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So I asked this question previously in the control theory section and got no responses. Ill try it here. Basically I need to find the ranges of k and \$\beta\$ such that the steady state error will be less than 10% for unit step input.

I'm currently at the point where I know that the steady state error is given by:

$$ 9 < \beta k$$

Now I'm using the closed loop characteristic equation:

$$2s^4 +2s^3+ (2 + \beta)s^2 + (\beta - 10 + 2k)s + 1 +\beta k = 0$$

$$ \begin{matrix} s^4 & 2 & (2+\beta) & (1+\beta k) \\ s^3 & 2 & \beta-10+2k \\ s^2 &12-2k & 1+\beta k \\ s^1 & \frac{(12-2k)(\beta-10+2k)-(2)(1+\beta k)}{12-2k} \\ s^0 & 1+\beta k \end{matrix} $$

So at this point I'm not sure how to find the ranges for k and \$\beta\$ and apply it to the steady state error? Can I say that:

$$ 12 - 4k > 0$$ $$ - 4k > -12$$ $$ 4k < 12$$ $$ k < 3 $$

I'm just stuck as this point. Any thoughts? Thanks.

So I asked this question previously in the control theory section and got no responses. Ill try it here. Basically I need to find the ranges of k and \$\beta\$ such that the steady state error will be less than 10% for unit step input.

I'm currently at the point where I know that the steady state error is given by:

$$ 9 < \beta k$$

Now I'm using the closed loop characteristic equation:

$$2s^4 +2s^3+ (2 + \beta)s^2 + (\beta - 10 + 2k)s + 1 +\beta k = 0$$

$$ \begin{matrix} s^4 & 2 & (2+\beta) & (1+\beta k) \\ s^3 & 2 & \beta-10+2k \\ s^2 &12-2k & 1+\beta k \\ s^1 & \frac{(12-2k)(\beta-10+2k)-(2)(1+\beta k)}{12-2k} \\ s^0 & 1+\beta k \end{matrix} $$

So at this point I'm not sure how to find the ranges for k and \$\beta\$ and apply it to the steady state error? Can I say that:

$$ 12 - 2k > 0$$ $$ - 2k > -12$$ $$ 2k < 12$$ $$ k < 6$$

I'm just stuck as this point. Any thoughts? Thanks.

So I asked this question previously in the control theory section and got no responses. Ill try it here. Basically I need to find the ranges of k and \$\beta\$ such that the steady state error will be less than 10% for unit step input.

I'm currently at the point where I know that the steady state error is given by:

$$ 9 < \beta k$$

Now I'm using the closed loop characteristic equation:

$$2s^4 +2s^3+ (2 + \beta)s^2 + (\beta - 10 + 2k)s + 1 +\beta k = 0$$

$$ \begin{matrix} s^4 & 2 & (2+\beta) & (1+\beta k) \\ s^3 & 2 & \beta-10+2k \\ s^2 &12-4 k & 1+\beta k \\ s^1 & \frac{(12-4 k)(\beta-10+2k)-(2)(1+\beta k)}{12-4k} \\ s^0 & 1+\beta k \end{matrix} $$

So at this point I'm not sure how to find the ranges for k and \$\beta\$ and apply it to the steady state error? Can I say that:

$$ 12 - 4k > 0$$ $$ - 4k > -12$$ $$ 4k < 12$$ $$ k < 3 $$

I'm just stuck as this point. Any thoughts? Thanks.

So I asked this question previously in the control theory section and got no responses. Ill try it here. Basically I need to find the ranges of k and \$\beta\$ such that the steady state error will be less than 10% for unit step input.

I'm currently at the point where I know that the steady state error is given by:

$$ 9 < \beta k$$

Now I'm using the closed loop characteristic equation:

$$2s^4 +2s^3+ (2 + \beta)s^2 + (\beta - 10 + 2k)s + 1 +\beta k = 0$$

$$ \begin{matrix} s^4 & 2 & (2+\beta) & (1+\beta k) \\ s^3 & 2 & \beta-10+2k \\ s^2 &12-2k & 1+\beta k \\ s^1 & \frac{(12-2k)(\beta-10+2k)-(2)(1+\beta k)}{12-2k} \\ s^0 & 1+\beta k \end{matrix} $$

So at this point I'm not sure how to find the ranges for k and \$\beta\$ and apply it to the steady state error? Can I say that:

$$ 12 - 4k > 0$$ $$ - 4k > -12$$ $$ 4k < 12$$ $$ k < 3 $$

I'm just stuck as this point. Any thoughts? Thanks.

So I asked this question previously in the control theory section and got no responses. Ill try it here. Basically I need to find the ranges of k and \$\beta\$ such that the steady state error will be less than 10% for unit step input.

I'm currently at the point where I know that the steady state error is given by:

$$ 9 < \beta k$$

Now I'm using the closed loop characteristic equation:

$$2s^4 +2s^3+ (2 + \beta)s^2 + (\beta - 10 + 2k)s + 1 +\beta k = 0$$

$$ \begin{matrix} s^4 & 2 & (2+\beta) & (1+\beta k) \\ s^3 & 2 & \beta-10+2k \\ s^2 &12-4 k & 1+\beta k \\ s^1 & \frac{(12-4 k)(\beta-10+2k)-(2)(1+\beta k)}{12-4k} \\ s^0 & 1+\beta k \end{matrix} $$

So at this point I'm not sure how to find the ranges for k and $\beta$ and apply it to the steady state error? Can I say that:

$$ 12 - 4k > 0$$ $$ - 4k > -12$$ $$ 4k < 12$$ $$ k < 3 $$

I'm just stuck as this point. Any thoughts? Thanks.

So I asked this question previously in the control theory section and got no responses. Ill try it here. Basically I need to find the ranges of k and \$\beta\$ such that the steady state error will be less than 10% for unit step input.

I'm currently at the point where I know that the steady state error is given by:

$$ 9 < \beta k$$

Now I'm using the closed loop characteristic equation:

$$2s^4 +2s^3+ (2 + \beta)s^2 + (\beta - 10 + 2k)s + 1 +\beta k = 0$$

$$ \begin{matrix} s^4 & 2 & (2+\beta) & (1+\beta k) \\ s^3 & 2 & \beta-10+2k \\ s^2 &12-4 k & 1+\beta k \\ s^1 & \frac{(12-4 k)(\beta-10+2k)-(2)(1+\beta k)}{12-4k} \\ s^0 & 1+\beta k \end{matrix} $$

So at this point I'm not sure how to find the ranges for k and \$\beta\$ and apply it to the steady state error? Can I say that:

$$ 12 - 4k > 0$$ $$ - 4k > -12$$ $$ 4k < 12$$ $$ k < 3 $$

I'm just stuck as this point. Any thoughts? Thanks.