第十次作业参考答案

01第一题

第一小题中的求解除了（14）（15）小题之外，其他的各题都可以在MATLAB中使用MATLAB的符号计算帮助求解，一边检查求解的结果正确性。
使用MATLAB求解第一小题可以参见下面的链接：

利用MATLAB帮助求解作业中的Laplace变换和Z变换

求解：

（1）
L[1−e−at]=1s−1s+a=as(s+a)L\left[ {1 - e^{ - at} } \right] = {1 \over s} - {1 \over {s + a}} = {a \over {s\left( {s + a} \right)}}L[1−e−at]=s1−s+a1=s(s+a)a

（2）
L[sin⁡(t)+2cos⁡(t)]=1s2+1+2ss2+1=2s+1s2+1L\left[ {\sin \left( t \right) + 2\cos \left( t \right)} \right] = {1 \over {s^2 + 1}} + {{2s} \over {s^2 + 1}} = {{2s + 1} \over {s^2 + 1}}L[sin(t)+2cos(t)]=s2+11+s2+12s=s2+12s+1

（3）
L[t⋅e−2t]=1(s+2)2L\left[ {t \cdot e^{ - 2t} } \right] = {1 \over {\left( {s + 2} \right)^2 }}L[t⋅e−2t]=(s+2)21

（4） 由：
L[sin⁡(2t)]=2s4+4L\left[ {\sin \left( {2t} \right)} \right] = {2 \over {s^4 + 4}}L[sin(2t)]=s4+42
再根据s域平移性质，可以得到：

L[e−t⋅sin⁡(2t)]=2(s+1)2+4=2s2+2s+5L\left[ {e^{ - t} \cdot\sin \left( {2t} \right)} \right] = {2 \over {\left( {s + 1} \right)^2 + 4}} = {2 \over {s^2 + 2s + 5}}L[e−t⋅sin(2t)]=(s+1)2+42=s2+2s+52

（5） 因为
L[tn]=n!sn+1L\left[ {t^n } \right] = {{n!} \over {s^{n + 1} }}L[tn]=sn+1n!

所以：L[1+2t]=1s+2s2=s+2s2L\left[ {1 + 2t} \right] = {1 \over s} + {2 \over {s^2 }} = {{s + 2} \over {s^2 }}L[1+2t]=s1+s22=s2s+2

再由s域平移性质，可得：L[(1+2t)⋅e−t]=s+3(s+1)2L\left[ {\left( {1 + 2t} \right) \cdot e^{ - t} } \right] = {{s + 3} \over {\left( {s + 1} \right)^2 }}L[(1+2t)⋅e−t]=(s+1)2s+3

（6） 由L[1−cos⁡(αt)]=1s−ss2+α2L\left[ {1 - \cos \left( {\alpha t} \right)} \right] = {1 \over s} - {s \over {s^2 + \alpha ^2 }}L[1−cos(αt)]=s1−s2+α2s

再由s域平移性质，可得：L{[1−cos⁡(αt)]e−βt}=1s+β−s+β(s+β)2+α2L\left\{ {\left[ {1 - \cos \left( {\alpha t} \right)} \right]e^{ - \beta t} } \right\} = {1 \over {s + \beta }} - {{s + \beta } \over {\left( {s + \beta } \right)^2 + \alpha ^2 }}L{[1−cos(αt)]e−βt}=s+β1−(s+β)2+α2s+β

（7） L[t2+2t]=2s3+2s2=2s+2s3L\left[ {t^2 + 2t} \right] = {2 \over {s^3 }} + {2 \over {s^2 }} = {{2s + 2} \over {s^3 }}L[t2+2t]=s32+s22=s32s+2

（8） L[2δ(t)−3e−7t]=2−3s+7L\left[ {2\delta \left( t \right) - 3e^{ - 7t} } \right] = 2 - {3 \over {s + 7}}L[2δ(t)−3e−7t]=2−s+73

（9） 由L[sinh⁡(βt)]=βs2−β2L\left[ {\sinh \left( {\beta t} \right)} \right] = {\beta \over {s^2 - \beta ^2 }}L[sinh(βt)]=s2−β2β

再由s域的平移性质，可得：L[e−αtsinh⁡(βt)]=β(s+α)2−β2L\left[ {e^{ - \alpha t} \sinh \left( {\beta t} \right)} \right] = {\beta \over {\left( {s + \alpha } \right)^2 - \beta ^2 }}L[e−αtsinh(βt)]=(s+α)2−β2β

（10） 因为 cos⁡2(Ωt)=12+12cos⁡(2Ωt)\cos ^2 \left( {\Omega t} \right) = {1 \over 2} + {1 \over 2}\cos \left( {2\Omega t} \right)cos2(Ωt)=21+21cos(2Ωt)

所以：L[cos⁡2(Ωt)]=12s+12⋅ss2+4Ω2=12(1s+ss2+4Ω2)L\left[ {\cos ^2 \left( {\Omega t} \right)} \right] = {1 \over {2s}} + {1 \over 2} \cdot {s \over {s^2 + 4\Omega ^2 }} = {1 \over 2}\left( {{1 \over s} + {s \over {s^2 + 4\Omega ^2 }}} \right)L[cos2(Ωt)]=2s1+21⋅s2+4Ω2s=21(s1+s2+4Ω2s)

（11）
L[1β−α(e−αt−e−βt)]=1β−α(1s+α−1s+β)=1(s+α)(s+β)L\left[ {{1 \over {\beta - \alpha }}\left( {e^{ - \alpha t} - e^{ - \beta t} } \right)} \right] = {1 \over {\beta - \alpha }}\left( {{1 \over {s + \alpha }} - {1 \over {s + \beta }}} \right) = {1 \over {\left( {s + \alpha } \right)\left( {s + \beta } \right)}}L[β−α1(e−αt−e−βt)]=β−α1(s+α1−s+β1)=(s+α)(s+β)1

（12） 由于：L[e−tcos⁡(ωt)]=s+1(s+1)2+ω2L\left[ {e^{ - t} \cos \left( {\omega t} \right)} \right] = {{s + 1} \over {\left( {s + 1} \right)^2 + \omega ^2 }}L[e−tcos(ωt)]=(s+1)2+ω2s+1

所以：L[e−(t+a)⋅cos⁡(ωt)]=(s+1)e−a(s+1)2+ω2L\left[ {e^{ - \left( {t + a} \right)} \cdot \cos \left( {\omega t} \right)} \right] = {{\left( {s + 1} \right)e^{ - a} } \over {\left( {s + 1} \right)^2 + \omega ^2 }}L[e−(t+a)⋅cos(ωt)]=(s+1)2+ω2(s+1)e−a

