MA30061 Control Theory

Chapter 5 The transfer function

Definition 5.1. The Laplace transform of a function $f:[0,\infty )\to \mR ^{n_1\times n_2}$ is defined by

\[ \mL (f)=s\mapsto \int _0^\infty \e ^{-st}f(t)\,dt. \]

We will also use the notation $\hat {f}$ for $\mL (f)$.

Remark 5.2. If $\int _0^\infty \e ^{-\sigma t}|f(t)|\,dt<\infty $, then the Laplace transform is a function $\{s:\re (s)>\sigma \}\to \mC ^{n_1\times n_2}$. When Laplace transforming a function, we will always assume that such a $\sigma \in \mR $ exists.

Remark 5.3. The most important property for us of the Laplace transform is how it interacts with differentiation: for $\re (s)$ large enough
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\begin{equation} \label {eq:Laplacediff} \mL (f')(s)=s\mL (f)(s)-f(0). \end{equation}

This can be proven though integration by parts:

\[ \mL (f')(s) =\int _0^\infty \e ^{-st}f'(t)\,dt =\left [\e ^{-st}f(t)\right ]_{t=0}^\infty +s\int _0^\infty \e ^{-st}f(t)\,dt =-f(0)+s\mL (f)(s), \]

where we have used that $\lim _{t\to \infty }\e ^{-st}f(t)=0$ for $\re (s)$ large enough.

Applying (5.1) to $f'$ rather than $f$ gives $\mL (f'')(s)=s\mL (f')-f'(0)$ and using (5.1) then gives
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\begin{equation} \label {eq:Laplacediff2} \mL (f'')(s)=s^2\mL (f)(s)-sf(0)-f'(0). \end{equation}

Laplace transforming (3.1) with $x(0)=0$ gives

\[ s\hat {x}(s)=A\hat {x}(s)+B\hat {u}(s),\qquad \hat {y}(s)=C\hat {x}(s)+D\hat {u}(s). \]

Solving this gives

\[ \hat {x}(s)=(sI-A)^{-1}B\hat {u}(s),\qquad \hat {y}(s)=\left (C(sI-A)^{-1}B+D\right )\hat {u}(s). \]

Definition 5.4. The transfer function of (3.1) is the $\mC ^{p\times m}$ valued function

\[ G(s)=C(sI-A)^{-1}B+D. \]

Remark 5.5. A second interpretation of the transfer function is as follows. We seek a solution of (3.1) of the form $u(t)=\e ^{st}u_0$, $x(t)=\e ^{st}x_0$, $y(t)=\e ^{st}y_0$. Substituting this in (3.1) gives

\[ s\e ^{st}x_0=A\e ^{st}x_0+B\e ^{st}u_0,\qquad \e ^{st}y_0=C\e ^{st}x_0+D\e ^{st}u_0. \]

Here $\e ^{st}$ cancels out and we obtain

\[ sx_0=Ax_0+Bu_0,\qquad y_0=Cx_0+Du_0. \]

Solving this gives

\[ x_0=(sI-A)^{-1}Bu_0,\qquad y_0=\left [C(sI-A)^{-1}B+D\right ]u_0. \]

Therefore the transfer function is such that for the input $u(t)=\e ^{st}u_0$ and the initial condition $x^0:=(sI-A)^{-1}Bu_0$ we have $y(t)=\e ^{st}G(s)u_0$.

In the above it makes more sense to view $t\in \mR $ and consider the “initial condition” $x^0$ as the value at an intermediate time. That value for $x(0)$ is special in the sense that the output is again a constant times $\e ^{st}$.

