具体步骤如下:
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原始表达式:
4 + 3 e − j n ( 1 − 0.25 e − j n ) ( 1 − 0.5 e − j n ) ( 1 − 0.75 e − j n ) \frac{4 + 3e^{-jn}}{(1 - 0.25e^{-jn})(1 - 0.5e^{-jn})(1 - 0.75e^{-jn})} (1−0.25e−jn)(1−0.5e−jn)(1−0.75e−jn)4+3e−jn -
部分分式展开:
4 + 3 e − j n ( 1 − 0.25 e − j n ) ( 1 − 0.5 e − j n ) ( 1 − 0.75 e − j n ) = a 1 − 0.25 e − j n + b 1 − 0.5 e − j n + c 1 − 0.75 e − j n \frac{4 + 3e^{-jn}}{(1 - 0.25e^{-jn})(1 - 0.5e^{-jn})(1 - 0.75e^{-jn})} = \frac{a}{1 - 0.25e^{-jn}} + \frac{b}{1 - 0.5e^{-jn}} + \frac{c}{1 - 0.75e^{-jn}} (1−0.25e−jn)(1−0.5e−jn)(1−0.75e−jn)4+3e−jn=1−0.25e−jna+1−0.5e−jnb+1−0.75e−jnc -
求解系数 (a):
将等式两边乘以 (1 - 0.25e^{-jn}),然后令 (e^{-jn} = 4) 以消去其他项:
( 1 − 0.25 e − j n ) ⋅ 4 + 3 e − j n ( 1 − 0.25 e − j n ) ( 1 − 0.5 e − j n ) ( 1 − 0.75 e − j n ) = a ⋅ 1 − 0.25 e − j n 1 − 0.25 e − j n + b ⋅ 1 − 0.25 e − j n 1 − 0.5 e − j n + c ⋅ 1 − 0.25 e − j n 1 − 0.75 e − j n (1 - 0.25e^{-jn}) \cdot \frac{4 + 3e^{-jn}}{(1 - 0.25e^{-jn})(1 - 0.5e^{-jn})(1 - 0.75e^{-jn})} = a \cdot \frac{1 - 0.25e^{-jn}}{1 - 0.25e^{-jn}} + b \cdot \frac{1 - 0.25e^{-jn}}{1 - 0.5e^{-jn}} + c \cdot \frac{1 - 0.25e^{-jn}}{1 - 0.75e^{-jn}} (1−0.25e−jn)⋅(1−0.25e−jn)(1−0.5e−jn)(1−0.75e−jn)4+3e−jn=a⋅1−0.25e−jn1−0.25e−jn+b⋅1−0.5e−jn1−0.25e−jn+c⋅1−0.75e−jn1−0.25e−jn
当 (e^{-jn} = 4) 时,((1 - 0.25 \cdot 4) = 0):
4 + 3 ⋅ 4 ( 1 − 0.5 ⋅ 4 ) ( 1 − 0.75 ⋅ 4 ) = a \frac{4 + 3 \cdot 4}{(1 - 0.5 \cdot 4)(1 - 0.75 \cdot 4)} = a (1−0.5⋅4)(1−0.75⋅4)4+3⋅4=a
4 + 12 ( 1 − 2 ) ( 1 − 3 ) = a \frac{4 + 12}{(1 - 2)(1 - 3)} = a (1−2)(1−3)4+12=a
16 ( − 1 ) ( − 2 ) = a \frac{16}{(-1)(-2)} = a (−1)(−2)16=a
16 2 = a \frac{16}{2} = a 216=a
a = 8 a = 8 a=8 -
求解系数 (b):
类似地,将等式两边乘以 (1 - 0.5e^{-jn}),然后令 (e^{-jn} = 2) 以消去其他项:
