Question

A cascade control strategy is shown in the figure. The transfer function between the output \((y)\) and the secondary disturbance \((d_2)\) is defined as \(G_{d2}(s)=\dfrac{y(s)}{d_2(s)}\). Which one of the following is the CORRECT expression for \(G_{d2}(s)\)?

• A. \(\dfrac{1}{(11s+21)(0.1s+1)}\)
• B. \(\dfrac{1}{(s+1)(0.1s+1)}\)
• C. \(\dfrac{s+1}{(s+2)(0.1s+1)}\)
• D. \(\dfrac{s+1}{(s+1)(0.1s+1)}\)

Hint:

For disturbance transfer functions, set the set-point input to zero and keep only the disturbance path.
If the disturbance enters inside a feedback loop, include the loop gain in the denominator \(1+LG\).
After obtaining a polynomial denominator, factor it to compare with given options.

Answer 1

The Correct Option is A

Step 1: Loop reduction up to the point where \(d_2\) enters
Let \(G_1(s)=\dfrac{1}{0.1s+1}\) and \(G_2(s)=\dfrac{1}{s+1}\). From the diagram, the signal entering the \(d_2\)-summer comes from a unity gain followed by a controller of gain \(10\), and there is an outer unity feedback from \(y\) to the first summer (error \(=y_{\text{set}}-y\)). With \(y_{\text{set}}=0\) (for disturbance transfer) the path from \(y\) back to the \(d_2\)-summer through the controller forms a feedback around the plant. The equivalent forward block from the \(d_2\)-summer to \(y\) is \(G_1G_2\) with a feedback path of gain \(10\). 

Step 2: Closed-loop due to the controller feedback
Writing the closed-loop relation at the \(d_2\)-injection node (call its signal \(u\)): \[ y \;=\; G_1G_2\big(d_2 - 10\,y\big) \quad\Rightarrow\quad y\big(1+10G_1G_2\big)=G_1G_2\,d_2 . \] Hence \[ \frac{y}{d_2}=\frac{G_1G_2}{1+10G_1G_2} = \frac{1}{(0.1s+1)(s+1)+10}. \] 

Step 3: Rearrangement to the factored form
\[ (0.1s+1)(s+1)+10 =0.1s^2+1.1s+11 = (0.1s+1)(11s+21). \] Therefore \[ G_{d2}(s)=\frac{1}{(11s+21)(0.1s+1)}. \] This matches option (A).