求周期信号x(t)=0.5cos10t 0.2cos(100t-45°)通过传递函数H（s）=1/(0.005s 1)装置后的稳态响应 求一电路题
\u6c42\u5468\u671f\u4fe1\u53f7x(t)=0.5cos10t+0.2cos(100t-45\u00b0)\u901a\u8fc7\u4f20\u9012\u51fd\u6570H\uff08s\uff09=1/(0.005s+1)\u88c5\u7f6e\u540e\u7684\u7a33\u6001\u54cd\u5e94\u7531\u4e00\u8282\u7cfb\u7edf\u54cd\u5e94\u8f93\u51fa\u89c4\u5f8b\u5f97\uff0c
y\uff08t\uff09=A1COS(10t+a1)+A2cos(100t+a2) \u7531A\uff08w\uff09=1/(\u6839\u53f7[1+\uff080.005*10)^2])=A1/0.5
\u5f97A1=0.50 \u540c\u7406\uff0cA2=0.18
\u7531a\uff08w\uff09=-arctan\uff080.005*10\uff09=a1-0 \u5f97a1=-2.9\u00b0 \u540c\u7406a2=-71.6\u00b0
\u6240\u4ee5\uff0cy(t)=0.50cos(10t-2.9\u00b0)+0.18cos(100t-71.6\u00b0)
\u7535\u8def\u5206\u6790 \u8bd5\u9898\uff08\u2160\uff09
\u4e8c. \u586b\u7a7a \uff08\u6bcf\u98981\u5206\uff0c\u517110\u5206\uff09
1\uff0eKVL\u4f53\u73b0\u4e86\u7535\u8def\u4e2d \u5b88\u6052\u7684\u6cd5\u5219\u3002
2\uff0e\u7535\u8def\u4e2d\uff0c\u67d0\u5143\u4ef6\u5f00\u8def\uff0c\u5219\u6d41\u8fc7\u5b83\u7684\u7535\u6d41\u5fc5\u4e3a \u3002
3\uff0e\u82e5\u7535\u8def\u7684\u652f\u8def\u6570\u4e3ab\uff0c\u8282\u70b9\u6570\u4e3an\uff0c\u5219\u72ec\u7acb\u7684KCL\u65b9\u7a0b\u6570\u4e3a \u3002
4\uff0e\u5728\u7ebf\u6027\u7535\u8def\u53e0\u52a0\u5b9a\u7406\u5206\u6790\u4e2d\uff0c\u4e0d\u4f5c\u7528\u7684\u72ec\u7acb\u7535\u538b\u6e90\u5e94\u5c06\u5176 \u3002
5\uff0e\u5982\u679c\u4e24\u4e2a\u5355\u53e3\u7f51\u7edc\u7aef\u53e3\u7684 \u5b8c\u5168\u76f8\u540c\uff0c\u5219\u8fd9\u4e24\u4e2a\u5355\u53e3\u7f51\u7edc
\u7b49\u6548\u3002
6\uff0e\u82e5\u4e00\u9636\u7535\u8def\u7535\u5bb9\u7535\u538b\u7684\u5b8c\u5168\u54cd\u5e94\u4e3auc(t)= 8 - 3e-10t V,\u5219\u7535\u5bb9\u7535\u538b
\u7684\u96f6\u8f93\u5165\u54cd\u5e94\u4e3a \u3002
7\uff0e\u82e5\u4e00\u4e2a\u6b63\u5f26\u7535\u538b\u7684\u77ac\u65f6\u8868\u8fbe\u5f0f\u4e3a10cos(100\u03c0t+45\u00b0)V,\u5219\u5b83\u7684\u5468\u671fT
\u4e3a \u3002
8\uff0e\u6b63\u5f26\u7535\u538bu1(t)=220cos(10t+45\u00b0)V, u2(t)=220sin(10t+120\u00b0)V,
\u5219\u76f8\u4f4d\u5dee\u03c612 \uff1d \u3002
9\uff0e\u82e5\u7535\u611fL\uff1d2H\u7684\u7535\u6d41i =2 cos(10t+30\u00b0)A (\u8bbeu , \u4e3a\u5173\u8054\u53c2\u8003