（13） 因为：
t⋅e−(t−2)u(t−1)=e⋅[(t−1)⋅e−(t−1)+e−(t−1)]⋅u(t−1)t \cdot e^{ - \left( {t - 2} \right)} u\left( {t - 1} \right) = e \cdot \left[ {\left( {t - 1} \right) \cdot e^{ - \left( {t - 1} \right)} + e^{ - \left( {t - 1} \right)} } \right] \cdot u\left( {t - 1} \right)t⋅e−(t−2)u(t−1)=e⋅[(t−1)⋅e−(t−1)+e−(t−1)]⋅u(t−1)

且：L[(t−1)e−(t−1)u(t−1)]=e−s(s+1)2L\left[ {\left( {t - 1} \right)e^{ - \left( {t - 1} \right)} u\left( {t - 1} \right)} \right] = {{e^{ - s} } \over {\left( {s + 1} \right)^2 }}L[(t−1)e−(t−1)u(t−1)]=(s+1)2e−sL[e−(t−1)u(t−1)]=e−ss+1L\left[ {e^{ - \left( {t - 1} \right)} u\left( {t - 1} \right)} \right] = {{e^{ - s} } \over {s + 1}}L[e−(t−1)u(t−1)]=s+1e−s

所以：
L[t⋅e−(t−2)u(t−1)]=e⋅[1(s+1)2+1s+1]⋅e−s=(s+2)e−(s−1)(s+1)2L\left[ {t \cdot e^{ - \left( {t - 2} \right)} u\left( {t - 1} \right)} \right] = e \cdot \left[ {{1 \over {\left( {s + 1} \right)^2 }} + {1 \over {s + 1}}} \right] \cdot e^{ - s} = {{\left( {s + 2} \right)e^{ - \left( {s - 1} \right)} } \over {\left( {s + 1} \right)^2 }}L[t⋅e−(t−2)u(t−1)]=e⋅[(s+1)21+s+11]⋅e−s=(s+1)2(s+2)e−(s−1)

（14） 由拉普拉斯变换的s域平移特性：L[e−tf(t)]=F(s+1)L\left[ {e^{ - t} f\left( t \right)} \right] = F\left( {s + 1} \right)L[e−tf(t)]=F(s+1)
由尺度变换性质，得到：
L[e−taf(ta)]=aF(as+1)L\left[ {e^{ - {t \over a}} f\left( {{t \over a}} \right)} \right] = aF\left( {as + 1} \right)L[e−atf(at)]=aF(as+1)

（15） 由拉普拉斯变换的尺度特性：L[f(ta)]=aF(as)L\left[ {f\left( {{t \over a}} \right)} \right] = aF\left( {as} \right)L[f(at)]=aF(as)

再由s域平移性质，可得：L[e−atf(ta)]=aF[a(s+a)]=aF(as+a2)L\left[ {e^{ - at} f\left( {{t \over a}} \right)} \right] = aF\left[ {a\left( {s + a} \right)} \right] = aF\left( {as + a^2 } \right)L[e−atf(at)]=aF[a(s+a)]=aF(as+a2)

（16） 根据：cos⁡3(3t)=cos⁡(3t)⋅1+cos⁡(6t)2=14cos⁡(9t)+34cos⁡(t)\cos ^3 \left( {3t} \right) = \cos \left( {3t} \right)\cdot{{1 + \cos \left( {6t} \right)} \over 2} = {1 \over 4}\cos \left( {9t} \right) + {3 \over 4}\cos \left( t \right)cos3(3t)=cos(3t)⋅21+cos(6t)=41cos(9t)+43cos(t)

所以：L[cos⁡3(3t)]=14⋅ss2+81+34⋅ss2+9L\left[ {\cos ^3 \left( {3t} \right)} \right] = {1 \over 4} \cdot {s \over {s^2 + 81}} + {3 \over 4} \cdot {s \over {s^2 + 9}}L[cos3(3t)]=41⋅s2+81s+43⋅s2+9s

再根据s域的微分性质，可得：L[t⋅cos⁡3(3t)]=−dds(14ss2+81+34⋅ss2+9)=14[s2−81(s2+81)2+3s2−27(s2+9)2]L\left[ {t \cdot \cos ^3 \left( {3t} \right)} \right] = - {d \over {ds}}\left( {{1 \over 4}{s \over {s^2 + 81}} + {3 \over 4} \cdot {s \over {s^2 + 9}}} \right) = {1 \over 4}\left[ {{{s^2 - 81} \over {\left( {s^2 + 81} \right)^2 }} + {{3s^2 - 27} \over {\left( {s^2 + 9} \right)^2 }}} \right]L[t⋅cos3(3t)]=−dsd(41s2+81s+43⋅s2+9s)=41[(s2+81)2s2−81+(s2+9)23s2−27]
（17） 根据：L[cos⁡(2t)]=ss2+4L\left[ {\cos \left( {2t} \right)} \right] = {s \over {s^2 + 4}}L[cos(2t)]=s2+4s

连续两次应用s域微分性质，可得：L[t⋅cos⁡(2t)]=s2−4(s2+4)2L\left[ {t \cdot \cos \left( {2t} \right)} \right] = {{s^2 - 4} \over {\left( {s^2 + 4} \right)^2 }}L[t⋅cos(2t)]=(s2+4)2s2−4

L[t2⋅cos⁡(2t)]=2s2−24s(s2+4)3L\left[ {t^2 \cdot \cos \left( {2t} \right)} \right] = {{2s^2 - 24s} \over {\left( {s^2 + 4} \right)^3 }}L[t2⋅cos(2t)]=(s2+4)32s2−24s

（18） 根据
L[1−e−αt]=1s−1s+αL\left[ {1 - e^{ - \alpha t} } \right] = {1 \over s} - {1 \over {s + \alpha }}L[1−e−αt]=s1−s+α1

再由s域的积分性质，可得：
L[1t(1−e−αt)]=∫s+∞(1s1−1s1+α)ds1=ln⁡(s+α)−ln⁡s=−ln⁡(ss+α)L\left[ {{1 \over t}\left( {1 - e^{ - \alpha t} } \right)} \right] = \int_s^{ + \infty } {\left( {{1 \over {s_1 }} - {1 \over {s_1 + \alpha }}} \right)ds_1 } = \ln \left( {s + \alpha } \right) - \ln s = - \ln \left( {{s \over {s + \alpha }}} \right)L[t1(1−e−αt)]=∫s+∞(s11−s1+α1)ds1=ln(s+α)−lns=−ln(s+αs)

