Q:

Transfer function of boost converter

When I have gone through some books I have found the expression of the DC Gain of the the boost converter in current mode is given by 
H0=(RL/Ri)*1/(2*M+(RL*Tsw)/(L*M^2)*(1/2+Sa/Sn))

But I did not find out what is the expression of the M, please can you give me the expression of this M and from where does it come??
Here is attached the final expression given by OnSemiconductor

A:

The answer of Verbal Kint makes most sense.  
If you multiply the DC gain on slide 35 by \$ \frac{R_L}{R_L} \$ 
$$H_0 = \frac{k_0-k_i}{g_f - (g_0+g_i) - g_r - \frac{1}{R_L} } \frac{R_L}{R_L} $$ 
and substitute the equations given in slide 27 into the term \$k_0-k_i\$,  
then \$H_0\$ evaluates to
$$H_0 = R_L \frac{\frac{1-D}{R_i}}{g_f R_L - (g_0+g_i) R_L - g_r R_L - 1 } = \frac{R_L}{R_i} \frac{1-D}{g_f R_L - (g_0+g_i) R_L - g_r R_L - 1 } $$ 
Using \$M=\frac{1}{1-D}\$ it looks more and more to the DC gain \$H_0\$ given on slide 50:
$$H_0 = \frac{R_L}{R_i} \frac{1}{M (g_f R_L - (g_0+g_i) R_L - g_r R_L - 1) } $$ 

