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Assuming the closed-loop system is stable, find the steady-state error if

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Answer #1

Inner altitude Loop:

There is another loop inside this:

Forward gain of this loop is:

$G(s) = \frac{1}{Js}$

Feedback gain:

H(s) = Ku

closed loop transfer function:

P()+GH

$P(s) = \frac{\frac{1}{Js}K_w}{1+\frac{1}{Js}K_w}$

P(s) =

Now consider the whole inner altitude loop:

Forward gain of this loop is:

$P(s)*\frac{1}{s}*k_{\theta} = \frac{K_w}{Js+K_w}*\frac{1}{s}*k_{\theta}$

Feedback gain is 1

closed loop transfer function:

$Q(s) =\frac{\theta(s)}{\theta_r(s)}= \frac{ \frac{K_w}{Js+K_w}*\frac{1}{s}*k_{\theta}}{1+ \frac{K_w}{Js+K_w}*\frac{1}{s}*k_{\theta}}$

θ(s)

0(s)

Now consider the loop on the right side:

Forward gain of this loop is:

F(s) =

Feedback gain is

B(s) = b

closed loop transfer function:

F B

$E(s) = \frac{\frac{1}{s}*b}{1+\frac{1}{s}*b}$

$E(s) = \frac{b}{s+b}$

Now the block diagram can be reduced as:

D(s) X(s) Q(s)

Let the whole part circled in blue be:

Tis

$\small X(s) = ((X_r-X(s))*I(s)+D(s))E(s)*(1/s)$

But if, ,

$\small X(s) = (-X(s)*I(s)+D(s))E(s)*(1/s)$

$\small X(s) = -X(s)*I(s)*E(s)*(1/s)+D(s)*E(s)*(1/s)$

$\small X(s)(1+I(s)*E(s)*(1/s))=D(s)*E(s)*(1/s)$

$\small \Rightarrow X(s)=\frac{D(s)*E(s)}{s+I(s)*E(s)}$

$E(s) = \frac{b}{s+b}$  

Disturbance is a step function: $D(s) = \frac{1}{s}$

$\small \Rightarrow X(s)=\frac{\frac{1}{s}*\frac{b}{s+b}}{s+I(s)*\frac{b}{s+b}}$

Tis

$Q(s) = \frac{K_wk_{\theta}}{Js^2+K_ws+ K_wk_{\theta}}$

$\small I(s)=K_p(1+T_ds+\frac{1}{T_is})* \frac{K_wk_{\theta}}{Js^2+K_ws+ K_wk_{\theta}}*g$

$\small I(s)=K_p(\frac{T_is+T_iT_ds^2+1}{T_is})* \frac{K_wk_{\theta}}{Js^2+K_ws+ K_wk_{\theta}}*g$

I(s) ak

$\small \Rightarrow X(s)=\frac{\frac{1}{s}*\frac{b}{s+b}}{s+I(s)*\frac{b}{s+b}}$

$\small \Rightarrow X(s)=\frac{\frac{1}{s}*\frac{b}{s+b}}{s+\frac{(T_is+T_iT_ds^2+1)K_pK_wk_{\theta}}{(T_is)(Js^2+K_ws+ K_wk_{\theta)}}*g*\frac{b}{s+b}}$

$\small \Rightarrow X(s)=\frac{\frac{1}{s}b*(T_is)(Js^2+K_ws+ K_wk_{\theta})}{s*(s+b)(T_is)(Js^2+K_ws+ K_wk_{\theta})+(T_is+T_i T_d s^2+1)K_pK_wk_{\theta}*g*b}$

By final value theorem:

$\small X_{ss} = \lim_{s\rightarrow 0}sX(s)$

$\small \Rightarrow X_{ss}=\lim_{s\rightarrow 0}s*\frac{\frac{1}{s}b*(T_is)(Js^2+K_ws+ K_wk_{\theta})}{s*(s+b)(T_is)(Js^2+K_ws+ K_wk_{\theta})+(T_is+T_i T_d s^2+1)K_pK_wk_{\theta}*g*b}$

$\small \Rightarrow X_{ss}=\lim_{s\rightarrow 0}\frac{b*(T_is)(Js^2+K_ws+ K_wk_{\theta})}{s*(s+b)(T_is)(Js^2+K_ws+ K_wk_{\theta})+(T_is+T_i T_d s^2+1)K_pK_wk_{\theta}*g*b}$

