Question 5 of 6: Identifying a first-order-plus-dead-time plant from a sampled step test

Question 5: Identifying a first-order-plus-dead-time plant from a sampled step test (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. A nine-sample record of a unit step test on an unknown plant driven through a zero-order hold, and the assumed model structure.

Find. A procedure for identifying the discrete model $G(z)$ together with its numerical parameters, and from those the three continuous parameters $K$, $\tau$ and $T$.

Approach. Assume the first-order-plus-delay difference equation, extract the delay from the response gap less the hold’s own unit delay, read the two remaining parameters from the first two moving samples, validate by propagating the recursion, then invert the ZOH mapping to recover the continuous parameters.

1. Part (a) — state the procedure before touching the numbers. For a first-order-plus-dead-time plant sampled behind a zero-order hold the discrete model has the structure$$G(z)=\frac{b}{z-a}\,z^{-N},\qquad\text{i.e.}\qquad y(k)=a\,y(k-1)+b\,u(k-1-N)$$The identification is then four steps: (i) locate the sample at which the input steps and the sample at which the output first moves, and take the delay from the gap; (ii) read $b$ from the first moving output sample; (iii) read $a$ from the next one; (iv) propagate the recursion through the whole record as an independent check on the assumed structure.
2. Determine the delay order $N$ — and subtract the hold’s own sample. The input steps at $k=1$ and the output first moves at $k=3$, a gap of two samples. But the ZOH equivalent of ANY strictly proper plant already carries one inherent $z^{-1}$: a control applied at $k$ cannot affect $y(k)$, only $y(k+1)$. Hence$$N=\left(3-1\right)-1=\boxed{1}$$Charging the whole two-sample gap to transport delay is the single most common error on this question type and would double the identified dead time.
3. Read the gain parameter $b$ from the first moving sample. At $k=3$ the recursion reads $y(3)=a\,y(2)+b\,u(1)$, and $y(2)=0$, so the $a$-term drops out entirely:$$b=\frac{y(3)}{u(1)}=\frac{1.000}{1}=\boxed{1.000}$$
4. Read the pole parameter $a$ from the next sample. At $k=4$, $y(4)=a\,y(3)+b\,u(2)$ with $u(2)=1$, so$$a=\frac{y(4)-b}{y(3)}=\frac{1.750-1.000}{1.000}=\boxed{0.750}$$$$G(z)=\frac{1}{z-0.75}\,z^{-1}=\frac{1}{z\,(z-0.75)}$$
5. Validate the model against the rest of the record. Propagating $y(k)=0.75\,y(k-1)+1$ from $y(3)=1$ gives $1.750,\ 2.3125,\ 2.7344,\ 3.0508,\ 3.2881$ against the measured $1.750,\ 2.312,\ 2.734,\ 3.051,\ 3.288$ — agreement to the last printed digit at every sample. An even sharper test is that the residuals from the final value, $y(\infty)-y(k)$, fall in a constant ratio of $0.750$; a constant residual ratio IS the evidence that a single real pole describes the data, and if it drifted, the first-order assumption would have to be abandoned.
6. Part (b) — invert the ZOH mapping for the pole. The zero-order-hold equivalent of $K/(\tau s+1)$ is $K\left(1-e^{-h/\tau}\right)\big/\left(z-e^{-h/\tau}\right)$, so the discrete pole is the exponential of the continuous one:$$a=e^{-h/\tau}\quad\Longrightarrow\quad \tau=\frac{-h}{\ln a}=\frac{-1}{\ln 0.750}=\boxed{3.476\ \text{s}}$$
7. Recover the DC gain and the dead time. Matching numerators, $b=K\left(1-e^{-h/\tau}\right)=K(1-a)$, hence$$K=\frac{b}{1-a}=\frac{1.000}{0.250}=\boxed{4.000}$$and the transport delay is the delay order times the sample period, $T=N h=\boxed{1.000\ \text{s}}$. The identified plant is therefore$$P(s)=\frac{4\,e^{-s}}{3.476\,s+1}$$Two independent sanity checks close the answer: $K$ equals the discrete DC gain $b/(1-a)=4$, which is where the record is visibly heading, and re-sampling this $P(s)$ returns $a=0.750$ and $b=1.000$ exactly.