Questions by admin / GO Electronics

1 votes

0 answers

1801

GATE ECE 2011 | Question: 11
In the circuit shown below, the value of $\mathrm{R}_{\mathrm{L}}$ such that the power transferred to $\mathrm{R}_{\mathrm{L}}$ is maximum is $5\; \Omega$ $10\; \Omega$ $15\; \Omega$ $20\; \Omega$

In the circuit shown below, the value of $\mathrm{R}_{\mathrm{L}}$ such that the power transferred to $\mathrm{R}_{\mathrm{L}}$ is maximum is$5\; \Omega$$10\; \Omega$$15\...

112 views

asked Sep 3, 2022

1 votes

0 answers

1802

GATE ECE 2011 | Question: 12
The value of the integral $\oint_c \frac{-3 z+4}{\left(z^2+4 z+5\right)} d z$ where $c$ is the circle $|z|=1$ is given by $0$ $1 / 10$ $4 / 5$ $1$

The value of the integral $\oint_c \frac{-3 z+4}{\left(z^2+4 z+5\right)} d z$ where $c$ is the circle $|z|=1$ is given by$0$$1 / 10$$4 / 5$$1$

100 views

asked Sep 3, 2022

1 votes

0 answers

1803

GATE ECE 2011 | Question: 13
A transmission line of characteristic impedance $50\; \Omega$ is terminated by a $50\; \Omega$ load. When excited by a sinusoidal voltage source at $10\; \mathrm{GHz}$, the phase difference between two points spaced $2 \mathrm{~mm}$ ... $1.6 \times 10^8 \mathrm{~m} / \mathrm{s}$ $3 \times 10^8 \mathrm{~m} / \mathrm{s}$

A transmission line of characteristic impedance $50\; \Omega$ is terminated by a $50\; \Omega$ load. When excited by a sinusoidal voltage source at $10\; \mathrm{GHz}$, t...

103 views

asked Sep 3, 2022

1 votes

0 answers

1804

GATE ECE 2011 | Question: 14
Consider the following statements regarding the complex Poynting vector $\vec{P}$ for the power radiated by a point source in an infinite homogeneous and lossless medium. $\operatorname{Re}(\vec{P})$ denotes the real part of $\vec{P}, S$ denotes a ... decreases with increasing radial distance from the source

Consider the following statements regarding the complex Poynting vector $\vec{P}$ for the power radiated by a point source in an infinite homogeneous and lossless medium....

100 views

asked Sep 3, 2022

1 votes

0 answers

1805

GATE ECE 2011 | Question: 15
An analog signal is band-limited to $4 \; \mathrm{kHz}$, sampled at the Nyquist rate and the samples are quantized into $4$ levels. The quantized levels are assumed to be independent and equally probable. If we transmit two quantized samples per second, the information rate is ... $2 \; \text{bits/sec}$ $3 \; \mathrm{bits/sec}$ $4 \; \mathrm{bits/sec}$

An analog signal is band-limited to $4 \; \mathrm{kHz}$, sampled at the Nyquist rate and the samples are quantized into $4$ levels. The quantized levels are assumed to be...

91 views

asked Sep 3, 2022

1 votes

1 answer

1806

GATE ECE 2011 | Question: 16
The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by $G(s) H(s)=k \frac{s(s+1)}{(s+2)(s+3)}$ $G(s) H(s)=k \frac{(s+1)}{s(s+2)(s+3)^2}$ $G(s) H(s)=k \frac{1}{s(s-1)(s+2)(s+3)}$ $G(s) H(s)=k \frac{(s+1)}{s(s+2)(s+3)}$

The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by$G(s) H(s)=k \frac{s(s+1)}{(s+2)(s+3)}$$G(s) H(s)=k...

298 views

asked Sep 3, 2022

1 votes

0 answers

1807

GATE ECE 2011 | Question: 17
A system is defined by its impulse response $h(n)=2^n u(n-2)$. The system is stable and causal causal but not stable stable but not causal unstable and noncausal

A system is defined by its impulse response $h(n)=2^n u(n-2)$. The system isstable and causalcausal but not stablestable but not causalunstable and noncausal

98 views

asked Sep 3, 2022

1 votes

1 answer

1808

GATE ECE 2011 | Question: 18
If the unit step response of a network is $\left(1-e^{-\alpha t}\right)$, then its unit impulse response is $\alpha e^{-\alpha t}$ $\alpha^{-1} e^{-\alpha t}$ $\left(1-\alpha^{-1}\right) e^{-\alpha t}$ $(1-\alpha) e^{-\alpha t}$

If the unit step response of a network is $\left(1-e^{-\alpha t}\right)$, then its unit impulse response is$\alpha e^{-\alpha t}$$\alpha^{-1} e^{-\alpha t}$$\left(1-\alph...