（19） 根据L[e−3t−e−5t]=1s+3−1s+5L\left[ {e^{ - 3t} - e^{ - 5t} } \right] = {1 \over {s + 3}} - {1 \over {s + 5}}L[e−3t−e−5t]=s+31−s+51

再由s域的积分性质可得：
L[e−3t−e−5tt]=∫s+∞(1s1+3−1s1+5)ds1=ln⁡(s+5s+3)L\left[ {{{e^{ - 3t} - e^{ - 5t} } \over t}} \right] = \int_s^{ + \infty } {\left( {{1 \over {s_1 + 3}} - {1 \over {s_1 + 5}}} \right)ds_1 } = \ln \left( {{{s + 5} \over {s + 3}}} \right)L[te−3t−e−5t]=∫s+∞(s1+31−s1+51)ds1=ln(s+3s+5)

（20） 根据：L[sin⁡(αt)]=αs2+α2L\left[ {\sin \left( {\alpha t} \right)} \right] = {\alpha \over {s^2 + \alpha ^2 }}L[sin(αt)]=s2+α2α

再由s域的积分性质可得：
L[sin⁡(αt)t]=∫s+∞αs12+α2ds1=∫s+∞1(s1α)2+1d(s1α)=π2−arctan⁡(sα)=arctan⁡(αs)L\left[ {{{\sin \left( {\alpha t} \right)} \over t}} \right] = \int_s^{ + \infty } {{\alpha \over {s_1^2 + \alpha ^2 }}ds_1 } = \int_s^{ + \infty } {{1 \over {\left( {{{s_1 } \over \alpha }} \right)^2 + 1}}d\left( {{{s_1 } \over \alpha }} \right)} = {\pi \over 2} - \arctan \left( {{s \over \alpha }} \right) = \arctan \left( {{\alpha \over s}} \right)L[tsin(αt)]=∫s+∞s12+α2αds1=∫s+∞(αs1)2+11d(αs1)=2π−arctan(αs)=arctan(sα)

02第二题

求解：

（1）解答：
X(z)=∑n=0∞(14)nz−n=4z4z−1=zz−14,∣z∣>14X\left( z \right) = \sum\limits_{n = 0}^\infty {\left( {{1 \over 4}} \right)^n z^{ - n} } = {{4z} \over {4z - 1}} = {z \over {z - {1 \over 4}}},\,\,\left| z \right| > {1 \over 4}X(z)=n=0∑∞(41)nz−n=4z−14z=z−41z,∣z∣>41

（2）解答：
X(z)=∑n=0∞(−13)nz−n=zz+13=3z3z+1,∣z∣>13X\left( z \right) = \sum\limits_{n = 0}^\infty {\left( { - {1 \over 3}} \right)^n z^{ - n} } = {z \over {z + {1 \over 3}}} = {{3z} \over {3z + 1}},\;\;\;\;\left| z \right| > {1 \over 3}X(z)=n=0∑∞(−31)nz−n=z+31z=3z+13z,∣z∣>31

（3）解答：
X(z)=∑n=−∞0(12)nz−n=11−2z,∣z∣<12X\left( z \right) = \sum\limits_{n = - \infty }^0 {\left( {{1 \over 2}} \right)^n z^{ - n} } = {1 \over {1 - 2z}},\,\,\,\,\left| z \right| < {1 \over 2}X(z)=n=−∞∑0(21)nz−n=1−2z1,∣z∣<21

（4）解答：
X(z)=∑n=−∞0(−12)nz−n=11+2z,∣z∣<12X\left( z \right) = \sum\limits_{n = - \infty }^0 {\left( { - {1 \over 2}} \right)^n z^{ - n} } = {1 \over {1 + 2z}},\,\,\,\,\,\left| z \right| < {1 \over 2}X(z)=n=−∞∑0(−21)nz−n=1+2z1,∣z∣<21

（5）解答：
X(z)=∑n=−∞−1−(16)n−1z−nu[−n−1]=36z6z−1,∣z∣<16X\left( z \right) = \sum\limits_{n = - \infty }^{ - 1} { - \left( {{1 \over 6}} \right)^{n - 1} z^{ - n} u\left[ { - n - 1} \right]} = {{36z} \over {6z - 1}},\,\,\,\,\left| z \right| < {1 \over 6}X(z)=n=−∞∑−1−(61)n−1z−nu[−n−1]=6z−136z,∣z∣<61

（6）解答：
X(z)=1z−1,∣z∣>1X\left( z \right) = {1 \over {z - 1}},\,\,\,\,\left| z \right| > 1X(z)=z−11,∣z∣>1

（7）解答：
X(z)=∑n=03(110)n⋅z−n=1−(110z)31−110z=(10z)3−1(10z)2(10z−1),∣z∣>0X\left( z \right) = \sum\limits_{n = 0}^3 {\left( {{1 \over {10}}} \right)^n \cdot z^{ - n} = {{1 - \left( {{1 \over {10z}}} \right)^3 } \over {1 - {1 \over {10z}}}} = {{\left( {10z} \right)^3 - 1} \over {\left( {10z} \right)^2 \left( {10z - 1} \right)}},\,\,\,\,\,\left| z \right| > 0}X(z)=n=0∑3(101)n⋅z−n=1−10z11−(10z1)3=(10z)2(10z−1)(10z)3−1,∣z∣>0

（8）解答：
X(z)=zz−15+zz−16=z(60z−11)(5z−1)⋅(6z−1),∣z∣>15X\left( z \right) = {z \over {z - {1 \over 5}}} + {z \over {z - {1 \over 6}}} = {{z\left( {60z - 11} \right)} \over {\left( {5z - 1} \right) \cdot \left( {6z - 1} \right)}},\,\,\,\,\left| z \right| > {1 \over 5}X(z)=z−51z+z−61z=(5z−1)⋅(6z−1)z(60z−11),∣z∣>51

（9）解答：
X(z)=1−12z−8=2z8−12z8,∣z∣>0X\left( z \right) = 1 - {1 \over 2}z^{ - 8} = {{2z^8 - 1} \over {2z^8 }},\,\,\,\,\left| z \right| > 0X(z)=1−21z−8=2z82z8−1,∣z∣>0

03第三题

求双边序列x[n]=(1a)∣n∣,∣a∣>1x\left[ n \right] = \left( {{1 \over a}} \right)^{\left| n \right|} ,\,\,\,\,\left| a \right| > 1x[n]=(a1)∣n∣,∣a∣>1
的z变换，并表明收敛于及绘制出零极点图。

