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1、《智能控制技术》第二章习题答案22-2解:(1)语气算子”很”λ=2即μw(a)=µr(a)00≤a≤25“很年轻”模糊集隶属度函数μw(a)=2−1[a−25)]1+(25≤a≤2005(2)“不老也不年轻”V=O∩Y1−[1+(a−25)2]−`125≤a<51.95μv(a)=1−[1+(a−50)2]−151.9≤a≤20052-3解:(1)μ0000.30.400ze(e)∩μps(e)=−30+−20+−10+0+10+20+30(2)μ(e)∪μ(e)=0+0+0.4+1+1.0+0.3+0zeps−30
2、−40−100102030(0.6∧0.5)∨(0.9∧0.1)(0.6∧0.7)∨(0.9∧0.4)0.50.62-4解:(1)PoQ==(0.2∧0.5)∨(0.7∧0.1)(0.2∧0.7)∨(0.7∧0.4)0.20.40.60.6PoQoR=0.40.40.60.90.60.5(2)P∪Q=P∪QoS=0.20.70.60.50.60.5(3)(PoQ)∪(QoS)=(P∪Q)oS=0.60.52-5解:先求关系矩阵R=A×B000000.20.20
3、.20.10R=A×B=0.50.50.40.10A′=[0.10.610.60.1]0.80.80.40.101.00.80.40.10则方向角的变化B′=A′oR=[0.60.60.40.10]2-6解:B′=非常重=0.008+0.064+0.216+0.512+1V1V2V3V4V5B″=不非常重=B′=0.992+0.936+0.784+0.488+0V1V2V3V4V51关系矩阵R=(A×B)∪(A×B″)0.20.40.60.810.20.40.60.80.8A×B=0.20.40.60
4、.60.60.20.40.40.40.40.20.20.20.20.2000000.20.20.20.20A×B″=0.40.40.40.400.60.60.60.48800.80.80.7840.48800.20.40.60.810.20.20.60.80.8R=(A×B)∪(A×B″)=0.40.40.60.60.60.60.60.60.4880.40.80.80.7840.4880.2A=非常轻=1+0.512+0.216+0.064+0.0081U1
5、U2U3U4U5A=重=0+0.2+0.4+0.6+0.82U1U2U3U4U5则对应的B1=[0.2160.40.60.81]B2=[0.80.80.7840.4880.4]2-7解:将方程按合成运算展开(0.3∧x)∨(0.2∧x)∨(0∧x)=0.2(1)123(0.5∧x)∨(0∧x)∨(0.6∧x)=0.4(2)123(0.2∧x)∨(0.4∧x)∨(0.1∧x)=0.2(3)123对于式(1)有:[x]=0.2[x]=[0.2,1][x]=Φ123(x)=[0,0.2](x)=[0,1](x)=[0,1]123对于式(2)有:
6、[x]=0.4[x]=Φ[x]=0.4123(x)=[0,0.4](x)=[0,1](x)=[0,0.4]123对于式(3)有:[x]=[0.2,1][x]=0.2[x]=Φ1232(x)=[0,1](x)=[0,0.2](x)=[0,1]123则有1R=(0.2,[0,1],[0,1])U([0,0.2],[0.2,1],[0,1])U([0,0.2],[0,1],Φ)2R=(0.4,[0,1],[0,0.4])U([0,0.4],Φ,[0,0.4])U([0,0.4],[0,1],0.4)3R=([0.2,1],[0,0.2],[0,
7、1])U([0,1],0.2,[0,1])123所以方程的解为R=RURUR0.10.50.52-8解:D=A×B=0.110.60.10.10.10.10.10.10.50.40.50.50.40.50.10.10.1关系矩阵R=DT×C=1×[0.41]=0.410.60.40.60.10.10.10.10.10.10.10.10.10.10.51又因为D′=A′×B′=0.10.50.5
8、0.10.10.1则C′=[0.10.510.10.50.50.10.10.1]oR=[0.40.5]即C′=0.4+0.5W1W233-1模糊逻辑控制器由哪几部分组成?各完成什么功能?答