
matlab中函数y=exp(j*x)在区间[-1,1]上变化时,能求出最大值最小值是怎么回事啊?
1个回答
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>> x=-1:0.1:1;
>> y=x.*exp(j*x)
y =
Columns 1 through 7
-0.5403 + 0.8415i -0.5594 + 0.7050i -0.5574 + 0.5739i -0.5354 + 0.4510i -0.4952 + 0.3388i -0.4388 + 0.2397i -0.3684 + 0.1558i
Columns 8 through 14
-0.2866 + 0.0887i -0.1960 + 0.0397i -0.0995 + 0.0100i 0 0.0995 + 0.0100i 0.1960 + 0.0397i 0.2866 + 0.0887i
Columns 15 through 21
0.3684 + 0.1558i 0.4388 + 0.2397i 0.4952 + 0.3388i 0.5354 + 0.4510i 0.5574 + 0.5739i 0.5594 + 0.7050i 0.5403 + 0.8415i
>> b=min(y)
b =
0
>> a=max(y)
a =
-0.5403 + 0.8415i
复数不能比大小,而在matlab中,当数据为复数时,求最大最小值是按其绝对值处理,即
max(abs(y)),min(abs(y))。
>> y=x.*exp(j*x)
y =
Columns 1 through 7
-0.5403 + 0.8415i -0.5594 + 0.7050i -0.5574 + 0.5739i -0.5354 + 0.4510i -0.4952 + 0.3388i -0.4388 + 0.2397i -0.3684 + 0.1558i
Columns 8 through 14
-0.2866 + 0.0887i -0.1960 + 0.0397i -0.0995 + 0.0100i 0 0.0995 + 0.0100i 0.1960 + 0.0397i 0.2866 + 0.0887i
Columns 15 through 21
0.3684 + 0.1558i 0.4388 + 0.2397i 0.4952 + 0.3388i 0.5354 + 0.4510i 0.5574 + 0.5739i 0.5594 + 0.7050i 0.5403 + 0.8415i
>> b=min(y)
b =
0
>> a=max(y)
a =
-0.5403 + 0.8415i
复数不能比大小,而在matlab中,当数据为复数时,求最大最小值是按其绝对值处理,即
max(abs(y)),min(abs(y))。
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