Block-Diagram Algebra with a Feedforward Path: 23-Chem-A6 Process Dynamics and Control, May 2013, Q8

Problem 8: Block-Diagram Algebra with a Feedforward Path (20%)

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 feedback loop with a parallel path $G_2$ around $G_1$, a disturbance $D$ entering between $G_1$ and $G_3$, and unity feedback:

Block	Transfer function
$G_1$	$10$
$G_2$ (parallel path from controller output)	$2/(3s+1)$
$G_3$	$1/(s-1)$ (open-loop unstable)
Controller	$K_c$ (proportional)

Find. (a) $Y/D$ and the characteristic equation; (b) the stabilising range of $K_c$.

Problem 8: the block diagram. The controller $K_c$ feeds $G_1$ (then the disturbance $D$ is added ahead of $G_3$) and, in parallel, $G_2$ direct to the output summer; unity feedback closes the loop. Effective forward path $G_1G_3+G_2$; disturbance path $G_3$.

Approach. Trace the diagram to express $Y$ in terms of the controller output and the disturbance, substitute the proportional law $u=K_c(Y_{sp}-Y)$, solve for $Y$, set $Y_{sp}=0$ to isolate $Y/D$, then apply the Routh test to the characteristic polynomial.

Assemble the output. The controller output $u=K_c(Y_{sp}-Y)$ drives two paths: through $G_1$ then (with $D$ added) through $G_3$, and directly through $G_2$ to the final summer. Thus $$Y=G_3\big(G_1u+D\big)+G_2u=(G_1G_3+G_2)\,u+G_3D.$$
Close the loop. Substituting $u=K_c(Y_{sp}-Y)$ and collecting $Y$: $Y\big[1+K_c(G_1G_3+G_2)\big]=K_c(G_1G_3+G_2)Y_{sp}+G_3D$. Setting $Y_{sp}=0$ gives $$\dfrac{Y}{D}=\dfrac{G_3}{1+K_c\big(G_1G_3+G_2\big)}.$$
Substitute the blocks. $G_1G_3+G_2=\dfrac{10}{s-1}+\dfrac{2}{3s+1}=\dfrac{10(3s+1)+2(s-1)}{(s-1)(3s+1)}=\dfrac{32s+8}{(s-1)(3s+1)}$. Hence $$\boxed{\dfrac{Y}{D}=\dfrac{3s+1}{3s^2+(32K_c-2)s+(8K_c-1)}}$$ (the $(s-1)$ from $G_3$ cancels against the denominator’s factor after clearing fractions).
Characteristic equation. The denominator gives $$\boxed{3s^2+(32K_c-2)s+(8K_c-1)=0}$$ obtained equivalently from $1+K_c(G_1G_3+G_2)=0\Rightarrow(s-1)(3s+1)+K_c(32s+8)=0$.
(b) Routh stability. For the quadratic $3s^2+(32K_c-2)s+(8K_c-1)$ all coefficients must be positive: $32K_c-2>0\Rightarrow K_c>1/16$ and $8K_c-1>0\Rightarrow K_c>1/8$. The binding condition is $$\boxed{K_c>\tfrac18=0.125}.$$ Below this the open-loop-unstable $G_3$ is not stabilised (a positive real root persists); above it both roots move into the LHP.

Result	Expression
$Y/D$	$(3s+1)/[3s^2+(32K_c-2)s+(8K_c-1)]$
Characteristic equation	$3s^2+(32K_c-2)s+(8K_c-1)=0$
Routh conditions	$K_c>1/16$ and $K_c>1/8$
Stability range	$K_c>1/8=0.125$

Concept & Theory

Topic. Block-diagram reduction with a parallel (feedforward) branch and a mid-loop disturbance; stabilising an open-loop-unstable element by feedback.

Key relations. For a single loop the closed-loop transfer to a disturbance is (direct path from $D$ to $Y$)$/(1+L)$ with loop gain $L=K_c(G_1G_3+G_2)$; Routh: a quadratic is stable iff all coefficients share one sign.

Why this works. The parallel $G_2$ adds to the forward path, so the effective process is $G_1G_3+G_2$; the characteristic equation $1+L=0$ governs every input-output pair. Because $G_3$ has a RHP pole, only sufficient loop gain ($K_c>1/8$) shifts both closed-loop roots into the LHP — another instance of gain stabilising an unstable plant.

Common pitfall. Omitting the $G_2$ term from the loop gain, or forgetting that the disturbance enters after $G_1$ so only $G_3$ (not $G_1G_3$) multiplies $D$ in the numerator.

Source. Coughanowr & LeBlanc, Ch. 11–13 (block-diagram algebra, closed-loop transfer functions); Stephanopoulos, Ch. 14–15 (feedback loop analysis, Routh test).

Question 8 of 8: Block-Diagram Algebra with a Feedforward Path

Problem 8: Block-Diagram Algebra with a Feedforward Path (20%)