How does total load (load + line) becomes inductive and capacitive as it moves along the line?

Discussion in 'General Electronics Chat' started by Dong-gyu Jang, Sep 16, 2015.

  1. Dong-gyu Jang

    Thread Starter Member

    Jun 26, 2015

    I'm now studying Smith chart and have some wired feeling about this.

    Let's say transmission line connects to the inductor as a load.

    According to the chart, as it moves from the load to RF source, total impedance (load + transmission line from the load to the measuring point) becomes periodically inductive and capacitive as movement on the chart is circulation.

    I'm really curious how inductive load + just line becomes capacitive and even purely resistive?

    Is there physically some sort of resonance to inductance and capacitance of the cable?
  2. Papabravo


    Feb 24, 2006
    A transmission line and a load are described by a differential equation. Depending on the geometry and boundary conditions the solution may consist of a standing wave, which is a function of position only, and a traveling wave which is a function of time only. As the two solutions interact at a given point the transmission line and the load can exhibit different characteristics. It will exhibit these differences as a function of frequency as well. You'll notice that a Smith Chart plotter might make several passes around the normalized impedance Z = 1 from inductive to capacitative and back again.
    Dong-gyu Jang likes this.
  3. KL7AJ

    Senior Member

    Nov 4, 2008
    Hi Dong0gyu:

    Smith favorite topic!

    Actually, if the load is a PURE inductance, there will never be a pure resistance at any location on the chart. The impedance will rotate around the outer perimeter of the chart. The closest thing you will get to a pure resistances is either 0 ohms at the left of infinity at the right.

    As far as how this can result in a capacitive situation, it's useful to look at a very mechanical view of a transmission line. There is a finite amount of capacitance between the two conductors....this is "real" capacitance, not just a mathematical abstraction. In addition, there is real inductance along the cable....a straight piece of wire does have some inductance. The full model of a transmission line, however, is best described as a ladder network, consisting of a very large number of tiny series inductors, and a very small shunt capacitor between each of them. So, yes, there is real inductance and capacitance to work with in a transmission line!

    If you do have a RESISTIVE load, things get even more interesting. A 1/4 wave length of transmission line can transform a pure resistive load to another pure resistive load of a different value. Such "Q-sections" are commonly used as impedance matching transformer in radio frequency circuits.

    I highly recommend the ARRL Antenna book for more details on the Smith Chart.

    Good luck!
    Dong-gyu Jang likes this.
  4. daviddeakin

    Active Member

    Aug 6, 2009
    Whether an impedance is inductive or capacitive is simply a matter of the phase shift between the voltage and current at (OK across) that point in the circuit. If you send a signal down a cable and it reflects back off the far end, then at every point along the cable you have a mixture (sum) of the forward-going signal, and the reflected signal, each of which has a particular amount of phase shift. As the two signals sum together, the phase of the resultant signal (voltage/current) defines whether that point on the cable 'looks' inductive or capacitive. Of course, the exact point on the cable that you are most interested in is the point where you inject the signal, since this is the bit that loads the source, but the principle still holds for any point along the cable.
    Dong-gyu Jang likes this.