262 views

asked Sep 3, 2022

3 votes

2 answers

1809

GATE ECE 2011 | Question: 19
The output $\mathrm{Y}$ in the circuit below is always $\text{“1"}$ when two or more of the inputs $\mathrm{P, Q, R}$ are $\text{“0"}$ two or more of the inputs $\mathrm{P, Q, R}$ are $\text{“1"}$ any odd number of the inputs $\mathrm{P, Q, R}$ is $\text{“0"}$ any odd number of the inputs $\mathrm{P, Q, R}$ is $\text{“1"}$

The output $\mathrm{Y}$ in the circuit below is always $\text{“1"}$ whentwo or more of the inputs $\mathrm{P, Q, R}$ are $\text{“0"}$two or more of the inputs $\mathr...

526 views

asked Sep 3, 2022

1 votes

0 answers

1810

GATE ECE 2011 | Question: 20
In the circuit shown below, capacitors $\text{C}_1$ and $\text{C}_2$ are very large and are shorts at the input frequency. $\mathrm{v}_{\mathrm{i}}$ is a small signal input. The gain magnitude $\left|\mathrm{v}_{\mathrm{o}} / \mathrm{v}_{\mathrm{i}}\right|$ at $10 \; \mathrm{Mrad} / \mathrm{s}$ is maximum minimum unity zero

In the circuit shown below, capacitors $\text{C}_1$ and $\text{C}_2$ are very large and are shorts at the input frequency. $\mathrm{v}_{\mathrm{i}}$ is a small signal inp...

98 views

asked Sep 3, 2022

1 votes

0 answers

1811

GATE ECE 2011 | Question: 21
Drift current in semiconductors depends upon only the electric field only the carrier concentration gradient both the electric field and the carrier concentration both the electric field and the carrier concentration gradient

Drift current in semiconductors depends upononly the electric fieldonly the carrier concentration gradientboth the electric field and the carrier concentrationboth the el...

102 views

asked Sep 3, 2022

1 votes

0 answers

1812

GATE ECE 2011 | Question: 22
A Zener diode, when used in voltage stabilization circuits, is biased in reverse bias region below the breakdown voltage reverse breakdown region forward bias region forward bias constant current mode

A Zener diode, when used in voltage stabilization circuits, is biased in reverse bias region below the breakdown voltagereverse breakdown regionforward bias regionforward...

125 views

asked Sep 3, 2022

1 votes

0 answers

1813

GATE ECE 2011 | Question: 23
The circuit shown below is driven by a sinusoidal input $\mathrm{v}_{\mathrm{i}}=\mathrm{V}_{\mathrm{P}} \cos (\mathrm{t} / \mathrm{RC})$. The steady state output $\text{v}_\text{o}$ is $\left(\mathrm{V}_p / 3\right) \cos (\mathrm{t / R C})$ ... $\left(\mathrm{V}_p / 2\right) \sin (\mathrm{t / R C})$

The circuit shown below is driven by a sinusoidal input $\mathrm{v}_{\mathrm{i}}=\mathrm{V}_{\mathrm{P}} \cos (\mathrm{t} / \mathrm{RC})$. The steady state output $\text{...

117 views

asked Sep 3, 2022

1 votes

0 answers

1814

GATE ECE 2011 | Question: 24
Consider a closed surface $S$ surrounding a volume V. If $\vec{r}$ is the position vector of a point inside $S$, with $\hat{n}$ the unit normal on $S$, the value of the integral $\unicode{x222F}_S \; 5 \vec{r} . \hat{n} d S$ is $3 \mathrm{V}$ $5 \mathrm{V}$ $10 \mathrm{V}$ $15 \mathrm{V}$

Consider a closed surface $S$ surrounding a volume V. If $\vec{r}$ is the position vector of a point inside $S$, with $\hat{n}$ the unit normal on $S$, the value of the i...

