![]() There you can read off the resistance value (a little over 200 Ohms in this case), and you can note the distance of traversal either in degrees in the reflection coefficient or in units of lambda. Once you hit the x-axis, you know the impedance at that distance is real. As t_n_k said, the magnitude of the reflection coefficient does not change, which is why the circle is the correct shape. One starts by marking the terminating impedance on the plot, and then one can move away from the load (toward the generator) by moving on a circle clockwise. Overlayed on top of this is a mapping of the impedance that goes with that reflection coefficient (impedance is normalized relative to Zo). The plot is a polar plot in which the radius represents the magnitude of the reflection coefficient and the angle represents the angle of the reflection coefficient. The Smith chart may look complicated, but it is simple in concept. ![]() I've attached a drawing of the Smith chart that goes along with t_n_k's description. How did he find this insight? I'm guessing from a familiarity with the theory and particularly with the visualization of that theory as enabled by using a Smith chart. T_n_k gave a nice intuitive description on a method to solve this. However, since we've made the recommendation to consider Smith charts, we would be remiss if we didn't make some attempt to let you see their value. A wavelength corresponds to 360° so that's a line length of 32.8°/360° or 0.091 wavelengths.Ĭlick to expand.It looks like you solved it. If the reflected wave is in phase with the input voltage then the total return phase change is 65.59° or 32.8° one way. In terms of phase relationship the incident wave must travel from the source to the load and the reflected wave must travel back to the source. The coefficient phase angle must be zero at the source. If the line input impedance seen at the source is purely resistive then the complex reflection coefficient is purely real.Only the phase changes as the wave travels along the line. The magnitude of the complex reflection coefficient is the same anywhere on the line.To solve your problem you simply need to know two things - given this is a loss-less line example.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |