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1361
GATE ECE 2007 | Question: 16
If the Laplace transform of a signal $y(t)$ is $Y(s)=\dfrac{1}{s(s-1)},$ then its final value is $-1$ $0$ $1$ Unbounded
If the Laplace transform of a signal $y(t)$ is $Y(s)=\dfrac{1}{s(s-1)},$ then its final value is$-1$$0$$1$Unbounded
1 votes
0 answers
1362
GATE ECE 2007 | Question: 17
If $R(\tau)$ is the autocorrelation function of a real, wide-sense stationary random process, then which of the following is $\text{NOT}$ true? $R(\tau)=R(-\tau)$ $|R(\tau)| \leq R(0)$ $R(\tau)=-R(-\tau)$ The mean square value of the process is $R(0)$
If $R(\tau)$ is the autocorrelation function of a real, wide-sense stationary random process, then which of the following is $\text{NOT}$ true?$R(\tau)=R(-\tau)$$|R(\tau)...
1 votes
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1363
GATE ECE 2007 | Question: 18
If $S(f)$ is the power spectral density of a real, wide-sense stationary random process, then which of the following is $\text{ALWAYS}$ true? $S(0) \geq S(f)$ $S(f) \geq 0$ $S(-f)=-S(f)$ $\displaystyle{}\int_{-\infty}^{\infty} S(f) d f=0$
If $S(f)$ is the power spectral density of a real, wide-sense stationary random process, then which of the following is $\text{ALWAYS}$ true?$S(0) \geq S(f)$$S(f) \geq 0$...
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1365
GATE ECE 2007 | Question: 20
If $\mathrm{C}$ is a closed curve enclosing a surface $S$, then the magnetic field intensity $\vec{H}$, the current density $\vec{J}$ and the electric flux density $\vec{D}$ ...
If $\mathrm{C}$ is a closed curve enclosing a surface $S$, then the magnetic field intensity $\vec{H}$, the current density $\vec{J}$ and the electric flux density $\vec{...
1 votes
0 answers
1366
GATE ECE 2007 | Question: 21
It is given that $X_{1}, X_{2}, \cdots ,X_{M}$ are $M$ non-zero, orthogonal vectors. The dimension of the vector space spanned by the $2 M$ vectors $X_{1}, X_{2}, \cdots, X_{M},-X_{1},-X_{2}, \cdots,-X_{M}$ is $2 M$ $M+1$ $M$ dependent on the choice of $X_{1}, X_{2}, \cdots, X_{M}$
It is given that $X_{1}, X_{2}, \cdots ,X_{M}$ are $M$ non-zero, orthogonal vectors. The dimension of the vector space spanned by the $2 M$ vectors $X_{1}, X_{2}, \cdots,...
1 votes
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1367
GATE ECE 2007 | Question: 22
Consider the function $f(x)=x^{2}-x-2$. The maximum value of $f(x)$ in the closed interval $[-4,4]$ is $18$ $10$ $-2.25$ indeterminate
Consider the function $f(x)=x^{2}-x-2$. The maximum value of $f(x)$ in the closed interval $[-4,4]$ is$18$$10$$-2.25$indeterminate
1 votes
0 answers
1369
GATE ECE 2007 | Question: 24
The solution of the differential equation $k^{2} \dfrac{d^{2} y}{d x^{2}}=y-y_{2}$ under the boundary conditions (i) $y=y_{1}$ at $x=0$ and (ii) $y=y_{2}$ at $x=\infty$, where $k, y_{1}$ and $y_{2}$ ... $y=\left(y_{1}-y_{2}\right) \sinh (x / k)+y_{1}$ $y=\left(y_{1}-y_{2}\right) \exp (-x / k)+y_{2}$
The solution of the differential equation $k^{2} \dfrac{d^{2} y}{d x^{2}}=y-y_{2}$ under the boundary conditions (i) $y=y_{1}$ at $x=0$ and (ii) $y=y_{2}$ at $x=\infty$, ...