92 views

asked Sep 3, 2022

1 votes

0 answers

1815

GATE ECE 2011 | Question: 25
The solution of the differential equation $\frac{d y}{d x}=k y, y(0)=c$ is $x=c e^{-k y}$ $x=k e^{c y}$ $y=c e^{k x}$ $y=c e^{-k x}$

The solution of the differential equation $\frac{d y}{d x}=k y, y(0)=c$ is$x=c e^{-k y}$$x=k e^{c y}$$y=c e^{k x}$$y=c e^{-k x}$

112 views

asked Sep 3, 2022

1 votes

0 answers

1816

GATE ECE 2011 | Question: 26
The electric and magnetic fields for a $\text{TEM}$ wave of frequency $14 \; \mathrm{GHz}$ in a homogeneous medium of relative permittivity $\varepsilon_r$ and relative permeability $\mu_r=1$ ... $\varepsilon_r=3, E_p=360 \pi$ $\varepsilon_r=9, E_p=360 \pi$ $\varepsilon_r=9, E_p=120 \pi$

The electric and magnetic fields for a $\text{TEM}$ wave of frequency $14 \; \mathrm{GHz}$ in a homogeneous medium of relative permittivity $\varepsilon_r$ and relative p...

105 views

asked Sep 3, 2022

1 votes

0 answers

1817

GATE ECE 2011 | Question: 27
A message signal $m(t)=\cos 2000 \pi t+4 \cos 4000 \pi t$ modulates the carrier $c(t)=\cos 2 \pi f_c t$ where $f_c=1 \; \mathrm{MHz}$ to produce an $\text{AM}$ signal. For demodulating the generated $\text{AM}$ signal using an envelope detector, the time constant ... $\mathrm{RC} <<1 \;\mu \mathrm{s}$ $\mathrm{RC} >> 0.5 \mathrm{~ms}$

A message signal $m(t)=\cos 2000 \pi t+4 \cos 4000 \pi t$ modulates the carrier $c(t)=\cos 2 \pi f_c t$ where $f_c=1 \; \mathrm{MHz}$ to produce an $\text{AM}$ signal. Fo...

101 views

asked Sep 3, 2022

1 votes

0 answers

1818

GATE ECE 2011 | Question: 28
The block diagram of a system with one input $u$ and two outputs $y_1$ and $y_2$ is given below. A state space model of the above system in terms of the state vector $\underline{x}$ ...

The block diagram of a system with one input $u$ and two outputs $y_1$ and $y_2$ is given below.A state space model of the above system in terms of the state vector $\und...

96 views

asked Sep 3, 2022

1 votes

0 answers

1819

GATE ECE 2011 | Question: 29
Two systems $H_1(z)$ and $H_2(z)$ are connected in cascade as shown below. The overall output $y(n)$ is the same as the input $x(n)$ with a one unit delay. The transfer function of the second system $\mathrm{H}_2(z)$ ... $\frac{\left(1-0.4 z^{-1}\right)}{z^{-1}\left(1-0.6 z^{-1}\right)}$

Two systems $H_1(z)$ and $H_2(z)$ are connected in cascade as shown below. The overall output $y(n)$ is the same as the input $x(n)$ with a one unit delay. The transfer f...

106 views

asked Sep 3, 2022

1 votes

0 answers

1820

GATE ECE 2011 | Question: 30
An $8085$ ... $8 \mathrm{CH}$ $64 \mathrm{H}$ $23 \mathrm{H}$ $15 \mathrm{H}$

An $8085$ assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is$$\...

103 views

asked Sep 3, 2022

1 votes

0 answers

1821

GATE ECE 2011 | Question: 31
The first six points of the $8$-point $\text{DFT}$ of a real valued sequence are $5,1-\mathrm{j} 3,0,3-\mathrm{j} 4,0$ and $3+\mathrm{j} 4$. The last two points of the $\text{DFT}$ are respectively $0,1-\mathrm{j} 3$ $0,1+\mathrm{j} 3$ $1+\mathrm{j} 3,5$ $1-\mathrm{j} 3,5$

The first six points of the $8$-point $\text{DFT}$ of a real valued sequence are $5,1-\mathrm{j} 3,0,3-\mathrm{j} 4,0$ and $3+\mathrm{j} 4$. The last two points of the $\...

109 views

asked Sep 3, 2022

1 votes

0 answers

1822

GATE ECE 2011 | Question: 32
For the $\text{BJT Q}_1$ in the circuit shown below, $\beta=\infty, \mathrm{V}_{\mathrm{BEon}}=0.7 \mathrm{V}, \mathrm{V}_{\mathrm{CEsat}}=0.7 \mathrm{V}$. The switch is initially closed. At time $t=0$, the switch is opened. The time $t$ at which $\mathrm{Q}_1$ leaves the active region is $10 \mathrm{~ms}$ $25 \mathrm{~ms}$ $50 \mathrm{~ms}$ $100 \mathrm{~ms}$

For the $\text{BJT Q}_1$ in the circuit shown below, $\beta=\infty, \mathrm{V}_{\mathrm{BEon}}=0.7 \mathrm{V}, \mathrm{V}_{\mathrm{CEsat}}=0.7 \mathrm{V}$. The switch is ...

