# Homework Help: Poles of transfer function and stability

1. Sep 24, 2015

### Maylis

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
part a

$$\vec {x} = \left [ \begin{array}{ccc} X-X_{s} \\ S-S_{s} \end{array} \right ]$$

$$\vec {u} = \left [ \begin{array}{ccc} S_{0}-S_{0s} \\ \frac {F_{0}}{V}- \big (\frac {F_{0}}{V} \big )_{s} \end{array} \right ]$$
$$\vec {y} = \left [ \begin{array}{ccc} X-X_{s} \end{array} \right ]$$
part b
Manipulated input
$$u_{M} = S_{0}-S_{0s}$$

Disturbance input
$$u_{D} = \frac {F_{0}}{V}- \big (\frac {F_{0}}{V} \big )_{s}$$
part c
The differential equations
$$\frac {dX}{dt} = \mu (S) X - \frac {F_{0}}{V}X = f_{1}$$
$$\frac {dS}{dt} = - \mu(S) \frac {X}{Y} + \frac {F_{0}}{V}(S_{0}-S) =f_{2}$$
are linearized into the form
$$\dot{\vec{x}} = A \vec {x} + B \vec {u}$$
Where $A_{ij} = \frac {\partial f_{i}}{\partial x_{j}}$ and $B_{ij} = \frac {\partial f_{i}}{\partial u_{j}}$ For example, $A_{11} = \frac {\partial f_{1}}{\partial X} = \mu(S)-\frac {F_{0}}{V}$ and $B_{11} = \frac {\partial f_{1}}{\partial S_{0}}=0$

The matrices $A$ and $B$ are solved symbolically and the differential equations are linearized below

$$\left [ \begin{array}{ccc} \frac {d(X-X_{s})}{dt} \\ \frac {d(S-S_{s})}{dt} \end{array} \right ] = \left [ \begin{array}{ccc} \frac {\mu_{m}S_{s}}{K_{s}+S_{s}}- \big ( \frac {F_{0}}{V} \big )_{s} & \frac {\mu_{m}K_{s}}{(K_{s}+S_{s})^{2}}X_{s} \\ - \frac {\mu_{m}S_{s}}{K_{s}+S_{s}} \cdot \frac {1}{Y} & - \frac {\mu_{m}K_{s}}{(K_{s}+S_{s})^{2}} \frac {X_{s}}{Y} - \big ( \frac {F_{0}}{V} \big )_{s} \end{array} \right ] \left [ \begin{array}{ccc} X-X_{s} \\ S-S_{s} \end{array} \right ] + \left [ \begin{array}{ccc} 0 & -X_{s} \\ \big (\frac {F_{0}}{V} \big )_{s} & S_{0s}-S_{s} \end{array} \right ] \left [ \begin{array}{ccc} S_{0}-S_{0s} \\ \big ( \frac {F_{0}}{V} \big ) - \big ( \frac {F_{0}}{V} \big )_{s} \end{array} \right ]$$

Using the steady state values and parameters given, the equations are linearized and tabulated in
the matrices $A$ and $B$

$$\left [ \begin{array}{ccc} \frac {d(X-X_{s})}{dt} \\ \frac {d(S-S_{s})}{dt} \end{array} \right ] = \left [ \begin{array}{ccc} 7.11 \times 10^{-5} & 0.267 \\ -0.303 & -0.909 \end{array} \right ] \left [ \begin{array}{ccc} X - 1.877 \\ S-0.563 \end{array} \right ] + \left [ \begin{array}{ccc} 0 & -1.877 \\ 0.1 & 5.687 \end{array} \right ] \left [ \begin{array}{ccc} S_{0}-6.25 \\ \frac {F_{0}}{V} - 0.1 \end {array} \right ]$$
part d
Recasting the matrices as two differential equations, where $X = x_{1}$ and $S = x_{2}$,

$$\frac {dx_{1}}{dt} = 7.11 \times 10^{-5}x_{1} + 0.267 x_{2} - 1.877u_{D} \hspace {2 in} (1)$$

$$\frac {dx_{2}}{dt} = -0.303x_{1} - 0.909x_{2} + 0.1u_{M}+5.687u_{D} \hspace {1.7 in} (2)$$

Taking the Laplace transform of both differential equations,

$$sX_{1}(s)= 7.11 \times 10^{-5}X_{1}(s) + 0.267 X_{2}(s) - 1.877U_{D}(s) \hspace {1.25 in} (3)$$

$$sX_{2}(s) = -0.303X_{1}(s) - 0.909X_{2}(s) + 0.1U_{M}(s)+5.687U_{D}(s) \hspace {0.80 in} (4)$$

Using eq. (3) to solve for$X_{2}(s)= f(X_{1}(s),U_{D}(s))$, we get

$$X_{2}(s)= 3.745X_{1}(s)(s-7.11 \times 10^{-5})+7.030U_{D}(s)$$

Plug in the expression for $X_{2}(s)$ into equation (4) to eliminate $X_{2}(s)$. We isolate $X_{1}(s)$, and the final expression for $X_{1}(s)= Y(s) = f(U_{D}(s),U_{M}(s))$ is

$$Y(s)= \frac {-7.030(s+0.909)+5.687}{12.360(s-7.11 \times 10^{-5})(s+0.909)+1} U_{D}(s)+ \frac {0.1}{12.360(s-7.11 \times 10^{-5})(s+0.909)+1} U_{M}(s)$$
part e
If the disturbance input, $\frac {F_{0}}{V}$, were to have a step change of magnitude 0.5, then the transfer function derived in part (d) would have an extra term for $U_{D}(s)$, which is $\frac {0.5}{s}$.

$$Y(s)= \frac {-7.030(s+0.909)+5.687}{12.360(s-7.11 \times 10^{-5})(s+0.909)+1} \cdot \frac {0.5}{s} + \frac {0.1}{12.360(s-7.11 \times 10^{-5})(s+0.909)+1} U_{M}(s)$$

The poles are $7.11 \times 10^{-5}$, $-0.909$, and $0$. With a pole at the origin, and one barely positive, the system is an integrating system and never settles to a steady state when a step input change is made.
part f

Working with only the disturbance input,

$$Y(s)= \frac {-3.515(s+0.909)+2.844}{12.360s(s-7.11 \times 10^{-5})(s+0.909)+s}$$

I am stuck because my transfer function looks funky, it has the + constant term in the denominator, and I don't think it is right. There should be some (s + a) kind of term, this + s all by itself is something I haven't seen before. The substitution is really ugly for this problem, but I don't think I did it incorrectly. I can't do partial fraction decomposition with that term sticking out there. When I try and solve for $s = -0.909$ or $s = 7.11 \times 10^{-5}$, I can't get rid of these terms from my transfer function, and then the answer is trivially zero for the constants.

Last edited: Sep 24, 2015
2. Sep 26, 2015

### Hesch

I don't quite understand why you get stucked.

Just multiply

12.360s(s−7.11×10−5)(s+0.909) = As3 + Bs2 + Cs + D

then add the last s, so that you will get ( C+1 ) instead of C.

Now find the roots in

As3 + Bs2 + (C+1)s + D

rewrite the denominator ( with the roots changed ) and do the partial fraction decomposition.

3. Sep 26, 2015

### Maylis

Yes, I realized it later. I think I was too tired and hungry and had a headache so I didn't notice

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