已知i1=2sin(wt+30 )A,i2= 4sin(wt-45 )A,求i=i1+i2的最大值,要详细过程啊!~~急!
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![已知i1=2sin(wt+30 )A,i2= 4sin(wt-45 )A,求i=i1+i2的最大值,要详细过程啊!~~急!](/uploads/image/z/9933582-30-2.jpg?t=%E5%B7%B2%E7%9F%A5i1%3D2sin%28wt%2B30+%29A%2Ci2%3D+4sin%28wt-45+%29A%2C%E6%B1%82i%3Di1%2Bi2%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%2C%E8%A6%81%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B%E5%95%8A%21%7E%7E%E6%80%A5%21)
已知i1=2sin(wt+30 )A,i2= 4sin(wt-45 )A,求i=i1+i2的最大值,要详细过程啊!~~急!
已知i1=2sin(wt+30 )A,i2= 4sin(wt-45 )A,求i=i1+i2的最大值,要详细过程啊!~~急!
已知i1=2sin(wt+30 )A,i2= 4sin(wt-45 )A,求i=i1+i2的最大值,要详细过程啊!~~急!
i=i1+i2=2sin(wt+30º)+4sin(wt-45º)
=2(sinwtcos30º+sin30coswt)+4(sinwtcos45º-sin45ºcoswt)
=√3sinwt+coswt+2√2sinwt-2√2coswt
=(√3+2√2)sinwt+(1-2√2)coswt
=√[(√3+2√2)²+(1-2√2)²]sin(wt+φ) ( 辅助角公式)
=√(3+8+4√6+1+8-4√2)sin(wt+φ)
=√(20+4√6-4√2)sin(wt+φ)
≤√(20+4√6-4√2)
所以i=i1+i2的最大值为√(20+4√6-4√2).
i=i1+i2=2sin(wt+30º)+4sin(wt-45º)
=2(sinwtcos30º+sin30coswt)+4(sinwtcos45º-sin45ºcoswt)
=√3sinwt+coswt+2√2sinwt-2√2coswt
=(√3+2√2)sinwt+(1-2√2)coswt
=√[(√...
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i=i1+i2=2sin(wt+30º)+4sin(wt-45º)
=2(sinwtcos30º+sin30coswt)+4(sinwtcos45º-sin45ºcoswt)
=√3sinwt+coswt+2√2sinwt-2√2coswt
=(√3+2√2)sinwt+(1-2√2)coswt
=√[(√3+2√2)²+(1-2√2)²]sin(wt+φ) ( 辅助角公式)
=√(3+8+4√6+1+8-4√2)sin(wt+φ)
=√(20+4√6-4√2)sin(wt+φ)
≤√(20+4√6-4√2)
所以i=i1+i2的最大值为√(20+4√6-4√2)。
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