Electronics

Question 1
If the channel is band limited to 6 kHz and signal to noise ratio is 16, what would be the capacity of channel?
A
16.15kbps
B
23.24 kbps
C
40.12 kbps
D
24.74 kbps
Question 1 Explanation: 
→ Shannon Capacity (Noisy Channel) In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the maximum data rate capacity.
C=B log​ 2​ (1+S/N).
where S and N are the signal and noise power, respectively, at the output of the channel.
→ This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel. =6 * log​ 2​ (1+16)
=6 * 4.08746284125034
=24.74 kbps
Question 2
At 100% modulation, the power in each sideband is__ of that of arrier.
A
50%
B
40%
C
60%
D
25%
Question 2 Explanation: 
→ Power in sidebands may be calculated as modulation index.
→ Each sideband is 25%. Total modulation index=1.
Question 3
The capacity relationship is given by
A
C=Wlog​ 2​ (1+S/N)
B
C=2Wlog​ 2​ (1+S/N)
C
C=Wlog​ 2​ (1-S/N)
D
C=Wlog​ 10​ (1+S/N)
Question 3 Explanation: 
→ Shannon Capacity (Noisy Channel) In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the maximum data rate capacity.
C=W log​ 2​ (1+S/N).
where S and N are the signal and noise power, respectively, at the output of the channel.
→ This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel.
Question 4
Twelve 1 ​ Ω ​ resistances are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is
A
5/6 ​ Ω
B
1/6 ​ Ω
C
6/5 ​ Ω
D
3/2 ​ Ω
Question 4 Explanation: 
Question 5
The final value theorem is used to find the
A
Steady state value of the system output
B
Initial value of the system output
C
Transient behaviour of the system output
D
None of these
Question 5 Explanation: 
The final value theorem is used to find the steady state value of the system output
Question 6
For the discrete signal x[n]=a​ n​ u[n], a>0 the z-transform is
A
(z+a)/z
B
(z-a)/z
C
z/(z-a)
D
z/(z+a)
Question 6 Explanation: 
Let x[n] be causal signal given by x[n] = a​ n​ u[n]
→ The Z-Transform of x[n] is given by

The ROC is defined by | az −1 | < 1 or | z| > | a| .
Region of Convergence(RoC)
Region of Convergence for a discrete time signal x[n] is dened as a continuous region in z plane where the Z-Transform converges. In order to determine RoC, it is convenient to represent the Z-Transform as a is rational X(z)=P(z)/Q(z)
1. The roots of the equation P(z) = 0 correspond to the ’zeros’ of X(z)
2. The roots of the equation Q(z) = 0 correspond to the ’poles’ of X(z)
3. The RoC of the Z-transform depends on the convergence of the polynomials P(z) and Q(z).
Question 7
​ For a periodic signal v(t)=30 sin 100t+10cos 300t +6 sin (500t + ​ π ​ /4), the fundamental frequency in rad/s
A
100
B
300
C
500
D
None of the above
Question 7 Explanation: 
Given, the signal
V (t) = 3 0 sin 100t + 10 cos 300t + 6 sin (500t + π/4)
So, we have
ω 1 = 100 rads
ω 2 = 300 rads
ω 3 = 500 rads
∴ The respective time periods are
T 1 = 2π/ω1 = 2π/100 sec
T 2 = 2π/ω2 = 2π/300 sec
T 3 = 2π/ω3 = 2π/500 sec
So, the fundamental time period of the signal is
LCM (T 1 , T 2 , T 3 ) =LCM (2π,2π,2π)/HCF (100,300,500)
as T 0 =2π/100
∴ The fundamental frequency, ω0 =2π/T 0 = 100 rad/s
Question 8
​ If the number of bits per sample in a PCM system is increased from a n to n+1, the improvement in signal to quantization noise ratio will be
A
3dB
B
6db
C
2n dB
D
n dB
Question 8 Explanation: 
(S/N q)dB
= (1.76 + 6 n) dB
(SQNR) 1 = 1 .76 + 6 n
(SQNR) 2 = 1 .76 + 6 n(n + 1 ) = 1 .76 + 6 n + 6
(SQNR) 2 − (SQNR) 1 = 1 .76 + 6 n + 6 − 1 .76 − 6 n = 6 dB
So for every one bit increase in bits per sample will result is 6 dB improvement in signal to quantization ratio.
