2003|Electronics and Comm (GATE Exam)-Previous Question Paper Solution

Description: GATE Exam Previous Year Question Paper Solution Electronics and Communication (ECE) - 2003
Number of Questions: 90
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Tags: Network Graphs Signals and System Electronic Devices Analog Circuits Digital circuits Digital Circuits
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A 4 bit ripple counter and a 4 bit synchronous counter are made using flip-flops having a propagation delay of 10 ns each. If the worst case delay in the ripple counter and the synchronous counter be R and S respectively, then

  1. R = 10 ns, S = 40 ns

  2. R = 40 ns, S = 10 ns

  3. R = 10 ns, S = 30 ns

  4. R = 30 ns, S = 10 ns


Correct Option: B
Explanation:

The output of the 74 series of TTL gates is taken from a BJT in

  1. totem pole and common collector configuration

  2. either totem pole or open collector configuration

  3. common base configuration

  4. common collector configuration


Correct Option: B
Explanation:

When output of the 74 series gate of TTL gates is taken from BJT then the configuration is either totem pole or open collector configuration.

The circuit shown in figure has 4 boxes each described by inputs P, Q, R and outputs Y, Z with Y = P $\oplus$Q $\oplus$R Z = RQ + $\bar P R + Q\bar P$

The circuit acts as a

  1. 4 bit adder giving P + Q

  2. 4 bit subtractor-giving P - Q

  3. 4 bit subtractor-giving Q - P

  4. 4 bit adder giving P + Q + R


Correct Option: B
Explanation:

The minimum number of comparators required to build an 8 it flash ADC is

  1. 8

  2. 63

  3. 255

  4. 256


Correct Option: C
Explanation:

In the flash analog to digital converter, the no. of comparators is equal to $2^{n-1}$, where n is no. of bits

So, $2^{n-1} = 2^8 - 1 = 255$

The circuit shown in figure converts

  1. BCD to binary code

  2. Binary to excess - 3 code

  3. Excess - 3 to Gray code

  4. Gray to Binary code


Correct Option: D
Explanation:

The number of distinct Boolean expression of 4 variables is

  1. 16

  2. 256

  3. 1024

  4. 65536


Correct Option: D
Explanation:

The number of distinct boolean expression of n variable is $2^{2n}.$ Thus $2^{2^4} = 2^{16} = 65536$

The circuit shown in figure is a 4-bit DAC The input bits 0 and 1 are represented by 0 and 5 V respectively. The OP AMP is ideal, but all the resistances and the 5V inputs have a tolerance of $\pm$10%. The specification (rounded to the nearest multiple of 5%) for the tolerance of the DAC is

  1. $\pm$35%

  2. $\pm$20%

  3. $\pm$10%

  4. $\pm$5%


Correct Option: A
Explanation:

Without any additional circuitry, an 8 : 1 MUX can be used to obtain

  1. some but not all Boolean functions of 3 variables

  2. all function of 3 variables but none of 4 variables

  3. all functions of 3 variables and some but not all of 4 variables

  4. all functions of 4 variables


Correct Option: D
Explanation:

A $2^n.1$ MUX can implement all logic functions of (n+1) variable without and y additional circuitry. Here n = 3. Thus a 8:1 MUX can implement all logic functions of 4 variable.

In the circuit shown in Figure, A is a parallel in, parallel-out 4-bit register, which loads at the rising edge of the clock C. The input lines are connected to a 4-bit bus, W. Its output acts as the input to a 16$\times$4 ROM whose output is floating when the enable input E is 0. A partial table of the contents of the ROM is as follows:

$$ \begin{array}{c|c} \ Address & 0 & 2 & 4 & 6 & 8 & 10 & 11 & 14 \\ Data & 0011 & 1111 & 0100 & 1010 & 1011 & 1000 & 0010 & 1000 \end{array} $$

The clock to the register is shown, and the data on the W bus at time t1 is 0110. The data on the bus at time t2 is

  1. 1111

  2. 1011

  3. 1000

  4. 0010


Correct Option: C
Explanation:

After $t = t_1$, at first rising edge of clock, the output of shift register is 0110, which in input to address line of ROM. At 0110 is applied to register. So at this time data stored in ROM at 1010 (10), 1000 will be on bus.

When W has the data 0110 and it is 6 in decimal, and it's data value at that add is 1010 then 1010 i.e. 10 is acting as odd, at time $t_2$ and data at that movement is 1000.