88 views

asked Sep 3, 2022

1 votes

0 answers

1823

GATE ECE 2011 | Question: 33
In the circuit shown below, the network $\mathrm{N}$ is described by the following $Y$ matrix: $Y=\left[\begin{array}{cc}0.1 \mathrm{~S} & -0.01 \mathrm{~S} \\ 0.01 \mathrm{~S} & 0.1 \mathrm{~S}\end{array}\right]$. The voltage gain $\dfrac{V_2}{V_1}$ is $1 / 90$ $ – 1 / 90$ $ – 1 / 99$ $ – 1 / 11$

In the circuit shown below, the network $\mathrm{N}$ is described by the following $Y$ matrix:$Y=\left[\begin{array}{cc}0.1 \mathrm{~S} & -0.01 \mathrm{~S} \\ 0.01 \mathr...

120 views

asked Sep 3, 2022

1 votes

0 answers

1824

GATE ECE 2011 | Question: 34
In the circuit shown below, the initial charge on the capacitor is $2.5 \mathrm{~mC}$, with the voltage polarity as indicated. The switch is closed at time $t=0$. The current $i(t)$ at a time $t$ after the switch is closed is $i(t)=15 \exp (-2 \times 10^3 \mathrm{t) \;A}$ ... $i(t)=10 \exp (-2 \times 10^3 \mathrm{t) \;A}$ $i(t)=-5 \exp (-2 \times 10^3 \mathrm{t) \;A}$

In the circuit shown below, the initial charge on the capacitor is $2.5 \mathrm{~mC}$, with the voltage polarity as indicated. The switch is closed at time $t=0$. The cur...

118 views

asked Sep 3, 2022

2 votes

1 answer

1825

GATE ECE 2011 | Question: 35
The system of equations $ \begin{aligned} &x+y+z=6 \\ &x+4 y+6 z=20 \\ &x+4 y+\lambda z=\mu \end{aligned} $ has NO solution for values of $\lambda$ and $\mu$ given by $\lambda=6, \mu=20$ $\lambda=6, \mu \neq 20$ $\lambda \neq 6, \mu=20$ $\lambda \neq 6, \mu \neq 20$

The system of equations $$ \begin{aligned} &x+y+z=6 \\ &x+4 y+6 z=20 \\ &x+4 y+\lambda z=\mu \end{aligned} $$ has NO solution for values of $\lambda$ and $\mu$ given by$\...

221 views

asked Sep 3, 2022

1 votes

0 answers

1826

GATE ECE 2011 | Question: 36
A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is $2 / 36$ $2 / 6$ $5 / 12$ $1 / 2$

A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is$2 / 36$$2 / 6$$5 / 12$$1 / 2$

61 views

asked Sep 3, 2022

1 votes

0 answers

1827

GATE ECE 2011 | Question: 37
A current sheet $\vec{J}=10 \hat{u}_y \; A / m$ lies on the dielectric interface $x=0$ between two dielectric media with $\varepsilon_{r 1}=5, \mu_{r 1}=1$ in Region-$1 \; (x<0)$ and $\varepsilon_{r 2}=2, \mu_{r 2}=2$ in Region-$2 \; (x>0)$. If the magnetic field ... $\vec{H}_2=3 \hat{u}_x+30 \hat{u}_y+10 \hat{u}_z \; A / m$

A current sheet $\vec{J}=10 \hat{u}_y \; A / m$ lies on the dielectric interface $x=0$ between two dielectric media with $\varepsilon_{r 1}=5, \mu_{r 1}=1$ in Region-$1 \...

113 views

asked Sep 3, 2022

1 votes

0 answers

1828

GATE ECE 2011 | Question: 38
A transmission line of characteristic impedance $50 \; \Omega$ is terminated in a load impedance $\text{Z}_\text{L} .$ The $\text{VSWR}$ of the line is measured as $5$ and the first of the voltage maxima in the line is observed at a distance of $\lambda / 4$ from the load. The ... is $10 \; \Omega$ $250 \; \Omega$ $(19.23+j 46.15) \; \Omega$ $(19.23-j 46.15) \; \Omega$

A transmission line of characteristic impedance $50 \; \Omega$ is terminated in a load impedance $\text{Z}_\text{L} .$ The $\text{VSWR}$ of the line is measured as $5$ an...

