Electronics and Communication Engineering (ECE)
The nominal quiescent collector current of a transistor is 1.2 mA. If the range of β for this transistor is 80 ≤ β ≤ 120 and if the quiescent collector current changes by +-10 percent, the range in value for rπ is- a)1.73 kΩ < rπ < 2.59 kΩ
- b)1.93 kΩ < rπ < 2.59 kΩ
- c)1.73 kΩ < rπ < 2.59 kΩ
- d)1.56 kΩ < rπ < 2.88 kΩ
Correct answer is option 'D'. Can you explain this answer?
The nominal quiescent collector current of a transistor is 1.2 mA. If the range of β for this transistor is 80 ≤ β ≤ 120 and if the quiescent collector current changes by +-10 percent, the range in value for rπ is
a)
1.73 kΩ < rπ < 2.59 kΩ
b)
1.93 kΩ < rπ < 2.59 kΩ
c)
1.73 kΩ < rπ < 2.59 kΩ
d)
1.56 kΩ < rπ < 2.88 kΩ
Tanya Satija answered • 3 hours ago
to - a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
to a)
b)
c)
d)
Pooja Yadav answered • 3 hours ago
The circuit shown in the figure contains a dependent current source between A and B terminals. The thevenin’s equivalent resistance in kW between the terminals C and D is
- a)10 kW
- b)20 kW
- c)0 kW
- d)15 kW
Correct answer is option 'B'. Can you explain this answer?
a)
10 kW
b)
20 kW
c)
0 kW
d)
15 kW
Shoarib Quershi answered • 3 hours agoThevenin’s equivalent circuit is made by connecting 1A current source at the output and short circuiting the battery.
At node A,
At node A,
then find 
rad/sec. The value of 
in the volts is.
volts 
volts 
A frequency modulator has output 
. Where 
= 100 Hz/V and m (t) is shown in the figure. According Carson’s rule the bandwidth of s(t) is
- a)700 Hz
- b)1400 Hz
- c)2800 Hz
- d)200 Hz
Correct answer is option 'B'. Can you explain this answer?
. Where = 100 Hz/V and m (t) is shown in the figure. According Carson’s rule the bandwidth of s(t) is a)
700 Hz
b)
1400 Hz
c)
2800 Hz
d)
200 Hz
Harshit Ahir answered • 3 hours ago
= 500 Hz,So Bandwidth = 2
2(500 + 200)
= 1400 Hz
In a twin-wire transmission line in air, the adjacent voltage maxima are at 12 cm and 18 cm. The operating frequency is.- a)2.5 GHz
- b)1.25 GHz
- c)5 GHz
- d)10 GHz
Correct answer is option 'A'. Can you explain this answer?
a)
2.5 GHz
b)
1.25 GHz
c)
5 GHz
d)
10 GHz
Md Shaik Junaid answered • 3 hours agoAdjacent maxima appear with a distance of 
f = 2.5 GHZ
f = 2.5 GHZ
The steady state output 
is
For an unit step input determine the time required so that response will settle in 2% of its final value. 
The value of 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The value of
a)
b)
c)
d)
Anamika Kadyan answered • 3 hours ago
The current 12 for the circuit shown below is______________ mA •
Correct answer is '7'. Can you explain this answer?
The current 12 for the circuit shown below is______________ mA •
Srinidhi Akella answered • 12 hours agoJust apply nodal analysis
let the nodal voltage be V
(V-30)/2k + V/1k = 9mA
i.e V-30+2V=18. i.e V=16volts
so I2=(30-V)/2k= 7mA
let the nodal voltage be V
(V-30)/2k + V/1k = 9mA
i.e V-30+2V=18. i.e V=16volts
so I2=(30-V)/2k= 7mA
The energy of a causal exponential pulse shown in figure is
- a)1/2α
- b)-α
- c)- 1/α
- d)2α
Correct answer is option 'A'. Can you explain this answer?
The energy of a causal exponential pulse shown in figure is
a)
1/2α
b)
-α
c)
- 1/α
d)
2α
Geet Naman answered • 14 hours agoGiven, 
∴ Normalized energy of given signal is
∴ Normalized energy of given signal is
The process of transmitting two or more information signals simultaneously over the same channel is called- a)multiplexing
- b)telemetry
- c)detection
- d)modulation
Correct answer is option 'A'. Can you explain this answer?