1 votes
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1370
GATE ECE 2007 | Question: 25
The equation $x^{3}-x^{2}+4 x-4=0$ is to be solved using the Newton-Raphson method. If $x=2$ is taken as the initial approximation of the solution, then the next approximation using this method will be $\frac{2}{3}$ $\frac{4}{3}$ $1$ $\frac{3}{2}$
The equation $x^{3}-x^{2}+4 x-4=0$ is to be solved using the Newton-Raphson method. If $x=2$ is taken as the initial approximation of the solution, then the next approxim...
1 votes
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1372
GATE ECE 2007 | Question: 27
If the semi-circular contour $D$ of radius $2$ is as shown in the figure, then the value of the integral $\displaystyle{}\oint_{D} \dfrac{1}{\left(s^{2}-1\right)} d s$ is $j \pi$ $-j \pi$ $-\pi$ $\pi$
If the semi-circular contour $D$ of radius $2$ is as shown in the figure, then the value of the integral $\displaystyle{}\oint_{D} \dfrac{1}{\left(s^{2}-1\right)} d s$ is...
1 votes
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1373
GATE ECE 2007 | Question: 28
Two series resonant filters are as shown in the figure. Let the $\text{3-dB}$ bandwidth of Filter $1$ be $B_{1}$ and that of Filter $2$ be $B_{2}$. The value of $\dfrac{B_{1}}{B_{2}}$ is $4$ $1$ $\frac{1}{2}$ $\frac{1}{4}$
Two series resonant filters are as shown in the figure. Let the $\text{3-dB}$ bandwidth of Filter $1$ be $B_{1}$ and that of Filter $2$ be $B_{2}$. The value of $\dfrac{B...
1 votes
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1374
GATE ECE 2007 | Question: 29
For the circuit shown in the figure, the Thevenin voltage and resistance looking into $\mathrm{X}-\mathrm{Y}$ are $4 / 3 \mathrm{~V}, 2 \; \Omega$ $4 \mathrm{~V}, 2 / 3 \; \Omega$ $4 / 3 \mathrm{~V}, 2 / 3 \; \Omega$ $4 \mathrm{~V}, 2 \; \Omega$
For the circuit shown in the figure, the Thevenin voltage and resistance looking into $\mathrm{X}-\mathrm{Y}$ are$4 / 3 \mathrm{~V}, 2 \; \Omega$$4 \mathrm{~V}, 2 / 3 \; ...
1 votes
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1375
GATE ECE 2007 | Question: 30
In the circuit shown, $V_{C}$ is $0$ volts at $t=0 \; \mathrm{sec}$. For $t>0$, the capacitor current $i_{c}(t)$, where $t$ is in seconds, is given by $0.50 \exp (-25 t) \; \mathrm{mA}$ $0.25 \exp (-25 t) \; \mathrm{mA}$ $0.50 \exp (-12.5 t) \; \mathrm{mA}$ $0.25 \exp (-6.25 t) \; \mathrm{mA}$
In the circuit shown, $V_{C}$ is $0$ volts at $t=0 \; \mathrm{sec}$. For $t>0$, the capacitor current $i_{c}(t)$, where $t$ is in seconds, is given by$0.50 \exp (-25 t) \...
1 votes
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1376
GATE ECE 2007 | Question: 31
In the $\text{AC}$ network shown in the figure, the phasor voltage $\text{V}_{\text{AB}}$ (in Volts) is $0$ $5 \angle 30^{\circ}$ $12.5 \angle 30^{\circ}$ $17 \angle 30^{\circ}$
In the $\text{AC}$ network shown in the figure, the phasor voltage $\text{V}_{\text{AB}}$ (in Volts) is$0$$5 \angle 30^{\circ}$$12.5 \angle 30^{\circ}$$17 \angle 30^{\cir...
1 votes
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1378
GATE ECE 2007 | Question: 33
Group I lists four types of $p-n$ ... $\mathrm{P}-2, \mathrm{Q}-1, \mathrm{R}-2, \mathrm{S}-2$
Group I lists four types of $p-n$ junction diodes. Match each device in Group I with one of the options in Group II to indicate the bias condition of that device in its n...
1 votes
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1379
GATE ECE 2007 | Question: 34
The $\text{DC}$ current gain ($\beta$ ) of a $\text{BJT}$ is $50.$ Assuming that the emitter injection efficiency is $0.995$, the base transport factor is $0.980$ $0.985$ $0.990$ $0.995$
The $\text{DC}$ current gain ($\beta$ ) of a $\text{BJT}$ is $50.$ Assuming that the emitter injection efficiency is $0.995$, the base transport factor is$0.980$$0.985$$0...
