ECE 486 Control Systems
Lecture 4
More on System Dynamic Response
Recall from last time, we talked about impulse reponses, transfer functions and frequency responses. We started by investigating system output due to a simple unit impulse then we had slow build-up from there. At the end of last lecture, we concluded that given a sinusoidal input
\[ u(t) = A \cos(\omega t), \]
an LTI system with zero initial conditions \(x(0) = 0\) will respond with a system output
\[ y(t) = A M(\omega) \cos\big(\omega t + \varphi(\omega) \big), \] where \(M(\omega) = |H(j \omega)|\) is the modulus of the transfer function \(H(s)\) and \(\varphi(\omega)\) is the argument of \(H(j \omega)\).
Recall the set-up we had last time.
Problem set-up: Consider the system as in Figure 1, compute the response \(y\) to a given input \(u\) under a given set of initial conditions.
Figure 1: Problem set-up
Note: Our aforementioned case of zero initial conditions shall be included in the above general situations. That is to say, when we do not necessarily have the knowledge of zero initial conditions, we may expect both the transient response due to nonzero initial conditions, and the steady-state response as time goes to infinity, i.e., the effect of the initial conditions “washes away”.
More on Laplace Transform with Examples
Recall one-sided (or unilateral) Laplace transform
\begin{align*} \LL \{ f(t)\} \triangleq F(s) = \int^\infty_0 f(t)e^{-st}\d t. \qquad \text{(lower limit is actually $0^-$)} \end{align*}The following examples are used to show how to calculate Laplace transform by hand given a simple function \(f(t)\). If you need to refresh your memory, please refer to our textbook FPE, Appendix A.
Example 1: Compute the Laplace transform of unit step function,
\begin{align*} f(t) &= 1(t) = \begin{cases} 1, & t \ge 0 \\ 0, & t < 0\end{cases} \end{align*}Solution:
\begin{align*} \LL \{1(t)\} &= \int^\infty_0 e^{-st} \cdot 1 \d t \\ &= -\frac{1}{s}e^{-st}\Big|^\infty_0 \\ &= \frac{1}{s}. \qquad \text{(pole at $s=0$)} \end{align*}We need \({\rm Re}(s) > 0\), so that \(e^{-st} \to 0\) as \(t \to + \infty\).
Example 2: Compute the Laplace transform of cosine function \(f(t) = \cos t\).
Solution: First we use the definition for complex cosine function,
\begin{align*} \LL \{\cos t\} &= \LL \left\{ \frac{1}{2}e^{jt} + \frac{1}{2}e^{-jt}\right\} \\ \tag{1} \label{d4_eq1} &= \frac{1}{2} \LL \{e^{jt}\} + \frac{1}{2}\LL\{e^{-jt}\}. &\qquad \text{(by linearity)} \end{align*}Next we compute the Laplace transforms separately. As for the first term,
\begin{align*} \LL\{e^{jt}\} &= \int^\infty_0 e^{jt}e^{-st}\d t \\ &= \int^\infty_0 e^{(j-s)t}\d t \\ &= \frac{1}{j-s}e^{(j-s)t} \Big|^\infty_0 \\ &= - \frac{1}{j-s}. \qquad \text{(pole at $s=j$)} \end{align*}In both cases, we require \({\rm Re}(s) > 0\), i.e., \(s\) must lie in the right half-plane (RHP). Continue with Equation \eqref{d4_eq1}, we have
\begin{align*}\require{cancel} \LL \{\cos t\} &= \frac{1}{2} \LL \{e^{jt}\} + \frac{1}{2}\LL\{e^{-jt}\} \\ &= \frac{1}{2}\left(-\frac{1}{j-s} + \frac{1}{j+s} \right) \\ &= \frac{1}{2}\left( \frac{-\cancel{j}-s+\cancel{j}-s}{(j-s)(j+s)}\right) \\ &= \frac{1}{2}\left( \frac{-2s}{-1+\cancel{js}-\cancel{js}-s^2}\right) \\ &= \frac{s}{s^2+1}, \qquad \text{(poles at $s = \pm j$)} \end{align*}where \({\rm Re}(s) > 0\).