The process of transmitting two or more information signals simultaneously over the same channel is called
a)
multiplexing
b)
telemetry
c)
detection
d)
modulation
Priya Tanwar answered • 14 hours agoMultiplexing may be defined as the process of combining several message signals together and send them over the same communication channel, i.e. FDM, TDM and CDM.
Tanishka Rao asked • 22 hours agoDelhi averages three murder per week and their occurrences follow a poission distribution.Que:How many weeds per year (average) can the Delhi expect the number of murders per week to equal orexceed the average number per week ?- a)15
- b)20
- c)25
- d)30
Correct answer is option 'D'. Can you explain this answer?
Delhi averages three murder per week and their occurrences follow a poission distribution.
Que:How many weeds per year (average) can the Delhi expect the number of murders per week to equal orexceed the average number per week ?
a)
15
b)
20
c)
25
d)
30
Direction flag is used with- a)string instructions
- b)stack instructions
- c)arithmetic instructions
- d)branch instructions
Correct answer is option 'A'. Can you explain this answer?
Direction flag is used with
a)
string instructions
b)
stack instructions
c)
arithmetic instructions
d)
branch instructions
Saryu Bakshi answered • 14 hours agoDirection flag is used only with string instructions.
Ramya Srivalli asked • 22 hours agoIn an unbiased P-N Junction zero current implies that- a) The potential biases has disappeared
- b)No. of holes diffusing from P-side to N-side equals the no. of electrons diffusing from N-side to P-side
- c)Total current crossing Junction from P-side to N-side equals the total current crossing Junction from N-side top-side
- d)No carriers cross the Junction
Correct answer is option 'C'. Can you explain this answer?
In an unbiased P-N Junction zero current implies that
a)
The potential biases has disappeared
b)
No. of holes diffusing from P-side to N-side equals the no. of electrons diffusing from N-side to P-side
c)
Total current crossing Junction from P-side to N-side equals the total current crossing Junction from N-side top-side
d)
No carriers cross the Junction
The amount of frequency shift in FM is- a)directly proportional to amplitude of the modulating signal.
- b)inversely proportional to amplitude of the modulating signal.
- c)independent of amplitude of the modulating signal.
- d)proportional to frequency deviation.
Correct answer is option 'A'. Can you explain this answer?
The amount of frequency shift in FM is
a)
directly proportional to amplitude of the modulating signal.
b)
inversely proportional to amplitude of the modulating signal.
c)
independent of amplitude of the modulating signal.
d)
proportional to frequency deviation.
Lakshmi Priya Ghildiyal answered • 14 hours agoFor FM, the frequency changes with the change in the amplitude of modulating signal value.
Anumala Krishna Vamsi asked • yesterdayA binary symmetric channel (BSC) has a transition probability of 1/8. If the
binary symbol X is such that P(X = 0) = 9/10, then the probability of error for
an optimum receiver will be Correct answer will be uploaded soon - a)a
- b)b
- c)c
- d)d
Correct answer is option 'A'. Can you explain this answer?
A binary symmetric channel (BSC) has a transition probability of 1/8. If the
binary symbol X is such that P(X = 0) = 9/10, then the probability of error for
an optimum receiver will be
binary symbol X is such that P(X = 0) = 9/10, then the probability of error for
an optimum receiver will be
Correct answer will be uploaded soon
a)
a
b)
b
c)
c
d)
d
Answer the question with the help of the following table
Q.If in the next year, Crackle and Dairy Mill decided to join hands and increase their market share from their
combined existing share by 20%. What will be their combined production if the mark size remains the same?- a)26251
- b)28520
- c)31780
- d)Indeterminable
Correct answer is option 'A'. Can you explain this answer?
Answer the question with the help of the following table
Q.
If in the next year, Crackle and Dairy Mill decided to join hands and increase their market share from their
combined existing share by 20%. What will be their combined production if the mark size remains the same?
combined existing share by 20%. What will be their combined production if the mark size remains the same?
a)
26251
b)
28520
c)
31780
d)
Indeterminable
Diya Ahuja answered • 14 hours ago
A linear time invariant (LTI) system with the transfer function
is connected in unity feedback configuration as shown in the figure.
For the closed loop system shown, the root locus for 0 < K < ∞ intersects the imaginary axis for K = 1.5. The closed loop system is stable for- a)K > 1.5
- b)1 < K < 1.5
- c)0 < K < 1
- d)no positive value of K
Correct answer is option 'A'. Can you explain this answer?