1 votes
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1380
GATE ECE 2007 | Question: 35
Group I lists four different semiconductor devices. Match each device in Group I with its characteristic property in Group II. ... $\text{P-3, Q-4, R-1, S-2}$ $\text{P-3, Q-2, R-1, S-4}$
Group I lists four different semiconductor devices. Match each device in Group I with its characteristic property in Group II.$$\begin{array}{ll} \textbf{Group I} & \qqua...
1 votes
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1381
GATE ECE 2007 | Question: 36
For the $\text{Op-Amp}$ circuit shown in the figure, $V_{o}$ is $-2 \mathrm{~V}$ $-1 \mathrm{~V}$ $-0.5 \mathrm{~V}$ $0.5 \mathrm{~V}$
For the $\text{Op-Amp}$ circuit shown in the figure, $V_{o}$ is$-2 \mathrm{~V}$$-1 \mathrm{~V}$$-0.5 \mathrm{~V}$$0.5 \mathrm{~V}$
1 votes
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1382
GATE ECE 2007 | Question: 37
For the $\text{BJT}$ circuit shown, assume that the $\beta$ of the transistor is very large and $V_{B E}=0.7 \mathrm{~V}$. The mode of operation of the $\text{BJT}$ is cut-off saturation normal active reverse active
For the $\text{BJT}$ circuit shown, assume that the $\beta$ of the transistor is very large and $V_{B E}=0.7 \mathrm{~V}$. The mode of operation of the $\text{BJT}$ iscut...
1 votes
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1386
GATE ECE 2007 | Question: 41
The Boolean expression $Y=\overline{A} \;\overline{B}\; \overline{C} D+\overline{A} B C \overline{D}+A \overline{B}\; \overline{C} D+A B \overline{C}\; \overline{D}$ ... $Y=\overline{A} B C \overline{D}+\overline{B}\; \overline{C} D+A B \overline{C} \;\overline{D}$
The Boolean expression $Y=\overline{A} \;\overline{B}\; \overline{C} D+\overline{A} B C \overline{D}+A \overline{B}\; \overline{C} D+A B \overline{C}\; \overline{D}$ can ...
1 votes
1 answer
1389
GATE ECE 2007 | Question: 44
The following binary values were applied to the $\mathrm{X}$ and $\mathrm{Y}$ inputs of the NAND latch shown in the figure in the sequence indicated below: \[ \mathrm{X}=0, \mathrm{Y}=1 ; \quad \mathrm{X}=0, \mathrm{Y}=0 ; \quad \mathrm{X}=1, \mathrm{Y}=1 \ ...
The following binary values were applied to the $\mathrm{X}$ and $\mathrm{Y}$ inputs of the NAND latch shown in the figure in the sequence indicated below:\[ \mathrm{X}=0...
1 votes
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1390
GATE ECE 2007 | Question: 45
For the circuit shown, the counter state $\left(\mathrm{Q}_{1} \mathrm{Q}_{0}\right)$ follows the sequence $00,01,10,11,00 \ldots$ $00,01,10,00,01 \ldots$ $00,01,11,00,01 \ldots$ $00,10,11,00,10 \ldots$
For the circuit shown, the counter state $\left(\mathrm{Q}_{1} \mathrm{Q}_{0}\right)$ follows the sequence$00,01,10,11,00 \ldots$$00,01,10,00,01 \ldots$$00,01,11,00,01 \l...
1 votes
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1392
GATE ECE 2007 | Question: 47
The $\text{3-dB}$ bandwidth of the low-pass signal $e^{-t} u(t)$, where $u(t)$ is the unit step function, is given by $\frac{1}{2 \pi} \mathrm{Hz}$ $\frac{1}{2 \pi} \sqrt{\sqrt{2}-1} \mathrm{~Hz}$ $\infty$ $1 \mathrm{~Hz}$
The $\text{3-dB}$ bandwidth of the low-pass signal $e^{-t} u(t)$, where $u(t)$ is the unit step function, is given by$\frac{1}{2 \pi} \mathrm{Hz}$$\frac{1}{2 \pi} \sqrt{\...