\u65b9\u5411)\uff0c \u5219\u5b83\u7684\u7535\u538bu\u4e3a \u3002
10\uff0e\u6b63\u5f26\u7a33\u6001\u7535\u8def\u4e2d,\u82e5\u65e0\u6e90\u5355\u53e3\u7f51\u7edc\u5438\u6536\u7684\u590d\u529f\u7387 \uff1d80\uff0bj60 VA,\u5219
\u529f\u7387\u56e0\u6570\u03bb\uff1d \u3002
*11\uff0eL1=5H, L2=2H, M=1H \u7684\u8026\u5408\u7535\u611f\u53cd\u63a5\u4e32\u8054\u7684\u7b49\u6548\u7535\u611f\u4e3a \u3002
\u4e09\uff0e\u6c42\u4e0b\u56fe\u5355\u53e3\u7f51\u7edc\u7684\u8bfa\u987f\u7b49\u6548\u7535\u8def\uff0c\u5e76\u753b\u7b49\u6548\u7535\u8def\u56fe\u3002(15\u5206)
a
b
\u56db\uff0e\u7528\u7ed3\u70b9\u5206\u6790\u6cd5\uff0c\u6c42\u5404\u7ed3\u70b9\u7535\u4f4d\u548c\u7535\u538b\u6e90\u529f\u7387\u3002(15\u5206)
1 2
\u4e94\uff0e\u4e00\u9636\u7535\u8def\u5982\u56fe\uff0ct = 0\u5f00\u5173\u65ad\u5f00\uff0c\u65ad\u5f00\u524d\u7535\u8def\u4e3a\u7a33\u6001\uff0c\u6c42t \u2265 0\u7535\u611f
\u7535\u6d41 L(t) ,\u5e76\u753b\u51fa\u6ce2\u5f62\u3002(15\u5206)
\u516d\uff0e\u542b\u7406\u60f3\u53d8\u538b\u5668\u6b63\u5f26\u7a33\u6001\u76f8\u91cf\u6a21\u578b\u7535\u8def\u5982\u56fe\uff0c \uff1d100\u22200\u00b0V\uff0c
\u6c42 3 \u3002(15\u5206)
*\u4e03\uff0e \u542b\u7a7a\u5fc3\u53d8\u538b\u5668\u6b63\u5f26\u7a33\u6001\u7535\u8def\u5982\u56fe\uff0cuS(t)=10 cos ( 5t + 15\u00b0)V,
\u6c42\u7535\u6d41 1(t), 2(t)\u3002(15\u5206)
\u7535\u8def\u5206\u6790 \u8bd5\u9898\uff08\u2161\uff09
\u4e00\uff0e\u5355\u9879\u9009
D\uff0e10\u2220180\u00b0V
\u4e8c. \u586b\u7a7a \uff08\u6bcf\u98981\u5206\uff0c\u517110\u5206\uff09
1\uff0e\u7535\u8def\u7684\u4e24\u7c7b\u7ea6\u675f\u662f \u3002
2\uff0e\u4e00\u53ea100\u03a9\uff0c1w\u7684\u7535\u963b\u5668\uff0c\u4f7f\u7528\u65f6\u7535\u963b\u4e0a\u7684\u7535\u538b\u4e0d\u5f97\u8d85\u8fc7 V\u3002
3\uff0e\u542bUS\u548cIS \u4e24\u76f4\u6d41\u7535\u6e90\u7684\u7ebf\u6027\u975e\u65f6\u53d8\u7535\u963b\u7535\u8def\uff0c\u82e5IS\u5355\u72ec\u4f5c\u7528\u65f6\uff0c
R \u4e0a\u7684\u7535\u6d41\u4e3aI\u2032\uff0c\u5f53US\u5355\u72ec\u4f5c\u7528\u65f6\uff0cR\u4e0a\u7684\u7535\u6d41\u4e3aI"\uff0c\uff08I\u2032\u4e0eI"