A linear time invariant (LTI) system with the transfer function
is connected in unity feedback configuration as shown in the figure.
For the closed loop system shown, the root locus for 0 < K < ∞ intersects the imaginary axis for K = 1.5. The closed loop system is stable for
is connected in unity feedback configuration as shown in the figure.
For the closed loop system shown, the root locus for 0 < K < ∞ intersects the imaginary axis for K = 1.5. The closed loop system is stable for
a)
K > 1.5
b)
1 < K < 1.5
c)
0 < K < 1
d)
no positive value of K
Rachita Sahoo answered • 14 hours ago
If closed loop system to be stable all coefficients to positive
The state diagram of a finite state machine (FSM) designed to detect an overlapping sequence of three bits is shown in the figure. The FSM has an input ‘In’ and an output ‘Out’. The initial state of the FSM is S0.
Correct answer is '4 to 4'. Can you explain this answer?
The state diagram of a finite state machine (FSM) designed to detect an overlapping sequence of three bits is shown in the figure. The FSM has an input ‘In’ and an output ‘Out’. The initial state of the FSM is S0.
Shilpa Jaiswal answered • 14 hours agoFrom the state diagram, let us obtain the transition of states and out when IN channel.
Initial state is So, the input sea is 10101101001101
→ The ticketed mark now corresponding to output = 1.
So output will be 1 ‘4’ times.
Initial state is So, the input sea is 10101101001101
→ The ticketed mark now corresponding to output = 1.
So output will be 1 ‘4’ times.
For a (-)ve feedback second order control system in its step response, the maximum value of the output response in time domain form is given by
Q. The damping factor (damping coefficient) of this system is- a)0.84 per sec.
- b)0.45 per sec.
- c)4.2 per sec.
- d)5 per sec.
Correct answer is option 'C'. Can you explain this answer?
For a (-)ve feedback second order control system in its step response, the maximum value of the output response in time domain form is given by
Q. The damping factor (damping coefficient) of this system is
a)
0.84 per sec.
b)
0.45 per sec.
c)
4.2 per sec.
d)
5 per sec.
Sanjukta Pal answered • 14 hours agoFrom the conventional approach we have,
Give that C(t)Imax = 1.75
∴ Mp = Maximum overshoot = 1.75 - 1.00 = 0.75
A single tone signal 14Cos(40π x103t) is used to modulate a carrier Accoswct. Then determine the number of channels available in very high frequency (VHF) band for this signal, if amplitude modulation and frequency modulation scheme employs respectively {Assume sensitivity of FM modulator = 10KHz/V}- a)33,750 and 4218
- b)3375 and 421
- c)67,500 and 8437
- d)6,750 and 844
Correct answer is option 'D'. Can you explain this answer?
A single tone signal 14Cos(40π x103t) is used to modulate a carrier Accoswct. Then determine the number of channels available in very high frequency (VHF) band for this signal, if amplitude modulation and frequency modulation scheme employs respectively {Assume sensitivity of FM modulator = 10KHz/V}
a)
33,750 and 4218
b)
3375 and 421
c)
67,500 and 8437
d)
6,750 and 844
Srinidhi Akella answered • 21 hours agoThe VHF band is 30M-300Mhz
now if am modulation is done...the bw of each signal is 2fm=2(20khz)=40khz
total bw of VHF band is (300-30)MHz=270Mhz
assuming guard band to be zero, the number of signals each with bw 40khz occupying VHF band is
270Mhz/40khz=6750
similarly for FM, md=(Kf)(Am)/fm=10k(14)/(20k)=7
so bw of each signal is 2(mf+1)fm=2(8)(20k)=320khz
so numb... more
now if am modulation is done...the bw of each signal is 2fm=2(20khz)=40khz
total bw of VHF band is (300-30)MHz=270Mhz
assuming guard band to be zero, the number of signals each with bw 40khz occupying VHF band is
270Mhz/40khz=6750
similarly for FM, md=(Kf)(Am)/fm=10k(14)/(20k)=7
so bw of each signal is 2(mf+1)fm=2(8)(20k)=320khz
so numb... more
What will be sensitivity of modulator shown below, if its depth of modulation is0.4 and x(t) = cos (340nt) ?
- a)0.68 KHz/V
- b)68 Hz/V
- c)36 Hz/V
- d)52 Hz/V
Correct answer is option 'B'. Can you explain this answer?
