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Homework5Solutions

# Homework5Solutions - So"I ‘ iii"\cirr’ui’...

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Unformatted text preview: So "I ‘ iii," \cirr’ui’: HW#5 Solution I Page 1 31 BM E 232\$ BME 31 l Page 2 a=xlsread(’?ressure time data.xls‘); ____ t=a(:,l); pr=a(:,2); i=pr./(7.32); plot(t,i) >% A=£4.% -0w125 0.125 —%.§;1 —1 2 —E:0 0.125 -0.325 0.2; 0.1 0 0x2 -0.55] A = 4.4000 —0.1250 0.1250 —4«§000 1.0000 —1_0000 2.0000 -2,0000 0 0.1350 —0.3250 0,2000 0.1000 0 0.2000 -0,5500 }} E=f050S—5;-3.5] E = 0 0 —5‘0000 —2‘5000 }> X=inv§AjKE X: 9?.5000 272.5000 172(5000 85.0000 S‘Vﬁmg Vlzz7z\$ Var“ W15 Va}: 4:35 V5" 3 {rs} Paww éiﬁswmﬁ 5% W WWW mm scum; F; H = avéxﬁwvx): ﬁvrm «(w/1):: - 593751441 5. PSpice Keys: 11: RS 270 V1=3.104V; V2=3.298V; V3=2.529V; V4=3.104V; V5=4.273V; V6=-2.72V; V7=4.273V; V8=4.273V Total dissipated:279.7mW 12: V1=—2.328V; V2=—O.432V; V3=O.l45V; V4=-4.368V; V5=0.877V; V6=OV; V7=O.877V; V8=0.877V Total dissipated:lS7.3mW Superposition: V123,1O4V+ —2.328V + 3.393V+ —S.09V =»0i921V; V2=3.298V+ -O.432V +5‘119V+ —7.678V =0307V; V3=2.529V+ 0.14SV + —1.725V+ 2.588V =3.537V; V4=3.104V+ -4.368V +3.393V+ —5.09V =-2.961V; V5:-2.72V+ 0.877V +7.697V + -11.55V =1‘302V; V6:-2.72V+ 0V + 0V + 0V:—2.72V; V7=4.273V+ 0.877V + -4.303V + 6.455V =7.302V; V8=4.273V+ 0.877V + 7.697V +6‘455V =19.3V [127880.9mW; I22-667i9mW,‘ V1:664.5mW; V2:263.3mW Total dissipated:620.973mW:279.7mW+157.3mW+455mW+1573mW Thus voltage obey superposition theory, while the dissipation power does not. LR ...
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