rex_toms
春芽
共回答了14个问题采纳率:85.7% 举报
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).
1年前
10