What will be sensitivity of modulator shown below, if its depth of modulation is
0.4 and x(t) = cos (340nt) ?
a)
0.68 KHz/V
b)
68 Hz/V
c)
36 Hz/V
d)
52 Hz/V
Srinidhi Akella answered • 21 hours agoThe modulation index for FM is md=(Kf)(Am)/fm
given x(t)=cos(340Ï€t). ...i.e Am=1volt and fm=170hz and given mf=0.4
so substituting we get 0.4=Kf(1)/170. i.e Kf=68hz/v
given x(t)=cos(340Ï€t). ...i.e Am=1volt and fm=170hz and given mf=0.4
so substituting we get 0.4=Kf(1)/170. i.e Kf=68hz/v
The open-loop transfer function of unity feedback system is G(s) = 
. Thes (s + 2) (s + 10)range of k for which closed-loop system is stable.- a)k < 240
- b)k > 240
- c)k < 340
- d)k > 340
Correct answer is option 'A'. Can you explain this answer?
The open-loop transfer function of unity feedback system is G(s) = . Thes (s + 2) (s + 10)range of k for which closed-loop system is stable.
. Thes (s + 2) (s + 10)range of k for which closed-loop system is stable.a)
k < 240
b)
k > 240
c)
k < 340
d)
k > 340
Shrinjini answered • yesterdayIt is type-1 and order —3 system so the Nyquist plot is
For stability:
An EM wave travels in free space with the electric field component, E. = 100 ei(a866 + a5z)ax V/m and ω = 3 x 108 rad/sec. The time average power in the wave is- a)(1.33 ay - 2.3 a2) W/m2
- b)(- 2.63 a2 + 5.6 ay) W/m2
- c)(11.5 ay + 6.6 az) W/m2
- d)(0.866 ay + 0.5 az) W/m2
Correct answer is option 'C'. Can you explain this answer?
An EM wave travels in free space with the electric field component, E. = 100 ei(a866 + a5z)ax V/m and ω = 3 x 108 rad/sec. The time average power in the wave is
a)
(1.33 ay - 2.3 a2) W/m2
b)
(- 2.63 a2 + 5.6 ay) W/m2
c)
(11.5 ay + 6.6 az) W/m2
d)
(0.866 ay + 0.5 az) W/m2
Ashi Choudhary answered • yesterdayStandard equation is,
Consider a cascaded block of an LTI system
The impulse response of S1. is h1. t = δ t -2 The impulse response of 52 is h2 t = t.e-ru tThe input x(t) is wide sense stationary random process with Rxx 0 = 3 and variance = 1Q. The mean of the random process w(t) is- a)W(t) will not be wide sense stationary so mean will not a be a constant.
- b)
- c)1
- d)2
Correct answer is option 'B'. Can you explain this answer?
Consider a cascaded block of an LTI system
The impulse response of S1. is h1. t = δ t -2 The impulse response of 52 is h2 t = t.e-ru t
The input x(t) is wide sense stationary random process with Rxx 0 = 3 and variance = 1
Q. The mean of the random process w(t) is
a)
W(t) will not be wide sense stationary so mean will not a be a constant.
b)
c)
1
d)
2
Mridul Gupta answered • yesterday
then causal sequence f(n) is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
then causal sequence f(n) is
a)
b)
c)
d)
Swati Srivastava answered • yesterday
On equating both sides we get. A 4- C = 3.
4A- 2B + C D = 5,
48 + D = 0
For the continuous signal x(t) shown in the fig draw The following signals
Q.What is the signal x(-t)- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
For the continuous signal x(t) shown in the fig draw The following signals
Q.
What is the signal x(-t)
a)
b)
c)
d)
Monika Singh answered • yesterdayThe signal Is a folded signal. so
A circuit using an op - AMP is shown in the given figure it has
- a) Voltage series feed back
- b) Voltage shunt feed back
- c) Current shunt feed back
- d) Current series feed back
Correct answer is option 'B'. Can you explain this answer?
A circuit using an op - AMP is shown in the given figure it has
a)
Voltage series feed back
b)
Voltage shunt feed back
c)
Current shunt feed back
d)
Current series feed back
Shabhan Qurishi answered • yesterday
Since the feedback current is proportional to the output voltage, it is a voltage shunt feedback amplifier.