Question 9
A carrier A​ c​ Cos(w​ C​ )t is frequency modulated by a signal E​ m​ Cos(w​ m​ )t. The modulation index is m​ f​ . the expression for the resulting FM signal is
A
A​ c,​ Cos[w​ c​ t+ m​ f​ Sin(w​ m​ )t]
B
A​ c​ Cos[w​ c​ t+ m​ f​ Cos(w​ m​ )t]
C
A​ c​ Cos[w​ c​ t+ ​ π ​ m​ f​ Sin w​ m​ t]
D
A​ c​ Cos[w​ c​ t+ 2 ​ π ​ m​ f​ E​ m​ Cos(w​ m​ )t/w​ m​ ]
Question 9 Explanation: 
Question 10
The open loop transfer function of a feedback control system is G(s).H(s)=1/(S+1)​ 3​ . The gain margin of the system is
A
2
B
4
C
8
D
16
Question 10 Explanation: 
Given data,
The open loop transfer function of a feedback control system is G(s).H(s)=1/(S+1)​ 3
Gain margin=?
→ Here, the formula is -3tan​ -1​ (W​ pc​ ) = -180​ 0
tan​ -1​ (W​ pc​ ) = -180​ 0​ /3
= -60​ 0
W​ pc​ =√3 rad/sec
G(s).H(s)=1/(S+1)​ 3
|G(s).H(s)|=X= 1/(1+W​ pc2​ )
=1⁄8
G(s).H(s)=1/X
= 8
Question 11
​ An ideal op-amp is an ideal
A
Voltage controlled current source
B
voltage controlled voltage source
C
current controlled current source
D
current controlled voltage source
Question 11 Explanation: 
An ideal op-amp lead us to an op-amp with input resistance is infinite, so no current flows into either input terminal (the “current rule”) and that the differential input offset voltage is zero (the “voltage rule”).
Ideal operational amplification will have
➝ High input impedance ( R in = ∞ ) and low output impedance ( R out = 0 )

➝ R in = ∞ , so zero input current
➝ Infinite voltage gain
➝ So, it is a voltage controlled voltage source device.
Question 12
A solution for the differential equation x'(t) + 2x(t) =​ δ ​ (t) with initial condition x(~0)=0
A
e​ -2t​ u(t)
B
e​ 2t​ u(t)
C
e​ -t​ u(t)
D
e​ t​ u(t)
Question 13
In MOSFET fabrication, the channel length is delayed during the process of
A
Isolation oxide growth
B
Channel stop implantation
C
Ply-Silicon gate patterning
D
Lithography step leading to the contact pad
Question 13 Explanation: 
In MOSFET fabrication channel length is defined during Poly-Silicon gate patterning process
Question 14
The open loop transfer function of a feedback control system is G(s).H(s)=1/(S+1)​ 3​ . The gain margin of the system is
A
2
B
4
C
8
D
16
Question 14 Explanation: 
The open loop transfer function of a feedback control system is G(s).H(s)=1/(S+1)​ 3
Gain margin=?
