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\begin{gather}
\frac{\theta_m(s)}{E_a(s)}=\frac{K(s)}{(s+a)} \\ 
\nonumber \text{The equivalent inertia $J_m$ is} \\
J_m=J_a+J_L)\frac{25}{20})^2=0.02+1\frac{1}{100}=0.03 \\ 
\nonumber \text{where $J_L-1$ is the load intertial at $\theta_o$. The equivalent viscous damping $D_m$ at the armature is}\\
D_m=D_a+D_L(\frac{25}{250})^2=0.01+\frac{1}{100}=0.02 \quad \text{where $D_L$ is viscous damping at $\theta_o$} \\
\nonumber \text{from the problem, $K_t=0.5N-m/A$ and $K_b=0.5V-s/rad$ and the armature resistance $R_a=8$ ohms} \\
\nonumber \text{quantities along with $J_m$ and $D_m$ are substituted with below formula} \\
\frac{\theta_m(s)}{E_a(s)}=\frac{K_t(R_a*J_m)}{s[s+\frac{1}{J_m(D_m+\frac{K_tK_b}{R_a}}}=\frac{2.083}{s(s+1.71)}\\\frac{\theta_m(s)}{E_a(s)}=\frac{K(s)}{(s+a)} \\ \nonumber \text{The equivalent inertia $J_m$ is} 
\nonumber{To calculate transfer function entirely, there is one job left to finish.This is about multiplying gear ratio with the arrive}\\
\nonumber{at the transfer function relating load displacement to armature voltage} \\
\frac{\theta_u(s)}{E_a{s)}=0.1\frac{\theta_m(s)}{E_a(s)}=\frac{0.2083}{s(s+1.71)}
\end{gather}
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