If the functions W, X, Y and Z are as follows: W = R + $\bar P Q + \bar RS$ X = PQ$\bar R \bar S + \bar P \bar Q \bar R \bar S + P \bar Q \bar R \bar S$ Y = RS + $\overline{PR + P \bar Q + \bar P \bar Q}$ Z = R + S + $\overline{PQ + \bar P \bar Q \bar R + P \bar Q \bar S}$ then

  1. W = Z, X = $\mathcal{Z}$

  2. W = Z, X = Y

  3. W = Y

  4. W = Y = $\mathcal{Z}$


Correct Option: A
Explanation:

A 0 to 6 counter consists of 3 flip flops and a combination circuit of 2 input gate(s). The combination circuit consists of

  1. one AND gate

  2. one OR gate

  3. one AND gate and one OR gate

  4. two AND gates


Correct Option: D
Explanation:

Counter must be reset when it count 111. This can be implemented by following circuitry:

The DTL, TTL, ECL and CMOS families of digital ICs are compared in the following 4 columns:

(P) (Q) (R) (S) Fanout is minimum DTL DTL TTL CMOS Power consumption is minimum TTL CMOS ECL DTL Propagation delay is minimum CMOS ECL TTL TTL
The correct column is

  1. P

  2. Q

  3. R

  4. S


Correct Option: B
Explanation:

The DTL has minimum fan out and CMOS has minimum power consumption. Propagation delay is minimum in ECL.

In an 8085 microprocessor, the instruction CMP B has been executed while the content of the accumulator is less than that of register B. As a result,

  1. Carry flag will be set but Zero flag will be reset

  2. Carry flag will be reset but Zero flag will be set

  3. both Carry flag and Zero flag will be reset

  4. both Carry flag and Zero flag will be set


Correct Option: A
Explanation:

CMP B $\Rightarrow$ Compare the accumulator content with context of Register B

If A $\lt$ R CY is set and zero flag will be reset.

The unit of $\nabla$$\times$ H is

  1. Ampere

  2. Ampere/metre

  3. Ampere/metre2

  4. Ampere-metre


Correct Option: C
Explanation:

Medium 1 has the electrical permittivity $\xi_1$= 1.5 $\xi_0$ farad/m and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permittivity $\xi_2$ = 2.5 $\xi_0$ farad/m and occupies the region to the right of x = 0 plane. If E1 in medium 1 is E1 = (2ux -3uy + 1uz) volt/m, then E2 in medium 2 is

  1. (2.0ux - 7.5uy + 2.5uz) volt/m

  2. (2.0ux - 2.0uy + 0.6uz) volt/m

  3. (1.2ux - 3.0uy + 1.0uz) volt/m

  4. (1.2ux - 2.0uy + 0.6uz) volt/m


Correct Option: C
Explanation:

The bandgap of silicon at 300 K is

  1. 1.36 eV

  2. 1.10 eV

  3. 0.80 eV

  4. 0.67 eV


Correct Option: B
Explanation:

A particular green LED emits light of wavelength 5490°A. The energy bandgap of the semiconductor material used there is (Planck's constant = 6.626$\times$10-34J -s)

  1. 2.26 eV

  2. 1.98 eV

  3. 1.17 eV

  4. 0.74 eV


Correct Option: A
Explanation:

If the electric field intensity is given by E = (xux + yuy + zuz) volt/m, the potential difference between X(20,0) and Y(1,2,3) is

  1. +1 volt

  2. -1 volt

  3. +5 volt

  4. +6 volt


Correct Option: C
Explanation:

A uniform plane wave travelling in air is incident on the plane boundary between air and another dielectric medium with $\epsilon_r$= 4. The reflection coefficient for the normal incidence, is

  1. zero

  2. 0.5$\angle$180�

  3. 0.333$\angle$0�

  4. 0.333$\angle$180�


Correct Option: D
Explanation:

The depth of penetration of electromagnetic wave in a medium having conductivity $\sigma$ at a frequency of 1 MHz is 25 cm. The depth of penetration at a frequency of 4 MHz will be

  1. 6.25 cm

  2. 12.50 cm

  3. 50.00 cm

  4. 100.00 cm


Correct Option: B
Explanation:

A rectangular metal wave-guide filled with a dielectric material of relative permittivity $\epsilon_r$ = 4 has the inside dimensions 3.0 cm $\times$ 1.2 cm. The cut-off frequency for the dominant mode is

  1. 2.5 GHz

  2. 5.0 GHz

  3. 10.0 GHz

  4. 12.5 GHz


Correct Option: A
Explanation:

Two identical antennas are placed in the $\theta$= $\dfrac{\pi}{2}$ plane as shown in figure. The elements have equal amplitude excitation with 180° polarity difference, operating at wavelength $\lambda$. The correct value of the magnitude of the far-zone resultant electric field strength normalized with that of a single element, both computed for $\phi$= 0, is

  1. 2 cos $\left( \dfrac{2\pi s}{\lambda} \right)$

  2. 2 sin $\left( \dfrac{2\pi s}{\lambda} \right)$

  3. 2 cos $\left( \dfrac{\pi s}{\lambda} \right)$

  4. 2 sin $\left( \dfrac{\pi s}{\lambda} \right)$


Correct Option: D
Explanation:

The electron concentration in a sample of uniformly doped n-type silicon at 300 K varies linearly from 1017/cm3 at x = 0 to 6 $\times$1016/cm3 at x = 2 $\mu$m. Assume a situation that electrons are supplied to keep this concentration gradient constant with time. If electronic charge is 1.6$\times$10-19 coulomb and the diffusion constant Dn = 35 cm2/s, the current density in the silicon, if no electric field is present, is

  1. zero

  2. -112 A/cm2

  3. +1120 A/cm2

  4. -1120 A/cm2


Correct Option: D
Explanation:

If P is Passivation, Q is n-well implant, R is metallization and S is soruce/drain diffusion, then the order in which they are carried out in a standard n-well CMOS fabrication process, is

  1. P-Q-R-S

  2. Q-S-R-P

  3. R-P-S-Q

  4. S-R-Q-P


Correct Option: B
Explanation:

At 300 K, for a diode current of 1 mA, a certain germanium diode requires a forward bias of 0.1435V, whereas a certain silicon diode requires a forward bias of 0.718 V. Under the conditions stated above, the closest approximation of the ratio of reverse saturation current in germanium diode to that in silicon diode is

  1. 1

  2. 5

  3. 4 $\times$103

  4. 8 $\times$103


Correct Option: C
Explanation:

n-type silicon is obtained by doping silicon with

  1. germanium

  2. aluminum

  3. boron

  4. phosphorus


Correct Option: D
Explanation:

 .

For an n-channel enhancement type MOSFET, if the source is connected at a higher potential than that of the bulk, (i.e. VSB > 0) the threshold voltage VT of the MOSFET

  1. remains unchanged

  2. decreases

  3. changes polarity

  4. increases


Correct Option: D
Explanation:

 Option (4) is correct.

When the gate-to-source voltage (VGS) of a MOSFET with threshold voltage of 400 mV, working in saturation is 900 mV, the drain current in observed to be 1 mA. Neglecting the channel width modulation effect and assuming that the MOSFET is operating at saturation, the drain current for an applied VGS of 1400 mV is

  1. 0.5 mA

  2. 2.0 mA

  3. 3.5 mA

  4. 4.0 mA


Correct Option: D
Explanation:

The action of a JFET in its equivalent circuit can best be represented as a

  1. Current Controlled Current Source

  2. Current Controlled Voltage Source

  3. Voltage Controlled Voltage Source

  4. Voltage Controlled Current Source


Correct Option: D
Explanation:

The input to a coherent detector is DSB-SC signal plus noise. The noise at the detector output is

  1. the in-phase component

  2. the quadrature component

  3. zero

  4. the envelope


Correct Option: A
Explanation:

The input is a coherent detector is DSB - SC signal plus noise. The noise at the detector output is the in-phase component as the quadrature component $n_0(t)$ of the noise $n(t)$ is completely rejected by the detector.

Let x(t) = 2 cos (800$\pi$t) + cos (1400$\pi$t). x(t) is sampled with the rectangular pulse train shown in figure. The only spectral components (in kHz) present in the sampled signal in the frequency range 2.5 kHz to 3.5 kHz are

  1. 2.7, 3.4

  2. 3.3, 3.6

  3. 2.6, 2.7, 3.3, 3.4, 3.6

  4. 2.7, 3.3


Correct Option: D
Explanation:

Choose the correct one from among the alternatives A, B, C, D after matching an item in Group 1 with the most appropriate item in Group 2.||| |---|---| | Group 1| Group 2| | P Ring modulator| 1 Clock recovery| | Q VCO| 2 Demodulation of FM| | R Foster-Seely discriminator| 3 Frequency conversion| | S Mixer| 4 Summing the two inputs| | | 5 Generation of FM| | | 6 Generation of DSB-Sc|

  1. P - 1, Q - 3, R - 2, S - 4

  2. P - 6, Q - 5, R - 2, S - 3

  3. P - 6, Q - 1, R - 3, S - 2

  4. P - 5, Q - 6, R - 1, S - 3


Correct Option: B
Explanation:

Let X and Y be two statistically independent random variables uniformly distributed in the ranges (-1,1) and (-2,1) respectively. Let Z = X + Y, then the probability that [Z$\le$-2] is

  1. zero

  2. $\dfrac{1}{6}$

  3. $\dfrac{1}{3}$

  4. $\dfrac{1}{12}$


Correct Option: B
Explanation:

Let Y and Z be the random variables obtained by sampling X(t) at t = 2 and t = 4 respectively. Let W = Y - Z. The variance of W is

  1. 13.36

  2. 9.36

  3. 2.64

  4. 8.00


Correct Option: C
Explanation:

The noise at the input to an ideal frequency detector is white. The detector is operating above threshold. The power spectral density of the noise at the output is

  1. raised cosine

  2. flat

  3. parabolic

  4. Gaussian


Correct Option: C
Explanation:

The noise at the input to an ideal frequency detector is white. The PSD of noise at the output is parabolic.