Recall from last time convolution in time domain is “equivalent” to multiplication in frequency domain by Laplace transform
\[ \LL \{f \star g\} = \LL \{f\} \cdot \LL \{g\}. \] Therefore, a quick application of this to our system input, transfer function and system output leads to \(Y(s) = H(s)U(s)\).
Example 3: Consider system dynamics \[\dot{y}(t) = -y(t) + u(t),\, y(0) = 0.\] Compute the system output response in time domain due to cosine input \(u(t) = \cos t\).
Solution: From the example of last lecture, we know the system transfer function \(H(s) = \frac{1}{s+1}\). (Set \(a = 1\) in this case.) We also computed in Example 2
\begin{align*} U(s) &= \LL \{ \cos t \}\\ &=\frac{s}{s^2+1}. \end{align*}The Laplace transform of the system output \(Y(s)\) is
\begin{align*} Y(s)&= H(s)U(s) \\ &= \frac{s}{(s+1)(s^2+1)}. \end{align*}Time domain response can be obtained by applying inverse Laplace transform,
\begin{align*} y(t) &= \LL^{-1}\{Y(s)\} \\ &= \LL^{-1}\left\{ \dfrac{s}{(s+1)(s^2+1)} \right\}. \end{align*}Here we want to further apply partial fraction decomposition. The three relevant Laplace transform entries in the table are
\begin{align*} \frac{1}{s+1} & \qquad \LL^{-1}\left\{\frac{1}{s+1}\right\} = e^{-t}, & \qquad (\# 7) \\ \frac{1}{s^2+1} & \qquad \LL^{-1}\left\{\frac{1}{s^2+1}\right\} = \sin t, & \qquad (\# 17) \\ \frac{s}{s^2+1} & \qquad \LL^{-1}\left\{\frac{s}{s^2+1}\right\} = \cos t. & \qquad (\# 18) \end{align*}Since
\begin{align*} Y(s) = \dfrac{s}{(s+1)(s^2+1)}, \end{align*}we can break \(Y(s)\) into two terms, splitting the factors of the denominator of \(Y(s)\) and each term of the sum is proper,
\begin{align*} \tag{2} \label{d4_eq2} Y(s) &= \frac{a}{s+1} + \frac{bs + c}{s^2+1}. \quad \text{(need $bs+c$ so that ${\rm deg(num)} = {\rm deg(den)}-1$)} \end{align*}Then the problem becomes how to compute \(a\), \(b\) and \(c\) in the above decomposition.
To find \(a\), multiply both sides of Equation \eqref{d4_eq2} by \(s+1\) to isolate \(a\),
\begin{align*} (s+1)Y(s) &= \frac{s}{s^2 + 1} \\ &= a + \frac{(s+1)(bs+c)}{(s^2+1)}. \tag{3} \label{d4_eq3} \end{align*}If we set \(s = -1\), then the second term on the right hand side of Equation \eqref{d4_eq3} is gone. Thus,
\begin{align*} a &= (s+1)Y(s)\Big|_{s=-1} \\ &= -\frac{1}2. \end{align*}By the similar token, we can find \(b\) and \(c\). Multiply both sides of Equation \eqref{d4_eq2} by \(s^2+1\) to isolate \(bs + c\),
\begin{align*} (s^2+1)Y(s) &= \frac{s}{s+1} \\ &= \frac{a(s^2+1)}{s+1} + bs + c. \tag{4} \label{d4_eq4} \end{align*}We set \(s = j\) to kill off the first term on the right hand side of Equation \eqref{d4_eq4}. \[ bj+c = (s^2+1)Y(s)\Big|_{s=j} = \frac{j}{1+j}. \] We then match \({\rm Re}(\cdot)\) and \({\rm Im}(\cdot)\) parts on both sides, \(c + bj = \dfrac{j}{1+j} = \dfrac{j(1-j)}{(1+j)(1-j)} = \dfrac{1}{2} + \dfrac{j}{2}\). Hence \(b = c = \frac{1}{2}\).