5.1 Examples

5.1.1 First order systems

The transfer function of the first order system

\[ T\dot {y}(t)+y(t)=Ku(t), \]

\[ G(s)=\frac {K}{Ts+1}. \]

5.1.2 Second order systems (low pass)

The transfer function of the second order (low pass) system

\[ T^2\ddot {y}(t)+2\zeta T\dot {y}(t)+y(t)=Ku(t), \]

\[ G(s)=\frac {K}{T^2s^2+2\zeta Ts+1}=\frac {K\omega _0^2}{s^2+2\zeta \omega _0s+\omega _0^2}. \]

The easiest way to obtain this formula is by Laplace transforming the equations using (5.1) and (5.2) rather than by going through the state. We have that

\[ \mL \left (T^2\ddot {y}+2\zeta T\dot {y}+y\right )=\mL \left (Ku\right ), \]

which using linearity, (5.2) and the fact that $y(0)=\dot {y}(0)=0$ (which comes from the assumption of zero initial condition $x(0)$) gives

\[ T^2s^2\hat {y}+2\zeta Ts\hat {y}+\hat {y}=K\hat {u}, \]

which gives

\[ (T^2s^2+2\zeta Ts+1)\hat {y}=K\hat {u}, \]

which gives

\[ \hat {y}=\frac {K}{T^2s^2+2\zeta Ts+1}\hat {u}. \]

Since the relation $\hat {y}(s)=G(s)\hat {u}(s)$ characterizes the transfer function, we must have $G(s)=\frac {K}{T^2s^2+2\zeta Ts+1}$.

5.1.3 Second order systems (band pass)

The transfer function of the second order (band pass) system

\[ T^2\ddot {q}(t)+2\zeta T\dot {q}(t)+q(t)=Ku(t),\qquad y(t)=\dot {q}(t), \]

\[ G(s)=\frac {Ks}{T^2s^2+2\zeta Ts+1}=\frac {K\omega _0^2s}{s^2+2\zeta \omega _0s+\omega _0^2}. \]

5.2 Case study: control of a tape drive

We compute the transfer function from the disturbance $v_e$ to each of the state variables $v_1$, $v_2$ and $T$. Laplace transforming the differential equations with zero initial condition (and $u_1=0$ and $u_2=0$) gives

\begin{align*} M_1s\hat {v}_1+d_1\hat {v}_1-\hat {T}&=0,\\ M_2s\hat {v}_2+d_2\hat {v}_2+\hat {T}&=0,\\ s\hat {T}+k(\hat {v}_1-\hat {v}_2)&=k\hat {v}_e. \end{align*} From the first equation we obtain $\hat {v}_1=\frac {1}{M_1s+d_1}\hat {T}$ and from the second equation we obtain $\hat {v}_2=\frac {-1}{M_2s+d_2}\hat {T}$. Substituting this into the third equation gives

\[ \left (s+\frac {k}{M_1s+d_1}+\frac {k}{M_2s+d_2}\right )\hat {T}=k\hat {v}_e. \]

Combining the terms in brackets gives

\[ \frac {s(M_1s+d_1)(M_2s+d_2)+k(M_2s+d_2)+k(M_1s+d_1)}{(M_1s+d_1)(M_2s+d_2)}\hat {T}=k\hat {v}_e, \]

from which we deduce that

\[ \hat {T}=\frac {k(M_1s+d_1)(M_2s+d_2)}{s(M_1s+d_1)(M_2s+d_2)+k(M_2s+d_2)+k(M_1s+d_1)}\hat {v}_e, \]

so that the transfer function from $v_e$ to $T$ is

\[ G_{Tv_e}(s)=\frac {k(M_1s+d_1)(M_2s+d_2)}{s(M_1s+d_1)(M_2s+d_2)+k(M_2s+d_2)+k(M_1s+d_1)}. \]

Using the earlier equations $\hat {v}_1=\frac {1}{M_1s+d_1}\hat {T}$ and $\hat {v}_2=\frac {-1}{M_2s+d_2}\hat {T}$, we then obtain

\begin{align*} G_{v_1v_e}(s)&=\frac {k(M_2s+d_2)}{s(M_1s+d_1)(M_2s+d_2)+k(M_2s+d_2)+k(M_1s+d_1)}, \\ G_{v_2v_e}(s)&=\frac {-k(M_1s+d_1)}{s(M_1s+d_1)(M_2s+d_2)+k(M_2s+d_2)+k(M_1s+d_1)}. \end{align*} Note that in each case the denominator equals

\[ M_1M_2s^3+(d_1M_2+d_2M_1)s^2+(d_1d_2+kM_1+kM_2)s+(d_1+d_2)k, \]

which we already saw in Section 2.3.