→ Here, the formula is -3tan​ -1​ (W​ pc​ ) = -180​ 0 tan​ -1​ (W​ pc​ ) = -180​ 0​ /3
= -60​ 0
W​ pc​ =√3 rad/sec
G(s).H(s)=1/(S+1)​ 3
|G(s).H(s)|=X= 1/(1+W​ pc2​ )
=1⁄8
G(s).H(s)=1/X
= 8
Question 15
​ The final value theorem is used to find the
A
Steady state value of the system output
B
Initial value of the system output
C
Transient behaviour of the system output
D
None of these
Question 15 Explanation: 
The final value theorem is used to find the steady state value of the system output
Question 16
​ For the discrete signal x[n]=a​ n​ u[n], a>0 the z-transform is
A
(z+a)/z
B
(z-z)/z
C
z/(z-a)
D
z/(z+a)
Question 16 Explanation: 
Let x[n] be causal signal given by x[n] = a​ n​ u[n]
→ The Z-Transform of x[n] is given by

Region of Convergence(RoC)
Region of Convergence for a discrete time signal x[n] is dened as a continuous region in z plane where the Z-Transform converges. In order to determine RoC, it is convenient to represent the Z-Transform as a is rational X(z)=P(z)/Q(z)
1. The roots of the equation P(z) = 0 correspond to the ’zeros’ of X(z)
2. The roots of the equation Q(z) = 0 correspond to the ’poles’ of X(z)
3. The RoC of the Z-transform depends on the convergence of the polynomials P(z) and Q(z).
Question 17
​ For a periodic signal v(t)=30 sin 100t+10cos 300t +6 sin (500t + ​ π ​ /4), the fundamental frequency in rad/s
A
100
B
300
C
500
D
None of the above
Question 17 Explanation: 
Given, the signal
V (t) = 3 0 sin 100t + 10 cos 300t + 6 sin (500t + π/4)
So, we have
ω 1 = 100 rads
ω 2 = 300 rads
ω 3 = 500 rads
∴ The respective time periods are
T 1 = 2π/ω1 = 2π/100 sec
T 2 = 2π/ω2 = 2π/300 sec
T 3 = 2π/ω3 = 2π/500 sec
So, the fundamental time period of the signal is
LCM (T 1 , T 2 , T 3 ) =LCM (2π,2π,2π)/HCF (100,300,500)
as T 0 =2π/100
∴ The fundamental frequency, ω0 =2π/T 0 = 100 rad/s
Question 18
​ If the number of bits per sample in a PCM system is increased from a n to n+1, the improvement in signal to quantization noise ratio will be
A
3 dB
B
6 db
C
2n dB
D
n dB
Question 18 Explanation: 
(S/N q)dB
= (1.76 + 6 n) dB
(SQNR) 1 = 1 .76 + 6 n
(SQNR) 2 = 1 .76 + 6 n(n + 1 ) = 1 .76 + 6 n + 6
(SQNR) 2 − (SQNR) 1 = 1 .76 + 6 n + 6 − 1 .76 − 6 n = 6 dB
So for every one bit increase in bits per sample will result is 6 dB improvement in signal to quantization ratio.
Question 19
At 100% modulation, the power in each sideband is__ of that of carrier.
A
50%
B
40%
C
60%
D
25%
Question 19 Explanation: 
Power in sidebands may be calculated as modulation index.
→ Each sideband is 25%. Total modulation index=1.
Question 20
Twelve 1 ​ Ω ​ resistances are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is
A
5/6 ​ Ω
B
1/6 ​ Ω
C
6/5 ​ Ω
D
3/2 ​ Ω
Question 20 Explanation: 
Apply Ohms law, the resistance is equal to the voltage divided by the current. The resulting equivalent resistance is 5/6 Ω.
R=E/I
R=(2*(1/2)V)/3A
= ⅚
Note: This is the cube structure consisting of 12 resistors electrically connected between the 8 vertices. Each resistor is 1 Ω, but any value can be used so long as they are all the same.
