Question 10.3: APPLICATION TO A WILKINSON POWER DIVIDER Find the noise figu...
APPLICATION TO A WILKINSON POWER DIVIDER
Find the noise figure of a Wilkinson power divider when one of the output ports is terminated in a matched load. Assume an insertion loss factor of L from the input to either output port.
From Chapter 7 the scattering matrix of a Wilkinson divider is given as
[S]=\frac{-j}{\sqrt{2L} }\left[\begin{matrix} 0 & 1 & 1 \\ 1 & 0 &0 \\ 1 & 0 & 0 \end{matrix} \right]where the factor L ≥ 1 accounts for the dissipative loss from port 1 to port 2 or 3 (note that dissipative loss is distinct from the −3 dB power division ratio). To evaluate the noise figure of the Wilkinson divider, we first terminate port 3 with a matched load; this converts the three-port device into a two-port device. If we assume a matched source at port 1, we have Γ_{s} = 0. Equation (10.26) then gives Γ_{out} = S_{22} = 0, and so the available gain can be calculated from (10.25) as
G_{21}=\frac{\text{power available from network} }{\text{power available from source }} = \frac{\left|S_{21}\right|^2\left(1-\left|\Gamma _S\right|^2 \right) }{\left|1-S_{11}\Gamma _S\right| ^2(1-\left|\Gamma _{out}\right| ^2)} (10.25)
\Gamma _{out}= S_{22}+ \frac{S_{12}S_{21}\Gamma _S}{1-S_{11}\Gamma _S} (10.26)
G_{21}=\mid S_{21} \mid ^{2} =\frac{1}{2L},The equivalent noise temperature of the Wilkinson divider is, from (10.28),
T_{e}=\frac{1-G_{21}}{G_{21}}T=(2L-1)T ,where T is the physical temperature of the divider. Using (10.11) gives the noise figure as
F= 1+\frac{T_{e}}{T_{0}}=1+(2L-1)\frac{T}{T_{0}} ,Observe that if the divider is at room temperature, then T = T_{0} and the above reduces to F = 2L. If the divider is at room temperature and lossless, this reduces to F = 2 = 3 dB. In this case the source of the noise power is the isolation resistor contained in the Wilkinson divider circuit.
Because the network is matched at its input and output, it is easy to obtain these same results using the thermodynamic argument directly. Thus, if we apply an input noise power of kTB to port 1 of the matched divider at temperature T , the system will be in thermal equilibrium and the output noise power must be kTB.
We can also express the output noise power as the sum of the input power times the gain of the divider, and N_{added}, the noise power added by the divider itself (referenced to the input to the divider):
Solving for N_{added} gives N_{added} = kTB(2L − 1), so the equivalent noise temperature is
T_{e}=\frac{N_{added}}{kB}=(2L-1)T,in agreement with the above.