求解：

将双边序列分成右边和左边序列，分别求出各自的Z变换。
xR[n]=(1a)n,n≥0,xL[n]=(1a)−n=an,n<0x_R \left[ n \right] = \left( {{1 \over a}} \right)^n ,\,\,\,\,\,n \ge 0,\,\,\,\,\,\,x_L \left[ n \right] = \left( {{1 \over a}} \right)^{ - n} = a^n ,\,\,\,\,\,n < 0xR[n]=(a1)n,n≥0,xL[n]=(a1)−n=an,n<0

Z{xR[n]}=zz−1a,∣z∣>1aZ\left\{ {x_R \left[ n \right]} \right\} = {z \over {z - {1 \over a}}},\,\,\,\,\left| z \right| > {1 \over a}Z{xR[n]}=z−a1z,∣z∣>a1

Z{xL[n]}=∑n=−∞−1anz−n=∑n=1∞zna−nZ\left\{ {x_L \left[ n \right]} \right\} = \sum\limits_{n = - \infty }^{ - 1} {a^n z^{ - n} } = \sum\limits_{n = 1}^\infty {z^n a^{ - n} }Z{xL[n]}=n=−∞∑−1anz−n=n=1∑∞zna−n=∑n=0∞zna−n−1=11−za−1−1=−zz−a= \sum\limits_{n = 0}^\infty {z^n a^{ - n} } - 1 = {1 \over {1 - za^{ - 1} }} - 1 = {{ - z} \over {z - a}}=n=0∑∞zna−n−1=1−za−11−1=z−a−z

Z{x[n]}=Z{xR[n]+xL[n]}Z\left\{ {x\left[ n \right]} \right\} = Z\left\{ {x_R \left[ n \right] + x_L \left[ n \right]} \right\}Z{x[n]}=Z{xR[n]+xL[n]}=zz−1a+−zz−a=(1−a2)zaz2−(1+a2)z+a= {z \over {z - {1 \over a}}} + {{ - z} \over {z - a}} = {{\left( {1 - a^2 } \right)z} \over {az^2 - \left( {1 + a^2 } \right)z + a}}=z−a1z+z−a−z=az2−(1+a2)z+a(1−a2)z

04第四题

求解：

（1）解答

L−1(1s+2)=e−2t,t≥0L^{ - 1} \left( {{1 \over {s + 2}}} \right) = e^{ - 2t} ,\,\,\,\,t \ge 0L−1(s+21)=e−2t,t≥0

（2）解答：
L−1(42s+5)=L−1(2s+52)=2⋅e−52tt≥1L^{ - 1} \left( {{4 \over {2s + 5}}} \right) = L^{ - 1} \left( {{2 \over {s + {5 \over 2}}}} \right) = 2 \cdot e^{ - {5 \over 2}t} \,\,\,\,t \ge 1L−1(2s+54)=L−1(s+252)=2⋅e−25tt≥1

（3）解答：
2s(2s+3)=23s+−23s+32{2 \over {s\left( {2s + 3} \right)}} = {{{2 \over 3}} \over s} + {{ - {2 \over 3}} \over {s + {3 \over 2}}}s(2s+3)2=s32+s+23−32

L−1(23s+−23s+32)=23u(t)−23e−32t.u(t)L^{ - 1} \left( {{{{2 \over 3}} \over s} + {{ - {2 \over 3}} \over {s + {3 \over 2}}}} \right) = {2 \over 3}u\left( t \right) - {2 \over 3}e^{ - {3 \over 2}t} .u\left( t \right)L−1(s32+s+23−32)=32u(t)−32e−23t.u(t)

（4）解答：
1s(s2+3)=13s+−13ss2+3{1 \over {s\left( {s^2 + 3} \right)}} = {{{1 \over 3}} \over s} + {{ - {1 \over 3}s} \over {s^2 + 3}}s(s2+3)1=s31+s2+3−31s

L−1(13s−13ss2+3)=13u(t)−13cos⁡3t⋅u(t)L^{ - 1} \left( {{{{1 \over 3}} \over s} - {{{1 \over 3}s} \over {s^2 + 3}}} \right) = {1 \over 3}u\left( t \right) - {1 \over 3}\cos \sqrt 3 t \cdot u\left( t \right)L−1(s31−s2+331s)=31u(t)−31cos3t⋅u(t)

（5）解答：

3(s+4)(s+3)=3s+3−3s+4{3 \over {\left( {s + 4} \right)\left( {s + 3} \right)}} = {3 \over {s + 3}} - {3 \over {s + 4}}(s+4)(s+3)3=s+33−s+43

L−1[3s+3−3s+4]=3e−3t−3e−4t,t≥0L^{ - 1} \left[ {{3 \over {s + 3}} - {3 \over {s + 4}}} \right] = 3e^{ - 3t} - 3e^{ - 4t} ,\,\,\,\,t \ge 0L−1[s+33−s+43]=3e−3t−3e−4t,t≥0

（6）解答：
3s(s+4)(s+3)=−9s+3+12s+4{{3s} \over {\left( {s + 4} \right)\left( {s + 3} \right)}} = {{ - 9} \over {s + 3}} + {{12} \over {s + 4}}(s+4)(s+3)3s=s+3−9+s+412

L−1(−9s+3+12s+4)=−9e−3t+12e−4t,t≥0L^{ - 1} \left( {{{ - 9} \over {s + 3}} + {{12} \over {s + 4}}} \right) = - 9e^{ - 3t} + 12e^{ - 4t} ,\,\,\,\,t \ge 0L−1(s+3−9+s+412)=−9e−3t+12e−4t,t≥0

（7）解答：
L−1(1s2+1+1)=sin⁡t+δ(t),t≥0L^{ - 1} \left( {{1 \over {s^2 + 1}} + 1} \right) = \sin t + \delta \left( t \right),\,\,\,\,t \ge 0L−1(s2+11+1)=sint+δ(t),t≥0

（8）解答：
1s2−3s+2=−1s−1+1s−2{1 \over {s^2 - 3s + 2}} = {{ - 1} \over {s - 1}} + {1 \over {s - 2}}s2−3s+21=s−1−1+s−21

L−1(−1s−1+1s−2)=−et+e2t,t≥0L^{ - 1} \left( {{{ - 1} \over {s - 1}} + {1 \over {s - 2}}} \right) = - e^t + e^{2t} ,\,\,\,\,t \ge 0L−1(s−1−1+s−21)=−et+e2t,t≥0