X(t) is a random process with a constant mean value of 2 and the autocorrelation function Rx ($\tau$) = 4$\lfloor e^{-0.2 | d} + 1 \rfloor$.

Let X be the Gaussian random variable obtained by sampling the process at t = ti and let Q ($\alpha$) = $\int_\infty^\infty \dfrac{1}{\sqrt {2\pi}} e^{\dfrac{-y^2}{2}} dy$ The probability that [ x $\le$ 1 ] is

  1. 1 - Q(0.5)

  2. Q(0.5)

  3. Q $\dfrac{1}{2 \sqrt 2}$

  4. 1-Q $\left( \dfrac{1}{2 \sqrt 2} \right)$


Correct Option: D
Explanation:

c(t) and m(t) are used to generate an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term cos[2$\pi$(1008 $\times$103t)] in the FM signal (in terms of the Bessel coefficients) is

  1. 5j4(3)

  2. $\dfrac{5}{2} j_\theta (3)$

  3. $\dfrac{5}{2} j_\theta (3)$

  4. 5j4(6)


Correct Option: D
Explanation:

Let m (t) = cos [ (4$\pi$ $\times$ 103) t] be the message signal and c(t) = 5 cos[(2$\pi$ $\times$ 106)t] be the carrier. c(t) and m(t) are used to generate an AM signal. The modulation index of the generated AM signal is 0.5. Then the quantity $\dfrac{Total sideband power}{Carrier power}$ is

  1. $\dfrac{1}{2}$

  2. $\dfrac{1}{4}$

  3. $\dfrac{1}{3}$

  4. $\dfrac{1}{8}$


Correct Option: D
Explanation:

At a given probability of error, the binary coherent FSK is inferior to binary coherent PSK by

  1. 6 dB

  2. 3 dB

  3. 2 dB

  4. 0 dB


Correct Option: B
Explanation:

We have $P_c = \dfrac{1}{2} erfc \left( \sqrt {\dfrac{E_d}{2\eta}} \right)$

Since $P_\epsilon$ of Binary FSK is 3 dB inferior to binary PSK

A DSB-SC signal is to be generated with a carrier frequency fc = 1 MHz using a nonlinear device with the input-output characteristic V0 = a0vi + a1vi3 where a0 and a1 are constants. The output of the nonlinear device can be filtered by an appropriate band-pass filter. Let vi = A'c cos (2$\pi$f'c t) + m(t) where m(t) is the message signal. The value of f'c (in MHz) is

  1. 1.0

  2. 0.333

  3. 0.5

  4. 3.0


Correct Option: C
Explanation:

A sinusoidal signal with peak-to-peak amplitude of 1.536 V is quantised into 128 levels using a mid-rise uniform quantiser. The quantisation noise power is

  1. 0.768 V2

  2. 48 $\times$10-6 V2

  3. 12 $\times$10-6 V2

  4. 3.072 V2


Correct Option: C
Explanation:

If Eb, the energy per bit of a binary digital signal, is 10-6 watt-sec and the onesided power spectral density of the white noise, N0 = 10-5 W/Hz, then the output SNR of the matched filter is

  1. 26 dB

  2. 10 dB

  3. 20 dB

  4. 13 dB


Correct Option: D
Explanation:

The input to a linear delta modulator having a step-size $\Delta$ = 0.628 is a sine wave with frequency fm and peak amplitude Em. If the sampling frequency fs = 40 kHz, the combination of the sine wave frequency and the peak amplitude, where slope overload will take place, is

  1. Em = 0.3 V, fm = 8 kHz

  2. Em = 1.5 V, fm = 4 kHz

  3. Em = 1.5 V, fm = 2 kHz

  4. Em = 3.0 V, fm = 1 kHz


Correct Option: B
Explanation:

 

A super heterodyne receiver is to operate in the frequency range 550 kHz - 1650 kHz, with the intermediate frequency of 450 kHz. Let R = $\dfrac{C_{max}}{C_{min}}$ denote the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, then

  1. R = 4.41, I = 1600

  2. R = 2.10, I = 1150

  3. R = 3.0, I = 1600

  4. R = 9.0, I = 1150


Correct Option: A
Explanation:

If S represents the carrier synchronization at the receiver and $\rho$ represents the bandwidth efficiency, then the correct statement for the coherent binary PSK is

  1. $\rho$ = 0.5, S is required

  2. $\rho$ = 1.0, S is required

  3. $\rho$ = 0.5, S is not required

  4. $\rho$ = 1.0, S is not required


Correct Option: A
Explanation:

A signal is sampled at 8 kHz and is quantized using 8-bit uniform quantizer. Assuming SNRq for a sinusoidal signal, the correct statement for PCM signal with a bit rate of R is

  1. R = 32 kbps, SNRq = 25.8 dB

  2. R = 64 kbps, SNRq = 49.8 dB

  3. R = 64 kbps, SNRq = 55.8 dB

  4. R = 32 kbps, SNRq = 49.8 dB


Correct Option: B
Explanation:

The Fourier series expansion of a real periodic signal with fundamental frequency f0 is given by gp (t) = $\displaystyle \sum_{n=-\infty}^\omega C_n e^{j^2 x n f_0 t}$it is given that C3 = 3 + j5. Then C-3 is

  1. 5+j3

  2. -3-j5

  3. -5+j3

  4. 3-j5


Correct Option: D
Explanation:

The Laplace transform of i(t) is given by I(s) = $\dfrac{2}{s(1+s)}$ As t $\rightarrow$$\infty$, the value of i(t) tends to

  1. 0

  2. 1

  3. 2

  4. $\infty$


Correct Option: C
Explanation:

A sequence x(n) with the z-transform X(z) = z4 + z2 -2z + 2-3z-4 is applied as an input to a linear, time-invariant system with the impulse response h(n) = 2$\delta$(n-3) where $\delta$(n) = $ \begin{cases} 1, n = 0 \\ 0, otherwise \end{cases}

$ The output at n = 4 is

  1. -6

  2. zero

  3. 2

  4. -4


Correct Option: B
Explanation:

The system under consideration is an RC low-pass filter (RC-LPF) with R = 1.0 k$\Omega$and C = 1.0 $\delta$F.

Let tg(f) be the group delay function of the given RC-LPF and f2 = 100 Hz. Then tg(f2) in ms, is

  1. 0.717

  2. 7.17

  3. 71.7

  4. 4.505


Correct Option: A
Explanation:

The system under consideration is an RC low-pass filter (RC-LPF) with R = 1.0 k$\Omega$and C = 1.0$\mu$F.

Let H(f) denote the frequency response of the RC-LPF. Let f1 be the highest frequency such that 0$\le$|f| $\le$f1, $\dfrac{ | H(f_1) | }{H(0)}$$\ge$ 0.95. Then f1 (in Hz) is

  1. 327.8

  2. 163.9

  3. 52.2

  4. 104.4


Correct Option: C
Explanation:

Let x(t) be the input to a linear, time-invariant system. The required output is 4x (t-2). The transfer function of the system should be

  1. 4 ej4$\pi$f

  2. 2 e-j8$\pi$f

  3. 4 e-j4$\pi$f

  4. 2 ej8$\pi$f


Correct Option: C
Explanation:

Let P be linearity, Q be time-invariance, R be causality and S be stability. A discrete time system has the input-output relationship, Y(n) = $ \begin{cases} x(n), & n \ge 1 \\ 0 & n=0 \\ x(n+1), & n \le -1 \end{cases}

$ where x(n) is the input and y(n) is the output. The above system has the properties

  1. P, S but not Q, R

  2. P, Q, S but not R

  3. P, Q, R, S

  4. Q, R, S but not P


Correct Option: A
Explanation:

A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100. If each R, L and C is doubled from its original value, the new Q of the circuit is

  1. 25

  2. 50

  3. 100

  4. 200


Correct Option: B
Explanation:

The driving point impedance Z(s) of a network has the pole-zero locations as shown in the figure below. If Z(0) = 3, then Z(s) is

  1. $\dfrac{3(s+3)}{s^2 + 2s + 3}$

  2. $\dfrac{2(s+3)}{s^2 + 2s + 2}$

  3. $\dfrac{3(s-3)}{s^2 - 2s - 2}$

  4. $\dfrac{2(s-3)}{s^2 - 2s - 2}$


Correct Option: B
Explanation:

The minimum number of equations required to analyse the circuit shown in the figure below is

  1. 3

  2. 4

  3. 6

  4. 7


Correct Option: B
Explanation:

Number of loops = b n + 1 = minimum number of equations Number of branches = b = 8 Number of nodes = 5 Minimum number of equations = 8 − 5 + 1 = 4  Hence, (2) is the correct option.

The current flowing through the resistance R in the circuit in figure has the form P cos 4t, where P is

  1. (0.18 + j0.72)

  2. (0.46 + j1.90)

  3. -(0.18 + j1.90)

  4. -(0.192 + j0.144)


Correct Option: A

A source of angular frequency 1 rad/sec has a source impedance consisting of 1 $\Omega$resistance in series with 1 H inductance. The load that will obtain the maximum power transfer is

  1. 1 $\Omega$resistance

  2. 1 $\Omega$resistance in parallel with 1 H inductance

  3. 1 $\Omega$resistance in series with 1 F capacitor

  4. 1 $\Omega$ resistance in parallel with 1 F capacitor


Correct Option: C
Explanation:

Assume that the switch S is in position 1 for a long time and thrown to position 2 at t = 0.

