武大电气电路仿真实验报告 - 图文(5)

自强、弘毅、求是、拓新

10.8angle(H)0.60.40.200.511.52ww2.533.540angle(H)-0.5-1-1.500.511.52ww2.533.54

(a)线性频率响

0-10dB-20-3010ww0-0.5-1-1.50angle(H)10ww0

(b)对数频率响

2、频率响应：二阶低通电路

令H0=1，画出Q=1/3,1/2,1/√2,1,2,5的幅频相频响应，当Q=1/√2时，成为最平幅度特性，即在通带内其幅频特性最为平坦。

clear,formatcompact

for Q=[1/3,1/2,1/sqrt(2),1,2,5] ww=logspace(-1,1,50); H=1./(1+j*ww/Q+(j*ww).^2); figure(1)

subplot(2,1,1),plot(ww,abs(H)),hold on

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自强、弘毅、求是、拓新

subplot(2,1,2),plot(ww,angle(H)),hold on figure(2)

subplot(2,1,1),semilogx(ww,20*log10(abs(H))),hold on subplot(2,1,2),semilogx(ww,angle(H)),hold on end

figure(1),subplot(2,1,1),grid,xlabel('w'),ylabel('abs(H)') subplot(2,1,2),grid,xlabel('w'),ylabel('angle(H))') figure(2),subplot(2,1,1),grid,xlabel('w'),ylabel('abs(H)') subplot(2,1,2),grid,xlabel('w'),ylabel('angle(H)')

6abs(H)420012345w6789100-1angle(H))-2-3-4012345w678910

(a)线性频率响

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自强、弘毅、求是、拓新

200abs(H)-20-40-60-1100110w100-1angle(H)-2-3-4-1100110w10

(b)对数频率响

3、频率响应：二阶带通电路

clear,formatcompact H0=1;wn=1;

for Q=[5,10,20,50,100] w=logspace(-1,1,50); H=H0./(1+j*Q*(w./wn-wn./w)); figure(1)

subplot(2,1,1),plot(w,abs(H)),grid,holdon subplot(2,1,2),plot(w,angle(H)),grid,holdon figure(2)

subplot(2,1,1),semilogx(w,20*log10(abs(H))),grid,holdon subplot(2,1,2),semilogx(w,angle(H)),grid,holdon end

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自强、弘毅、求是、拓新

10.80.60.40.20012345678910210-1-2012345678910

0-20-40-60-110210-1-2-110100101100101

4、复杂谐振电路的计算

clear,formatcompact

R1=2;R2=3;L1=0.75e-3;L2=0.25e-3;C=1000e-12;Rs=28200; L=L1+L2;R=R1+R2; Rse=Rs*(L/L1)^2 f0=1/(2*pi*sqrt(C*L)) Q0=sqrt(L/C)/R,R0=L/C/R; Re=R0*Rse/(R0+Rse) Q=Q0*Re/R0,B=f0/Q s=log10(f0);

f=logspace(s-.1,s+.1,501);w=2*pi*f;

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自强、弘毅、求是、拓新

z1e=R1+j*w*L;z2e=R2+1./(j*w*C); ze=1./(1./z1e+1./z2e+1./Rse); subplot(2,1,1),loglog(w,abs(ze)),grid

axis([min(w),max(w),0.9*min(abs(ze)),1.1*max(abs(ze))]) subplot(2,1,2),semilogx(w,angle(ze)*180/pi) axis([min(w),max(w),-100,100]),grid

fh=w(find(abs(1./(1./z1e+1./z2e))>50000))/2/pi; fhmin=min(fh),fhmax=max(fh)

仿真结果： Rse =

5.0133e+004 f0 =

1.5915e+005 Q0 = 200 Re =

4.0085e+004 Q =

40.0853 B =

3.9704e+003 fhmin =

1.5770e+005 fhmax =

1.6063e+005

实验总结

1学会MATLAB的频率计算

2用MATLAB 中的abs(H)和angle(H)语句可直接计算幅频响应，而且其图像的频率坐标可以是线性的（用plot），也可以用时半对数的（用semilogx）。

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