5.3 Case study: a suspension system

It is possible (though tedious) to compute the transfer functions of the fixed structure suspension system as given in Section 1.2.1. These are

\begin{align*} G_{\rm handling}(s)&= -r_1~\frac {m_s m_{us}s^3 + d (m_s+m_{us}) s^2 + k_s (m_s+m_{us}) s}{m_s m_{us} s^4+ d (m_s+m_{us}) s^3 + \left [k_s (m_s+m_{us}) + k_{us} m_s \right ] s^2 + d k_{us} s + k_s k_{us}}, \\ G_{\rm stroke}(s)&= -r_2~\frac {k_{us}m_ss}{m_s m_{us} s^4+ d (m_s+m_{us}) s^3 + \left [k_s (m_s+m_{us}) + k_{us} m_s \right ] s^2 + d k_{us} s + k_s k_{us}}, \\ G_{\rm comfort}(s)&= -\frac {d k_{us} s^2 + k_s k_{us} s}{m_s m_{us} s^4+ d (m_s+m_{us}) s^3 + \left [k_s (m_s+m_{us}) + k_{us} m_s \right ] s^2 + d k_{us} s + k_s k_{us}}. \end{align*}

5.4 Problems

Let $M,k>0$ and $d\geq 0$. Consider as in Section 1.4

\[ \dot {q}=v-v_e,\qquad M\dot {v}+dv+kq=F_e. \]

In the context of this chapter we have the input $u:=\sbm {F_e\\v_e}$ and the output $y:=\sbm {q\\v}$. Determine the transfer function of the above system.

Solution. Laplace transforming with zero initial conditions gives
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\begin{equation} \label {eq:problems:transferfunction1} s\hat {q}=\hat {v}-\hat {v_e},\qquad (Ms+d)\hat {v}+k\hat {q}=\hat {F}_e.\tag {$\dagger $} \end{equation}

Multiplying the second equation by $s$ and substituting the first equation gives

\[ (Ms^2+ds+k)\hat {v}=s\hat {F}_e+k\hat {v}_e. \]

Hence

\[ \hat {v}=\frac {s}{Ms^2+ds+k}\hat {F}_e+\frac {k}{Ms^2+ds+k}\hat {v}_e. \]

It follows that
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\begin{equation} \label {eq:problems:transferfunction2} \hat {v}-\hat {v}_e=\frac {s}{Ms^2+ds+k}\hat {F}_e+\frac {-Ms^2-ds}{Ms^2+ds+k}\hat {v}_e\tag {$\ast $} \end{equation}

This gives

\[ G_{vF_e}(s)=\frac {s}{Ms^2+ds+k},\qquad G_{vv_e}(s)=\frac {k}{Ms^2+ds+k}. \]

Substituting ($\ast $) into the first equation in ($\dagger $) gives

\[ \hat {q}=\frac {1}{Ms^2+ds+k}\hat {F}_e+\frac {-Ms-d}{Ms^2+ds+k}\hat {v}_e. \]

Hence

\[ G_{qF_e}(s)=\frac {1}{Ms^2+ds+k},\quad G_{qv_e}(s)=-\frac {Ms+d}{Ms^2+ds+k}. \]

Therefore the transfer function is

\[ G=\bbm { \frac {1}{Ms^2+ds+k}&-\frac {Ms+d}{Ms^2+ds+k}\\ \frac {s}{Ms^2+ds+k}&\frac {k}{Ms^2+ds+k} }. \]

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