Question

The transfer function from D(s) to C(s) is _________. Given \(G(s) = \frac{2{3s+1}\) and \(H(s) = 3\)}

• A. \(2 / (3s+7)\)
• B. \(2 / (3s+1)\)
• C. \(6 / (3s+7)\)
• D. \(3 / (3s+1)\)

Hint:

For any standard closed-loop block diagram, the transfer function from any input to the output can be found using this simple shorthand rule: \[ \text{Transfer Function} = \frac{\text{Forward Path Gain between that input and output}}{1 + \text{Loop Gain}} \] - Here, the forward path from $D(s)$ to $C(s)$ passes only through $G(s)$, so the forward gain is $G(s)$. - The total loop path passes through $G(s)$ and $H(s)$, so the loop gain is $G(s)H(s)$. This immediately gives $\frac{G(s)}{1+G(s)H(s)}$.

Answer 1

The Correct Option is A

Concept: In control systems engineering, the transfer function between an input signal and an output signal is determined by analyzing the block diagram layout. When evaluating the system's response to an internal disturbance or alternative input $D(s)$, we find the transfer function by setting all other external inputs (like the primary reference input $R(s)$) to zero.

Step 1: Analyzing the Block Diagram Structure
Let us trace the signals through the provided feedback loop diagram:
• The system output is $C(s)$.
• This output passes through the feedback block $H(s) = 3$ and enters the primary summing junction with a negative sign.
• The block in the forward path between the reference input and the second summing junction has a gain of $1$.
• The disturbance signal $D(s)$ enters the system at the second summing junction with a positive sign.
• The output of this second summing junction passes through the main plant block $G(s) = \frac{2}{3s+1}$ to produce the final output $C(s)$.

Step 2: Setting the Primary Input $R(s)$ to Zero
To isolate the transfer function from $D(s)$ to $C(s)$, set the reference input $R(s) = 0$. Looking at the loop:
• The feedback signal leaving block $H(s)$ is $C(s) \cdot H(s)$.
• Since $R(s) = 0$, the output of the first summing junction is simply $-C(s)H(s)$.
• This signal passes through the block with gain $1$, so it arrives at the second summing junction unchanged as $-C(s)H(s)$.
• At the second summing junction, this feedback signal is added to the disturbance input $D(s)$, giving an output of: $D(s) - C(s)H(s)$.

Step 3: Setting up the Loop Equation
This combined signal now passes through the plant block $G(s)$ to produce the output $C(s)$: \[ C(s) = [D(s) - C(s)H(s)] \cdot G(s) \] Expand the terms: \[ C(s) = D(s)G(s) - C(s)G(s)H(s) \] Move all terms containing $C(s)$ to the left side of the equation: \[ C(s) + C(s)G(s)H(s) = D(s)G(s) \] Factor out $C(s)$: \[ C(s)[1 + G(s)H(s)] = D(s)G(s) \] Isolate the transfer function $\frac{C(s)}{D(s)}$: \[ \frac{C(s)}{D(s)} = \frac{G(s)}{1 + G(s)H(s)} \]

Step 4: Substituting the Given Expressions
We are given $G(s) = \frac{2}{3s+1}$ and $H(s) = 3$. Substitute these values into the transfer function: \[ \frac{C(s)}{D(s)} = \frac{\frac{2}{3s+1}}{1 + \left(\frac{2}{3s+1}\right) \cdot 3} \] Simplify the expression in the denominator: \[ \frac{C(s)}{D(s)} = \frac{\frac{2}{3s+1}}{1 + \frac{6}{3s+1}} = \frac{\frac{2}{3s+1}}{\frac{(3s+1) + 6}{3s+1}} \] Cancel out the common $(3s+1)$ terms in the denominators: \[ \frac{C(s)}{D(s)} = \frac{2}{(3s+1) + 6} = \frac{2}{3s+7} \] The calculated transfer function is exactly $\frac{2}{3s+7}$.