\u53c2\u8003\u65b9\u5411\u76f8\u540c\uff09\uff0c\u5219\u5f53US\u548cIS \u5171\u540c\u4f5c\u7528\u65f6\uff0cR\u4e0a\u7684\u529f\u7387\u5e94\u4e3a \u3002
4\uff0e\u82e5\u7535\u963b\u4e0a\u7535\u538bu\u4e0e\u7535\u6d41i\u4e3a\u975e\u5173\u8054\u53c2\u8003\u65b9\u5411\uff0c\u5219\u7535\u5bfcG\u7684\u8868\u8fbe\u5f0f
\u4e3a \u3002
5\uff0e\u5b9e\u9645\u7535\u538b\u6e90\u4e0e\u7406\u60f3\u7535\u538b\u6e90\u7684\u533a\u522b\u5728\u4e8e\u5b9e\u9645\u7535\u538b\u6e90\u7684\u5185\u963b \u3002
6\uff0e\u7535\u611f\u5143\u4ef6\u80fd\u5b58\u50a8 \u80fd \u3002
7\uff0e\u82e5\u7535\u5bb9\u4e0a\u7535\u538bu\u4e0e\u7535\u6d41i\u4e3a\u975e\u5173\u8054\u53c2\u8003\u65b9\u5411\uff0c\u5219u \uff0ci \u7684\u77ac\u65f6VCR
\u8868\u8fbe\u5f0f\u4e3a \u3002
* 8\uff0eR = 2 , L = 1H , C = 1F \u7684\u4e32\u8054\u4e8c\u9636\u7535\u8def\uff0c\u96f6\u8f93\u5165\u54cd\u5e94\u7684\u7c7b\u578b
\u662f \u3002
9\uff0e\u6b63\u5f26\u7a33\u6001\u7535\u8def\u4e2d, \u67d0\u7535\u611f\u4e24\u7aef\u7535\u538b\u6709\u6548\u503c\u4e3a20V\uff0c\u6d41\u8fc7\u7535\u6d41\u6709\u6548\u503c
\u4e3a2A\uff0c\u6b63\u5f26\u91cf\u5468\u671fT =\u03c0S , \u5219\u7535\u611f\u7684\u7535\u611f\u91cfL\uff1d \u3002
10\uff0e\u6b63\u5f26\u7a33\u6001L\uff0cC\u4e32\u8054\u7535\u8def\u4e2d, \u7535\u5bb9\u7535\u538b\u6709\u6548\u503c\u4e3a8V , \u7535\u611f\u7535\u538b\u6709\u6548\u503c
\u4e3a12V , \u5219\u603b\u7535\u538b\u6709\u6548\u503c\u4e3a \u3002
11\uff0e\u6b63\u5f26\u7a33\u6001\u7535\u8def\u4e2d, \u4e00\u4e2a\u65e0\u6e90\u5355\u53e3\u7f51\u7edc\u7684\u529f\u7387\u56e0\u6570\u4e3a0. 5 , \u7aef\u53e3\u7535\u538b
u(t) \uff1d10cos (100t +\u03c8u) V\uff0c\u7aef\u53e3\u7535\u6d41 (t) = 3 cos(100t - 10\u00b0)A (u,i\u4e3a
\u5173\u8054\u53c2\u8003\u65b9\u5411),\u5219\u7535\u538b\u7684\u521d\u76f8\u03c8u\u4e3a \u3002
*\u4e09\uff0e\u6c42\u4e0b\u56fe\u7535\u8def\u4e2d\uff0c\u8d1f\u8f7d\u7535\u963bRL\u83b7\u5f97\u7684\u6700\u5927\u529f\u7387\u3002(15\u5206)
\u56db\uff0e\u7535\u8def\u5982\u56fe\uff0c\uff081\uff09\u6c42a,b \u4e24\u7aef\u7535\u538bUab \u3002(2) \u82e5a,b \u7528\u5bfc\u7ebf\u77ed\u8def\uff0c\u6c42\u5bfc
