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1641
GATE ECE 2010 | Question: 8
In the silicon $\text{BJT}$ circuit shown below, assume that the emitter area of transistor $\text{Q1}$ is half that of transistor $\text{Q2.}$ The value of current $\text{I}_0$ is approximately $0.5 \mathrm{~mA}$ $2 \mathrm{~mA}$ $9.3 \mathrm{~mA}$ $15 \mathrm{~mA}$
In the silicon $\text{BJT}$ circuit shown below, assume that the emitter area of transistor $\text{Q1}$ is half that of transistor $\text{Q2.}$The value of current $\text...
1 votes
0 answers
1643
GATE ECE 2010 | Question: 10
Assuming the $\text{OP-AMP}$ to be ideal. the voltage gain of the amplifier shown below is $-\frac{\mathrm{R}_2}{\mathrm{R}_1}$ $-\frac{\mathrm{R}_3}{\mathrm{R}_1}$ $-\left(\frac{\mathrm{R}_2 \| \mathrm{R}_3}{\mathrm{R}_1}\right)$ $-\left(\frac{\mathrm{R}_2+\mathrm{R}_3}{\mathrm{R}_1}\right)$
Assuming the $\text{OP-AMP}$ to be ideal. the voltage gain of the amplifier shown below is$-\frac{\mathrm{R}_2}{\mathrm{R}_1}$$-\frac{\mathrm{R}_3}{\mathrm{R}_1}$$-\left(...
2 votes
1 answer
1644
GATE ECE 2010 | Question: 11
Match the logic gates in Column A with their equivalents in Column B. $\text{P-2, Q-4, R-1, S-3}$ $\text{P-4, Q-2, R-1, S-3}$ $\text{P-2, Q-4, R-3, S-1}$ $\text{P-4, Q-2, R-3, S-1}$
Match the logic gates in Column A with their equivalents in Column B.$\text{P-2, Q-4, R-1, S-3}$$\text{P-4, Q-2, R-1, S-3}$$\text{P-2, Q-4, R-3, S-1}$$\text{P-4, Q-2, R-3...
2 votes
1 answer
1645
GATE ECE 2010 | Question: 12
For the output $\text{F}$ to be $1$ in the logic circuit shown, the input combination should be $\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0$ $\text{A = 1, B = 0, C = 0}$ $\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=0$ $\text{A = 0, B = 0, C = 1}$
For the output $\text{F}$ to be $1$ in the logic circuit shown, the input combination should be$\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0$$\text{A = 1, B = 0, C = 0}$$\ma...
1 votes
0 answers
1646
GATE ECE 2010 | Question: 13
In the circuit shown, the device connected to $\text{Y5}$ can have address in the range $2000 – \mathrm{20FF}$ $\mathrm{2D00} – \mathrm{2DEF}$ $\mathrm{2E00} – \mathrm{2EFF}$ $\text{FD00 - FDFF}$
In the circuit shown, the device connected to $\text{Y5}$ can have address in the range$2000 – \mathrm{20FF}$$\mathrm{2D00} – \mathrm{2DEF}$$\mathrm{2E00} – \mathr...
1 votes
0 answers
1647
GATE ECE 2010 | Question: 14
Consider the $z$-transform $X(z)=5 z^2+4 z^{-1}+3 ; 0<|z|<\infty$. The inverse $z$-transform $x[n]$ is $5\; \delta[n+2]+3\; \delta[n]+4\; \delta[n-1]$ $5\; \delta[n-2]+3\; \delta[n]+4\; \delta[n+1]$ $5\; u[n+2]+3\; u[n]+4\; u[n-1]$ $5\; u[n-2]+3\; u[n]+4\; u[n+1]$
Consider the $z$-transform $X(z)=5 z^2+4 z^{-1}+3 ; 0<|z|<\infty$. The inverse $z$-transform $x[n]$ is$5\; \delta[n+2]+3\; \delta[n]+4\; \delta[n-1]$$5\; \delta[n-2]+3\; ...
1 votes
0 answers
1648
GATE ECE 2010 | Question: 15
Two discrete time systems with impulse responses $h_t[n]=\delta[n-1]$ and $h_2[n]=\delta[n-2]$ are connected in cascade. The overall impulse response of the cascaded system is $\delta[n-1]+ \delta[n-2]$ $\delta[n-4]$ $\delta[n-3]$ $\delta[n-1] \delta[n-2]$
Two discrete time systems with impulse responses $h_t[n]=\delta[n-1]$ and $h_2[n]=\delta[n-2]$ are connected in cascade. The overall impulse response of the cascaded syst...
