Question

For the block diagram shown in the figure, the transfer function \(\dfrac{Y(s)}{R(s)}\) is: \begin{center} \includegraphics[width=0.5\textwidth]{04.jpeg} \end{center}

• A. \(\dfrac{2s+3}{s+1}\)
• B. \(\dfrac{3s+2}{s-1}\)
• C. \(\dfrac{s+1}{3s+2}\)
• D. \(\dfrac{3s+2}{s+1}\)

Hint:

Always define intermediate signals at each summing node when solving block diagrams. Substituting step-by-step avoids confusion with feedback paths.

Answer 1

The Correct Option is D

Step 1: Define signals.
Let input be \(R(s)\), output be \(Y(s)\). From diagram: - First summing node input = \(2R(s)\) (top branch), plus feedback from output of integrator. - Output of first node enters integrator \(\frac{1}{s}\).

Step 2: First summing junction.
Let its output be \(X(s)\). \[ X(s) = 2R(s) + Y(s) \]

Step 3: Output of integrator.
\[ \frac{1}{s} X(s) \] This goes into second summing junction along with \(3R(s)\).

Step 4: Second summing junction.
\[ Y(s) = \frac{1}{s}X(s) + 3R(s) \]

Step 5: Substitute \(X(s)\).
\[ Y(s) = \frac{1}{s}[2R(s) + Y(s)] + 3R(s) \]

Step 6: Simplify.
Multiply through by \(s\): \[ sY(s) = 2R(s) + Y(s) + 3sR(s) \] \[ sY(s) - Y(s) = (2 + 3s)R(s) \] \[ (s-1)Y(s) = (3s+2)R(s) \]

Step 7: Transfer function.
\[ \frac{Y(s)}{R(s)} = \frac{3s+2}{s-1} \] Wait! Let's carefully check — did we misinterpret the feedback path? From the diagram: The block "1" feeds forward from \(Y(s)\) back into the first summing junction. So indeed: \[ X(s) = 2R(s) + (1)\cdot Y(s) \] Yes, that's consistent. After simplifying, we got \(\dfrac{3s+2}{s-1}\), which corresponds to option (B).

Final Answer: \[ \boxed{\frac{3s+2}{s-1}} \]