\u7ebf\u4e2d\u7535\u6d41Iab \u3002(15\u5206)
\u4e94\uff0e\u7528\u7f51\u5b54\u5206\u6790\u6cd5\uff0c\u6c42\u56fe\u793a\u7535\u8def\u7f51\u5b54\u7535\u6d41I1\uff0cI2\u53ca4\u03a9\u7535\u963b\u7684\u529f\u7387\u3002(15\u5206)
\u516d\uff0e\u4e00\u9636\u7535\u8def\u5982\u56fe\uff0ct = 0\u5f00\u5173\u65ad\u5f00\uff0c\u65ad\u5f00\u524d\u7535\u8def\u4e3a\u7a33\u6001\uff0c\u6c42t \u2265 0\u7535\u5bb9
\u7535\u538buC(t) \uff0c\u5e76\u753b\u51fa\u6ce2\u5f62\u3002(15\u5206)
\u4e03\uff0e\u56fe\u793a\u7535\u8def\u4e2d\uff0c\u6b63\u5f26\u7535\u538b\u6e90uS(t)= 4 cos t V, \u76f4\u6d41\u7535\u6d41\u6e90IS\uff1d6A\uff0c
\u6c42\u7535\u6d41 1(t), 2(t)\uff0c 3(t) \u3002(15\u5206)
\u7535\u8def\u5206\u6790 \u8bd5\u9898\uff08\u2162\uff09
\u4e8c. \u586b\u7a7a \uff08\u6bcf\u98981\u5206\uff0c\u517110\u5206\uff09
1\uff0eKCL\u4f53\u73b0\u4e86\u7535\u8def\u4e2d \u5b88\u6052\u7684\u6cd5\u5219\u3002
2\uff0e\u7535\u8def\u4e2d\uff0c\u67d0\u5143\u4ef6\u77ed\u8def\uff0c\u5219\u5b83\u4e24\u7aef\u7535\u538b\u5fc5\u4e3a \u3002
3\uff0e\u82e5\u7535\u8def\u7684\u652f\u8def\u6570\u4e3ab\uff0c\u8282\u70b9\u6570\u4e3an\uff0c\u5176\u7f51\u5b54\u6570\u4e3a \u3002
4\uff0e\u7f51\u5b54\u5206\u6790\u6cd5\u53ea\u9002\u7528\u4e8e \u7535\u8def\u3002
5\uff0e\u4e00\u4e2a\u6709\u6e90\u5355\u53e3\u7f51\u7edc\u7aef\u53e3\u5f00\u8def\u7535\u538bUOC\uff1d12V\uff0c\u7aef\u53e3\u77ed\u8def\u7535\u6d41
ISC\uff1d3A\uff0c\u5219\u5355\u53e3\u7f51\u7edc\u5bf9\u5916\u8f93\u51fa\u6700\u5927\u529f\u7387PLmax\u662f W \u3002
6\uff0e\u82e5\u7535\u611fL\u7684\u7535\u538bu\u4e0e\u7535\u6d41i\u4e3a\u5173\u8054\u53c2\u8003\u65b9\u5411\uff0c\u5219u \uff0ci \u7684\u77ac\u65f6VCR
\u8868\u8fbe\u5f0f\u4e3a \u3002
7\uff0e\u6b63\u5f26\u7535\u538bu(t) =100cos(10t - 45\u00b0)V,\u5219\u5b83\u5bf9\u5e94\u7684\u76f8\u91cf \u7684\u6781\u5750\u6807\u5f0f
\u4e3a \u3002
8\uff0e\u6b63\u5f26\u7535\u6d41\u7684\u6709\u6548\u503c10A\uff0c\u521d\u76f830\u00b0\uff0c\u5468\u671f20ms\uff0c\u5199\u51fa\u7528cos\u8868\u793a