1 votes
0 answers
1650
GATE ECE 2010 | Question: 17
The transfer function $\text{Y(s) / R(s)}$ of the system shown is $0$ $\frac{1}{\text{s}+1}$ $\frac{2}{\text{s}+1}$ $\frac{2}{\text{s}+3}$
The transfer function $\text{Y(s) / R(s)}$ of the system shown is$0$$\frac{1}{\text{s}+1}$$\frac{2}{\text{s}+1}$$\frac{2}{\text{s}+3}$
1 votes
0 answers
1651
GATE ECE 2010 | Question: 18
A system with the transfer function $\frac{Y(s)}{X(s)}=\frac{s}{s+p}$ has an output $y(t)=\cos \left(2 t-\frac{\pi}{3}\right)$ for the input signal $x(t)=p \cos \left(2 t-\frac{\pi}{2}\right)$. Then, the system parameter $’p’$ is $\sqrt{3}$ $\frac{2}{\sqrt{3}}$ $1$ $\frac{\sqrt{3}}{2}$
A system with the transfer function $\frac{Y(s)}{X(s)}=\frac{s}{s+p}$ has an output $y(t)=\cos \left(2 t-\frac{\pi}{3}\right)$ for the input signal $x(t)=p \cos \left(2 t...
1 votes
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1652
GATE ECE 2010 | Question: 19
For the asymptotic Bode magnitude plot shown below, the system transfer function can be $\frac{10 \text{s}+1}{0.1 \text{s}+1}$ $\frac{100 \text{s}+1}{0.1 \text{s}+1}$ $\frac{100 \mathrm{s}}{10 \mathrm{s}+1}$ $\frac{0.1 \text{s}+1}{10 \text{s}+1}$
For the asymptotic Bode magnitude plot shown below, the system transfer function can be$\frac{10 \text{s}+1}{0.1 \text{s}+1}$$\frac{100 \text{s}+1}{0.1 \text{s}+1}$$\frac...
1 votes
0 answers
1653
GATE ECE 2010 | Question: 20
Suppose that the modulating signal is $m(t)=2 \cos \left(2 \pi f_{m} t\right)$ and the carrier signal is $x_{c}(t)=A_{C} \cos \left(2 \pi f_{c}t\right)$. Which one of the following is a conventional $\text{AM}$ ...
Suppose that the modulating signal is $m(t)=2 \cos \left(2 \pi f_{m} t\right)$ and the carrier signal is $x_{c}(t)=A_{C} \cos \left(2 \pi f_{c}t\right)$. Which one of the...
1 votes
0 answers
1654
GATE ECE 2010 | Question: 21
Consider an angle modulated signal $x(t)=6 \cos \left[2 \pi \times 10^{6} t+2 \sin (8000 \pi t)+4 \cos (8000 \pi t)\right] \mathrm{V}$. The average power of $x(t)$ is $10 \mathrm{~W}$ $18 \mathrm{~W}$ $20 \mathrm{~W}$ $28 \mathrm{~W}$
Consider an angle modulated signal $x(t)=6 \cos \left[2 \pi \times 10^{6} t+2 \sin (8000 \pi t)+4 \cos (8000 \pi t)\right] \mathrm{V}$. The average power of $x(t)$ is$10 ...
1 votes
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1656
GATE ECE 2010 | Question: 23
A transmission line has a characteristic impedance of $50 \; \Omega$ and a resistance of $0.1 \; \Omega / \mathrm{m}$. If the line is distortionless, the attenuation constant (in $\mathrm{Np/m})$ is $500$ $5$ $0.014$ $0.002$
A transmission line has a characteristic impedance of $50 \; \Omega$ and a resistance of $0.1 \; \Omega / \mathrm{m}$. If the line is distortionless, the attenuation cons...
1 votes
0 answers
1657
GATE ECE 2010 | Question: 24
Consider the pulse shape $s(t)$ as shown. The impulse response $h(t)$ of the filter matched to this pulse is ​​​​​​​
Consider the pulse shape $s(t)$ as shown. The impulse response $h(t)$ of the filter matched to this pulse is ​​​​​​​
1 votes
0 answers
1658
GATE ECE 2010 | Question: 25
The electric field component of a time harmonic plane $\text{EM}$ wave traveling in a nonmagnetic lossless dielectric medium has an amplitude of $1 \; \mathrm{V/m}$. If the relative permittivity of the medium is $4$ ... is $\frac{1}{30 \pi}$ $\frac{1}{60 \pi}$ $\frac{1}{120 \pi}$ $\frac{1}{240 \pi}$
The electric field component of a time harmonic plane $\text{EM}$ wave traveling in a nonmagnetic lossless dielectric medium has an amplitude of $1 \; \mathrm{V/m}$. If t...