So we just found that
Now we can use linearity and tables:
\begin{align*} y(t) &= \LL^{-1}\left\{ -\frac{1}{2(s+1)} + \frac{s}{2(s^2+1)} + \frac{1}{2(s^2+1)} \right\} \\ &= -\frac{1}{2}\LL^{-1}\left\{\frac{1}{s+1}\right\} + \frac{1}{2}\LL^{-1}\left\{\frac{s}{s^2+1}\right\} + \frac{1}{2}\LL^{-1}\left\{\frac{1}{s^2+1}\right\} \\ &= -\frac{1}{2}e^{-t} + \frac{1}{2}\cos t + \frac{1}{2}\sin t \quad \text{(by transform tables)}\\ &= -\frac{1}{2}e^{-t} + \frac{1}{\sqrt{2}}\cos(t-\pi/4). \quad \text{(by $\cos(a-b)=\cos a\cos b + \sin a \sin b)$} \end{align*}The system time domain response in Example 3 consists of two parts,
\begin{align*} u(t) = \cos t \quad \longrightarrow \quad y(t) = \underbrace{-\frac{1}{2}e^{-t}}_{\text{transient} \atop \text{response}} + \underbrace{\frac{1}{\sqrt{2}}\cos(t-\pi/4)}_{\text{steady-state} \atop \text{response}}, \end{align*}of which the transient response vanishes as \(t \to \infty\). (We shall see later why.)
Recall the example we did last time, we may also compute the system time domain response using frequency response formula,
\begin{align*} H(s) &= \frac{1}{s+1} \qquad \Longrightarrow \qquad H(j\omega) = \frac{1}{j\omega+1}. \end{align*}In the case \(u(t) = \cos t\) we have \(A=1\) and \(\omega = 1\), then
\begin{align*} y(t) &= M(1) \cos\big(t + \varphi(1)\big) \\ &= \frac{1}{\sqrt{2}}\cos\big(t-\pi/4\big). \end{align*}So we only recover the steady-state component.
To recap, considering the system \[\dot{y}(t) = -y(t) + u(t), \qquad y(0) = 0,\] we did computation of system response in time domain in two different ways,
- Using partial fraction decomposition, we got
- Using frequency response formula, we got
Since they look different, and one of them is missing the transient response, we have to decide which one is the correct solution.
For a quick sanity check, we verify whether the initial condition \(y(0) = 0\) is satisfied in both cases. At \(t=0\), \(\dfrac{1}{\sqrt{2}}\cos(t-\pi/4) = \frac{1}{2} \neq 0\), which is inconsistent with the initial condition \(y(0)=0\). The term \(-\frac{1}{2}e^{-t}\Big|_{t=0} = -\frac{1}{2}\) cancels the steady-state term, so indeed \(y(0)=0\). Therefore, the first method gives the correct system output \(y(t)\).
Upshot: The frequency response formula only gives the steady-state part of the response. However the inverse Laplace transform gives the whole response including the transient part.
System Response Due to Input with Nonzero Initial Conditions
Laplace Transforms and Differentiation
Given a differentiable function \(f(t)\), the Laplace transform of its time derivative \(f'(t)\) is given by
\begin{align*} \LL\{ f'(t)\} &= \int^\infty_0 f'(t)e^{-st}\d t \\ &= f(t)e^{-st}\Big|^\infty_0 + s \int^\infty_{0} e^{-st} f(t) \d t \qquad \text{(by integration by parts)} \\ &= -f(0) + s F(s). \hspace{4.5cm} \text{(Assume $f(t)e^{-st} \to 0$ as $t \to \infty$)} \end{align*}Laplace transform of first derivative of \(f(t)\): \(\LL \{ f'(t) \} = s F(s) - f(0)\).
Similarly, we have
\begin{align*} \LL \{ f''(t) \} &= \LL \{ (f'(t))' \} = s \LL \{ f'(t) \} - f'(0) \\ &= s^2 F(s) - s f(0) - f'(0). \end{align*}When we have nonzero initial conditions, i.e., nonzero \(f(0)\), \(f'(0)\) etc., we need to take them into account according to the above formulae of Laplace transform of derivatives.
Example 4: Consider the system
\begin{align*} \ddot{y}(t) + 3\dot{y}(t) + 2y(t) = u(t), \mbox{ where } y(0) = \dot{y}(0) = 0. \end{align*}Compute system transfer function \(H(s) = \dfrac{Y(s)}{U(s)}\).