\u6b64\u7535\u6d41\u7684\u77ac\u65f6\u5f0f\u4e3a \u3002
9\uff0e\u6b63\u5f26\u7a33\u6001\u7535\u8def\u4e2d\uff0c\u7535\u6d41\u8d85\u524d\u7535\u538b\uff0d90\u00b0\u7684\u5143\u4ef6\u662f \u3002
10\uff0e\u6b63\u5f26\u7a33\u6001\u7535\u8def\u4e2d,\u7535\u5bb9\u7684\u7535\u538b\u6709\u6548\u503c\u4e3a10V\uff0c\u7535\u6d41\u6709\u6548\u503c\u4e3a2A\uff0c
\u7535\u5bb9\u5438\u6536\u7684\u5e73\u5747\u529f\u7387\u662f \u3002
*11\uff0e\u82e5\u6b63\u5f26\u7a33\u6001\u65e0\u6e90\u5355\u53e3\u7f51\u7edc\u7aef\u53e3\u7535\u538bu(t)=100 \uff0b100cos t V,\u7aef\u53e3\u7535\u6d41
i (t)= 1\uff0b10 cos( t \uff0d60\u00b0)\uff0b50cos 2t A\uff0c(\u8bbeu , \u4e3a\u5173\u8054\u53c2\u8003\u65b9\u5411)
\u5219\u7f51\u7edc\u5438\u6536\u7684\u5e73\u5747\u529f\u7387\u662f \u3002
\u4e09\uff0e\u7535\u8def\u5982\u56fe\uff0c\u82e5\u7535\u4f4d\u5668ab\u95f4\u7535\u963b\u4e3a80\u03a9\uff0c\u95ee\u7535\u538b\u8868\u8bfb\u6570\u662f\u591a\u5c11\uff1f\u5e76\u6807
\u51fa\u7535\u538b\u8868\u6781\u6027\uff0c\u518d\u6c42a\u70b9\u7535\u4f4dUa \u3002\uff08\u8bbe\u7535\u538b\u8868\u5185\u963b\u65e0\u7a77\u5927\uff09(15\u5206)
*\u56db\uff0e\u5df2\u77e5UR\uff1d2V\uff0c\u6c42R\u7684\u503c\u53ca\u53d7\u63a7\u6e90\u5438\u6536\u7684\u529f\u7387\u3002(15\u5206)
\u4e94\uff0e\u7535\u8def\u5982\u56fe\uff0c\u7528\u53e0\u52a0\u5b9a\u7406\u6c42\u5404\u652f\u8def\u7535\u6d41\u53ca\u7535\u6d41\u6e90\u529f\u7387\u3002(15\u5206)
\u516d\uff0e\u4e00\u9636\u7535\u8def\u5982\u56fe\uff0ct = 0\u5f00\u5173\u95ed\u5408\uff0c\u95ed\u5408\u524d\u7535\u8def\u4e3a\u7a33\u6001\uff0c\u6c42t \u2265 0\u7535\u6d41
L(t) \u3001 C(t)\u3001 (t) \u3002(15\u5206)
\u4e03\uff0e \u6b63\u5f26\u7a33\u6001\u7535\u8def\u5982\u56fe\uff0cuS(t)=100 cos 10t V,\u6c42\u7535\u6d41 (t)\u3001 1(t)\u3001
2(t) \u3002\u753b\u51fa\u4e09\u4e2a\u7535\u6d41\u7684\u76f8\u91cf\u56fe\u3002(15\u5206)
\u7535\u8def\u5206\u6790\uff08\u2160\uff09\u53c2\u8003\u7b54\u6848
\u4e00\uff0e\u5355\u9879\u9009\u62e9\u9898
1\uff0eC 2. B 3. A 4. D 5. B
6. D 7. A 8. C 9. D 10. D
11. A 12. C 13. A 14. D 15. A
16. D
\u4e8c\uff0e\u586b\u7a7a