1 votes
0 answers
1659
GATE ECE 2010 | Question: 26
If $e^{y}=x^{\frac{1}{x}}$, then $y$ has a maximum at $x=e$ minimum at $x=e$ maximum at $x=e^{-1}$ minimum at $x=e^{-1}$
If $e^{y}=x^{\frac{1}{x}}$, then $y$ has amaximum at $x=e$minimum at $x=e$maximum at $x=e^{-1}$minimum at $x=e^{-1}$
1 votes
0 answers
1660
GATE ECE 2010 | Question: 27
A fair coin is tossed independently four times. The probability of the event "the number of times heads show up is more than the number of times tails show up" is $\frac{1}{16}$ $\frac{1}{8}$ $\frac{1}{4}$ $\frac{5}{16}$
A fair coin is tossed independently four times. The probability of the event "the number of times heads show up is more than the number of times tails show up" is$\frac{1...
1 votes
0 answers
1661
GATE ECE 2010 | Question: 28
If $\vec{A}=x y \hat{a}_{x}+x^{2} \hat{a}_{y}$, then $\oint_{c} \vec{A} \cdot d \vec{l}$ over the path shown in the figure is $0$ $\frac{2}{\sqrt{3}}$ $1$ $2 \sqrt{3}$
If $\vec{A}=x y \hat{a}_{x}+x^{2} \hat{a}_{y}$, then $\oint_{c} \vec{A} \cdot d \vec{l}$ over the path shown in the figure is$0$$\frac{2}{\sqrt{3}}$$1$$2 \sqrt{3}$
1 votes
0 answers
1662
GATE ECE 2010 | Question: 29
The residues of a complex function $X(z)=\dfrac{1-2 z}{z(z-1)(z-2)}$ at its poles are $\frac{1}{2},-\frac{1}{2}$ and $1$ $\frac{1}{2}, \frac{1}{2}$ and $-1$ $\frac{1}{2}, 1$ and $-\frac{3}{2}$ $\frac{1}{2},-1$ and $\frac{3}{2}$
The residues of a complex function $X(z)=\dfrac{1-2 z}{z(z-1)(z-2)}$ at its poles are$\frac{1}{2},-\frac{1}{2}$ and $1$$\frac{1}{2}, \frac{1}{2}$ and $-1$$\frac{1}{2}, 1$...
1 votes
0 answers
1663
GATE ECE 2010 | Question: 30
Consider a differential equation $\dfrac{d y(x)}{d x}-y(x)=x$ with the initial condition $y(0)=0$. Using Euler's first order method with a step size of $0.1$, the value of $y(0.3)$ is $0.01$ $0.031$ $0.0631$ $0.1$
Consider a differential equation $\dfrac{d y(x)}{d x}-y(x)=x$ with the initial condition $y(0)=0$. Using Euler's first order method with a step size of $0.1$, the value o...
1 votes
0 answers
1664
GATE ECE 2010 | Question: 31
Given $f(t)=\mathscr{L}^{-1}\left[\dfrac{3 s+1}{s^{3}+4 s^{2}+(K-3) s}\right]$. If $\displaystyle{}\lim _{t \rightarrow \infty} f(t)=1$, then the value of $K$ is $1$ $2$ $3$ $4$
Given $f(t)=\mathscr{L}^{-1}\left[\dfrac{3 s+1}{s^{3}+4 s^{2}+(K-3) s}\right]$. If $\displaystyle{}\lim _{t \rightarrow \infty} f(t)=1$, then the value of $K$ is$1$$2$$3$...
1 votes
0 answers
1665
GATE ECE 2010 | Question: 32
In the circuit shown, the switch $S$ is open for a long time and is closed at $t=0$. The current $i(t)$ for $t \geq 0$ is $i(t)=0.5-0.125 e^{-1000t} \mathrm{~A}$ $i(t)=1.5-0.125 e^{-1000t} \mathrm{~A}$ $i(t)=0.5-0.5 e^{-1000t} \mathrm{~A}$ $i(t)=0.375 e^{- 1000t} \mathrm{~A}$
In the circuit shown, the switch $S$ is open for a long time and is closed at $t=0$. The current $i(t)$ for $t \geq 0$ is $i(t)=0.5-0.125 e^{-1000t} \mathrm{~A}$$i(t)=1.5...
