EC GATE 2010 - Online Test
Q1. A continuous time LTI system is described by
Assuming zero initial conditions, the response y(t) of the above system for the input 
is given by
is given by
Answer : Option B
Explaination / Solution:
System is described as
Taking laplace transform on both side of given equation
Transfer function of the system
By Partial fraction
Taking inverse laplace transform
Q2.
A unity negative feedback closed loop system has a plant with the transfer function 
and a controller GC (s) in the
and a controller GC (s) in thefeed forward path. For a unit set input, the transfer function of the controller that gives minimum steady state error is
Answer : Option D
Explaination / Solution:
Steady state error is given as
ess will be minimum if 
is maximum In option (D)
is maximum In option (D)
Q3. In the circuit shown, the power supplied by the voltage source is
Answer : Option A
Explaination / Solution:
Applying nodal analysis
Current from voltage source is
Since current through voltage source is zero, therefore power delivered is zero.
Q4. The signal flow graph of a system is shown below:
The state variable representation of the system can be
Answer : Option D
Explaination / Solution:
Q5. Assign output of each integrator by a state variable
The transfer function of the system is
Answer : Option C
Explaination / Solution:
By masson’s gain formula
Q6. The eigen values of a skew-symmetric matrix are
Answer : Option C
Explaination / Solution:
Eigen value of a Skew-symmetric matrix are either zero or pure imaginary in conjugate pairs.
Q7. The trigonometric Fourier series for the waveform f (t) shown below contains
Answer : Option C
Explaination / Solution:
For a function x(t) trigonometric fourier series is
For an even function x(t),Bn = 0
Since given function is even function so coefficient Bn = 0, only cosine and constant
terms are present in its fourier series representation.
Constant term :
Constant term is negative.
Q8. A function n(x) satisfied the differential equation 
where L is a constant. The boundary conditions are :n(0) = K and n(∞) = 0. The solution to this equation is
where L is a constant. The boundary conditions are :n(0) = K and n(∞) = 0. The solution to this equation is
Answer : Option D
Explaination / Solution:
Given differential equation
Q9. Consider the z -transform 
The inverse z -transform x[n] is
The inverse z -transform x[n] is
Answer : Option A
Explaination / Solution:
Q10. If ey = x1/x , then y has a
Answer : Option A
Explaination / Solution:
