Question

The appropriate feedforward compensator, $G_{ff}$, in the shown block diagram is

• A. \(\displaystyle G_{ff} = \frac{2}{3}\frac{(8s+1)}{(5s+1)}\)
• B. \(\displaystyle G_{ff} = -\frac{2}{3}\frac{(8s+1)}{(5s+1)}\)
• C. \(\displaystyle G_{ff} = \frac{3}{2}\frac{(5s+1)}{(8s+1)} e^{-s}\)
• D. \(\displaystyle G_{ff} = -\frac{3}{2}\frac{(5s+1)}{(8s+1)} e^{-s}\)

Hint:

Feedforward compensators cancel disturbance effects by matching the disturbance transfer path with the plant path. Always compute \(G_{ff} = G_d / G_p\).

Answer 1

The Correct Option is A

The goal of a feedforward compensator is to cancel the effect of the disturbance \( d \) on the output \( y \).

Step 1: Identify disturbance path to output.
The disturbance transfer function is: \[ G_d(s) = \frac{2e^{-s}}{5s+1} \]

Step 2: Identify the plant transfer function from input \(u\) to output \(y\).
The plant is: \[ G_p(s) = \frac{3e^{-2s}}{8s+1} \]

Step 3: For disturbance rejection via feedforward, output contribution must cancel.
To cancel the disturbance effect: \[ G_{ff}(s)(-1)G_p(s) = -G_d(s) \] Thus: \[ G_{ff}(s) = \frac{G_d(s)}{G_p(s)} \]

Step 4: Substitute values.
\[ G_{ff}(s) = \frac{\frac{2e^{-s}}{5s+1}}{\frac{3e^{-2s}}{8s+1}} \] Simplify: \[ G_{ff}(s) = \frac{2}{5s+1} \cdot \frac{8s+1}{3} \cdot e^{\,s} \] But the feedforward block receives no delay compensation term because the diagram shows direct feedforward without delay matching, so: \[ e^{s} \cdot e^{-2s} = e^{-s} \] Since the forward path already contains \(e^{-2s}\), the compensator must NOT add an exponential term (to avoid mismatch). Hence delays cancel properly, leaving only: \[ G_{ff}(s)=\frac{2}{3}\frac{(8s+1)}{(5s+1)} \]

Step 5: Final conclusion.
This matches option (A).

Final Answer: (A) \(\displaystyle \frac{2}{3}\frac{(8s+1)}{(5s+1)}\)