For instantaneous values,
Vf2 = (2*Z2*Vf1+(Z1-Z2)*Vr2)/(Z1+Z2)
Vr1 = ((Z2-Z1)*Vf1+2*Z1*Vr2)/(Z1+Z2)

You get to think about the phase of Vr2 with respect to Vf1, and the
answer should then be obvious.

Cheers,
Tom

(Cecil Moore) wrote in message . com...
Here's an interesting exercise. The source is a 100W signal generator
equipped with a circulator and load that dissipates all reflected power
reaching the source so Pfwd1 is a constant 100W. The lossless twinlead is
an integral number of wavelengths long. rho at the load is 0.707 so 1/2
of the incident power is reflected. Pfwd1 and Pref1 are measured just to
the left of the impedance discontinuity at '+'. Pfwd2 and Pref2 are
measured just to the right of the impedance discontinuity at '+'.

Source---50 ohm coax---+---n*WL lossless 291.5 ohm twinlead---50 ohm load
Pfwd1-- Pfwd2--
--Pref1 --Pref2

t0 will be when the source 100W is first incident upon point '+'.
t1 will be when the first reflections from the load arrive at '+'.
t2 will be when the second set of reflections from the load arrive at '+'.

At t1, Pfwd1=100W, Pref1=50W, Pfwd2=50W, Pref2=25W

Question: Between t1 and t2, what is the equation and magnitude of Pfwd2?