??? 05/06/06 02:53 Modified: 05/06/06 02:59 Read: times |
#115742 - Vector-diagram Responding to: ???'s previous message |
Mike said:
The thing that keeps throwing me is the fact that the real component of the input impedance varies with frequency. Do you know what the "vector diagram" is? (Someone shall correct me, if my english is wrong!) The vector diagram shows the Û- and the Î-vector rotating counter clock wise. The angle between the two vectors represents the phase shift of Û relative to Î. You can also draw the Z-vector in this diagram: Z is the quotient of Û and Î, where Û and Î means the amplitudes of U(t) and I(t), represented by the lengths of vectors. Z is the complex impedance of the circuit of course. You can also draw the Z-vector, as follows: Chose a length that is equal to Û/Î and draw it under an angle to the real-axis which is equal to the phase shift of Û relative to Î. So, if the phase angle between Û and Î is zero, then the Z-vector will coincide with the real-axis. The projection of Z-vector onto the real-axis will give you "R" and the projection of Z-vector onto the imaginary-axis will give you "X"! How can you deduce Z from the "R"- and "X"-plots from the picture of your exercise then? What must Z be at very low and very high frequencies? What kind of circuit can you clearly deduce from that? Kai |
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