over 2\alpha}\omega-{\pi(1-\alpha) \over 2\alpha}]e^{j\omega t}d\omega \\ &= {\Large{ {\cos[{ T_b \over 2\alpha}\omega-{\pi(1-\alpha) \over 2\alpha}] \over jt}e^{j\omega t} - { { T_b \over 2\alpha} \sin[{ T_b \over 2\alpha}\omega-{\pi(1-\alpha) \over 2\alpha}] \over t^2}e^{j\omega t} \over 1-{({ T_b \over 2\alpha})^2 \over t^2}}} {\Big |}_{\large {1-\alpha \over T_b}\pi}^{\large {1+\alpha \over T_b}\pi} \\ &= { 1 \over {\large 1 - {T_b^2 \over 4\alpha^2 t^2}}}[-{e^{j ({\large {1+\alpha \over T_b}\pi}) t} \over jt} -{e^{j ({\large {1-\alpha \over T_b}\pi}) t} \over jt}] \end{aligned}

### 借助三角函数的和差化积公式：

{ sin ⁡ ( α + β ) + sin ⁡ ( α − β ) = 2 sin ⁡ α cos ⁡ β ⇒ sin ⁡ α + sin ⁡ β = 2 sin ⁡ α + β 2 cos ⁡ α − β 2 \large \begin{cases} \sin(\alpha+\beta)+\sin(\alpha-\beta) = 2\sin\alpha\cos\beta \\ \Rightarrow \sin\alpha + \sin\beta = 2\sin{\alpha + \beta \over 2}\cos{\alpha - \beta \over 2} \end{cases}

### 推出：(5)

( 3 ) + ( 4 ) ⇒ ( 5 ) : (3) + (4) \Rightarrow (5) :

∫ − 1 + α T b π − 1 − α T b π cos ⁡ [ T b 2 α ( − ω − π ( 1 − α ) T b ) ] e j ω t d ω + ∫ 1 − α T b π 1 + α T b π cos ⁡ [ T b 2 α ( ω − π ( 1 − α ) T b ) ] e j ω t d ω = 1 1 − T b 2 4 α 2 t 2 [ e j ( − 1 − α T b π ) t j t + e j ( − 1 + α T b π ) t j t ] + 1 1 − T b 2 4 α 2 t 2 [ − e j ( 1 + α T b π ) t j t − e j ( 1 − α T b π ) t j t ] = 1 1 − T b 2 4 α 2 t 2 [ − 2 sin ⁡ ( 1 + α T b π ) t − 2 sin ⁡ ( 1 − α T b π ) t t ] = 1 1 − T b 2 4 α 2 t 2 ( − 4 sin ⁡ π T b t cos ⁡ π α T b t t ) \begin{aligned} & \int_{\large -{1+\alpha \over T_b}\pi}^{\large -{1-\alpha \over T_b}\pi}\cos[{ T_b \over 2\alpha}(-\omega-{\pi(1-\alpha) \over T_b})]e^{j\omega t}d\omega+\int_{\large {1-\alpha \over T_b}\pi}^{\large {1+\alpha \over T_b}\pi}\cos[{ T_b \over 2\alpha}(\omega-{\pi(1-\alpha) \over T_b})]e^{j\omega t}d\omega \\ &= { 1 \over {\large 1 - {T_b^2 \over 4\alpha^2 t^2}}}[{e^{j ({\large -{1-\alpha \over T_b}\pi}) t} \over jt} + {e^{j ({\large -{1+\alpha \over T_b}\pi}) t} \over jt}] + { 1 \over {\large 1 - {T_b^2 \over 4\alpha^2 t^2}}}[-{e^{j ({\large {1+\alpha \over T_b}\pi}) t} \over jt} -{e^{j ({\large {1-\alpha \over T_b}\pi}) t} \over jt}] \\ &= { 1 \over {\large 1 - {T_b^2 \over 4\alpha^2 t^2}}} [{-2\sin({1+\alpha \over T_b}\pi) t -2\sin({1-\alpha \over T_b}\pi)t \over t}] \\ &= { 1 \over {\large 1 - {T_b^2 \over 4\alpha^2 t^2}}} ({-4\sin{\pi \over T_b}t\cos{\pi\alpha \over T_b}t \over t}) \end{aligned}

