A control system is shown in the Figure. Which option represents the correct transfer function of the system?
Show Hint
Check exactly where the second block's input line is tapped from. It shares its input with the first block, so its output equals \(C(s)\), which makes the inner feedback cancel to zero.
Step 1: Label the signals in the diagram. Let \(E(s)\) be the output of the first summing junction, so \(E(s)\) is also the signal that enters the first block \(1/(s+4)\). Reading the diagram, the branch feeding the second block \(1/(s+4)\) is tapped from this very same line, so the second block's input is also \(E(s)\), not \(C(s)\).
Step 2: Write the two block outputs. \[ C(s)=E(s)\cdot\frac{1}{s+4} \] \[ \text{Second block output}=E(s)\cdot\frac{1}{s+4} \] Since both blocks are identical and both receive exactly the same input \(E(s)\), their outputs are identical: \[ \text{Second block output}=C(s) \]
Step 3: Evaluate the second summing junction. The second summing junction adds \(C(s)\), coming off the main output line on its "+" input, and subtracts the second block's output on its "-" input. Since these two signals are equal, \[ F(s)=C(s)-C(s)=0 \] Here \(F(s)\) is the signal this second summing junction sends back to the first summing junction's negative input.
Step 4: Apply this to the first summing junction. \[ E(s)=R(s)-F(s)=R(s)-0=R(s) \] The feedback contributes nothing, for every value of \(s\), so the error signal simply equals the reference input.
Step 5: Find the overall transfer function. \[ C(s)=E(s)\cdot\frac{1}{s+4}=\frac{R(s)}{s+4} \] \[ \frac{C(s)}{R(s)}=\frac{1}{s+4} \]
Step 6: Explain why the other options are wrong. Option (A), \(1/(s+4)^2\), and option (D), \(1/(s^2+8s+17)\), would appear if the second block were wrongly assumed to be fed by \(C(s)\) instead of \(E(s)\), leading to a non-zero inner loop. Option (C), \(2/(s+4)\), would follow from wrongly adding the two block outputs instead of subtracting them at the second summing junction. Careful tracing of the actual tap point shows the inner loop cancels exactly, leaving the simple cascade result.