In a Vidicon camera tube- a)grid G1 accelerates electrons
- b)grid G2 accelerates electrons and grid G3, slows down electrons
- c)grid G3 accelerates electrons and grid G2 slows down electrons
- d)both grids G2 and G3 accelerate electrons
Correct answer is option 'B'. Can you explain this answer?
In a Vidicon camera tube
a)
grid G1 accelerates electrons
b)
grid G2 accelerates electrons and grid G3, slows down electrons
c)
grid G3 accelerates electrons and grid G2 slows down electrons
d)
both grids G2 and G3 accelerate electrons
Telecom Tuners answered • yesterdayGrid G2 accelerates the electrons. Grid G3, is focus grid and slows the electrons so that they fall perpendicularly on the target at a low speed without causing secondary emission.
The received signal frequency at any time of a super-heterodyne receiver having IF = 456 kHz is 1 MHz. The corresponding image signal is- a)within the medium band
- b)outside the medium band
- c)depend upon modulation index
- d)depends on Modulating frequency
Correct answer is option 'A'. Can you explain this answer?
The received signal frequency at any time of a super-heterodyne receiver having IF = 456 kHz is 1 MHz. The corresponding image signal is
a)
within the medium band
b)
outside the medium band
c)
depend upon modulation index
d)
depends on Modulating frequency
Gate Funda answered • yesterdayImage signal frequency = Received signal frequency - 2 x Intermediate frequency.
= 1 MHz - 2 x 456kHz
1 MHz - 912 kHz 
88 kHzwhich lies in medium bond.
Consider the following - Generation of SSB signals
- Design of minimum phase type filters
- Representation of band pass signals
Which of the above applications is Hilbert transform used?- a)1 only
- b)1, 2 and 3
- c)1, 2 only
- d)1 and 3
Correct answer is option 'B'. Can you explain this answer?
Consider the following
- Generation of SSB signals
- Design of minimum phase type filters
- Representation of band pass signals
a)
1 only
b)
1, 2 and 3
c)
1, 2 only
d)
1 and 3
L Narayan Reddy answered • yesterdayHilbert transform is used in all the applications.
Yash Deep Singh asked • 2 days agoThe frequency response of a microphone means- a)bandwidth of frequencies having output with in ± 1 dB of the output at 1000 Hz
- b)bandwidth of frequencies having output with in ± 3 dB of the output at 1000 Hz
- c)bandwidth of frequencies having output with in ± 1 dB of the output at 500 Hz
- d)bandwidth of frequencies having output with in ± 3 dB of the output at 500 Hz
Correct answer is option 'A'. Can you explain this answer?
The frequency response of a microphone means
a)
bandwidth of frequencies having output with in ± 1 dB of the output at 1000 Hz
b)
bandwidth of frequencies having output with in ± 3 dB of the output at 1000 Hz
c)
bandwidth of frequencies having output with in ± 1 dB of the output at 500 Hz
d)
bandwidth of frequencies having output with in ± 3 dB of the output at 500 Hz
Graphic equaliser system in stereophonic systems generally have- a)3 or 4 bands
- b)1 band
- c)more than 8 bands
- d)large number of bands
Correct answer is option 'A'. Can you explain this answer?
Graphic equaliser system in stereophonic systems generally have
a)
3 or 4 bands
b)
1 band
c)
more than 8 bands
d)
large number of bands
Deepanshu answered • yesterdayStereo systems have 3 or 4 bands at 300 Hz, 1000 Hz, 10000 Hz etc. in graphic equaliser system.
Consider the following - A signal f(t) and its Hilbert transform fh(t) have the same energy density spectrum
- A signal f(t) and its Hilbert transform, fh(t) have the same auto correlation function
- A signal f(t) and its Hilbert transform are mutually orthogonal
Which of the above are true?- a)1, 2 and 3
- b)1 and 2 only
- c)1 only
- d)2 and 3 only
Correct answer is option 'A'. Can you explain this answer?
Consider the following
- A signal f(t) and its Hilbert transform fh(t) have the same energy density spectrum
- A signal f(t) and its Hilbert transform, fh(t) have the same auto correlation function
- A signal f(t) and its Hilbert transform are mutually orthogonal
a)
1, 2 and 3
b)
1 and 2 only
c)
1 only
d)
2 and 3 only
Sachin Kaushal answered • yesterdayAll are correct.