At t = 0+, the current i1 is

  1. $\dfrac{-V}{2R}$

  2. $\dfrac{-V}{R}$

  3. $\dfrac{-V}{4R}$

  4. zero


Correct Option: A

The differential equation for the current i(t) in the circuit of figure given below is

  1. 2$\dfrac{d^2i}{dt^2} + 2 \dfrac{di}{dt} + i(t) = sin t$

  2. $\dfrac{d^2i}{dt^2} + 2 \dfrac{di}{dt} + 2i(t) = cos t$

  3. 2$\dfrac{d^2i}{dt^2} + 2 \dfrac{di}{dt} + i(t) = cos t$

  4. $\dfrac{d^2i}{dt^2} + 2 \dfrac{di}{dt} + 2i(t) = sin t$


Correct Option: C
Explanation:

Assume that the switch S is in position 1 for a long time and thrown to position 2 at t = 0.

I1 (s) and I2 (s) are the Laplace transforms of i1 (t) and i2 (t) respectively. The equation for the loop current I1 (s) and I2 (s) for the circuit shown in figure Q.33-34, after the switch is brought from position 1 to position 2 at t = 0, are

  1. $\left[ \begin{array} \ R+Ls+\dfrac{1}{Cs} & -Ls \\ -Ls & R+Ls+\dfrac{1}{Cs} \end{array} \right]

    \left[ \begin{array} \ I_1 & (s) \\ I_2 & (s) \end{array} \right]

    =

    \left[ \begin{array} -\dfrac{V}{S} \\ 0 \end{array} \right] $

  2. $\left[ \begin{array} \ R+Ls+\dfrac{1}{Cs} & -Ls \\ -Ls & R+\dfrac{1}{Cs} \end{array} \right]

    \left[ \begin{array} \ I_1 & (s) \\ I_2 & (s) \end{array} \right]

    =

    \left[ \begin{array} \ -\dfrac{V}{S} \\ 0 \end{array} \right] $

  3. $\left[ \begin{array} \ R+Ls+\dfrac{1}{Cs} & -Ls \\ -Ls & R+Ls+\dfrac{1}{Cs} \end{array} \right]

    \left[ \begin{array} \ I_1 & (s) \\ I_2 & (s) \end{array} \right]

    =

    \left[ \begin{array} \ \dfrac{V}{S} \\ 0 \end{array} \right] $

  4. $\left[ \begin{array} \ R+Ls+\dfrac{1}{Cs} & -Ls \\ -Ls & R+Ls+\dfrac{1}{Cs} \end{array} \right]

    \left[ \begin{array} \ I_1 & (s) \\ I_2 & (s) \end{array} \right]

    =

    \left[ \begin{array} \ -\dfrac{V}{S} \\ 0 \end{array} \right] $


Correct Option: A
Explanation:

 

The impedance parameters Z11 and Z12 of the two-port network in figure are

  1. Z11 = 2.75 $\Omega$and Z12 = 0.25 $\Omega$

  2. Z11 = 3 $\Omega$and Z12 = 0.5 $\Omega$

  3. Z11 = 3 $\Omega$and Z12 = 0.25 $\Omega$

  4. Z11 = 2.25 $\Omega$and Z12 = 0.5 $\Omega$


Correct Option: A
Explanation:

 

Twelve 1$\Omega$ resistances are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is

  1. $\dfrac{5}{6}$ $\Omega$

  2. $\dfrac{1}{6}$ $\Omega$

  3. $\dfrac{6}{5}$ $\Omega$

  4. $\dfrac{3}{2}$ $\Omega$


Correct Option: A
Explanation:

For Current i, there are 3 similar paths. So current will be divide in 3 paths.

An input voltage v(t) = 10$\sqrt 2$ cos(t+100) + 10$\sqrt 3$ cos (2t+10o)V is applied to a series combination of resistance R = 1 $\Omega$ and an inductance L = 1 H. The resulting steady state current i(t) in ampere is

  1. 10 cos (t + 550) + 10 cos (2t + 100 + tan-12)

  2. 10 cos (t + 550) + 10 $\sqrt{ \dfrac{3}{2} }$ cos (2t + 550)

  3. 10 cos (t - 350) + 10 cos ( 2t + 100 - tan-12)

  4. 10 cos (t - 350) + 10 $\sqrt{ \dfrac{3}{2} }$cos (2t - 350)


Correct Option: C
Explanation:

The circuit shown in figure is best described as a

  1. bridge rectifier

  2. ring modulator

  3. frequency discriminatory

  4. voltage doubler


Correct Option: D
Explanation:

This circuit having two diode and capacitor pair in parallel, works as voltage doubler.