Question 21
A solution for the differential equation x'(t) + 2x(t) =​ δ ​ (t) with initial condition x(~0)=0
A
e​ -2t​ u(t)
B
e​ 2t​ u(t)
C
e​ -t​ u(t)
D
e​ t​ u(t)
Question 22
The noise due to random behavior of charge carriers is:
A
Partition noise
B
Industrial noise
C
Shot noise
D
Flicker noise
Question 22 Explanation: 
Shot noise is caused due to random behaviour of charge carriers. Shot noise is generated due to random emission of electrons from cathodes in electron tubes. In semiconductor devices, shot noise is generated due to random generation and recombination of electron-hole pairs
Question 23
The addition of p-type impurity to intrinsic material creates allowable energy levels
A
slightly above conduction band
B
slightly below conduction band
C
slightly above valence band
D
slightly below valence band
Question 24
Measurement of Hall coefficient enables the determination of
A
temperature coefficient
B
recovery time of stored carrier
C
Fermi level
D
type of conductivity and concentration of charge carriers
Question 25
For input voltage of V​ m​ sin net, peak inverse voltage rating needed for diode in a half wave rectifier using capacitor filter is
A
V​ m
B
2V​ m
C
V m/√2
D
V m
E
None of Above
Question 26
A current signal is given by i (t) = 20e −50t . The initial and final value of current are
A
0 and 20
B
20 and 0
C
0 and 10
D
20 and 10
Question 27
Common emitter configuration is preferred when main criterion is
A
Voltage gain
B
Current gain
C
Both voltage gain and current gain
D
Input impedance
Question 28
Which of the following can be used for accurate and stable time base circuit in a digital frequency meter ?
A
Quartz
B
Carbon
C
Aluminium
D
Copper
Question 29
An electric machine will have high efficiency when
A
Input / output is low
B
Reactive power is more
C
KWh consumption is low
D
Losses are low
Question 30
Voltage across capacitor in RLC series circuit is maximum
A
At resonance
B
Just before resonance
C
Just after resonance
D
Much after resonance
Question 31
Which of the following is the most economical method of starting a single phase motor
A
Resistant start method
B
Inductance start method
C
Capacitance start method
D
Split phase method
Question 32
A 1 mA ammeter has a resistance of 100 ᘯ . It is to be converted to a 1 A ammeter. The value of shunt resistance is
A
0.001ᘯ
B
0.1001ᘯ
C
100000ᘯ
D
100ᘯ
Question 33
When two resistance are connected in parallel, they have
A
Same resistances value
B
Same voltage across them
C
Same current passing through them
D
Different resistance value
Question 34
Power consumed by purely inductive circuit is
A
VI
B
VIcosx
C
VI sinx
D
Zero
Question 35
The turn ratio of a transformer is 20:1. If a load of 40 ohm is connected across the secondary. What will be the effective resistance seen looking into the primary ?
A
800 ohm
B
4 k ohm
C
8 k ohm
D
16 k ohm
Question 36
The electric flux and field intensity inside a conducting sphere is
A
Minimum
B
Maximum
C
Uniform
D
Zero
Question 37
The effect of doping intrinsic semiconductor is to
A
Move the Fermi level away from the centre of forbidden band
B
Move the Fermi level towards the centre of forbidden band
C
Change the crystal structure of the semiconductor
D
To keep the Fermi level at the middle of the forbidden band
Question 38
A 50KW, 3-phase, 400V, 60H induction motor runs at 1140 rpm. The % slip will be
A
1%
B
3%
C
5%
D
24%
Question 39
Which of the following device can be used for controlling the speed of a dc motor ?
A
Thyratron
B
Thyristor
C
Transistor
D
Thermistor
Question 40
The impulse response of an RL circuit is a
A
Rising exponential function
B
Decaying exponential function
C
Step function
D
Parabolic function
Question 41
When the speed at which a conductor is moved through a magnetic field is increased, the induced voltage
A
Increases
B
Decreases
C
Remains constant
D
Reaches zero
Question 42
AB is a vertical pole. The end A is on the level ground, C is the middle point of AB and P is a point on the level ground. The portion CB subtends an angle β at P. If AP = nAB, then
A
tan β = (n+1)/(2n2+1)
B
tan β = n/(2n2+1)
C
t an β = (n-1)/(2n2+1)
D
tan β = n / (n2+1)
Question 43
There exist positive integers A, B and C, with no common factor greater than 1, such that, Alog 200 (5) + B log 200 (2) = C then what is the value of A + B + C ?
A
6
B
7
C
8
D
9
There are 43 questions to complete.

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