（9）解答：
1s(RCs+1)=1s−1s+1RC{1 \over {s\left( {RCs + 1} \right)}} = {1 \over s} - {1 \over {s + {1 \over {RC}}}}s(RCs+1)1=s1−s+RC11

L−1[1s−1s+1RC]=1−e−1RCt,t≥0L^{ - 1} \left[ {{1 \over s} - {1 \over {s + {1 \over {RC}}}}} \right] = 1 - e^{ - {1 \over {RC}}t} ,\,\,\,\,t \ge 0L−1[s1−s+RC11]=1−e−RC1t,t≥0

（10）解答：
1−RCss(1+RCs)=1s−2s+1RC{{1 - RCs} \over {s\left( {1 + RCs} \right)}} = {1 \over s} - {2 \over {s + {1 \over {RC}}}}s(1+RCs)1−RCs=s1−s+RC12

L−1(1s−2s+1RC)=1−2e−tRC,t≥0L^{ - 1} \left( {{1 \over s} - {2 \over {s + {1 \over {RC}}}}} \right) = 1 - 2e^{ - {t \over {RC}}} ,\,\,\,\,t \ge 0L−1(s1−s+RC12)=1−2e−RCt,t≥0

（11）解答：
ω(s2+ω2)⋅1(RCs+1){\omega \over {\left( {s^2 + \omega ^2 } \right)}} \cdot {1 \over {\left( {RCs + 1} \right)}}(s2+ω2)ω⋅(RCs+1)1

ωs2+ω2⋅1RCs+1RC=ms+ns2+ω2⋅ls+1RC{\omega \over {s^2 + \omega ^2 }} \cdot {{{1 \over {RC}}} \over {s + {1 \over {RC}}}} = {{ms + n} \over {s^2 + \omega ^2 }} \cdot {l \over {s + {1 \over {RC}}}}s2+ω2ω⋅s+RC1RC1=s2+ω2ms+n⋅s+RC1l

ms2+mRCs+ns+nRC+ls2+lω2(s2+ω2)(s+1RC)=ωRC(s2+ω2)(s+1RC){{ms^2 + {m \over {RC}}s + ns + {n \over {RC}} + ls^2 + l\omega ^2 } \over {\left( {s^2 + \omega ^2 } \right)\left( {s + {1 \over {RC}}} \right)}} = {{{\omega \over {RC}}} \over {\left( {s^2 + \omega ^2 } \right)\left( {s + {1 \over {RC}}} \right)}}(s2+ω2)(s+RC1)ms2+RCms+ns+RCn+ls2+lω2=(s2+ω2)(s+RC1)RCω

(m+l)s2+(n+mRC)s+lω2(s2+ω2)(s+1RC)=ωRC(s2+ω2)(s+1RC){{\left( {m + l} \right)s^2 + \left( {n + {m \over {RC}}} \right)s + l\omega ^2 } \over {\left( {s^2 + \omega ^2 } \right)\left( {s + {1 \over {RC}}} \right)}} = {{{\omega \over {RC}}} \over {\left( {s^2 + \omega ^2 } \right)\left( {s + {1 \over {RC}}} \right)}}(s2+ω2)(s+RC1)(m+l)s2+(n+RCm)s+lω2=(s2+ω2)(s+RC1)RCω

KaTeX parse error: Undefined control sequence: \matrix at position 10: \left\{ {\̲m̲a̲t̲r̲i̲x̲{ {m + l = 0}n+mRC=0{n + {m \over {RC}} = 0}n+RCm=0nRC+lω2=ωRC{{n \over {RC}} + l\omega ^2 = {\omega \over {RC}}}RCn+lω2=RCω

KaTeX parse error: Undefined control sequence: \matrix at position 10: \left\{ {\̲m̲a̲t̲r̲i̲x̲{ {m = {{ - R…n=ω1+(RCω)2{n = {\omega \over {1 + \left( {RC\omega } \right)^2 }}}n=1+(RCω)2ωl=RCω1+(RCω)2{l = {{RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}}l=1+(RCω)2RCω

ωs2+ω2⋅1RCs+1=−RCω1+(RCω)2s+ω1+(RCω)2s2+ω2+RCω1+(RCω)2s+1RC{\omega \over {s^2 + \omega ^2 }} \cdot {1 \over {RCs + 1}} = {{{{ - RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}s + {\omega \over {1 + \left( {RC\omega } \right)^2 }}} \over {s^2 + \omega ^2 }} + {{{{RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}} \over {s + {1 \over {RC}}}}s2+ω2ω⋅RCs+11=s2+ω21+(RCω)2−RCωs+1+(RCω)2ω+s+RC11+(RCω)2RCω

−RC1+(RCω)2cos⁡(ωt)+11+(RCω)2sin⁡(ωt)+RCω1+(RCω)2e−tRC,t≥0{{ - RC} \over {1 + \left( {RC\omega } \right)^2 }}\cos \left( {\omega t} \right) + {1 \over {1 + \left( {RC\omega } \right)^2 }}\sin \left( {\omega t} \right) + {{RC\omega } \over {1 + \left( {RC\omega } \right)^2 }}e^{ - {t \over {RC}}} ,\,\,\,\,\,t \ge 01+(RCω)2−RCcos(ωt)+1+(RCω)21sin(ωt)+1+(RCω)2RCωe−RCt,t≥0

（12）解答：
4s+5s2+5s+6=4s+5(s+2)(s+3)=−3s+2+7s+3{{4s + 5} \over {s^2 + 5s + 6}} = {{4s + 5} \over {\left( {s + 2} \right)\left( {s + 3} \right)}} = {{ - 3} \over {s + 2}} + {7 \over {s + 3}}s2+5s+64s+5=(s+2)(s+3)4s+5=s+2−3+s+37

L−1(4s+5s2+5s+6)=−e−2t+7e−3t,t≥0L^{ - 1} \left( {{{4s + 5} \over {s^2 + 5s + 6}}} \right) = - e^{ - 2t} + 7e^{ - 3t} ,\,\,\,\,t \ge 0L−1(s2+5s+64s+5)=−e−2t+7e−3t,t≥0

（13）解答：
100(s+50)s2+201s+200=4900199s+1+15000199s+200{{100\left( {s + 50} \right)} \over {s^2 + 201s + 200}} = {{{{4900} \over {199}}} \over {s + 1}} + {{{{15000} \over {199}}} \over {s + 200}}s2+201s+200100(s+50)=s+11994900+s+20019915000