Choose proper substitutes for X and Y to make the following statement correct. Tunnel diode and Avalanche photodiode are operated in X bias and Y bias respectively.

  1. X: reverse, Y: reverse

  2. X: reverse, Y: forward

  3. X: forward, Y: reverse

  4. X: forward, Y: forward


Correct Option: C
Explanation:

Tunnel diode shows the negative characteristics in forward bias. It is used in forward bias.

Avalanche photo diode in reverse bias.

If the differential voltage gain and the common mode voltage gain of a differential amplifier are 48 dB and 2 dB respectively, then its common mode rejection ratio is

  1. 23 dB

  2. 25 dB

  3. 46 dB

  4. 50 dB


Correct Option: C
Explanation:

Generally, the gain of a transistor amplifier falls at high frequencies due to the

  1. internal capacitances of the device

  2. coupling capacitor at the input

  3. skin effect

  4. coupling capacitor at the output


Correct Option: A
Explanation:

If the input to the ideal comparator shown in figure is a sinusoidal signal of 8V (peak to peak) without any DC component, then the output of the comparator has a duty cycle of

  1. $\dfrac{1}{2}$

  2. $\dfrac{1}{3}$

  3. $\dfrac{1}{6}$

  4. $\dfrac{9}{12}$


Correct Option: B
Explanation:

Choose the correct match for input resistance of various amplifier configurations shown below ||| |---|---| | Configuration| Input resistance| | CB: Common Base| O: Low| | CC: Common Collector| MO: Moderate| | CE: Common Emitter| HI: High|

  1. CB-LO, CC-MO, CE-HI

  2. CB-LO, CC-HI, CE-MO

  3. CB-MO, CC-HI, CE-LO

  4. CB-HI, CC-LO, CE-MO


Correct Option: B
Explanation:

A PD controller is used to compensate a system. Compared to the uncompensated system, the compensated system has

  1. a higher type number

  2. reduced damping

  3. higher noise amplification

  4. larger transient overshoot


Correct Option: C
Explanation:

PD controller may accentuate noise at higher frequency. It does not affect the type of system and it increases the damping. It also reduces the maximum overshoot.

The figure shows the Nyquist plot of the open-loop transfer function G(s)H(s) of a system. If G(s)H(s) has one right hand pole, the closed loop system is

  1. always stable

  2. unstable with one closed loop right hand pole

  3. unstable with two closed loop right hand poles

  4. unstable with three closed loop right hand poles


Correct Option: A
Explanation:

An n-type silicon bar 0.1 cm long and $\mu$m2 in cross-sectional area has a majority carrier concentration of 5 $\times$ 1020/m3 and the carrier mobility is 0.13 m2/V-s at 300K. If the charge of an electron is 1.6$\times$10-19 coulomb, then the resistance of the bar is

  1. 106 ohm

  2. 104 ohm

  3. 10-1 ohm

  4. 10-4 ohm


Correct Option: A
Explanation:

The intrinsic carrier concentration of silicon sample of 300 K is 1.5 $\times$ 1016/m3. If after doping, the number of majority carriers is 5 $\times$ 1020/m3, the minority carrier density is

  1. 4.50 $\times$1011/m3

  2. 3.33 $\times$104/m3

  3. 5.00 $\times$1020/m3

  4. 3.00$\times$10-5/m3


Correct Option: A
Explanation:

The output voltage of the regulated power supply shown in figure is

  1. 3V

  2. 6V

  3. 9V

  4. 12V


Correct Option: C
Explanation:

What is the frequency of radio waves radiated out by an oscillating circuit consisting of condenser of capacity 0.02 microfarad and inductance 8 microhenry?

  1. 3.98 x 105 Hz

  2. 4.56 x 105 Hz

  3. 6.21 x 104 Hz

  4. 7.12 x 104 Hz


Correct Option: A
Explanation:

C = 0.02 x 10-6 F L = 8 x 10-6 H $\nu = \dfrac{1}{2\pi \sqrt{LC}}$  After putting the values, we get the frequency = 3.98 x 105 Hz

Match items in Group 1 with items in Group 2, most suitably ||| |---|---| | Group 1| Group 2| | P LED| 1 Heavy doping| | Q Avalanche photodiode| 2 Coherent radiation| | R Tunnel diode| 3 Spontaneous emission| | S LASER| 4 Current gain|

  1. P - 1, Q - 2, R - 4, S - 3

  2. P - 2, Q - 3, R - 1, S - 4

  3. P - 3, Q - 4, R - 1, S - 2

  4. P - 2, Q - 1, R - 4, S - 3


Correct Option: C
Explanation:

If the op-amp in figure is ideal, the output voltage Vout will be equal to

  1. 1 V

  2. 6 V

  3. 14 V

  4. 17 V


Correct Option: B
Explanation:

 

In the amplifier circuit shown in figure, the values of R1 and R2 are such that the transistor is operating at VCE = 3V and IC = 1.5mA when its $\beta$is 150. For a transistor with $\beta$of 200, the operating point (VCE, IC) is

  1. (2V, 2 mA)

  2. (3V, 2 mA)

  3. (4V, 2 mA)

  4. (4V, 1 mA)


Correct Option: A
Explanation:

 

An amplifier without feedback has a voltage gain of 50, input resistance of 1 K$\Omega$ and output resistance of 2.5 K$\Omega$. The input resistance of the current-shunt negative feedback amplifier using the above amplifier with a feedback factor of 0.2, is

  1. $\dfrac{1}{11}$ K$\Omega$

  2. $\dfrac{1}{5}$ K$\Omega$

  3. 5 K$\Omega$

  4. 11 K$\Omega$


Correct Option: A
Explanation:

Three identical amplifiers with each one having a voltage gain of 50, input resistance of 1 K$\Omega$and output resistance of 250 $\Omega$, are cascaded. The open circuit voltage gain of the combined amplifier is

  1. 49 dB

  2. 51 dB

  3. 98 dB

  4. 102 dB


Correct Option: C
Explanation:

An ideal sawtooth voltage waveform of frequency 500 Hz and amplitude 3 V is generated by charging a capacitor of 2 $\mu$F in every cycle. The charging requires

  1. constant voltage source of 3 V for 1 ms

  2. constant voltage source of 3 V for 2 ms

  3. constant current source of 3 mA for 1 ms

  4. constant current source of 3 mA for 2 ms


Correct Option: D
Explanation:

The signal flow graph of a system is shown in figure. The transfer function $\dfrac{C(s)}{R(s)}$ of the system is

  1. $\dfrac{6}{s^2+29s+6}$

  2. $\dfrac{6s}{s^2+29s+6}$

  3. $\dfrac{s(s+2)}{s^2+29s+6}$

  4. $\dfrac{s(s+27)}{s^2+29s+6}$


Correct Option: D
Explanation:

The root locus of the system G (s) H(s) = $\dfrac{k}{s(s+2)(s+3)}$ has the break - away point located at

  1. (- 0.5,0)

  2. (- 2.548,0)

  3. (- 4,0)

  4. (- 0.784,0)


Correct Option: D
Explanation:

The gain margin and the phase margin of a feedback system with G (s) H(s) = $\dfrac{s}{(s+100)^3}$are

  1. 0 dB, 0°

  2. $\infty$, $\infty$

  3. $\infty$, 0°

  4. 88.5 dB, $\infty$


Correct Option: B
Explanation:

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The approximate Bode magnitude plot of a minimum-phase system is shown in figure. The transfer function of the system is

  1. 108 $\dfrac{(s+0.1)^3}{(s+10)^2 (s+100)}$

  2. 107 $\dfrac{(s+0.1)^3}{(s+10) (s+100)}$

  3. 108 $\dfrac{(s+0.1)^2}{(s+10)^2 (s+100)}$

  4. 109 $\dfrac{(s+0.1)^3}{(s+10)^2 (s+100)^2}$


Correct Option: A
Explanation:

 

A second-order system has the transfer function $\dfrac{C(s)}{R(s)}$ = $\dfrac{4}{s^2+4s+4}$ with r(t) as the unit-step function, the response c(t) of the system is represented by


Correct Option: D
Explanation:

The zero-input response of a system given by the state-space equation is

  1. $\left[ \begin{array} \ te^t \\ t \end{array} \right]$

  2. $\left[ \begin{array} \ e^t \\ t \end{array} \right]$

  3. $\left[ \begin{array} \ e^t \\ te^t \end{array} \right]$

  4. $\left[ \begin{array} \ t \\ te^t \end{array} \right]$* 1/2


Correct Option: C
Explanation:

A short-circuited stub is shunt connected to a transmission line as shown in figure. If Z0 = 50 ohm, the admittance Y seen at the junction of the stub and the transmission line is

  1. (0.01 - j0.02) ohm

  2. (0.02 - j0.01) ohm

  3. (0.04 - j0.02) ohm

  4. (0.02 + j0) ohm


Correct Option: A
Explanation:

If the electric field intensity associated with a uniform plane electromagnetic wave traveling in a perfect dielectric medium is given by E(z,t) = 10 cos (2$\pi$$\times$ 107t = 0.1$\pi$z) volt/m, then the velocity of the traveling wave is

  1. 3.00 $\times$ 108 m/sec

  2. 2.00 $\times$ 108 m/sec

  3. 6.28 $\times$ 107 m/sec

  4. 2.00 $\times$ 107 m/sec


Correct Option: B
Explanation:

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