Identification of a first-order plant with transport delay from step data: 22-Elec-B2 Advanced Control Systems, December 2013, Q5

Question 5: Identification of a first-order plant with transport delay from step data (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Seven samples of a step test on an unknown first-order plant with an unknown integer transport delay, taken at one-second intervals through a zero-order hold.

Given data — Question 5
Quantity	Value
Sample period	$h = 1$ s
Input	unit step, applied at $k=1$ ($u=0$ at $k=0$, $u=1$ thereafter)
Output samples	$0,\ 0,\ 0,\ 4.000,\ 4.800,\ 4.960,\ 4.992$ for $k=0\dots6$
Assumed structure	first order continuous, plus a transport delay of $Nh$ with $N$ an unknown integer

Find. (a) the discrete transfer function $G(z)$ from $u(kh)$ to $y(kh)$; (b) the continuous transfer function $P(s)$, including its gain, time constant and delay.

Figure Q5.1 — the tabulated measurements. Two samples elapse between the input step at k = 1 and the first output motion at k = 3; one of those is the unavoidable ZOH/discretisation delay, so the transport delay is N = 1 sample.

Approach. Take the steady-state value to fix the gain, take the ratio of successive residuals to fix the discrete pole, count samples to fix the delay, then invert the zero-order-hold mapping to recover the continuous parameters.

Read the steady-state gain. The output is converging on a limit; the increments $4.000 \to 4.800 \to 4.960 \to 4.992$ are shrinking geometrically toward $$y(\infty) = 5.000 \quad\Longrightarrow\quad K_{p} = \frac{y(\infty)}{\Delta u} = \frac{5.000}{1} = 5 .$$
Extract the discrete pole from the residuals. For a first-order discrete system the distance from the final value decays by a constant factor each sample. Forming $y(\infty)-y(kh)$: $$1.000,\quad 0.200,\quad 0.040,\quad 0.008 \qquad\text{for } k=3,4,5,6 .$$ Each is one fifth of the one before, so the ratio is constant and $$a = \frac{0.200}{1.000} = \frac{0.040}{0.200} = \frac{0.008}{0.040} = 0.2 .$$ The constancy of this ratio is itself the evidence that a first-order model is adequate; a second-order plant would give a ratio that drifts.
Count the delay. The input step is applied at $k=1$ but the output first moves at $k=3$, a gap of two samples. One of those is unavoidable: the zero-order-hold equivalent of any strictly proper continuous plant carries an inherent $z^{-1}$, because a control applied at $k$ cannot affect the output until $k+1$. The remaining one sample is the transport delay: $$\boxed{N = 1, \ \text{i.e. a dead time of } Nh = 1 \ \text{second}}$$
Assemble the discrete model. The ZOH equivalent of $K_{p}/(\tau s+1)$ is $K_{p}(1-a)/(z-a)$, and the transport delay multiplies it by $z^{-N}$: $$G(z) = z^{-1}\,\frac{K_{p}(1-a)}{z-a} = z^{-1}\,\frac{5(1-0.2)}{z-0.2} \quad\Longrightarrow\quad \boxed{G(z) = \frac{4}{z(z-0.2)} = \frac{4z^{-2}}{1-0.2z^{-1}}}$$ Equivalently, the difference equation is $y(k) = 0.2\,y(k-1) + 4\,u(k-2)$. Running that recursion forward from rest reproduces the table exactly: $y(3)=4.000$, $y(4)=0.2(4)+4=4.800$, $y(5)=0.2(4.8)+4=4.960$ and $y(6)=0.2(4.96)+4=4.992$.
Invert the sampling to recover the time constant. The discrete pole is the image of the continuous pole, $a = e^{-h/\tau}$, so $$\tau = \frac{-h}{\ln a} = \frac{-1}{\ln 0.2} = \frac{1}{1.6094} \quad\Longrightarrow\quad \boxed{\tau = 0.6213\ \text{s}}$$
Write the continuous plant. Combining the gain, the time constant and the one-second transport delay, $$\boxed{P(s) = \frac{5\,e^{-s}}{0.6213\,s + 1}}$$ As a round-trip check, discretising this model at $h=1$ s gives $e^{-1/0.6213} = 0.200$ and a numerator gain $5(1-0.200) = 4.000$, returning $G(z)$ exactly.

Check: the one-sample discretisation delay must not be charged to the plant. The two-sample gap between the input step and the first output motion is the sum of the inherent $z^{-1}$ of any strictly proper ZOH equivalent and the physical transport delay. Attributing all of it to transport gives $N=2$ and a plant $5e^{-2s}/(0.6213s+1)$, whose discrete model would predict the first output motion at $k=4$ — one sample later than the data show. The data therefore identify $N=1$ unambiguously. A non-integer dead time is also excluded here: a delay shorter than 1 s would make $y(2)$ non-zero, and a delay $d$ between 1 s and 2 s would leave a residual $5e^{-(2-d)/\tau}$ at $k=3$ smaller than the printed $1.000$, so only $d = 1$ s fits the table exactly. With noisy real data, however, sub-sample delay resolution would need a faster sampler.

Final results — Question 5
Quantity	Symbol	Value
Steady-state gain	$K_{p}$	5
Discrete pole	$a$	0.2
Transport delay	$N$	1 sample = 1.0 s
Discrete model	$G(z)$	$\dfrac{4}{z(z-0.2)}$
Difference equation	—	$y(k)=0.2y(k-1)+4u(k-2)$
Continuous time constant	$\tau$	0.6213 s
Continuous plant	$P(s)$	$\dfrac{5e^{-s}}{0.6213s+1}$

Topic. Non-parametric identification of a first-order-plus-dead-time model from a sampled step test, and inversion of the zero-order-hold mapping.

Key relations. $K_{p}=y(\infty)/\Delta u$; residual ratio $[y(\infty)-y(k+1)]/[y(\infty)-y(k)] = a$; $a=e^{-h/\tau}$ so $\tau=-h/\ln a$; ZOH equivalent $K_{p}(1-a)/(z-a)$; transport delay of $N$ samples multiplies by $z^{-N}$; a strictly proper plant always contributes one further $z^{-1}$.

Why this works. A first-order step response is completely determined by three numbers, and the sampled data expose all three directly: the limit gives the gain, the geometric decay of the residual gives the pole, and the count of dead samples gives the delay. Working with residuals rather than with the raw output is what makes the pole readable by inspection, because it converts an exponential approach into a pure geometric sequence. The last step is simply reading the ZOH pole map backwards; because that map is one-to-one for real poles, the continuous model is recovered uniquely.

Common pitfall. Counting the dead time wrongly — either by measuring from $k=0$ instead of from the instant the input actually steps, or by charging the inherent discretisation delay to the plant. Both errors shift $N$ by one and change the answer to part (b). A second pitfall is fitting the pole from a single pair of samples: taking all three residual ratios and confirming they agree is what justifies the first-order assumption rather than merely assuming it.

Source. Franklin, Powell & Workman, Digital Control of Dynamic Systems, 3rd ed., Ch. 4 (ZOH equivalents) and Ch. 12 (system identification); Åström & Hägglund, Advanced PID Control, ISA, Ch. 2 (step-response models with dead time).

Question 5 of 6: Identification of a first-order plant with transport delay from step data

Question 5: Identification of a first-order plant with transport delay from step data (25 marks)