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Re: The i operator


From: Ben Abbott
Subject: Re: The i operator
Date: Mon, 26 Jan 2009 07:02:47 -0500


On Jan 25, 2009, at 10:39 PM, Steve wrote:

Hi,

I’d appreciate some advice with Octave and use of the i imaginary operator.

I can get this to work:

octave-3.0.3.exe:27> (2 * (pi) * 10.7e6 * 1.62e-6i)

ans = 0.00000 + 108.91273i


And this works where I define the constant w as 2*pi*10.7e6:

octave-3.0.3.exe:28> w = 2 * (pi) * 10.7e6

w = 6.7230e+007

octave-3.0.3.exe:29> (w * 1.62e-6i)

ans = 0.00000 + 108.91273i

I’d like to also define the constant L and use it in the calculation,
but I get this error:

octave-3.0.3.exe:30> w = 2 * (pi) * 10.7e6

w = 6.7230e+007

octave-3.0.3.exe:31> L = 1.62e-6i

L = 0.0000e+000 + 1.6200e-006i

octave-3.0.3.exe:32> w * Li

error: `Li' undefined near line 32 column 5

error: evaluating binary operator `*' near line 32, column 3


How does one use the i operator with constants so that Octave recognizes
the i as the imaginary operator?

Thanks.

Steve

I assume  you are asking for help, and not reporting a bug?

For the first example, the following works

        (2 * (pi) * 10.7e6 * 1.62*10^(-6i))
        ans = 3.4432e+07 - 1.0333e+08i

For the last, it can be wise to use 1i in the event "i" has been used as a variable.

        w = 2 * (pi) * 10.7e6;
        L = 1.62*10^(-6i);
        w * L * 1i
        ans = 1.0333e+08 + 3.4431e+07i

Notice the answers for each of the above. The "i" is not part of the exponent.

Evaluate 1e1i in either Matlab or Octave gives the same result.

        1e1i
        ans = 0 + 10i

Thus, just like the answers above, 1e1i is interpreted as (1e1)*1i

Ben






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