Solution: With zero initial conditions, we take the Laplace transform of the system dynamics on both sides,
\begin{align*} &s^2 Y(s) + 3s Y(s) + 2 Y(s) = U(s) \\ \implies& H(s) = \frac{Y(s)}{U(s)} = \frac{1}{s^2 + 3s + 2}. \end{align*}Example 5: Consider the system
\begin{align*} \ddot{y}(t) + 3\dot{y}(t) + 2y(t) = u(t), \mbox{ where } y(0) = \alpha, \dot{y}(0) = \beta. \end{align*}Compute the step response of the system, i.e., time domain response due to step input \(u(t) = 1(t)\).
Important: As a comparison with Example 4, we notice \(Y(s) = H(s)U(s)\) no longer holds if \(\alpha \neq 0\) or \(\beta \neq 0\).
Solution: Again, take Laplace transforms of both sides, mind the I.C.’s this time,
\begin{align*} s^2 Y(s) - s{y(0)} - \dot{y}(0) + 3 sY(s) - 3y(0) + 2Y(s) &= U(s) \\ s^2 Y(s) - s\alpha - \beta + 3 sY(s) - 3\alpha + 2Y(s) &= U(s). \end{align*}Plug in \(U(s) = \LL\{1(t)\} = 1/s\) and simplify,
\begin{align*} s^2 Y(s) - s\alpha - \beta + 3sY(s) - 3\alpha + 2Y(s) = \frac{1}{s}, \end{align*} \begin{align*} \implies Y(s) &= \frac{\alpha s + (3\alpha + \beta) + \frac{1}{s}}{s^2 + 3s + 2} \\ &=\frac{\alpha s^2 + (3\alpha + \beta)s + 1}{s(s+1)(s+2)}. \end{align*}Notice if \(\alpha = \beta = 0\), then \(Y(s) = \dfrac{1}{s(s+1)(s+2)} = H(s)U(s)\).
In order to compute system response in time domain, apply inverse Laplace transform,
\begin{align*} y(t) &= \LL^{-1}\{ Y(s)\} \\ &= \LL^{-1}\left\{ \frac{\alpha s^2 + (3\alpha + \beta)s + 1}{s(s+1)(s+2)} \right\}. \end{align*}By partial fraction decomposition,
\begin{align*} \frac{\alpha s^2 + (3\alpha + \beta)s + 1}{s(s+1)(s+2)} &= \frac{a}{s} + \frac{b}{s+1} + \frac{c}{s+2}, \end{align*}This gives \(a=1/2,\,\, b = 2\alpha + \beta - 1,\,\, c = -\alpha -\beta + 1/2\).
Therefore,
\begin{align*} y(t) &= \LL^{-1}\{ Y(s)\} \\ &= \LL^{-1}\left\{ \frac{\alpha s^2 + (3\alpha + \beta)s + 1}{s(s+1)(s+2)} \right\} \\ &= \LL^{-1}\left\{ \frac{1}{2s} + (2\alpha + \beta -1)\frac{1}{s+1} + \frac{-\alpha - \beta + 1/2}{s+2} \right\} \\ &=\frac{1}{2}1(t) + (2\alpha + \beta -1)e^{-t} + (1/2-\alpha-\beta) e^{-2t} \end{align*}is the system step response in time domain. It has both the transient and steady-state components,
- The transient terms are \(e^{-t}\), \(e^{-2t}\). They decay to zero at exponential rates \(-1\) and \(-2\). Also note the poles of transfer function \(H(s) = \dfrac{1}{(s+1)(s+2)}\) are at \(s=-1\) and \(s=-2\); these two poles are called stable poles since both of them lie in LHP.
- The steady-state term is \(\frac{1}{2}1(t)\) which indicates the steady-state value of \(1/2\).
DC Gain, Steady-State Value and Final Value Theorem
DC Gain
The steady-state value of the unit step response of the system is called its DC gain. It is also the ratio of system output and input signals when transients die out.
\begin{align*} \text{DC gain} = y(\infty) = \lim_{t \to \infty} y(t) \,\, \text{for } u(t) = 1(t). \end{align*}In Example 5, the unit step response was \[ y(t) = \frac{1}{2}1(t) + (2\alpha + \beta -1)e^{-t} + (1/2-\alpha-\beta) e^{-2t}, \] therefore by definition, \(\text{DC gain} = y(\infty) = 1/2\).
