求具有滚降系数 a 的余弦滚降特性 H(f) 相应的冲激响应 h(t)

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。 αH(f)h(t)H(f)=Tb,0f((1α)/2Tb)2Tb{ 1+cos[απTb(f2Tb1α)]},2Tb1αf2Tb1+α0,f((1+α)/2Tb),0α1

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,t2τ0,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,0ωTbπ(1α)0,h1(t)=2Tbsin2πtTbπ(1α)t=TbsinπtTbπ(1α)t={ Tb,0f2Tb(1α)0,h1(t)=2Tbsin2πtTbπ(1α)t=TbsinπtTbπ(1α)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(f2Tb1α)]},2Tb1αf2Tb1+α=2Tb{ 1+cos[2αTb(ωTbπ(1α))]},Tbπ(1α)ωTbπ(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} h(t)2πh2(t)Tb4πh2(t)=2Tb{ 1+cos[2αTb(ωTbπ(1α))]}ejωtdω={ 1+cos[2αTb(ωTbπ(1α))]}ejωtdω=Tb1+απTb1απ{ 1+cos[2αTb(ωTbπ(1α))]}ejωtdω+Tb1απTb1+απ{ 1+cos[2αTb(ωTbπ(1α))]}ejωtdω=jtejωtTb1+απTb1απ+jtejωtTb1απTb1+απ+Tb1+απTb1απcos[2αTb(ωTbπ(1α))]ejωtdω+Tb1απTb1+απcos[2αTb(ωTbπ(1α))]ejωtdω=2[tsinTbπ(1+α)ttsinTbπ(1α)t]+Tb1+απTb1απcos[2αTb(ωTbπ(1α))]ejωtdω+Tb1απTb1+απcos[2αTb(ωTbπ(1α))]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)ejωtt2asin(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=1t2a2jtcos(aω±b)ejωtt2asin(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+απTb1απcos[2αTb(ωTbπ(1α))]ejωtdω=Tb1+απTb1απcos[2αTbω2απ(1α)]ejωtdω=1t2(2αTb)2jtcos[2αTbω2απ(1α)]ejωtt22αTbsin[2αTbω2απ(1α)]ejωtTb1+απTb1απ=14α2t2Tb21[jtej(Tb1απ)t(jtej(Tb1+απ)t)]=14α2t2Tb21[jtej(Tb1απ)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} Tb1απTb1+απcos[2αTb(ωTbπ(1α))]ejωtdω=Tb1+απTb1απcos[2αTbω2απ(1α)]ejωtdω=1t2(2αTb)2jtcos[2αTbω2απ(1α)]ejωtt22αTbsin[2αTbω2απ(1α)]ejωtTb1απTb1+απ=14α2t2Tb21[jtej(Tb1+απ)tjtej(Tb1απ)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(αβ)=2sinαcosβsinα+sinβ=2sin2α+βcos2αβ

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+απTb1απcos[2αTb(ωTbπ(1α))]ejωtdω+Tb1απTb1+απcos[2αTb(ωTbπ(1α))]ejωtdω=14α2t2Tb21[jtej(Tb1απ)t+jtej(Tb1+απ)t]+14α2t2Tb21[jtej(Tb1+απ)tjtej(Tb1απ)t]=14α2t2Tb21[t2sin(Tb1+απ)t2sin(Tb1απ)t]=14α2t2Tb21(t4sinTbπtcosTbπαt)

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)ttsinTbπ(1α)t]+14α2t2Tb21(t4sinTbπtcosTbπαt)

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)ttsinTbπ(1α)t]πTb14α2t2Tb21(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(1α)t+2πTb[tsinTbπ(1+α)ttsinTbπ(1α)t]πTb14α2t2Tb21(tsinTbπtcosTbπαt)=2πTb[tsinTbπ(1+α)t+tsinTbπ(1α)t]πTb14α2t2Tb21(tsinTbπtcosTbπαt)=πTb(tsinTbπtcosTbπαt)πTb14α2t2Tb21(tsinTbπtcosTbπαt)=4α2t2Tb214α2t2Tb2πtTb(sinTbπtcosTbπαt)=1Tb24α2t21πtTb(sinTbπtcosTbπαt)=Sa(πt/Tb)14α2t2/Tb2cos(παt/Tb),Sa(x)=xsinx

End

如果这篇文章有所帮助,那就太好了!

猜你喜欢

转载自blog.csdn.net/m0_53793870/article/details/122361695

相关文章