For the system shown in fig.the steady state error component due to unit step disturbance is 0.000012 and steady state error component due to unit ramp input is 0.003. The values of K1 and K2 are respectively
- a)16.4, 1684
- b)1250, 2.4
- c)125 x 103, 0.016
- d)463, 3981
Correct answer is option 'C'. Can you explain this answer?
For the system shown in fig.the steady state error component due to unit step disturbance is 0.000012 and steady state error component due to unit ramp input is 0.003. The values of K1 and K2 are respectively
a)
16.4, 1684
b)
1250, 2.4
c)
125 x 103, 0.016
d)
463, 3981
Piyush Vasuja answered • yesterdayIf R (s) = 0
Error in output due to disturbance
Error due to ramp input
Error in output due to disturbance
Error due to ramp input
In the voltage doubler circuit shown in the figure, the switch ‘S’ is closed at t = 0. Assuming diodes D1 and D2 to be ideal, load resistance to be infinite and initial capacitor voltages to be zero, the steady state voltage across capacitors C1 and C2 will be
- a)vc1 = 10V, vc2 = 5V
- b)vc1 = 10V, vc2 = –5V
- c)vc1 = 5V, vc2 = 10V
- d)vc1 = 5V, vc2 = –10V
Correct answer is option 'D'. Can you explain this answer?
In the voltage doubler circuit shown in the figure, the switch ‘S’ is closed at t = 0. Assuming diodes D1 and D2 to be ideal, load resistance to be infinite and initial capacitor voltages to be zero, the steady state voltage across capacitors C1 and C2 will be
a)
vc1 = 10V, vc2 = 5V
b)
vc1 = 10V, vc2 = –5V
c)
vc1 = 5V, vc2 = 10V
d)
vc1 = 5V, vc2 = –10V
Nitin Kumar answered • yesterdaySince RLOAD is infinite, we can replace with open circuit
C1 will be charged in first +Ve half cycle
VC1 = +5v
C2 will be charged in first –Ve half cycle
VC2 = -10v
C1 will be charged in first +Ve half cycle
VC1 = +5v
C2 will be charged in first –Ve half cycle
VC2 = -10v
A uniformly doped npn bipolar transistor has following parameters:NE = 1018 cm-3 NB = 5 x 1016 cm-3,
Nc = 2 x 1019 cm-3,
DE = 8 cm2 /s , DB = 15 cm2 /s , Dc = 14 cm2 /s
xE = 0.8 μm, xB = 0.7 μmThe emitter injection efficiency γ is - a)0.999
- b)0.977
- c)0.982
- d)0.934
Correct answer is option 'B'. Can you explain this answer?
A uniformly doped npn bipolar transistor has following parameters:
NE = 1018 cm-3 NB = 5 x 1016 cm-3,
Nc = 2 x 1019 cm-3,
DE = 8 cm2 /s , DB = 15 cm2 /s , Dc = 14 cm2 /s
xE = 0.8 μm, xB = 0.7 μm
Nc = 2 x 1019 cm-3,
DE = 8 cm2 /s , DB = 15 cm2 /s , Dc = 14 cm2 /s
xE = 0.8 μm, xB = 0.7 μm
The emitter injection efficiency γ is
a)
0.999
b)
0.977
c)
0.982
d)
0.934
Zeba Ahmed answered • yesterday
A half wavelength dipole is kept in the x-y plane and oriented along 45° from the x-axis. Determine the direction of null in the radiation pattern for 
. Here the angle 
is measured from the z-axis, and the angle 
is measured from the x-axis in the x-y plane.- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
A half wavelength dipole is kept in the x-y plane and oriented along 45° from the x-axis. Determine the direction of null in the radiation pattern for . Here the angle is measured from the z-axis, and the angle is measured from the x-axis in the x-y plane.
. Here the angle is measured from the z-axis, and the angle is measured from the x-axis in the x-y plane.a)
b)
c)
d)
Santosh Kumar answered • yesterdayThe null occurs along axis of the antenna which is
A silicon npn bipolar transistor has doping concentration of NE = 2 x 1018cm-3, NB =1017cm-3 and NC = 15 x 1016 cm-3. The area is 10-3 cm2 and neutral base width is 1 μm. The transistor is biased in the active region at VBE = 0.5 V. The collector current is
(DB = 20 cm2/s)- a)9 μA
- b)17μA
- c)22 μA
- d)11 μA
Correct answer is option 'B'. Can you explain this answer?