1 votes
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1393
GATE ECE 2007 | Question: 48
A Hilbert transformer is a non-linear system non-causal system time-varying system low-pass system
A Hilbert transformer is anon-linear systemnon-causal systemtime-varying systemlow-pass system
1 votes
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1394
GATE ECE 2007 | Question: 49
The frequency response of a linear, time-invariant system is given by $H(f)=\dfrac{5}{1+j 10 \pi f}$. The step response of the system is $5\left(1-e^{-5 t}\right) u(t)$ $5\left(1-e^{-\frac{t}{5}}\right) u(t)$ $\frac{1}{5}\left(1-e^{-5 t}\right) u(t)$ $\frac{1}{5}\left(1-e^{-\frac{t}{5}}\right) u(t)$
The frequency response of a linear, time-invariant system is given by $H(f)=\dfrac{5}{1+j 10 \pi f}$. The step response of the system is$5\left(1-e^{-5 t}\right) u(t)$$5\...
1 votes
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1396
GATE ECE 2007 | Question: 51
The $z$-transform $X[z]$ of a sequence $x[n]$ is given by $X[z]=\dfrac{0.5}{1-2 z^{-1}}$. It is given that the region of convergence of $X[z]$ includes the unit circle. The value of $x[0]$ is $-0.5$ $0$ $0.25$ $0.5$
The $z$-transform $X[z]$ of a sequence $x[n]$ is given by $X[z]=\dfrac{0.5}{1-2 z^{-1}}$. It is given that the region of convergence of $X[z]$ includes the unit circle. T...
1 votes
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1397
GATE ECE 2007 | Question: 52
A control system with a $\text{PD}$ controller is shown in the figure. If the velocity error constant $K_{v}=1000$ and the damping ratio $\zeta=0.5$, then the values of $K_{P}$ and $K_{D}$ are $K_{P}=100, K_{D}=0.09$ $K_{P}=100, K_{D}=0.9$ $K_{P}=10, K_{D}=0.09$ $K_{P}=10, K_{D}=0.9$
A control system with a $\text{PD}$ controller is shown in the figure. If the velocity error constant $K_{v}=1000$ and the damping ratio $\zeta=0.5$, then the values of $...
1 votes
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1398
GATE ECE 2007 | Question: 53
The transfer function of a plant is $T(s)=\dfrac{5}{(s+5)\left(s^{2}+s+1\right)}$. The second-order approximation of $T(s)$ using dominant pole concept is $\frac{1}{(s+5)(s+1)}$ $\frac{5}{(s+5)(s+1)}$ $\frac{5}{s^{2}+s+1}$ $\frac{1}{s^{2}+s+1}$
The transfer function of a plant is $T(s)=\dfrac{5}{(s+5)\left(s^{2}+s+1\right)}$. The second-order approximation of $T(s)$ using dominant pole concept is$\frac{1}{(s+5)(...
1 votes
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1399
GATE ECE 2007 | Question: 54
The open-loop transfer function of a plant is given as $G(s)=\dfrac{1}{s^{2}-1}$. If the plant is operated in a unity feedback configuration, then the lead compensator that can stabilize this control system is $\frac{10(s-1)}{s+2}$ $\frac{10(s+4)}{s+2}$ $\frac{10(s+2)}{s+10}$ $\frac{2(s+2)}{s+10}$
The open-loop transfer function of a plant is given as $G(s)=\dfrac{1}{s^{2}-1}$. If the plant is operated in a unity feedback configuration, then the lead compensator th...
1 votes
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1400
GATE ECE 2007 | Question: 55
A unity feedback control system has an open-loop transfer function $G(s)=\dfrac{K}{s\left(s^{2}+7 s+12\right)}$. The gain $K$ for which $s=-1+j 1$ will lie on the root locus of this system is $4$ $5.5$ $6.5$ $10$
A unity feedback control system has an open-loop transfer function $G(s)=\dfrac{K}{s\left(s^{2}+7 s+12\right)}$. The gain $K$ for which $s=-1+j 1$ will lie on the root lo...
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