L−1(100(s+50)s2+201s+200)=4900199e−t+15000199e−200t,t≥0L^{ - 1} \left( {{{100\left( {s + 50} \right)} \over {s^2 + 201s + 200}}} \right) = {{4900} \over {199}}e^{ - t} + {{15000} \over {199}}e^{ - 200t} ,\,\,\,\,\,t \ge 0L−1(s2+201s+200100(s+50))=1994900e−t+19915000e−200t,t≥0

（14）解答：
s+3(s+1)3⋅(s+2)=A(s+1)3+B(s+1)2+Cs+1+Ds+2{{s + 3} \over {\left( {s + 1} \right)^3 \cdot \left( {s + 2} \right)}} = {A \over {\left( {s + 1} \right)^3 }} + {B \over {\left( {s + 1} \right)^2 }} + {C \over {s + 1}} + {D \over {s + 2}}(s+1)3⋅(s+2)s+3=(s+1)3A+(s+1)2B+s+1C+s+2D

D=s+3(s+1)3∣s=−2=−1D = \left. {{{s + 3} \over {\left( {s + 1} \right)^3 }}} \right|_{s = - 2} = - 1D=(s+1)3s+3∣∣∣∣∣s=−2=−1

A=s+3s+2∣s=−1=2A = \left. {{{s + 3} \over {s + 2}}} \right|_{s = - 1} = 2A=s+2s+3∣∣∣∣s=−1=2

B=dds(s+3s+2)∣s=−1=1s+2−s+3(s+2)2∣s=−1=−1B = \left. {{d \over {ds}}\left( {{{s + 3} \over {s + 2}}} \right)} \right|_{s = - 1} = \left. {{1 \over {s + 2}} - {{s + 3} \over {\left( {s + 2} \right)^2 }}} \right|_{s = - 1} = - 1B=dsd(s+2s+3)∣∣∣∣s=−1=s+21−(s+2)2s+3∣∣∣∣∣s=−1=−1

C=12⋅d2ds2(s+3s+2)∣s=−1=122(s+3)(s+2)3−2(s+2)2∣s=−1=1C = \left. {{1 \over 2} \cdot {{d^2 } \over {ds^2 }}\left( {{{s + 3} \over {s + 2}}} \right)} \right|_{s = - 1} = \left. {{1 \over 2}{{2\left( {s + 3} \right)} \over {\left( {s + 2} \right)^3 }} - {2 \over {\left( {s + 2} \right)^2 }}} \right|_{s = - 1} = 1C=21⋅ds2d2(s+2s+3)∣∣∣∣s=−1=21(s+2)32(s+3)−(s+2)22∣∣∣∣∣s=−1=1

L−1(s+3(s+1)3(s+2))=t2e−t−te−t+e−t−e−2t,t≥0L^{ - 1} \left( {{{s + 3} \over {\left( {s + 1} \right)^3 \left( {s + 2} \right)}}} \right) = t^2 e^{ - t} - te^{ - t} + e^{ - t} - e^{ - 2t} ,\,\,\,\,\,t \ge 0L−1((s+1)3(s+2)s+3)=t2e−t−te−t+e−t−e−2t,t≥0

05第五题

求下列函数的拉普拉斯逆变换：
F(s)=ln⁡(ss+9)F\left( s \right) = \ln \left( {{s \over {s + 9}}} \right)F(s)=ln(s+9s)

求解：

本题应用到Laplace变换的一个性质，如下：
F(s)=∫0−∞f(t)e−stdtF\left( s \right) = \int_{0_ - }^\infty {f\left( t \right)e^{ - st} dt}F(s)=∫0−∞f(t)e−stdtF′(s)=dds∫0−∞f(t)e−stdt=∫0−∞−t⋅f(t)e−stdtF'\left( s \right) = {d \over {ds}}\int_{0_ - }^\infty {f\left( t \right)e^{ - st} dt} = \int_{0_ - }^\infty { - t \cdot f\left( t \right)e^{ - st} dt}F′(s)=dsd∫0−∞f(t)e−stdt=∫0−∞−t⋅f(t)e−stdtt⋅f(t)=−L−1[dF(s)ds]t \cdot f\left( t \right) = - L^{ - 1} \left[ {{{dF\left( s \right)} \over {ds}}} \right]t⋅f(t)=−L−1[dsdF(s)]

f(t)=L−1[ln⁡(ss+9)]f\left( t \right) = L^{ - 1} \left[ {\ln \left( {{s \over {s + 9}}} \right)} \right]f(t)=L−1[ln(s+9s)]

−t⋅f(t)=L−1[ddsln⁡(ss+9)]=L−1(−1s+9+1s)=1−e−9t,t≥0- t \cdot f\left( t \right) = L^{ - 1} \left[ {{d \over {ds}}\ln \left( {{s \over {s + 9}}} \right)} \right] = L^{ - 1} \left( { - {1 \over {s + 9}} + {1 \over s}} \right)\, = 1 - e^{ - 9t} ,\,\,\,\,t \ge 0−t⋅f(t)=L−1[dsdln(s+9s)]=L−1(−s+91+s1)=1−e−9t,t≥0

f(t)=−1t+1te−9t,t≥0f\left( t \right) = {{ - 1} \over t} + {1 \over t}e^{ - 9t} ,\,\,\,\,t \ge 0f(t)=t−1+t1e−9t,t≥0

如果使用MATLAB进行求解，结果如下：

>>ilaplace(log(s/(s+9)))
ans=(exp(-9*t)-1)/t

06第六题

求解：

（1）x[n]=δ[n]x\left[ n \right] = \delta \left[ n \right]x[n]=δ[n]
（2） x[n]=δ[n+3]x\left[ n \right] = \delta \left[ {n + 3} \right]x[n]=δ[n+3]
（3） x[n]=δ[n−1]x\left[ n \right] = \delta \left[ {n - 1} \right]x[n]=δ[n−1]
（4） x[n]=−2δ[n−2]+δ[n]+2δ[n+1]x\left[ n \right] = - 2\delta \left[ {n - 2} \right] + \delta \left[ n \right] + 2\delta \left[ {n + 1} \right]x[n]=−2δ[n−2]+δ[n]+2δ[n+1]
（5） x[n]=anu[n]x\left[ n \right] = a^n u\left[ n \right]x[n]=anu[n]
（6） x[n]=−anu[−n−1]x\left[ n \right] = - a^n u\left[ { - n - 1} \right]x[n]=−anu[−n−1]

07第七题

求解：

（1）求解：
X(z)=zz+0.5X\left( z \right) = {z \over {z + 0.5}}X(z)=z+0.5z

由于∣z∣>5\left| z \right| > 5∣z∣>5，所以序列是右边序列，因此序列为：x[n]=(−0.5)nu[n]x\left[ n \right] = \left( { - 0.5} \right)^n u\left[ n \right]x[n]=(−0.5)nu[n]