1. \u80fd\u91cf 2. 0 3. n-1 4. \u77ed\u8def 5. VCR
6. 5e-10t V 7.0.02 S 8. 15\u00b0 9. 40cos ( 10t + 120\u00b0)V
10. 0.8 11. 5H
\u4e09\uff0e\u89e3\uff1aab\u77ed\u8def\uff0c\u6c42\u77ed\u8def\u7535\u6d41Iab = Isc (\u7528\u53e0\u52a0\u5b9a\u7406)
Isc \uff1d
\u72ec\u7acb\u7535\u6d41\u6e90\u4e0d\u4f5c\u7528\uff0c\u6c42\u7b49\u6548\u7535\u963bRo
Ro \uff1d(6//6\uff0b2\uff0b5)//10 = 5\u03a9
\u56db\uff0e\u89e3\uff1a \u5217\u7ed3\u70b9\u65b9\u7a0b \u89e3\u5f97: U1 = 14V U2 = 8V
\u4e94\uff0e\u89e3\uff1a t < 0 , i L(0-) = = 4A
t > 0 , i L(0+) = i L(0-) = 4A
i L(\u221e) = (0-)\u7b49\u6548\u7535\u8def
\u03c4=
\u2234 i L(t) = i L(\u221e)+[i L(0+) - i L(\u221e)]
= 1 + 3 A t\u22650
\uff08\u221e\uff09 \u6c42Ro\u7b49\u6548\u7535\u8def
\uff08\u221e\uff09\u7b49\u6548\u7535\u8def
\u516d\uff0e\u89e3\uff1a\u5c06\u53d8\u538b\u5668\u6b21\u7ea7\u8d1f\u8f7d\u65ad\u5f00\uff0c\u6c42\u6b21\u7ea7\u7aef\u53e3\u5de6\u4fa7\u6234\u7ef4\u5357\u7b49\u6548\u7535\u8def\uff0c
\uff1d100\u22200\u00b0 \uff1d500\u22200\u00b0V \uff08\u6781\u6027\u4e3a\u4e0a\u201c\uff0b\u201d\u4e0b\u201c\uff0d\u201d\uff09
\u7531\u7b49\u6548\u7535\u8def\u5f97\uff1a V
\u4e03\uff0e\u89e3\uff1a\u753b\u51fa\u76f8\u91cf\u6a21\u578b \uff0c\u53ef\u5f97\uff1a
\u2234 1(t) = 5cos ( 5t - 30\u00b0)A
2(t) = 2.5 cos ( 5t - 165\u00b0)A
\u7535\u8def\u5206\u6790\uff08\u2161\uff09\u53c2\u8003\u7b54\u6848
\u4e00\uff0e\u5355\u9879\u9009\u62e9\u9898
1\uff0eA 2. D 3. C 4. A 5. D
6. B 7. B 8. B 9. D 10. C
11. A 12. B 13. B 14. A 15. C
16. B 17. D
\u4e8c\uff0e\u586b\u7a7a
1. \u62d3\u6251\u7ea6\u675f\u4e0e\u5143\u4ef6\u7ea6\u675f 2. 10 3. 4.\uff0d
5. \u4e0d\u4e3a0 6. \u78c1\u573a 7. 8. \u4e34\u754c\u963b\u5c3c 9. 5H
10. 4V 11. 50\u00b0(\u6216\uff0d70\u00b0)
\u4e09\uff0e\u89e3\uff1a\u65ad\u5f00RL\u6c42\u5de6\u4fa7\u5355\u53e3\u7f51\u7edc\u6234\u7ef4\u5b81\u7b49\u6548\u7535\u8def\uff0c
1\uff0e\u6c42UOC \uff1a\u2235 I = 0 \u2234 4I = 0
UOC = 2 4 + 8 = 16 V
2.\u6c42RO \uff1a \u5148\u6c42\u77ed\u8def\u7535\u6d41 Isc
I\uff1dIsc\uff0c I1 \uff1d 4\uff0dI \uff1d 4\uff0dIsc
4Isc \uff1d 2\uff084\uff0dIsc\uff09\uff0b 8