### 于是由 (2)、(5) 得 :

4 π T b h 2 ( t ) = 2 [ sin ⁡ π ( 1 + α ) T b t t − sin ⁡ π ( 1 − α ) T b t t ] + 1 1 − T b 2 4 α 2 t 2 ( − 4 sin ⁡ π T b t cos ⁡ π α T b t t ) {4\pi \over T_b} h_2(t) = 2[{\sin {\pi(1+\alpha) \over T_b}t \over t}- {\sin {\pi(1-\alpha) \over T_b}t \over t}] + { 1 \over {\large 1 - {T_b^2 \over 4\alpha^2 t^2}}} ({-4\sin{\pi \over T_b}t\cos{\pi\alpha \over T_b}t \over t})

h 2 ( t ) = T b 2 π [ sin ⁡ π ( 1 + α ) T b t t − sin ⁡ π ( 1 − α ) T b t t ] − T b π 1 1 − T b 2 4 α 2 t 2 ( sin ⁡ π T b t cos ⁡ π α T b t t ) h_2(t) = {T_b \over 2\pi}[{\sin {\pi(1+\alpha) \over T_b}t \over t}- {\sin {\pi(1-\alpha) \over T_b}t \over t}] - {T_b \over \pi}{ 1 \over {\large 1 - {T_b^2 \over 4\alpha^2 t^2}}} ({\sin{\pi \over T_b}t\cos{\pi\alpha \over T_b}t \over t})

## 最终h1(t)+h2(t)得出结果

h ( t ) = h 1 ( t ) + h 2 ( t ) = T b sin ⁡ π ( 1 − α ) t T b π t + T b 2 π [ sin ⁡ π ( 1 + α ) T b t t − sin ⁡ π ( 1 − α ) T b t t ] − T b π 1 1 − T b 2 4 α 2 t 2 ( sin ⁡ π T b t cos ⁡ π α T b t t ) = T b 2 π [ sin ⁡ π ( 1 + α ) T b t t + sin ⁡ π ( 1 − α ) T b t t ] − T b π 1 1 − T b 2 4 α 2 t 2 ( sin ⁡ π T b t cos ⁡ π α T b t t ) = T b π ( sin ⁡ π T b t cos ⁡ π α T b t t ) − T b π 1 1 − T b 2 4 α 2 t 2 ( sin ⁡ π T b t cos ⁡ π α T b t t ) = T b 2 4 α 2 t 2 T b 2 4 α 2 t 2 − 1 T b π t ( sin ⁡ π t T b cos ⁡ π α t T b ) = 1 1 − 4 α 2 t 2 T b 2 T b π t ( sin ⁡ π t T b cos ⁡ π α t T b ) = S a ( π t / T b ) cos ⁡ ( π α t / T b ) 1 − 4 α 2 t 2 / T b 2   , 其 中 S a ( x ) = sin ⁡ x x 。 \begin{aligned} h(t) &= h_1(t)+h_2(t) \\ &= T_b{\sin \large{ \pi (1-\alpha)t \over T_b } \over \pi t} + {T_b \over 2\pi}[{\sin {\pi(1+\alpha) \over T_b}t \over t}- {\sin {\pi(1-\alpha) \over T_b}t \over t}] - {T_b \over \pi}{ 1 \over {\large 1 - {T_b^2 \over 4\alpha^2 t^2}}} ({\sin{\pi \over T_b}t\cos{\pi\alpha \over T_b}t \over t}) \\ &= {T_b \over 2\pi}[{\sin {\pi(1+\alpha) \over T_b}t \over t}+{\sin {\pi(1-\alpha) \over T_b}t \over t}] - {T_b \over \pi}{ 1 \over {\large 1 - {T_b^2 \over 4\alpha^2 t^2}}} ({\sin{\pi \over T_b}t\cos{\pi\alpha \over T_b}t \over t}) \\ &= {T_b \over \pi}({\sin{\pi \over T_b}t\cos{\pi\alpha \over T_b}t \over t}) - {T_b \over \pi}{ 1 \over {\large 1 - {T_b^2 \over 4\alpha^2 t^2}}} ({\sin{\pi \over T_b}t\cos{\pi\alpha \over T_b}t \over t}) \\ &= { {T_b^2 \over 4\alpha^2 t^2} \over {\large {T_b^2 \over 4\alpha^2 t^2} - 1}} {T_b \over \pi t}(\sin{\pi t \over T_b}\cos{\pi\alpha t \over T_b}) \\ &= {1 \over {\large 1 - {4\alpha^2t^2 \over T_b^2}}} {T_b \over \pi t}(\sin{\pi t \over T_b}\cos{\pi\alpha t \over T_b}) \\ &= Sa(\pi t/T_b){\cos(\pi \alpha t /T_b) \over 1-4\alpha^2t^2/T_b^2}\quad\,,其中Sa(x)={\sin{x} \over x}。 \end{aligned}