Have you ever tried to connect two TV sets to a single down-lead by constructing a Y junction in coaxial cable? One can buy splitter boxes to do this: this section explains what may be inside such a box.
Clearly, a shunt connection of a coaxial cable of characteristic impedance Zo to two identical cables, both terminated, presents a load impedance of Zo/2 to the first cable. (There are two impedances Zo in parallel).
The reflection coefficient at this junction is -1/3 so 1/9 of the power is reflected and 8/9 transmitted, 4/9 in one of the shunt arms and 4/9 in the other. The transmission coefficient is therefore 2/3.
In terms of the three port scattering parameters for this junction we have s11 = -1/3 = s22 = s33 by symmetry, and s12=s13=s21=s31=s32=s23=2/3 also by symmetry.
There is thus a loss of 3.5dB into each arm, because of the unavoidable reflected power. This is not usually very significant of itself, but the reflection mismatch makes the down-lead into a resonant section of cable, particularly if the antenna end is not well matched. This can give rise to poor performance of the system on one or more of the channels because the antenna down-lead acts as a rejection filter.
Typically, at UHF, a down-lead of length say 10 metres may be upwards of 30 wavelengths long, so the dropouts in transmission occur at a set of frequencies spaced about 3% apart in a kind of comb structure.
Multiple reflections up and down the down-lead may also affect the quality of the received signal.
It is not possible to match a lossless reciprocal three port network on all ports; there must be a residual reflection (inherent to the junction) at at least one port. In other words, we cannot have s11, s22, and s33 all zero.
If we think of the columns of the S matrix as individual column vectors, labelled C1 C2 and C3, then the inner products (or dot products) of certain combinations have to satisfy the following relations in order that the total power in to the network is the same as the total power out of the network.
(C1).(C1*) = (C2).(C2*) = (C3).(C3*) = 1
(C1).(C2*) = (C2).(C3*) = (C3).(C1*) = 0
(C1*).(C2) = (C2*).(C3) = (C3*).(C1) = 0
There is a derivation of this relationship in my notes or in many microwave texts which deal with S matrices.
First of all, we consider a symmetrical parallel-connected junction of three 50 ohm cables, and let us call them A, B, and C.
If Vi volts is incident on the junction from arm A, what is the transmitted voltage on arms B and C?
There are a number of ways of looking at this problem. From the point of view of the wave travelling towards the junction along arm A, the junction presents an impedance (immediately, not counting any later reflections) of 25 ohms, which is the parallel combination of 50 ohms of B with 50 ohms of C.
Thus the reflection coefficient at the junction is (ZL-Zo)/(ZL+Zo) = (25 - 50)/(25 + 50) = -1/3 so the return voltage wave back along A has size (-1/3)Vi, which represents 1/9th of the incident power. Therefore, 8/9ths of the incident power must be transmitted in total into arms B and C, and as these are wholly equivalent, there must be 4/9ths of the power in each arm or 2/3Vi in each outgoing arm.
We notice that the total outgoing voltages add to 1 in units of Vi (-1/3+2/3+2/3) and the total outgoing powers add to one in units of the incident power (1/9+4/9+4/9).
Second, let us consider the situation when the two outgoing cables have characteristic impedance 75 ohms, but the incoming cable still has impedance 50 ohms.
The reflection coefficient is now (37.5 - 50)/(37.5 + 50) = -1/7 and so the reflected wave along arm A has size only (-1/7)Vi, so 1/49th of the incident power is returned along arm A and 24/49ths of the incident power goes out along arms B and C.
Allowing for the 75 ohm characteristic impedance of the outgoing arms B and C, the outgoing voltage wave amplitudes are sqrt[(3/2)(24/49)] = (6/7)Vi (because 75/50 = 3/2.)
Thus, adding voltages at the junction we find that in terms of Vi, the outgoing wave voltages (referred to the 50 ohms Zo of the incoming cable) sum to 1 since [-1/7 + (2/3){6/7+6/7}] = 1.
Also, the powers add to unity, in terms of the input power, as 1/49+24/49+24/49 = 1 and we note that there is very good coupling into the output cables with only about 2% of the power reflected.