（2）求解：
X(z)z=z−0.5(z+14)⋅(z+12)=−3z+14+4z+12{{X\left( z \right)} \over z} = {{z - 0.5} \over {\left( {z + {1 \over 4}} \right) \cdot \left( {z + {1 \over 2}} \right)}} = {{ - 3} \over {z + {1 \over 4}}} + {4 \over {z + {1 \over 2}}}zX(z)=(z+41)⋅(z+21)z−0.5=z+41−3+z+214
由于∣z∣>5\left| z \right| > 5∣z∣>5，所以序列是右边序列，因此序列为：x[n]=[−3(−14)n+4(−12)n]⋅u[n]x\left[ n \right] = \left[ { - 3\left( {{{ - 1} \over 4}} \right)^n + 4\left( {{{ - 1} \over 2}} \right)^n } \right] \cdot u\left[ n \right]x[n]=[−3(4−1)n+4(2−1)n]⋅u[n]

=(−14)n(−3+2n+2)⋅u[n]= \left( { - {1 \over 4}} \right)^n \left( { - 3 + 2^{n + 2} } \right) \cdot u\left[ n \right]=(−41)n(−3+2n+2)⋅u[n]

（3）求解：
X(z)=1−12z−11−14z−2=zz+12X\left( z \right) = {{1 - {1 \over 2}z^{ - 1} } \over {1 - {1 \over 4}z^{ - 2} }} = {z \over {z + {1 \over 2}}}X(z)=1−41z−21−21z−1=z+21z∣z∣>12\left| z \right| > {1 \over 2}\;\;\;∣z∣>21x[n]=(−12)n⋅u[n]x\left[ n \right] = \left( { - {1 \over 2}} \right)^n \cdot u\left[ n \right]x[n]=(−21)n⋅u[n]

（4）求解：
X(z)z=z−az(1−az)=−1a(z−a)z(z−1a)=−az+a−1az−1a{{X\left( z \right)} \over z} = {{z - a} \over {z\left( {1 - az} \right)}} = {{ - {1 \over a}\left( {z - a} \right)} \over {z\left( {z - {1 \over a}} \right)}} = {{ - a} \over z} + {{a - {1 \over a}} \over {z - {1 \over a}}}zX(z)=z(1−az)z−a=z(z−a1)−a1(z−a)=z−a+z−a1a−a1∣z∣>∣1a∣,\left| z \right| > \left| {{1 \over a}} \right|,∣z∣>∣∣∣∣a1∣∣∣∣,x[n]=−aδ[n]+(a−1a)⋅(1a)n⋅u[n]x\left[ n \right] = - a\delta \left[ n \right] + \left( {a - {1 \over a}} \right) \cdot \left( {{1 \over a}} \right)^n \cdot u\left[ n \right]x[n]=−aδ[n]+(a−a1)⋅(a1)n⋅u[n]

08第八题

利用三种方法求解下面X(z)X\left( z \right)X(z)的你变换x[n]x\left[ n \right]x[n]。

X(z)=10z(z−1)(z−2),(∣z∣>2)X\left( z \right) = {{10z} \over {\left( {z - 1} \right)\left( {z - 2} \right)}},\,\,\,\,\left( {\left| z \right| > 2} \right)X(z)=(z−1)(z−2)10z,(∣z∣>2)

求解：

方法1：围线积分方法：
因为∣z∣>2\left| z \right| > 2∣z∣>2，所以信号为右边序列。
x[n]=12πj∮CX(z)⋅zn−1dz=12πj∮C10zn(z−1)(z−2)dzx\left[ n \right] = {1 \over {2\pi j}}\oint\limits_C {X\left( z \right) \cdot z^{n - 1} dz = {1 \over {2\pi j}}\oint\limits_C {{{10z^n } \over {\left( {z - 1} \right)\left( {z - 2} \right)}}dz} }x[n]=2πj1C∮X(z)⋅zn−1dz=2πj1C∮(z−1)(z−2)10zndz

x[n]=∑mRes[10zn(z−1)(z−2)]z=zmx\left[ n \right] = \sum\limits_m^{} {{\mathop{\rm Re}\nolimits} s\left[ {{{10z^n } \over {\left( {z - 1} \right)\left( {z - 2} \right)}}} \right]} _{z = z_m }x[n]=m∑Res[(z−1)(z−2)10zn]z=zm

Res[10zn(z−1)(z−2)]z=1=−10{\mathop{\rm Re}\nolimits} s\left[ {{{10z^n } \over {\left( {z - 1} \right)\left( {z - 2} \right)}}} \right]_{z = 1} = - 10Res[(z−1)(z−2)10zn]z=1=−10Res[10zn(z−1)(z−2)]z=2=10⋅2n{\mathop{\rm Re}\nolimits} s\left[ {{{10z^n } \over {\left( {z - 1} \right)\left( {z - 2} \right)}}} \right]_{z = 2} = 10 \cdot 2^nRes[(z−1)(z−2)10zn]z=2=10⋅2n

x[n]=[−10+10⋅2n]⋅u[n]x\left[ n \right] = \left[ { - 10 + 10 \cdot 2^n } \right] \cdot u\left[ n \right]x[n]=[−10+10⋅2n]⋅u[n]

方法2：长除法：

x[n]=10z−1+30z−2+70z−3+⋯x\left[ n \right] = 10z^{ - 1} + 30z^{ - 2} + 70z^{ - 3} + \cdotsx[n]=10z−1+30z−2+70z−3+⋯=10(2n−1)⋅u[n]= 10\left( {2^n - 1} \right) \cdot u\left[ n \right]=10(2n−1)⋅u[n]

>> deconv([10,0,0,0,0,0,0,0,0],[1,-3,2])'
ans=10 30 70 150 310 630 1270

方法3：部分因式分解法：
X(z)z=10(z−1)(z−2)=−10z−1+10z−2{{X\left( z \right)} \over z} = {{10} \over {\left( {z - 1} \right)\left( {z - 2} \right)}} = {{ - 10} \over {z - 1}} + {{10} \over {z - 2}}zX(z)=(z−1)(z−2)10=z−1−10+z−210X(z)=−10zz−1+10zz−2X\left( z \right) = {{ - 10z} \over {z - 1}} + {{10z} \over {z - 2}}X(z)=z−1−10z+z−210z

x[n]=−10⋅u[n]+10⋅2nu[n]x\left[ n \right] = - 10 \cdot u\left[ n \right] + 10 \cdot 2^n u\left[ n \right]x[n]=−10⋅u[n]+10⋅2nu[n]

