Question

4. Consider the block diagram shown below where D(s) is a step disturbance input. D(s) Controller Plant R(s) + E(s) C(s) G2(s) Ideally you want your controller design to reject a step disturbance input at D(s). This means that in the steady state for D(s)-1, the value of Y(s) is unchanged (a) Ignoring the input R(s), what is the transfer function器in terms of Gi(s) and G2(s)? (b) For G1(s)Ks 2) and G2(s)0419 what is the steady state error resulting from step inputs R(s)-4 and D(s)B (c) For G1 (s) = (Kps+ Kpt 쓸) and G2(s) = 0419 what is the steady state error resulting from step inputs R(s)-4 and D(s)5 Hint: Its a linear system so you can consider the steady state error resulting from R(s) and the steady state error resulting from D(s) separately and then sum them!

Answer 1

From the block diagram, the error is given by

$E(s)=R(s)-C(s)=R(s)-G_2(s)(D(s)+G_1(s)E(s))$

$\Rightarrow E(s)+G_1(s)G_2(s)E(s)=R(s)-G_2(s)D(s)$

$\Rightarrow E(s)=\frac{1}{1+G_1(s)G_2(s)}R(s)-\frac{G_2(s)}{1+G_1(s)G_2(s)}D(s)$ ........Eq.1

Part-a

Ignoring the input , the transfer function $\frac{E(s)}{D(s)}$ can be obtained from Eq.1 as

$\frac{E(s)}{D(s)}=-\frac{G_2(s)}{1+G_1(s)G_2(s)}$

Part-b

Given $G_1(s)=K_C(s+Z)\ \&\ G_2(s)=\frac{0.419}{s^2}$

the inputs are $R(s)=\frac{A}{s}\ \&\ D(s)=\frac{B}{s}$

From Eq.1 the error is given as

$E(s)=\frac{1}{1+K_C(s+Z)\cdot \frac{0.419}{s^2}}\frac{A}{s}-\frac{\frac{0.419}{s^2}}{1+K_C(s+Z)\cdot \frac{0.419}{s^2}}\frac{B}{s}$

$\Rightarrow E(s)=\frac{As}{s^2+0.419K_C(s+Z)}-\frac{0.419B}{s^3+0.419K_Cs(s+Z)}$ .......Eq.2

According to the final value theorem, the steady state error is given as

$e_{ss}=\lim_{s\rightarrow0}sE(s)$ .......Eq.3

From Eq.2 and Eq.3, the steady state error is

$e_{ss} =\lim_{s\rightarrow0}sE(s)=\lim_{s\rightarrow0}s\left[ \frac{As}{s^2+0.419K_C(s+Z)}-\frac{0.419B}{s^3+0.419K_Cs(s+Z)}\right ]$

$\Rightarrow e_{ss} =\lim_{s\rightarrow0}\left[ \frac{As^2}{s^2+0.419K_C(s+Z)}-\frac{0.419B}{s^2+0.419K_C(s+Z)}\right ]$

$\Rightarrow e_{ss} =\left[ 0-\frac{0.419B}{0+0.419K_C(Z)}\right ]$

$\Rightarrow e_{ss} =\frac{-B}{K_CZ}$

So, we have a constant steady-state error of $\frac{-B}{K_CZ}$

Part-c

Given $G_1(s)=K_Ds+K_P+\frac{K_I}{s}\ \&\ G_2(s)=\frac{0.419}{s^2}$

the inputs are $R(s)=\frac{A}{s}\ \&\ D(s)=\frac{B}{s}$

From Eq.1 the error is given as

$E(s)=\frac{1}{1+\left(K_Ds+K_P+\frac{K_I}{s} \right )\cdot \frac{0.419}{s^2}}\frac{A}{s}-\frac{\frac{0.419}{s^2}}{1+\left(K_Ds+K_P+\frac{K_I}{s} \right )\cdot \frac{0.419}{s^2}}\frac{B}{s}$

$E(s)=\frac{As^2}{s^3+0.419(K_Ds^2+K_Ps+K_I)}-\frac{0.419B}{s^3+0.419(K_Ds^2+K_Ps+K_I)}$ .......Eq.4

From Eq.4 and Eq.3, the steady state error is

$e_{ss} =\lim_{s\rightarrow0}sE(s)=\lim_{s\rightarrow0}s\left[ \frac{As^2}{s^3+0.419(K_Ds^2+K_Ps+K_I)}-\frac{0.419B}{s^3+0.419(K_Ds^2+K_Ps+K_I)}\right ]$

$e_{ss} =\lim_{s\rightarrow0}\left[ \frac{As^3}{s^3+0.419(K_Ds^2+K_Ps+K_I)}-\frac{0.419Bs}{s^3+0.419(K_Ds^2+K_Ps+K_I)}\right ]$

$e_{ss} =\left[ 0-0\right ]$

So, we have a zero steady state error.