1
GATE ECE 2008
MCQ (Single Correct Answer)
+2
-0.6
A linear, time-invariant, causal continuous time system has a rational transfer function with simple poles at s=-2 and s=-4, and one simple zero at s=-1. A unit step u(t) is applied at the input of the system. At steady state, the output has constant value of 1. The impulse response of this system is
A
[exp(-2t) + exp(-4t)]u(t)
B
[-4exp(-2t) + 12exp(-4t) - exp(-t)]u(t)
C
[-4exp(-2t) + 12exp(-4t)]u(t)
D
[-0.5exp(-2t) + 1.5exp(-4t)]u(t)
2
GATE ECE 2008
MCQ (Single Correct Answer)
+2
-0.6
Group I lists a set of four transfer functions. Group II gives a list of possible step responses y(t). Match the step responses with the corresponding transfer functions

Group I $$$P=\frac{25}{s^2+25}\;\;\;\;\;Q=\frac{36}{s^2+20s+36}$$$ $$$R=\frac{36}{s^2+12s+36}\;\;\;\;\;S=\frac{49}{s^2+7s+49}$$$

Group II GATE ECE 2008 Control Systems - Time Response Analysis Question 11 English 1 GATE ECE 2008 Control Systems - Time Response Analysis Question 11 English 2 GATE ECE 2008 Control Systems - Time Response Analysis Question 11 English 3 GATE ECE 2008 Control Systems - Time Response Analysis Question 11 English 4

A
$$\begin{array}{l}P\;\;\;\;Q\;\;\;\;R\;\;\;\;S\\3\;\;\;\;\;1\;\;\;\;\;4\;\;\;\;\;2\end{array}$$
B
$$\begin{array}{l}P\;\;\;\;Q\;\;\;\;R\;\;\;\;S\\3\;\;\;\;\;2\;\;\;\;\;4\;\;\;\;\;1\end{array}$$
C
$$\begin{array}{l}P\;\;\;\;Q\;\;\;\;R\;\;\;\;S\\2\;\;\;\;\;1\;\;\;\;\;4\;\;\;\;\;3\end{array}$$
D
$$\begin{array}{l}P\;\;\;\;Q\;\;\;\;R\;\;\;\;S\\3\;\;\;\;\;4\;\;\;\;\;1\;\;\;\;\;2\end{array}$$
3
GATE ECE 2007
MCQ (Single Correct Answer)
+2
-0.6
The frequency response of a linear, time-invariant system is given by $$H\left(f\right)\;=\;\frac5{1\;+\;j10\mathrm{πf}}$$ .The step response of the system is:
A
$$5\left(1\;-\;e^{-5t}\right)u\left(t\right)$$
B
$$5\left(1\;-\;e^{-\frac15}\right)u\left(t\right)$$
C
$$\frac15\;\left(1\;-\;e^{-5t}\right)u\left(t\right)$$
D
$$\frac15\;\left(1\;-\;e^{-\frac15}\right)u\left(t\right)$$
4
GATE ECE 2007
MCQ (Single Correct Answer)
+2
-0.6
The transfer function of a plant is $$$T\left(s\right)=\frac5{\left(s+5\right)\left(s^2+s+1\right)}$$$ The second-order approximation of T (s) using dominant pole concept is:
A
$$\frac1{\left(s+5\right)\left(s+1\right)}$$
B
$$\frac5{\left(s+5\right)\left(s+1\right)}$$
C
$$\frac5{\left(s^2+s+1\right)}$$
D
$$\frac1{\left(s^2+s+1\right)}$$
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