使用MATLAB对应的变换命令：

>>iztrans(10*z/(z-1)/(z-2))
ans=10*2^n-10

09第九题

求解：

（1）求解：
X(z)z=10z(z−0.5)(z−0.25)=20z−0.5+−10z−0.25{{X\left( z \right)} \over z} = {{10z} \over {\left( {z - 0.5} \right)\left( {z - 0.25} \right)}} = {{20} \over {z - 0.5}} + {{ - 10} \over {z - 0.25}}zX(z)=(z−0.5)(z−0.25)10z=z−0.520+z−0.25−10X(z)=20zz−0.5+−10zz−0.25X\left( z \right) = {{20z} \over {z - 0.5}} + {{ - 10z} \over {z - 0.25}}X(z)=z−0.520z+z−0.25−10z∣z∣>0.5\left| z \right| > 0.5∣z∣>0.5x[n]=[20⋅(0.5)n−10⋅(0.25)n]⋅u[n]x\left[ n \right] = \left[ {20 \cdot \left( {0.5} \right)^n - 10 \cdot \left( {0.25} \right)^n } \right] \cdot u\left[ n \right]x[n]=[20⋅(0.5)n−10⋅(0.25)n]⋅u[n]

>>iztrans(10/(1-0.5/z)/(1-0.25/z))'
ans=20*(1/2)^n-10*(1/4)^n

（2）求解：
X(z)z=10z(z−1)(z+1)=5z−1+5z+1{{X\left( z \right)} \over z} = {{10z} \over {\left( {z - 1} \right)\left( {z + 1} \right)}} = {5 \over {z - 1}} + {5 \over {z + 1}}zX(z)=(z−1)(z+1)10z=z−15+z+15X(z)=5zz−1+5zz+1X\left( z \right) = {{5z} \over {z - 1}} + {{5z} \over {z + 1}}X(z)=z−15z+z+15z∣z∣>1\left| z \right| > 1∣z∣>1x[n]=5⋅[1+(−1)n]⋅u[n]x\left[ n \right] = 5 \cdot \left[ {1 + \left( { - 1} \right)^n } \right] \cdot u\left[ n \right]x[n]=5⋅[1+(−1)n]⋅u[n]

>>iztrans(10*z*z/(z-1)/(z+1))'
ans=5*(-1)^n+5

（3）求解： 根据正弦、余弦单边序列的z变换：
Z[cos⁡(ω0n)u[n]]=z(z−cos⁡ω0)z2−2zcos⁡ω0+1Z\left[ {\cos \left( {\omega _0 n} \right)u\left[ n \right]} \right] = {{z\left( {z - \cos \omega _0 } \right)} \over {z^2 - 2z\cos \omega _0 + 1}}Z[cos(ω0n)u[n]]=z2−2zcosω0+1z(z−cosω0)Z[sin⁡(ω0n)u[n]]=zsin⁡ω0z2−2zcos⁡ω0+1Z\left[ {\sin \left( {\omega _0 n} \right)u\left[ n \right]} \right] = {{z\sin \omega _0 } \over {z^2 - 2z\cos \omega _0 + 1}}Z[sin(ω0n)u[n]]=z2−2zcosω0+1zsinω0

X(z)=z2+zz2−2zcos⁡ω+1X\left( z \right) = {{z^2 + z} \over {z^2 - 2z\cos \omega + 1}}X(z)=z2−2zcosω+1z2+z=z(z−cos⁡ω)z2−2zcos⁡ω+1+1+cos⁡ωsin⁡ωzsin⁡ωz2−2zcos⁡ω+1= {{z\left( {z - \cos \omega } \right)} \over {z^2 - 2z\cos \omega + 1}} + {{1 + \cos \omega } \over {\sin \omega }}{{z\sin \omega } \over {z^2 - 2z\cos \omega + 1}}=z2−2zcosω+1z(z−cosω)+sinω1+cosωz2−2zcosω+1zsinω
x[n]=(cos⁡ωn+1+cos⁡ωsin⁡ωsin⁡ωn)⋅u[n]x\left[ n \right] = \left( {\cos \omega n + {{1 + \cos \omega } \over {\sin \omega }}\sin \omega n} \right) \cdot u\left[ n \right]x[n]=(cosωn+sinω1+cosωsinωn)⋅u[n]=sin⁡(nω)+sin⁡(n+1)ωsin⁡ω⋅u[n]= {{\sin \left( {n\omega } \right) + \sin \left( {n + 1} \right)\omega } \over {\sin \omega }} \cdot u\left[ n \right]=sinωsin(nω)+sin(n+1)ω⋅u[n]

>>iztrans((1+1/z)/(1-2*z^-1*cos(w)+z^-2))'
ans=(sin(n*w)*(cos(w)+1))/sin(w)-(cos(n*w)*(cos(w)+1))/cos(w)+(cos(n*w)*(2*cos(w)+1))/cos(w)

>>iztrans((z*z+z)/(z*z-2*z*cos(w)+1))'
ans=(sin(n*w)*(cos(w)+1))/sin(w)-(cos(n*w)*(cos(w)+1))/cos(w)+(cos(n*w)*(2*cos(w)+1))/cos(w)

sin⁡nω(cos⁡ω+1)sin⁡ω−cos⁡nω(cos⁡ω+1)cos⁡ω+cos⁡nω(2cos⁡+1)cos⁡ω{{\sin n\omega \left( {\cos \omega + 1} \right)} \over {\sin \omega }} - {{\cos n\omega \left( {\cos \omega + 1} \right)} \over {\cos \omega }} + {{\cos n\omega \left( {2\cos + 1} \right)} \over {\cos \omega }}sinωsinnω(cosω+1)−cosωcosnω(cosω+1)+cosωcosnω(2cos+1)
(sin⁡nω⋅cos⁡ω−sin⁡ω⋅cos⁡nω)⋅(cos⁡ω+1)+sin⁡ω⋅cos⁡nω⋅(2cos⁡ω+1)sin⁡ω⋅cos⁡ω{{\left( {\sin n\omega \cdot \cos \omega - \sin \omega \cdot \cos n\omega } \right) \cdot \left( {\cos \omega + 1} \right) + \sin \omega \cdot \cos n\omega \cdot (2\cos \omega + 1)} \over {\sin \omega \cdot \cos \omega }}sinω⋅cosω(sinnω⋅cosω−sinω⋅cosnω)⋅(cosω+1)+sinω⋅cosnω⋅(2cosω+1)

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