1)
ξ
ξ1
For critical damping 1=
Squaring on both sides 1=
2) Quality factor (Q) =
3)
− −7 =
C=
.
C
=
. =
.
8
= 300
-4
V = 14V
14 = 4 X R R=
= 3.5
Ω
4) At 3-dB frequencies current is multiplied by frequency.
√
of the current at resonant
5)
f 1000
kHz and given frequency is f = 995 kHz. k Hz. At f = 995 kHz the current
impedance is capacitive.
6) F=
F − +
10
+
−
10
+
10 + 0.9 10.9
7) In sigma delta ADC, high bit accuracy is achieved by over sampling.
8) After solving the given integral, its value is 1.
9) Y(n) = x(n-2)
− z (()) H(z)H(z) −.−.8 H(z) H(z) ((−.−.8)) H(z) −.−.8 Y(z) =
x(z)
=
=
=
=
= =
So correct option is B
10) x(n) =
{1,1 , 2 , 1, 1 }
By definition of z-transformation
∑()− =− ∑()− =−
x(z) =
=
= x(-1)z + x(0) +x(1)
z− z− + x(2)
z− z− Applying Kirchhoff’s law at point B. Let the potential at point B is V. = z+2-
+
11)
− + V V +
+
= - 6
12)
=0 = 6
Solve the circuit. So equivalentcapacitance acros ab wil be 0.2μF
13) The transfer function T(s) =
14)
I ∫(10asint)dt I =
a = 1.414
= 10 A.
+
is that of a high pass filter.
15) (QNP)α (())
=
= 2
(())
= 10 log( ) = - 3dB
–
It means quantisation noise decrease by 3dB and ve sign indicates decrease in quantisation noise.
16) In given circuit one Zener diode is forward biased and an d will behave as a normal diode. So voltage drop is 0.6 V and another zener diode is a reversed biased and voltage drop will be 6.3V. So total output (
V
) = (6.3 + 0.6)V = 6.9 V
17) Assuming that only X and Y logic inputs are available and their
X Y⨁
complements and
are not available,
sufficient to implement X
then 4 two input NAND gates are
Y.
18)
X XX …. X− X X X ….. X− ….. X− X− X +
19)
+
+
f t
+
=
n
=
= 100 nsec.
+
20) Conversion time follows the order as given below: Dual slope convertor > Successive approximation type > Flash type convertor 21) The resolution of D/A converter is approximately 0.4% of its full scale range, then it is a 8- Bit counter. 22) In a microprocessor, the resister which holds the address of the next instruction to be fetched is 23) In microprocessor, WAIT states are used to interface slow peripherals to the processor. 24)
A flip-flop is used to store 1 bit of information.
Race-around condition occurs in J-K flip-flops when both inputs are 1.
Master salve configuration is used in flip-flop to store 2 bit of info.
A transparent latch consists of a D-type flip flop.
25) 3 x 512 + 7 x 64 + 5 x 8 + 3
After solving this expres ion you wil find the number of 1’s in its equivalent –
representation are 9.
26) For emitter coupled logic, the switching speed is very high because, transmitter are saturated when conducting.
27) Output of divider is 1 Hz, Schmitt trigger will will not change the frequency and flip flop will half the output frequency, Hence output frequency is
⁄
Therefore T = T=
= 2 sec.
28) Gray code for 7: 0
1
1
1
0
1
0
0
29) 10 bit A/D converters, the quantisation error is 0.1 %
30) Expression for given circuit is A(BC + D)
31) The Boolean expression for the shaded area in the venn diagram is:
XY
Z+XY
Hz
32) Number of chips =
8
= 32
33) Binary representation of +19 in 8 bit is
and 2’s complement of 19 is representation is
-19 and binary representation and its binary
1 1 1 0 1 1 0 1
34) The function shown in figure will yield 7 terms when simplified.
35) Logic circuit given will convert the binary code into Gray Code
36) In synchronous counter clock is given to all the flip-flops simultaneously. So time required to change the state is equal to propagation delay time.
An electromagnetic wave incident on a perfect conductor is Entirely Reflected.
for a lossless line R =0 and G=0
Z R j ωL j ωL Z G jωC jωC Z =
=
=
39) Reflection coefficient is
|Γ| −+
Therefore VSWR is
− + +|−||| +− ⁄⁄ =
= =
=
=
=
=
= 2: 1
40) TE wave TM wave
All exist in i n waveguides
Both TE and TM waves
So TEM waves does not exist in waveguides.
41) Bandwidth of FM is given
by Carson’s Rule
(B.W)M 2(∆f f 2(∆f f ) 2(∆f f f f (∆f 10 kHkHz)z) (∆f 10 kHkHzz )
)
=
= 10 kHz
)
Therefore:
∆∆
= = 1 : 1
42) The Bandwidth required for QPSK modulated channel is
43) Magic T is Four Port junction.