( 1 − 0.5 e − j n ) ⋅ 4 + 3 e − j n ( 1 − 0.25 e − j n ) ( 1 − 0.5 e − j n ) ( 1 − 0.75 e − j n ) = b ⋅ 1 − 0.5 e − j n 1 − 0.5 e − j n + a ⋅ 1 − 0.5 e − j n 1 − 0.25 e − j n + c ⋅ 1 − 0.5 e − j n 1 − 0.75 e − j n (1 - 0.5e^{-jn}) \cdot \frac{4 + 3e^{-jn}}{(1 - 0.25e^{-jn})(1 - 0.5e^{-jn})(1 - 0.75e^{-jn})} = b \cdot \frac{1 - 0.5e^{-jn}}{1 - 0.5e^{-jn}} + a \cdot \frac{1 - 0.5e^{-jn}}{1 - 0.25e^{-jn}} + c \cdot \frac{1 - 0.5e^{-jn}}{1 - 0.75e^{-jn}} (1−0.5e−jn)⋅(1−0.25e−jn)(1−0.5e−jn)(1−0.75e−jn)4+3e−jn=b⋅1−0.5e−jn1−0.5e−jn+a⋅1−0.25e−jn1−0.5e−jn+c⋅1−0.75e−jn1−0.5e−jn
当 (e^{-jn} = 2) 时,((1 - 0.5 \cdot 2) = 0):
4 + 3 ⋅ 2 ( 1 − 0.25 ⋅ 2 ) ( 1 − 0.75 ⋅ 2 ) = b \frac{4 + 3 \cdot 2}{(1 - 0.25 \cdot 2)(1 - 0.75 \cdot 2)} = b (1−0.25⋅2)(1−0.75⋅2)4+3⋅2=b
4 + 6 ( 1 − 0.5 ) ( 1 − 1.5 ) = b \frac{4 + 6}{(1 - 0.5)(1 - 1.5)} = b (1−0.5)(1−1.5)4+6=b
10 ( 0.5 ) ( − 0.5 ) = b \frac{10}{(0.5)(-0.5)} = b (0.5)(−0.5)10=b
10 − 0.25 = b \frac{10}{-0.25} = b −0.2510=b
b = − 40 b = -40 b=−40 -
求解系数 (c):
将等式两边乘以 (1 - 0.75e^{-jn}),然后令 (e^{-jn} = \frac{4}{3}) 以消去其他项:
( 1 − 0.75 e − j n ) ⋅ 4 + 3 e − j n ( 1 − 0.25 e − j n ) ( 1 − 0.5 e − j n ) ( 1 − 0.75 e − j n ) = c ⋅ 1 − 0.75 e − j n 1 − 0.75 e − j n + a ⋅ 1 − 0.75 e − j n 1 − 0.25 e − j n + b ⋅ 1 − 0.75 e − j n 1 − 0.5 e − j n (1 - 0.75e^{-jn}) \cdot \frac{4 + 3e^{-jn}}{(1 - 0.25e^{-jn})(1 - 0.5e^{-jn})(1 - 0.75e^{-jn})} = c \cdot \frac{1 - 0.75e^{-jn}}{1 - 0.75e^{-jn}} + a \cdot \frac{1 - 0.75e^{-jn}}{1 - 0.25e^{-jn}} + b \cdot \frac{1 - 0.75e^{-jn}}{1 - 0.5e^{-jn}} (1−0.75e−jn)⋅(1−0.25e−jn)(1−0.5e−jn)(1−0.75e−jn)4+3e−jn=c⋅1−0.75e−jn1−0.75e−jn+a⋅1−0.25e−jn1−0.75e−jn+b⋅1−0.5e−jn1−0.75e−jn
当 (e^{-jn} = \frac{4}{3}) 时,((1 - 0.75 \cdot \frac{4}{3}) = 0):
4 + 3 ⋅ 4 3 ( 1 − 0.25 ⋅ 4 3 ) ( 1 − 0.5 ⋅ 4 3 ) = c \frac{4 + 3 \cdot \frac{4}{3}}{(1 - 0.25 \cdot \frac{4}{3})(1 - 0.5 \cdot \frac{4}{3})} = c (1−0.25⋅34)(1−0.5⋅34)4+3⋅34=c
4 + 4 ( 1 − 1 3 ) ( 1 − 2 3 ) = c \frac{4 + 4}{(1 - \frac{1}{3})(1 - \frac{2}{3})} = c (1−31)(1−32)4+4=c
8 ( 2 3 ) ( 1 3 ) = c \frac{8}{(\frac{2}{3})(\frac{1}{3})} = c (32)(31)8=c
8 2 9 = c \frac{8}{\frac{2}{9}} = c 928=c
c = 36 c = 36 c=36