Electronics Engineering - Online Test
Q1. The circuit shown below is driven by a sinusoidal input Vi = Vpcos(t/RC). The
steady state output vo
Answer : Option A
Explaination / Solution:
parallel combination of R and C equivalent impedance is
Transfer function can be written as
Q2. When the output Y in the circuit below is ‘1’, it implies that data has
Answer : Option A
Explaination / Solution:
No Explaination.
Q3. Match the inferences X, Y, and Z, about a system, to the corresponding properties of the elements of first column in Routh‟s Table of the system characteristic equation
X: The system is stable … P: … when all elements are positive
Y: The system is unstable … Q: … when any one element is zero
Z: The test breaks down … R: … when there is a change in sign of coefficients
Answer : Option D
Explaination / Solution:
No Explaination.
Q4. Consider the feedback system shown in the figure. The Nyquist plot of G (s) is
also shown. Which one of the following conclusions is correct ?
Answer : Option D
Explaination / Solution:
Given the feedback system and the Nyquist plot of G (s)is
For the given system, we have the open loop transfer function as
G (s) = KG (s)
Considering the open loop system G (s) is stable, we have no open loop poles in
right half plane
P = 0
From Nyquist theorem, we know that
N = P - Z
Where N is the number of encirclements of (-1 + j0), P is number of open loop
poles in right half plane, Z is number of closed loop poles in right half plane. For
stability, we must have
Z = 0
N = 0, if closed loop system is stable
N ≠ 0, if closed loop system is unstable
observing the Nyquist plot, we conclude that the plot of KG(s) encircles (-1 + j0)
if K> 1
Hence, N ≠ 0 for sufficient large and positive value of K . Thus, the closedsystem
is unstable for sufficiently large and positive K .
Q5. The transfer function 
Answer : Option D
Explaination / Solution:
For the given capacitance, 
in the circuit, we have the reactance.
in the circuit, we have the reactance.
Q6. The logic function implemented by the circuit below is (ground implies logic 0)
Answer : Option D
Explaination / Solution:
Q7. A first-order low-pass filter of time constant T is excited with different input signals (with zero
initial conditions up to t = 0). Match the excitation signals X, Y, Z with the corresponding time
responses for t ≥ 0:
X: Impulse P: 1 – e-t/T
Y: Unit step Q: t – T(1 – e(-t/T)
Z :Ramp R: e-t/T
Answer : Option C
Explaination / Solution:
In general the first order, L.P.F filter transfer function is 
because G(0) = k and G(∞) = ∞ if we take this transfer function as reference and give different input such as
s(t).r(t).u(t)
because G(0) = k and G(∞) = ∞ if we take this transfer function as reference and give different input such as
s(t).r(t).u(t)
Q8. The feedback configuration and the pole-zero locations of 
are shown below. The root locus for negative values of k , i.e. for -3 < k < 0, has breakaway/break-in points and angle of departure at pole P (with respect to the
positive real axis) equal to
are shown below. The root locus for negative values of k , i.e. for -3 < k < 0, has breakaway/break-in points and angle of departure at pole P (with respect to thepositive real axis) equal to
Answer : Option B
Explaination / Solution:
The characteristic equation is
Q9. A source 
has an internal impedance of 
If a purely resistive load connected to this source has to extract the maximum power out of the source, its value inΩ should be
has an internal impedance of If a purely resistive load connected to this source has to extract the maximum power out of the source, its value inΩ should be
Answer : Option C
Explaination / Solution:
For the purely resistive load, maximum average power is transferred when
where
is the equivalent thevinin (input) impedance of the circuit. Hence,
we obtain
where
is the equivalent thevinin (input) impedance of the circuit. Hence,we obtain
Q10. The output of a 3-stage Johnson (twisted ring) counter is fed to a digital-to
analog (D/A) converter as shown in the figure below. Assume all the states
of the counter to be unset initially. The waveform which represents the D/A
converter output v0 is
Answer : Option A
Explaination / Solution:
All the states of the counter are initially unset.
State Initially are shown below in table:
Q2Q1Q0
0 0 0 0
1 0 0 4
1 1 0 6
1 1 1 7
0 1 1 3
0 0 1 1
0 0 0 0
