例题设计满足下列技术指标的切比雪夫低通滤波器: 5-5a通带允许波纹:1dB002m×104rads; 阻带衰减:15dBo≥2mx2×104rads 解:通带截止频率O=2丌×10rd/s,最大允许衰减arm=ldB 阻带截止频率o,=4zx10ad/s,最小允许衰减amn=15B 归一化阻带截止频率=0=2ra/s ●求波动系数E: 0.1a 0.1×1 1=0.50885 求滤波器阶数n cosh (6-1. iooldan -1) cosh- (0.50885-1. 100. 1s -ly cosh(O) cosh(2) n=2.34≈3 求归一化增益K:KB21=05089×2049131
min 1 1 0.1 1 cosh ( 10 1) cosh ( ) s n − − − − 例题 5-5a max =1dB 4 2 10 / c = rad s = n 2.34 3 3 1 1 0.49131 0.50885 2 − = = 通带截止频率 阻带截止频率 K 设计满足下列技术指标的切比雪夫低通滤波器: 通带允许波纹: 1dB 0≤ω≤2π×104rad/s; 阻带衰减:15dB ω≥ 2π×2×104rad/s 4 4 10 / s = rad s ,最大允许衰减 ,最小允许衰减 min =15dB 归一化阻带截止频率 max 0.1 10 1 = − 2 / s s c rad s = = ⚫ 求波动系数ε: 0.1 1 10 1 0.50885 = − = ⚫ 求滤波器阶数n : 1 1 0.1 15 1 cosh (0.50885 10 1) cosh (2) − − − − = ⚫ 求归一化增益 : 1 1 2 K n − = 解:
例题5-5a 求归一化传递函数H(3): 查表5-3(2):b=0.49131,b1=123841,b2=0.98834 K 0.49131 H(S) s3+b2+bs+b33+0.9883432+1.238413+0.49131 求切比雪夫低通滤波器的传递函数H(s) 反归一化,将=代入 0.49131 H(S) )3+098834()2+1.23841(3)+049131 O.=2x×10 1.2187×10 s3+6,2104×104s2+48893×10°s+1.2187×10
⚫ 求切比雪夫低通滤波器的传递函数 H s( ) : 查表5-3(2): 例题5-5a 3 2 2 1 0 ( ) K H s s b s b s b = + + + 0 1 2 b b b = = = 0.49131, 1.23841, 0.98834 反归一化,将 代入: c s s = 3 2 0.49131 s s s 0.98834 1.23841 0.49131 = + + + ⚫ 求归一化传递函数 H s( ) : 3 2 4 0.49131 ( ) ( ) 0.98834( ) 1.23841( ) 0.49131 2 10 c c c c H s s s s = + + + = 4 3 4 2 9 4 1.2187 10 s s s 6.2104 10 4.8893 10 1.2187 10 = + + +