Isc \uff1d A
Ro \uff1d = 6\u03a9
RL\uff1d RO \uff1d 6\u03a9\u83b7\u6700\u5927\u529f\u7387
W
\u56db\uff0e\u89e3:\uff081\uff09Uab
= -10V
(2) Iab =
\u4e94\uff0e\u89e3\uff1a \uff083\uff0b2\uff0b4\uff09I1\uff0d4I2 \uff1d 17
\uff083+4\uff09I2 \uff0d 4I1 \uff1d \uff0d18
\u89e3\u5f97\uff1a I1 \uff1d 1A I2 \uff1d \uff0d2A
\u516d \uff0e\u89e3\uff1a t < 0 , u C(0-) = - 2V
t > 0 , u C (0+) = u C (0-) = -2V
u C (\u221e) = 10 \u2013 2 = 8V
\u03c4= (1 + 1) 0.25 = 0.5 S (0-)\u7b49\u6548\u7535\u8def
\u2234 u C (t) = u C (\u221e)+[u C (0+) - u C (\u221e)]
= 8 - 10 V t\u22650
(\u221e)\u7b49\u6548\u7535\u8def
\u4e03\uff0e\u89e3\uff1a6A\u5355\u72ec\u4f5c\u7528\u65f6\uff1a 1\u2032\uff1d 2\u2032\uff1d 6A\uff0c 3\u2032\uff1d 0
uS\u5355\u72ec\u4f5c\u7528\u65f6\uff0c\u753b\u51fa\u76f8\u91cf\u6a21\u578b \uff0c\u53ef\u5f97\uff1a
\u2234 1\u3003(t) = 0
2\u3003(t) \uff1d 4cos ( t - 45\u00b0)A
3\u3003 (t) = \uff0d4cos ( t - 45\u00b0) \uff1d 4cos ( t\uff0b135\u00b0) A
\u53e0\u52a0\uff1a 1\uff08t\uff09\uff1d 1\u2032\uff0b 1\u3003\uff1d 6A
2 (t) = 2\u2032+ 2\u3003 = 6 + 4cos ( t - 45\u00b0)A
3 (t) = 3\u2032+ 3\u3003= 4cos(t\uff0b135\u00b0) A
\u7535\u8def\u5206\u6790\uff08\u2162\uff09\u53c2\u8003\u7b54\u6848
\u4e00\uff0e\u5355\u9879\u9009\u62e9\u9898
1\uff0eC 2. A 3. C 4. D 5. A
6. D 7. B 8. A 9. C 10. B
11. D 12. B 13. A 14. D 15. C
16. A
\u4e8c\uff0e\u586b\u7a7a
1. \u7535\u8377 2. 0 3. b\u2013(n-1) 4. \u5e73\u9762 5. 9
6. 7.50 \u2220\uff0d45\u00b0V 8. 10 cos ( 100\u03c0t + 30\u00b0)A
9. \u7535\u611f 10. 0 11. 350 W
\u4e09\uff0e\u89e3\uff1a -12 =(120 + 60)I-30
I = = 0.1 A
U\u8868\uff1d 80I \u2013 30 = - 22 V
\u7535\u538b\u8868\u6781\u6027\u4e3a\u4e0b\u201c\uff0b\u201d\u3001\u4e0a\u201c\uff0d\u201d
Ua\uff1d - 22 + 60I = -16 V
\u56db\uff0e\u89e3\uff1a
\u4e94\uff0e\u89e3: \u7535\u6d41\u6e90\u5355\u72ec\u4f5c\u7528\uff0c
\uff1d 10A = = = 1A
\u7535\u538b\u6e90\u5355\u72ec\u4f5c\u7528
\uff1d\uff0d \uff1d2A
\uff1d \uff0d \uff1d \uff0d2\uff0d\uff08\uff0d3\uff09\uff1d1A
\u53e0\u52a0\uff1aI1 = + = 10 + 2 = 12A
I2 = + = 2 \u20132 = 0
I3 = + = 1 \u2013 3 = -2A