A silicon npn bipolar transistor has doping concentration of NE = 2 x 1018cm-3, NB =1017cm-3 and NC = 15 x 1016 cm-3. The area is 10-3 cm2 and neutral base width is 1 μm. The transistor is biased in the active region at VBE = 0.5 V. The collector current is
(DB = 20 cm2/s)
(DB = 20 cm2/s)
a)
9 μA
b)
17μA
c)
22 μA
d)
11 μA
N Ashish answered • 2 days ago
For the graph shown in fig.correct set is
- a)A
- b)B
- c)C
- d)D
Correct answer is option 'B'. Can you explain this answer?
For the graph shown in fig.correct set is
a)
A
b)
B
c)
C
d)
D
Annu Nagar answered • 2 days agoThere are 4 node and 6 branches.
t = n - 1 = 3, l = b - n + 1 = 3
The time taken for the output to settle within 2% of step input for the control system represented by 
is given by:- a)1.2 sec
- b)1.6 sec
- c)2 sec
- d)0.4 sec
Correct answer is option 'B'. Can you explain this answer?
The time taken for the output to settle within 2% of step input for the control system represented by is given by:
is given by:a)
1.2 sec
b)
1.6 sec
c)
2 sec
d)
0.4 sec
M Satish answered • 2 days agoComparing the above transfer function with the standard second order transfer function
The value of Z22 (Ω) for the circuit shown below is
- a) 4/11 Ω
- b) 9/4 Ω
- c)4/9 Ω
- d) 11/4 Ω
Correct answer is option 'A'. Can you explain this answer?
The value of Z22 (Ω) for the circuit shown below is
a)
4/11 Ω
b)
9/4 Ω
c)
4/9 Ω
d)
11/4 Ω
Yash Deep Singh answered • 2 days agoApplying KVL, we get
Putting the value of V1 from equation (ii) into equation (i), we get
The Theven's equivalent resistance to the left of AB in figure given below is given by
- a)1/2 Ω
- b)1 Ω
- c)3/2 Ω
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The Theven's equivalent resistance to the left of AB in figure given below is given by
a)
1/2 Ω
b)
1 Ω
c)
3/2 Ω
d)
None of these
Shiwan Gijai Deep answered • 2 days agoFor finding the Thevenin’s equivalent resistance, 8 A current source will be open circuited while V volt voltage source will be short circuited.
Hence
The initial state of the counter in the circuit shown below is 01, and resolution of DAC is 0.5 volt. Determine the output of DAC after three clock pulses.
- a)1.5 volt
- b) 2 volt
- c)1 volt
- d)0 volt
Correct answer is option 'D'. Can you explain this answer?
a)
1.5 volt
b)
2 volt
c)
1 volt
d)
0 volt
V Nagaindraja answered • 2 days agoThe state diagram of counter
So, output of DAC = 0 volt
So, output of DAC = 0 volt
.The angles of arrival of its root loci are 
in the given circuit below
Two pipes can fill a tank in 15 minutes and 25 minutes respectively. Both the pipes are opened together and after some time, the first pipe is closed and the tank is completely full in 15 minutes. For how many minutes is the first pipe opened?
- a)5 minutes
- b)6 minutes
- c)7 minutes
- d)8 minutes
Correct answer is option 'B'. Can you explain this answer?
a)
5 minutes
b)
6 minutes
c)
7 minutes
d)
8 minutes
Dilpreet Singh answered • 2 days ago
t = 6 min.
Find the value of
- a)
- b)
- c)
- d)3
Correct answer is option 'D'. Can you explain this answer?
a)
b)
c)
d)
3
Gargi Sharma answered • 2 days ago
y(y-3) = 0
y = 3
Consider the audio signal with spectral components limited to the frequency band of 300 to 3300 Hz. A PCM signal is generated with a sampling rate of 8000 samples/sec. The required output signal to quantizing noise ratio is 30 dB. The minimum number of bits per sample needed is :- a)4
- b)5
- c)6
- d)7
Correct answer is option 'B'. Can you explain this answer?
a)
4
b)
5
c)
6
d)
7
K Ragini Rao answered • 2 days ago(SNR) in dB = 1.76 + 20 log L
Where L is the number of levels
So minimum number of levels are 26, and minimum number of bits are 5.
Where L is the number of levels
So minimum number of levels are 26, and minimum number of bits are 5.
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