1 votes
0 answers
1666
GATE ECE 2010 | Question: 33
The current $\mathrm{I}$ in the circuit shown is $\text{-j1 A}$ $\text{j1 A}$ $0 \mathrm{~A}$ $20 \mathrm{~A}$
The current $\mathrm{I}$ in the circuit shown is $\text{-j1 A}$$\text{j1 A}$$0 \mathrm{~A}$$20 \mathrm{~A}$
1 votes
0 answers
1667
GATE ECE 2010 | Question: 34
In the circuit shown, the power supplied by the voltage source is $0 \mathrm{~W}$ $5 \mathrm{~W}$ $10 \mathrm{~W}$ $100 \mathrm{~W}$
In the circuit shown, the power supplied by the voltage source is$0 \mathrm{~W}$$5 \mathrm{~W}$$10 \mathrm{~W}$$100 \mathrm{~W}$
2 votes
1 answer
1670
GATE ECE 2010 | Question: 37
Assuming that all flip-flops are in reset condition initially, the count sequence observed at $\text{Q}_\text{A}$ in the circuit shown is $0010111 \ldots$ $0001011 \ldots$ $0101111 \ldots$ $0110100 \ldots$
Assuming that all flip-flops are in reset condition initially, the count sequence observed at $\text{Q}_\text{A}$ in the circuit shown is$0010111 \ldots$$0001011 \ldots$$...
1 votes
0 answers
1671
GATE ECE 2010 | Question: 38
The transfer characteristic for the precision rectifier circuit shown below is (assume ideal $\text{OP-AMP}$ and practical diodes)
The transfer characteristic for the precision rectifier circuit shown below is (assume ideal $\text{OP-AMP}$ and practical diodes)
2 votes
1 answer
1672
GATE ECE 2010 | Question: 39
The Boolean function realized by the logic circuit shown is $\text{F} = \sum_{\text{m}} (0, 1, 3, 5, 9, 10, 14)$ $\text{F}=\sum_{\text{m}}(2,3,5,7,8,12,13)$ $\text{F}=\sum_{\text{m}}(1,2,4,5,11, 14,15)$ $\text{F}= \sum_{\text{m}}(2,3,5,7,8,9,12)$
The Boolean function realized by the logic circuit shown is$\text{F} = \sum_{\text{m}} (0, 1, 3, 5, 9, 10, 14)$$\text{F}=\sum_{\text{m}}(2,3,5,7,8,12,13)$$\text{F}=\sum_{...
1 votes
0 answers
1673
GATE ECE 2010 | Question: 40
For the $8085$ ... $\text{00H}$ $45 \text{H}$ $\text{67H}$ $\text{E7H}$
For the $8085$ assembly language program given below, the content of the accumulator after the execution of the program is$$\begin{array}{|lll|}\hline 3000 & \text{MVI} &...
1 votes
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1676
GATE ECE 2010 | Question: 43
The Nyquist sampling rale for the signal $s(t)=\dfrac{\sin (500 \pi t)}{\pi t} \times \dfrac{\sin (700 \pi t)}{\pi t}$ is given by $400 \mathrm{~Hz}$ $600 \mathrm{~Hz}$ $1200 \mathrm{~Hz}$ $1400 \mathrm{~Hz}$
The Nyquist sampling rale for the signal $s(t)=\dfrac{\sin (500 \pi t)}{\pi t} \times \dfrac{\sin (700 \pi t)}{\pi t}$ is given by$400 \mathrm{~Hz}$$600 \mathrm{~Hz}$$120...
1 votes
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1677
GATE ECE 2010 | Question: 44
A unity negative feedback closed loop system has a plant with the transfer function $\text{G}(s)=\frac{1}{s^{2}+2 s+2}$ and a controller $\text{G}_{c}(s)$ ... $\text{G}_{c}(s)=1+\frac{2}{s}+3 s$
A unity negative feedback closed loop system has a plant with the transfer function $\text{G}(s)=\frac{1}{s^{2}+2 s+2}$ and a controller $\text{G}_{c}(s)$ in the feedforw...
1 votes
0 answers
1680
GATE ECE 2010 | Question: 47
In the circuit shown, all the Transmission line sections are lossless. The Voltage Standing Wave Ratio $\text{(VSWR)}$ on the $60 \Omega$ line is $1.00$ $1.64$ $2.50$ $3.00$
In the circuit shown, all the Transmission line sections are lossless. The Voltage Standing Wave Ratio $\text{(VSWR)}$ on the $60 \Omega$ line is$1.00$$1.64$$2.50$$3.00$...
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