I4 = + = 1 + 1 = 2A
P12A = - I1•1 12 = - 144W
\u516d\uff0e \u89e3\uff1a t < 0 , u C(0-) = 6 V i L(0-) = 0
t > 0 , \u4e3a\u4e24\u4e2a\u4e00\u9636\u7535\u8def
\u7535\u5bb9\u4e00\u9636\uff1au C(0\uff0b)\uff1du C(0-)\uff1d 6V
i C(0+) =
i C(\u221e) = 0
\u03c4= RC = 2 0.5 = 1S
\u2234 i C(t) = i C(0+) = - 3 A t\u22650
\u7535\u611f\u4e00\u9636\uff1ai L(0+) = i L(0-) = 0
i L(\u221e) =
\u03c4=
\u2234 i L(t) = i L(\u221e)\uff081\uff0d \uff09
= 3\uff081\uff0d \uff09A t\u22650
\u2234 i (t) = i L(t) \uff0di C(t) = 3\uff081\uff0d \uff09\uff0b3 A t\u22650
\u4e03\uff0e\u89e3\uff1a\u753b\u51fa\u76f8\u91cf\u6a21\u578b \uff0c\u53ef\u5f97\uff1a
\u2234 (t) = 10 cos ( 10 t - 45\u00b0)A
1(t) = 5 cos 10 t A
2(t) = 5 cos ( 10 t - 90\u00b0)A
\u4ee5\u4e0a\u56de\u7b54\u4f60\u6ee1\u610f\u4e48\uff1f
由一节系统响应输出规律得：
y（t）=A1COS(10t+a1)+A2cos(100t+a2) 由A（w）=1/(根号[1+（0.005*10)^2])=A1/0.5
得A1=0.50 同理，A2=0.18
由a（w）=-arctan（0.005*10）=a1-0 得a1=-2.9° 同理a2=-71.6°
所以，y(t)=0.50cos(10t-2.9°)+0.18cos(100t-71.6°)
一个信号既可以是模拟的也可以是数字的。如果它是连续时间和连续值，那么它就是一个模拟信号。如果它是离散时间和离散值，那么它就是一种数字信号。
除了这种区分外，信号也可以分为周期性的或非周期性的。周期性信号是一种经过一定时间重复本身的，而非周期性信号则不会重复。模拟和数字信号既可以是周期性的也可以是非周期性的。
扩展资料：
f(t)为两个子信号f1(t)=cos2t与f2(t)=sin3t的和，即f(t)=f1(t)+f2(t)，且f1(t)=f1(t-n1T1),f2(t)=f2(t-n2T2)，其中n1∈Z，n2∈Z。则当T1/T2=n2/n1(n1与n2必须为不可约的整数)时，f(t)即为周期信号，其周期T=n1T1=n2T2。
今子信号cos2t的周期为πs，子信号sin3t的周期为2π/3s。故有T1/T2=3/2。
由于3/2已为不能再约的整数比，故f(t)为周期信号，其周期T为2πs。
参考资料来源：百度百科--周期信号
楼主你好,此题属于频域法的概念问题.
目测你把加号漏了: x(t)=0.5*cos10t+0.2cos(100t-45°);H(s)=1/(0.005s+1)
首先指出,线性系统具有可线性叠加的性质,因此不妨将输入分解成
r1(t)=0.5cos10t;r2(t)=0.2cos(100t-45°)
频域法指出:对于线性系统,输入为某一频率的正弦波信号时,系统输出为同频率的正弦波
输出与输入的幅值比为传递函数的增益A(omg),相角变化为fai(omg)
这即是建立稳态输出与输入信号的桥梁
针对你给出的题目,H(s)=1/(0.005s+1),即G(jomg)=1/(0.005jomg+1)
有A(omg)=1/√((0.005omg)^2+1),fai(omg)=-arctan(0.005omg)
有A(10rad/s)=0.9988 A(100rad/s)=0.8944
fai(10rad/s)=-2.8624° fai(100rad/s)=-26.5651°
因此,对于r1(t)=0.5cos10t,有C1(t)=0.9988*0.5*cos(10t-2.8624°)
对r2(t)=0.2cos(100t-45°),有C2(t)=0.8944*0.2*cos(t-45°--26.5651°)
再由线性系统的性质,C(t)=C1(t)+C2(t)即可
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