44) Diplexer is made of
45) The gain G of an antenna of effective area A is G=
46)
z Z. Z √ 5 x 20 √ 100100 =
=
=
= 10
47) The input impedance of short circuited line of length l where capacitive.
< l <
is
48) Maximum coding gain occur in Turbo codes. 49) None of these 50) BCH codes belong to block code. 51)
m(t)
Frequency Modulation signal
PM
52) The Doppler shift observed at the ground station, when the satellite is over the head of the station is Maximum. 53)
According to Shannon’s channel capacity theorem: log(1SNR) log(115) log(16) C=B 4
4
= 4 x 4 = 16 kbps.
54)
P|ρ| P P P |ρ| |ρ| 2
Where
= Reverse Power
= input Power =
P = Reflection coefficient 2
= =
+− ||||
=
=
So VSWR =
=
+− ⁄⁄
=
⁄⁄ =
=1 : 5
55)
P log dB log = 10
30 = 10
10− W 10− P(dBm) log dB log dB dB
= 1000 P0 = 1000 x
= 1x
= 10
m
= 10
m= 0
m
W
56)
P()) (PG)dBL d BL G G (PG) L P()) – 10 = 10 log = 10
= 10 log(100) = 20 dB
Path loss (
) = 100dB
= (20 100)dBW = -80dBW
57) At condition of Equilibrium:
P⃗ ⃗Q R⃗ ⃗R ⃗P ⃗Q ⃗R ⃗22 ⃗33 ⃗33 ⃗44 ⃗2k ⃗2k ⃗2k +
= 0
= -(
)
= -(
= -(
)
)
(
)
58) Velocity
v⃗ ω⃗x r r ω⃗ ωk i j k v⃗ ω⃗ x r 0x 0y ωz ωyi ωxj i ∂ ∂j k∂ ωk ∂ωx ω∂y ∂0 ω⃗ ωk, therefore Curl v 2ω =
= xi + yj + zk, and
=
=
= -
Therefore, Curl v =
But as we know:
59)
sin2t (3si3sin2tsin6t) + + + + − (+− +)( +) (+)(+) = =
By taking Laplace transformation L(sin3 2t) = =
=
=
=
8 (+)(+)
= 2
60)
=
c o0sθ 10 si0nθ s i n θ 0 c o s θ cosθ cosθ sinθ sinθ θ θ (
– 0) + 0+
cos2 + sin2
( 0- (-
))
=1
61) Particular solution = = = = = =
D+ [− − ] [D+ e D+ cos2x] [ ′(D) cos2x] [ D cos2x] 8 ∫ cos2xdx 8
General solution: D2 + 4 =0 D = ±2j
=A cos2x + B sin2x
8 8 sin2x
Y = A cos2x + B sin2x + –
62) For concurrent, coefficient determinant should be equal to 0
24 13 13 3k2
=0
2(-6 + 3k) -1(-8+9) -1(4k-9) =0 2k – 4 = 0 k=2 63) The probability of both the balls being red is
x x =
64) f(x) = x3 + x sinx f(-x) = (-x3) + (-x) sin(-x) =-x3 + xsinx
f(x) + f(-x) = 2x sinx f(x) - f(-x) = 2x3
not an odd function.
not an even function.
So this is not an periodic function, because Periodic Function + Non-Periodic Function = Therefore given function is a constant function.
Non-Periodic Function
65)
− s e c ∫ √ − =
66)
λ [525 22 ] [0λ 0λ] [52 λ 22 λ] λ λ λ λ60 λ λ6
Characteristics equation is
A- I = 0 = 0 = 0
(5+ ) (2+ ) – 4 = 0 2 +
7
( +1) (
Therefore
67)
=
=
R Ω = 20 =
) =0
68) V = I XL 50 = 7.96 X
10− π 10 x L ≃ x 2 x
L = 99.97mH
100mH
69) In a capacitor, the electric charge is stored in metal plates.
70) Oscillator requires positive positive feedback.
71) Core loss loss in in transformer transformer varies varies significantly significantly with load.
72)
8 8 − + = =
So R = 36
Ω
+
=
=
73) Time constant of a series R-L circuit is equals to
Second
74) For RC circuit i(t) = at t =
∞
e−
(Steady State)
i=0
75) The given circuit is a High Pass Filter
76) At low frequency capacitor will be open circuited
A −− A (dB) log100 log10 40dB = =
= 20 =
= 100 = 20
77)
R C 2C and
Time constant = Rtotal Ctotal =
2C = RC
78)
Rtotal = + +
=
Ω
= 1
So I =
I=
= 2A
79) When L is doubled and C is halved, the resonance frequency of series tuned circuit becomes unchanged.
80) In a series resonant circuit, with the increase in L
Resonant frequency will decrease
Bandwidth will decreases
Q will increase
