Question:medium
For the control system shown in the Figure, the transfer function of a plant,
\[
G(s)=\frac{1}{(s+1)(s+2)}
\]
is connected in cascade with a compensator
\[
C(s)=K(s+\alpha),
\]
where \(K\) and \(\alpha\) are positive real valued constants. The compensator and plant are placed in the forward path of a unity negative feedback system with input \(R(s)\) and output \(Y(s)\).
Which of the following pairs \((K,\alpha)\) represent the correct values for the closed loop system to have poles at \(\left(-3\pm j\sqrt{5}\right)\)?
For the control system shown in the Figure, the transfer function of a plant,
\[ G(s)=\frac{1}{(s+1)(s+2)} \]
is connected in cascade with a compensator
\[ C(s)=K(s+\alpha), \]
where \(K\) and \(\alpha\) are positive real valued constants. The compensator and plant are placed in the forward path of a unity negative feedback system with input \(R(s)\) and output \(Y(s)\).
Which of the following pairs \((K,\alpha)\) represent the correct values for the closed loop system to have poles at \(\left(-3\pm j\sqrt{5}\right)\)?
\[ G(s)=\frac{1}{(s+1)(s+2)} \]
is connected in cascade with a compensator
\[ C(s)=K(s+\alpha), \]
where \(K\) and \(\alpha\) are positive real valued constants. The compensator and plant are placed in the forward path of a unity negative feedback system with input \(R(s)\) and output \(Y(s)\).
Which of the following pairs \((K,\alpha)\) represent the correct values for the closed loop system to have poles at \(\left(-3\pm j\sqrt{5}\right)\)?
Show Hint
Form the characteristic equation (s+1)(s+2)+K(s+alpha)=0 and match its sum and product of roots to the given complex pole pair.
Updated On: Jul 20, 2026
- \(2, 3\)
- \(3, 4\)
- \(2, 4\)
- \(3, 3\)
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Get the closed loop denominator.
With plant $G(s)=\dfrac{1}{(s+1)(s+2)}$ and compensator $C(s)=K(s+\alpha)$ in a unity feedback loop, the characteristic polynomial is
\[ (s+1)(s+2)+K(s+\alpha)=0 \]
Step 2: Build the quadratic directly from the desired poles.
If the poles must be $-3\pm j\sqrt5$, the quadratic with these roots is
\[ (s-(-3+j\sqrt5))(s-(-3-j\sqrt5))=(s+3)^2+5=s^2+6s+9+5=s^2+6s+14 \]
Step 3: Expand the actual characteristic equation.
\[ (s+1)(s+2)+K(s+\alpha)=s^2+(3+K)s+(2+K\alpha) \]
Step 4: Match coefficient by coefficient.
Comparing with $s^2+6s+14$:
\[ 3+K=6\ \Rightarrow\ K=3 \]
\[ 2+K\alpha=14\ \Rightarrow\ K\alpha=12\ \Rightarrow\ \alpha=\frac{12}{3}=4 \]
Step 5: Conclude.
\[ \boxed{(K,\alpha)=(3,4)} \]
With plant $G(s)=\dfrac{1}{(s+1)(s+2)}$ and compensator $C(s)=K(s+\alpha)$ in a unity feedback loop, the characteristic polynomial is
\[ (s+1)(s+2)+K(s+\alpha)=0 \]
Step 2: Build the quadratic directly from the desired poles.
If the poles must be $-3\pm j\sqrt5$, the quadratic with these roots is
\[ (s-(-3+j\sqrt5))(s-(-3-j\sqrt5))=(s+3)^2+5=s^2+6s+9+5=s^2+6s+14 \]
Step 3: Expand the actual characteristic equation.
\[ (s+1)(s+2)+K(s+\alpha)=s^2+(3+K)s+(2+K\alpha) \]
Step 4: Match coefficient by coefficient.
Comparing with $s^2+6s+14$:
\[ 3+K=6\ \Rightarrow\ K=3 \]
\[ 2+K\alpha=14\ \Rightarrow\ K\alpha=12\ \Rightarrow\ \alpha=\frac{12}{3}=4 \]
Step 5: Conclude.
\[ \boxed{(K,\alpha)=(3,4)} \]
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