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DC Wave Questions

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Andrew Holme

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wrote:
O.K. - now we're getting somewhere......you're saying the current and
voltage (and the implied impedance Z = V/I) of the "DC sine wave" is
the sum of the respective current and voltage of a +10V DC signal and
a -5V/+5V AC signal going into the same load.

Example:
DC +10V into load produces 1 Amp, therefore implied resistance = 10
ohm.
and
AC -5V/+5V (and given frequency) into load produces 0.5 amps,
therefore implied impedance = 20 ohms,

then what would the superposition prinicple predict as the resulting
combined current and impednace?

The current is simply the sum of the AC and DC components
e.g. 0.5 amps peak-to-peak AC with a 1 amp DC offset
Max. instantaneous current = 1.25 amps
Min. instantaneous current = 0.75 amps

Impedance can be represented as a complex number:
real part = reisitance = R = 10 ohms
imaginary part = reactance = X

Total impedance Z = R + jX

To work out the imaginary part, you have to do a vector addition because
current and voltage in a reactance are 90 degrees out of phase:

Ipk = Vpk / sqrt(X*X + R*R)

0.25 = 5 / sqrt(X*X + 10*10)

X = sqrt(300) = 17.3

i.e. Z = 10 + j*17.3

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