Imaginary signals (Re: Watt meters)

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br> 

Tesla list wrote:

 > Original poster: "John H. Couture" <couturejh-at-mgte-dot-com>
 >
 > All kidding aside, I find it interesting that something as real as an
 > electrical current can only be represented as something imaginary. Maybe it
 > is because Mathematics was invented by imperfect man but currents were made
 > by a perfect nature or God.

The imaginary signals are just a mathematical trick, because of the
equivalence that is valid for perpetual sinusoidal signals in a linear
circuit:

True behavior:
The current in a capacitor is i=C*dv/dt
The voltage over an inductor is v=L*di/dt
The voltage over a resistor is v=R*i
Phasor analysis equivalent:
The signals are represented by "phasors", that are complex values.
Cosinusoidal signals are real.
Sinusoidal signals are imaginary (negative).
The current in a capacitor is I=j*w*C*V
The voltage over an inductor is V=j*w*L*I
The voltage over a resistor is V=R*I

Or: to diferentiate a signal is equivalent to multiply its
phasor by j*w.

(v=V=voltage; i=I=current; w=frequency in rad/s; j=sqrt(-1))

To verify this, consider a signal A*cos(w*t) + B*sin(w*t)
Its equivalent phasor is A-j*B (sinusoids are treated as negative)
The derivative of this signal is -w*A*sin(w*t) + w*B*cos(w*t)
Multiplying (A-j*B)*(j*w) = j*w*A+w*B
That corresponds precisely to -w*A*sin(w*t) + w*B*cos(w*t)

The equivalent works for integrations too, that are equivalent to
divide the phasors by j*w (DC terms resulting from the integration
are ignored).

Just a trick to avoid some calculus.

Antonio Carlos M. de Queiroz