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Take positive volumetric flow rate in the direction of increasing $x$. For an arterial pressure wave, the characteristic admittance of an arterial wave givesThe backward wave contributes the opposite flow sign. Write $p_j$ for each junction's complex pressure amplitude, suppressing the common $e^{i\omega t}$ factor, and put $r=RY$, $C_{\beta }=\cos \beta$, $S_{\beta }=\sin \beta$. The topology is the two paths $K\to L\to N$ and $K\to M\to N$; their intermediate and common distal junctions each have a hydraulic resistance. The coordinate convention in the source puts the upstream end at $-\ell$ and the downstream end at zero in each segment. Consequently the wave transfer matrix of an arterial segment isEquivalently $q_u=Y(C_{\beta }{p}_u-p_d)/(i{S}_{\beta })$ and $q_d=Y(p_u-C_{\beta }{p}_d)/(i{S}_{\beta })$, with their continuous limits understood when $S_{\beta }=0$.

Symmetry gives $p_L=p_M=p_s$. Distal mass conservation and the resistive relation $Q_N=p_N/R$ giveAt $L$, the incoming flow supplies both its bed and segment $D$, soChoose the incident amplitude $P_A$ at $K$. If $P_{AR}$ is the reflected amplitude in vessel $A$, pressure continuity equation gives $p_K=P_A+P_{AR}$, while inlet flow is $2Y(P_A-P_{AR})=2Y(2P_A-p_K)$. Equating this to the two outgoing segment flows yieldsEliminating $p_s,p_K$ from these three relations gives the wave transmission through a resistively loaded arterial loop resultFor large $r$ at fixed phase, $p_N=2P_A{e}^{-2i\beta }+O(r^{-1})$ and $p_M=C_{\beta }{p}_N+O(r^{-1})$. HenceThese leading terms must not be used as nonzero denominators near a phase at which the normal intermediate flow vanishes.