$\small \Rightarrow X_{ss} = 0$

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Answer 1

Inner altitude Loop:

There is another loop inside this:

Forward gain of this loop is:

$G(s) = \frac{1}{Js}$

Feedback gain:

H(s) = Ku

closed loop transfer function:

P()+GH

$P(s) = \frac{\frac{1}{Js}K_w}{1+\frac{1}{Js}K_w}$

P(s) =

Now consider the whole inner altitude loop:

Forward gain of this loop is:

$P(s)*\frac{1}{s}*k_{\theta} = \frac{K_w}{Js+K_w}*\frac{1}{s}*k_{\theta}$

Feedback gain is 1

closed loop transfer function:

$Q(s) =\frac{\theta(s)}{\theta_r(s)}= \frac{ \frac{K_w}{Js+K_w}*\frac{1}{s}*k_{\theta}}{1+ \frac{K_w}{Js+K_w}*\frac{1}{s}*k_{\theta}}$

θ(s)

0(s)

Now consider the loop on the right side:

Forward gain of this loop is:

F(s) =

Feedback gain is

B(s) = b

closed loop transfer function:

F B

$E(s) = \frac{\frac{1}{s}*b}{1+\frac{1}{s}*b}$

$E(s) = \frac{b}{s+b}$

Now the block diagram can be reduced as:

D(s) X(s) Q(s)

Let the whole part circled in blue be:

Tis

$\small X(s) = ((X_r-X(s))*I(s)+D(s))E(s)*(1/s)$

But if, ,

$\small X(s) = (-X(s)*I(s)+D(s))E(s)*(1/s)$

$\small X(s) = -X(s)*I(s)*E(s)*(1/s)+D(s)*E(s)*(1/s)$

$\small X(s)(1+I(s)*E(s)*(1/s))=D(s)*E(s)*(1/s)$

$\small \Rightarrow X(s)=\frac{D(s)*E(s)}{s+I(s)*E(s)}$

$E(s) = \frac{b}{s+b}$  

Disturbance is a step function: $D(s) = \frac{1}{s}$

$\small \Rightarrow X(s)=\frac{\frac{1}{s}*\frac{b}{s+b}}{s+I(s)*\frac{b}{s+b}}$

Tis

$Q(s) = \frac{K_wk_{\theta}}{Js^2+K_ws+ K_wk_{\theta}}$

$\small I(s)=K_p(1+T_ds+\frac{1}{T_is})* \frac{K_wk_{\theta}}{Js^2+K_ws+ K_wk_{\theta}}*g$

$\small I(s)=K_p(\frac{T_is+T_iT_ds^2+1}{T_is})* \frac{K_wk_{\theta}}{Js^2+K_ws+ K_wk_{\theta}}*g$

I(s) ak

$\small \Rightarrow X(s)=\frac{\frac{1}{s}*\frac{b}{s+b}}{s+I(s)*\frac{b}{s+b}}$

$\small \Rightarrow X(s)=\frac{\frac{1}{s}*\frac{b}{s+b}}{s+\frac{(T_is+T_iT_ds^2+1)K_pK_wk_{\theta}}{(T_is)(Js^2+K_ws+ K_wk_{\theta)}}*g*\frac{b}{s+b}}$

$\small \Rightarrow X(s)=\frac{\frac{1}{s}b*(T_is)(Js^2+K_ws+ K_wk_{\theta})}{s*(s+b)(T_is)(Js^2+K_ws+ K_wk_{\theta})+(T_is+T_i T_d s^2+1)K_pK_wk_{\theta}*g*b}$

By final value theorem:

$\small X_{ss} = \lim_{s\rightarrow 0}sX(s)$

$\small \Rightarrow X_{ss}=\lim_{s\rightarrow 0}s*\frac{\frac{1}{s}b*(T_is)(Js^2+K_ws+ K_wk_{\theta})}{s*(s+b)(T_is)(Js^2+K_ws+ K_wk_{\theta})+(T_is+T_i T_d s^2+1)K_pK_wk_{\theta}*g*b}$

$\small \Rightarrow X_{ss}=\lim_{s\rightarrow 0}\frac{b*(T_is)(Js^2+K_ws+ K_wk_{\theta})}{s*(s+b)(T_is)(Js^2+K_ws+ K_wk_{\theta})+(T_is+T_i T_d s^2+1)K_pK_wk_{\theta}*g*b}$

$\small \Rightarrow X_{ss} = 0$