Question

Match the unit-step responses(1),(2) and (3) with transfer functions P(s), Q(s) and R(s), given below.

$P\left(s\right)=-\frac{1}{(s+1)};Q\left(s\right)=\frac{2(s-1)}{(s+10)(s+2)};R\left(s\right)=\frac{1}{(s+1{)}^2}$

• A. $P\left(s\right)-\left(3\right),Q\left(s\right)-\left(2\right),R\left(s\right)-\left(1\right)$
• B. $P\left(s\right)-\left(1\right),Q\left(s\right)-\left(2\right),R\left(s\right)-\left(3\right)$
• C. $P\left(s\right)-\left(2\right),Q\left(s\right)-\left(1\right),R\left(s\right)-\left(3\right)$
• D. $P\left(s\right)-\left(1\right),Q\left(s\right)-\left(3\right),R\left(s\right)-\left(2\right)$

Answer 1

The correct option is B $P\left(s\right)-\left(1\right),Q\left(s\right)-\left(2\right),R\left(s\right)-\left(3\right)$

$P\left(s\right)=\frac{-1}{s+1}$
Response,
$C\left(s\right)=P\left(s\right).u\left(s\right)=\frac{-1}{s+1}.\frac{1}{s}$

$\therefore c\left(\infty \right)=L{t}_{s\to 0}s.c\left(s\right)$

$=L{t}_{s\to 0}s.[\frac{-1}{s+1}.\frac{1}{s}]=-1$

$\therefore P\left(s\right)\to 1Match$

$Q\left(s\right)=\frac{2(s-1)}{(s+10)(s+2)}$

$C\left(s\right)=Q\left(s\right).u\left(s\right)$

$=\frac{2(s-1)}{(s+10)(s+2)}.\frac{1}{s}$

$c\left(\infty \right)=L{t}_{s\to 0}sC\left(s\right)$

$=L{t}_{s\to 0}s[\frac{2(s-1)}{(s+10)(s+2)}.\frac{1}{s}]=\frac{-1}{2}$

$Q\left(s\right)\to 2Match$

$R\left(s\right)=\frac{1}{(s+1{)}^2};C\left(s\right)=\frac{1}{(s+1{)}^2}.\frac{1}{s}$

$c\left(\infty \right)=L{t}_{s\to 0}C\left(s\right)=L{t}_{s\to 0}\frac{1}{(s+1{)}^2}.\frac{1}{s}=1$

$R\left(s\right)\Rightarrow 3Match$

$P\left(s\right)\to 1,Q\left(s\right)\to 2,R\left(s\right)\to 3$