文章目录
问题的提出
求 具 有 滚 降 系 数 α 的 余 弦 滚 降 特 性 H ( f ) 相 应 的 冲 激 响 应 h ( t ) 。 H ( f ) = { T b , 0 ≤ ∣ f ∣ ≤ ( ( 1 − α ) / 2 T b ) 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 0 , ∣ f ∣ ≤ ( ( 1 + α ) / 2 T b ) 其 中 , 0 ≤ α ≤ 1 。 求具有滚降系数 \alpha的余弦滚降特性H( f ) 相应的冲激响应 { h(t)}。\\ \\ \\ H(f)= \begin{cases} T_b\,,0 \leq |f| \leq ((1-\alpha)/2T_b) \\ \large {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} \\ 0 \,, |f| \leq ((1+\alpha)/2T_b) \end{cases} \\ \\ 其中\,,\, 0 \le \alpha \le 1。 求具有滚降系数α的余弦滚降特性H(f)相应的冲激响应h(t)。H(f)=⎩⎪⎨⎪⎧Tb,0≤∣f∣≤((1−α)/2Tb)2Tb{ 1+cos[απTb(∣f∣−2Tb1−α)]},2Tb1−α≤∣f∣≤2Tb1+α0,∣f∣≤((1+α)/2Tb)其中,0≤α≤1。
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)⇔F(ω)F(t)⇔2πf(−ω)
故由：
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} f(t)={ E,∣t∣≤2τ0,其他⇔F(ω)=2Eωsin2ωτ
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} H1(ω)⇒H1(f)={ Tb,0≤∣ω∣≤Tbπ(1−α)0,其他⇔h1(t)=2Tbsin2πtTbπ(1−α)t=TbsinπtTbπ(1−α)t={ Tb,0≤∣f∣≤2Tb(1−α)0,其他⇔h1(t)=2Tbsin2πtTbπ(1−α)t=TbsinπtTbπ(1−α)t
求解分段函数的第2部分对应的h2(t)
又由：
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} H2(f)⇒H2(ω)=2Tb{ 1+cos[απTb(∣f∣−2Tb1−α)]},2Tb1−α≤∣f∣≤2Tb1+α=2Tb{ 1+cos[2αTb(∣ω∣−Tbπ(1−α))]},Tbπ(1−α)≤∣ω∣≤Tbπ(1+α)
推出：(2)
对 应 的 h ( t ) 如 下 ： 2 π h 2 ( t ) = ∫ − ∞ ∞ T b 2 { 1 + cos [ T b 2 α ( ∣ ω ∣ − π ( 1 − α ) T b ) ] } e j ω t d ω ⇒ 4 π T b h 2 ( t ) = ∫ − ∞ ∞ { 1 + 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 { 1 + 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 { 1 + cos [ T b 2 α ( ω − π ( 1 − α ) T b ) ] } e j ω t d ω = e j ω t j t ∣ − 1 + α T b π − 1 − α T b π + e j ω t j t ∣ 1 − α T b π 1 + α T b π + ∫ − 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 = 2 [ sin π ( 1 + α ) T b t t − sin π ( 1 − α ) T b t t ] + ∫ − 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 对应的h(t)如下：\\ \begin{aligned} 2\pi h_2(t) &= \int_{-\infty}^{\infty}{T_b \over 2}\{1+\cos[{ T_b \over 2\alpha}(|\omega|-{\pi(1-\alpha) \over T_b})] \}e^{j\omega t}d\omega \\ \Rightarrow {4\pi \over T_b} h_2(t) &= \int_{-\infty}^{\infty}\{1+\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}\{1+\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}\{1+\cos[{ T_b \over 2\alpha}(\omega-{\pi(1-\alpha) \over T_b})] \}e^{j\omega t}d\omega \\ &= {e^{j\omega t} \over jt}|^{\large -{1-\alpha \over T_b}\pi}_{\large -{1+\alpha \over T_b}\pi}\,+ {e^{j\omega t} \over jt}|^{\large {1+\alpha \over T_b}\pi}_{\large {1-\alpha \over T_b}\pi} \,+ \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 \\ &= 2[{\sin {\pi(1+\alpha) \over T_b}t \over t}- {\sin {\pi(1-\alpha) \over T_b}t \over t}] \,+ \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 \end{aligned} 对应的h(t)如下：2πh2(t)⇒Tb4πh2(t)=∫−∞∞2Tb{ 1+cos[2αTb(∣ω∣−Tbπ(1−α))]}ejωtdω=∫−∞∞{ 1+cos[2αTb(∣ω∣−Tbπ(1−α))]}ejωtdω=∫−Tb1+απ−Tb1−απ{ 1+cos[2αTb(−ω−Tbπ(1−α))]}ejωtdω+∫Tb1−απTb1+απ{ 1+cos[2αTb(ω−Tbπ(1−α))]}ejωtdω=jtejωt∣−Tb1+απ−Tb1−απ+jtejωt∣Tb1−απTb1+απ+∫−Tb1+απ−Tb1−απcos[2αTb(−ω−Tbπ(1−α))]ejωtdω+∫Tb1−απTb1+απcos[2αTb(ω−Tbπ(1−α))]ejωtdω=2[tsinTbπ(1+α)t−tsinTbπ(1−α)t]+∫−Tb1+απ−Tb1−απcos[2αTb(−ω−Tbπ(1−α))]ejωtdω+∫Tb1−απTb1+απcos[2αTb(ω−Tbπ(1−α))]ejωtdω
借助以下公式：
由于：
∫ 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(aω±b)ejωtdω=jtcos(aω±b)ejωt+∫jtasin(aω±b)ejωtdω=jtcos(aω±b)ejωt−t2asin(aω±b)ejωt+∫t2a2cos(aω±b)ejωtdω
得：
∫ 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}}} ∫cos(aω±b)ejωtdω=1−t2a2jtcos(aω±b)ejωt−t2asin(aω±b)ejωt
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} ∫−Tb1+απ−Tb1−απcos[2αTb(−ω−Tbπ(1−α))]ejωtdω=∫−Tb1+απ−Tb1−απcos[−2αTbω−2απ(1−α)]ejωtdω=1−t2(−2αTb)2jtcos[−2αTbω−2απ(1−α)]ejωt−t2−2αTbsin[−2αTbω−2απ(1−α)]ejωt∣∣∣−Tb1+απ−Tb1−απ=1−4α2t2Tb21[jtej(−Tb1−απ)t−(−jtej(−Tb1+απ)t)]=1−4α2t2Tb21[jtej(−Tb1−απ)t+jtej(−Tb1+απ)t]
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} ∫Tb1−απTb1+απcos[2αTb(ω−Tbπ(1−α))]ejωtdω=∫−Tb1+απ−Tb1−απcos[2αTbω−2απ(1−α)]ejωtdω=1−t2(2αTb)2jtcos[2αTbω−2απ(1−α)]ejωt−t22αTbsin[2αTbω−2απ(1−α)]ejωt∣∣∣Tb1−απTb1+απ=1−4α2t2Tb21[−jtej(Tb1+απ)t−jtej(Tb1−απ)t]
借助三角函数的和差化积公式：
由：
{ 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} ⎩⎨⎧sin(α+β)+sin(α−β)=2sinαcosβ⇒sinα+sinβ=2sin2α+βcos2α−β
推出：(5)
( 3 ) + ( 4 ) ⇒ ( 5 ) : (3) + (4) \Rightarrow (5) : (3)+(4)⇒(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} ∫−Tb1+απ−Tb1−απcos[2αTb(−ω−Tbπ(1−α))]ejωtdω+∫Tb1−απTb1+απcos[2αTb(ω−Tbπ(1−α))]ejωtdω=1−4α2t2Tb21[jtej(−Tb1−απ)t+jtej(−Tb1+απ)t]+1−4α2t2Tb21[−jtej(Tb1+απ)t−jtej(Tb1−απ)t]=1−4α2t2Tb21[t−2sin(Tb1+απ)t−2sin(Tb1−απ)t]=1−4α2t2Tb21(t−4sinTbπtcosTbπαt)
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}) Tb4πh2(t)=2[tsinTbπ(1+α)t−tsinTbπ(1−α)t]+1−4α2t2Tb21(t−4sinTbπtcosTbπαt)
得：
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}) h2(t)=2πTb[tsinTbπ(1+α)t−tsinTbπ(1−α)t]−πTb1−4α2t2Tb21(tsinTbπtcosTbπαt)
最终h1(t)+h2(t)得出结果
则：
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} h(t)=h1(t)+h2(t)=TbsinπtTbπ(1−α)t+2πTb[tsinTbπ(1+α)t−tsinTbπ(1−α)t]−πTb1−4α2t2Tb21(tsinTbπtcosTbπαt)=2πTb[tsinTbπ(1+α)t+tsinTbπ(1−α)t]−πTb1−4α2t2Tb21(tsinTbπtcosTbπαt)=πTb(tsinTbπtcosTbπαt)−πTb1−4α2t2Tb21(tsinTbπtcosTbπαt)=4α2t2Tb2−14α2t2Tb2πtTb(sinTbπtcosTbπαt)=1−Tb24α2t21πtTb(sinTbπtcosTbπαt)=Sa(πt/Tb)1−4α2t2/Tb2cos(παt/Tb),其中Sa(x)=xsinx。
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