Online Electronics and Communication Engineering Test - Previous Exam Papers Test 5
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Emax - Emin = 0.5 x 1 kV = 0.5 kV
2Emax = 1.5 kVi, Emax = 0.75 kV
Emin = 0.25 kV.
Since value lies between - 1 and + 1. Therefore no real or complex solution exists.



.The damping coefficient for the given circuit is __________ .
Damping coefficient is given by,

Substituting R = 1 Ω, L = 1 H
∴
.

The given signal can be expressed as multiplication of x1(t) and x2(t) as shown below.
where A = 2, T/2 = 0.25 => T = 0.5
∴ x(t) = x1(t) x x2(t)
=> X(f) = X1(f) * X2(f)
Now X1(f) =
[δ(f - f0) + δ(f + f0)]
=
[sin c [T(f - f0)] + sin c[T(f + f0)]]
Now, A = 2, T = 0.5 and f0 =
= 1
=> X(f) = 0.5[sin c (0.5(f - 1)) + sin c (0.5(f + 1))].

A → 0100 0101
B → 0100 0101
Carry flag → 0
RAR will A → 0010 0010
XRA B 0110 0111
67.

The given circuit can be compared with Wheatstone's Bridge
∴Req = (10 + 10 + 10) || (10 + 10 + 10) = 30 || 30 = 15 Ω
.


Using Nodal analysis
...(i)
...(ii)
∴ V2 = 2Vx
∴ V2 = 2V ...(iii)



...(iv)
Put V2 = 2V


Put in equation (iii), Vs = 3V1 - V2
Vs = 3 x (+5V) - 2V
Vs = 15V - 2V = 13V.
has an output
for the input signal
. Then, the system parameter 'p' is|y(t)| = |M(ω)||x(t)|.

ω2 + p2 = ω2p2, 4 + p2 = 4p2
3p2 = 4,
.

P1 = 5 x 2 x 1 = 10
L1 = - 4
.
at x = p is given by


.

where region A corresponds to
Assertion (A): The expression E = - ΔV, where E is the electric field and V is the potential is not valid for time varying fields.
Reason (R): The curl of a gradient is identically zero.
- Addressing modes, CPU
- Instruction set, data formats
- Secondary memory, operating system
The klystron and travelling wave tube differ from each other,
- In TWT the microwave circuit is non- resonant.
- In klystron the microwave circuit is resonant.
- TWT uses attenuator.
- The wave is TWT is a non-propagating wave.
where F(s) is the Laplace transform of f(t). What is the steady-state value of f(t)?
What current distribution leads to this field?
[Hint: The algebra is trivial in cylindrical coordinates.]