The Final Value Theorem
Given the system set-up in Figure 1, we can write
\begin{align*} u(t) = 1(t) \qquad U(s) = \frac{1}{s} \qquad \Longrightarrow \qquad Y(s) = \frac{H(s)}{s}. \end{align*}Let’s look at several examples to see how to compute \(y(\infty)\) from \(Y(s)\).
- \(Y(s) = \dfrac{1}{s+a}, \,\, a > 0\), with a pole at \(s = -a < 0\), \(y(t) = e^{-at} \implies y(\infty) = 0\).
- \(Y(s) = \dfrac{1}{s+a}, \,\, a < 0\) with a pole at \(s = -a > 0\), \(y(t) = e^{-at} \implies y(\infty) = \infty\).
- \(Y(s) = \dfrac{1}{s^2+\omega^2}, \,\, \omega \in \RR\) with purely imaginary poles at \(s = \pm j\omega\), \(y(t) = \sin(\omega t)\implies y(\infty) \text{ does not exist}\).
- \(Y(s) = \dfrac{c}{s}\) with pole at the origin, \(s = 0\), \(y(t) = c1(t) \implies y(\infty) = c\).
In general, we have the Final Value Theorem (FVT), which states the following,
The Final Value Theorem (in Control): If all poles of \(sY(s)\) are strictly stable or lie in the open left half-plane (OLHP), i.e., have \({\rm Re}(s) < 0\), then \[ y(\infty) = \lim_{s \to 0} sY(s). \]
The Final Value Theorem (in Math): If \(\lim_{t\to \infty} f(t)\) exists, i.e, it has a finite limit, then \[ \lim_{t\to \infty} f(t) = \lim_{s \to 0} s F(s), \] where \(F(s)\) is the one-sided Laplace transform of \(f(t)\).
If we rerun the previous examples, hit \(Y(s)\) with an \(s\) then pass \(sY(s)\) to the limit,
- \(Y(s) = \dfrac{1}{s+a}\), \(sY(s) = \dfrac{s}{s+a}\). If \(a > 0\), then \(y(\infty) = 0\); if \(a < 0\), FVT does not hold since unstable poles make \(y(t)\) blow up as \(t \to \infty\). The limit \(y(\infty)\) does not exist in the first place.
- \(Y(s) = \dfrac{1}{s^2+\omega^2}\), \(sY(s) = \dfrac{s}{s^2+\omega^2}\), poles are purely imaginary (not in OLHP), FVT does not hold for the same reason that \(y(\infty)\) does not exist in the first place.
- \(Y(s) = \dfrac{c}{s}\), \(sY(s) = c\), no poles or poles at infinity, so \(y(\infty) = c\) is correctly given by FVT.
By Final Value Theorem, we may compute DC gain with ease when FVT holds. For example, system unit step response is nothing but
\begin{align*} Y(s) = \frac{H(s)}{s}. \end{align*}If all poles of \(sY(s) = H(s)\) are strictly stable, then by FVT
\begin{align*} y(\infty) = \lim_{s \to 0} H(s). \end{align*}Important: In general, we cannot pass the system transfer function \(H(s)\) to the limit and claim the resulting limit is the steady-state value. The special feature of the above formula was that \(Y(s)\) was the unit step response, where the input transform \(\LL \{ 1(t) \} = \frac{1}s\) would cancel \(s\) in the expression \(sY(s)\) of FVT. If \(Y(s)\) is not a unit step response, \(s\) will not be cancelled out by the Laplace transform of input \(U(s) \neq \frac{1}s\).
Example: Given system transfer function
\begin{align*} H(s) &= \frac{s^2 + 5s + 3}{s^3 + 4s^2 + 2s + 5}, \end{align*}compute the DC gain of the system.
Solution: All the poles of \(H(s)\) are strictly stable poles (we shall see this later using the Routh–Hurwitz criterion), so
\begin{align*} y(\infty) = H(s)\Big|_{s=0} = \frac{3}{5}. \end{align*}