smith chart logical problem question

[yefj]

Hello, in smith chart whole circle is half wavelength (lambda/2) so if we make full circle our occumulated phase is 0 degrees.
but if we go to simple phase of traveling wave exp(j*k*L)=exp(j*(2*pi/lambda*(lambda/2))=exp(pi)=-1 (180 phase)
where did i go wrong?
Thanks.

 

[G4BCH]

On a Smith chart 0 to 180 degrees is the same a 180 to 360 degrees, each half wave length travelled is the same, ignoring loss.

 

[yefj]

Hello G4BCH,so how does it explain
exp(j*k*L)=exp(j*(2*pi/lambda*(lambda/2))=exp(pi)=-1 (180 phase)
on smith chart full circle is half wavelength

 

[ferdem]

Consider that you are dealing with reflection coefficients on the Smith chart. Reflection coefficient involves forward and backward waves. If you go teta much away from a load, the phase of the reflection coefficient changes twice.

 

[mtwieg]

but if we go to simple phase of traveling wave exp(j*k*L)=exp(j*(2*pi/lambda*(lambda/2))=exp(pi)=-1 (180 phase)

Where does this equation come from?

 

[biff44]

let me try to explain.
the smith chart is a mapping of the REFLECTION COEFFICIENT PLANE.
to get reflection coefficient, physically, a signal has to start at a source, travel to a load, and then return to the source. you measure the return signal in relation to the original source signal.

So any result you get from a smith chart involves a wave travelling BOTH WAYS.
If the load is a half wave length away, the wave energy will pick up 180 degrees going from the source to the load, and another 180 degrees bouncing back from the load to the source.

the equations above are just saying that