Question 10.SP.14: If Rx is the inherent resistance of the inductor L for the b......

The circuit of Fig. 10-14(a) represents the circuit of Fig. 10-6 with Q ON and D OFF.  By KVL,
L\,\frac{d i_{L}}{d t} + {R_{x}}I_{L} = V_{1} \qquad 0 \le t \le D\,T_{s}      (1)

The circuit of Fig. 10-14(b) is valid for Q OFF and D ON.   Whence,
L\,\frac{d i_{L}}{d t} + R_{x}i_{L} = -v_{2}\;\;\;\;\;\;\;D T_{s} \leq t \leq T_{s}           (2)

In similar manner to the procedure of Problem 10.11, integrate (1) and (2), add the results, and divide by T_s to find
{\frac{L}{T_{s}}}\int_{i_{L}(0)}^{i_{L}(t)}\,d i_{L} + R_{x}\,{\frac{1}{T_{s}}}\int_{0}^{T_{s}}i_{L}\,d t = {\frac{V_{1}}{T_{s}}}\int_{0}^{DT_{s}}\,d t-{\frac{1}{T_{s}}}\int_{D T_{s}}^{t_{s}}\,v_{s}\,d t              (3)

For a periodic i_L, the first term of (3) must be zero. Recognize the average values of i_L and v_2, respectively, in the second term on each side of the equation to give
R_{x}I_{L} = D V_{1}  –  (1  –  D)V_{2}               (4)

From Fig. 10-14,
C \frac{d v_{2}}{d t} = -\frac{v_{2}}{R_{L}} \qquad 0 \le t \le D T_{s}          (5)
C \frac{d v_{2}}{d t} = i_{L}  –  \frac{v_{2}}{R_{L}}\;\;\;\;\;\;\;D T_{s} \leq t \leq T_{s}                  (6)

Integrate, add, and divide by T_s for (5) and (6).
\frac{C}{T_{s}}\int_{v_{2}(0)}^{v_2(T_{s})}d v_{2} = \frac{1}{T_{s}}\int_{D T_{s}}^{T_{s}}\,i_{L}\,d t  –  \frac{1}{R}\frac{1}{T_{s}}\int_{0}^{T_{s}}\,v_{2}\,d t            (7)

The first term of (7) must be zero for periodic v_2.   Owing to the straight-line segment description of i_L, the first term on the right-hand side of (7) can be written as (1  –  D)I_L.   Recognize the average value of v_2 in the last term.   Thus, (7) becomes

0 = (1  –  D)I_{L}  –  {\frac{V_{2}}{R_{L}}}                 (8)

Solve (8) for I_L, substitute the result into (4), and rearrange to yield

G_{V}^{\prime} = {\frac{V_{2}}{V_{1}}} = {\frac{D(1  –  D)R_{L}}{R_{x}  +  (1  –  D)^{2}R_{L}}}                 (9)