102 views

asked Sep 3, 2022

1 votes

0 answers

1829

GATE ECE 2011 | Question: 39
$\mathrm{X(t)}$ is a stationary random process with autocorrelation function $R_X(\tau)=\exp \left(-\pi \tau^2\right)$. This process is passed through the system shown below. The power spectral density of the output process $\mathrm{Y}(\mathrm{t})$ is $(4 \pi^2 f^2+1) \exp (\pi f^2)$ ... $(4 \pi^2 f^2+1) \exp (-\pi f)$ $(4 \pi^2 f^2-1) \exp (-\pi f)$

$\mathrm{X(t)}$ is a stationary random process with autocorrelation function $R_X(\tau)=\exp \left(-\pi \tau^2\right)$. This process is passed through the system shown be...

99 views

asked Sep 3, 2022

1 votes

0 answers

1830

GATE ECE 2011 | Question: 40
The output of a $3$-stage Johnson (twisted-ring) counter is fed to a digital-to-analog (D/A) converter as shown in the figure below. Assume all states of the counter to be unset initially. The waveform which represents the D/A converter output $\mathrm{V}_{\mathrm{o}}$ is

The output of a $3$-stage Johnson (twisted-ring) counter is fed to a digital-to-analog (D/A) converter as shown in the figure below. Assume all states of the counter to b...

99 views

asked Sep 3, 2022

1 votes

0 answers

1831

GATE ECE 2011 | Question: 41
Two $\text{D}$ flip-flops are connected as a synchronous counter that goes through the following $\mathrm{Q}_{\text{B}} \;\mathrm{Q}_{\mathrm{A}}$ sequence $00 \rightarrow 11 \rightarrow 01 \rightarrow 10 \rightarrow 00 \rightarrow \cdots$ The connections to the ...

Two $\text{D}$ flip-flops are connected as a synchronous counter that goes through the following $\mathrm{Q}_{\text{B}} \;\mathrm{Q}_{\mathrm{A}}$ sequence $00 \rightarro...

37 views

asked Sep 3, 2022

1 votes

0 answers

1832

GATE ECE 2011 | Question: 42
In the circuit shown below, for the $\text{MOS}$ transistors, $\mu_{\mathrm{n}} \mathrm{C}_{\mathrm{ox}}=100 \; \mu \mathrm{A} / \mathrm{V}^2$ and the threshold voltage $\mathrm{V}_{\mathrm{T}}=1 \mathrm{~V}$. The voltage $\mathrm{V}_{\mathrm{x}}$ at the source of the upper transistor is $1 \mathrm{~V}$ $2 \mathrm{~V}$ $3 \mathrm{~V}$ $3.67 \mathrm{~V}$

In the circuit shown below, for the $\text{MOS}$ transistors, $\mu_{\mathrm{n}} \mathrm{C}_{\mathrm{ox}}=100 \; \mu \mathrm{A} / \mathrm{V}^2$ and the threshold voltage $...

42 views

asked Sep 3, 2022

1 votes

0 answers

1833

GATE ECE 2011 | Question: 43
An input $\mathrm{x(t)}=\exp (-2 \mathrm{t)u(t})+\delta(\mathrm{t}-6)$ is applied to an LTI system with impulse response $\mathrm{h(t)=u(t})$. The output is $[1-\exp (-2 \mathrm{t)] u(t)+u(t}+6)$ $[1-\exp (-2 \mathrm{t)] u(t)+u(t}-6)$ $0.5[1-\exp (-2 \mathrm{t)] u(t)+u(t}+6)$ $0.5[1-\exp (-2 \mathrm{t)] u(t)+u(t}-6)$

An input $\mathrm{x(t)}=\exp (-2 \mathrm{t)u(t})+\delta(\mathrm{t}-6)$ is applied to an LTI system with impulse response $\mathrm{h(t)=u(t})$. The output is$[1-\exp (-2 \...

35 views

asked Sep 3, 2022

1 votes

0 answers

1834

GATE ECE 2011 | Question: 44
For a $\mathrm{BJT}$, the common-base current gain $\alpha=0.98$ and the collector base junction reverse bias saturation current $\mathrm{I}_{\mathrm{CO}}=0.6 \;\; \mu \mathrm{A}$. This $\mathrm{BJT}$ is connected in the common emitter mode and operated in the active ... mode of operation is $0.98 \mathrm{~mA}$ $0.99 \mathrm{~mA}$ $1.0 \mathrm{~mA}$ $1.01 \mathrm{~mA}$

For a $\mathrm{BJT}$, the common-base current gain $\alpha=0.98$ and the collector base junction reverse bias saturation current $\mathrm{I}_{\mathrm{CO}}=0.6 \;\; \mu \m...

33 views

asked Sep 3, 2022

1 votes

0 answers

1835

GATE ECE 2011 | Question: 45
If $F(s)=L[f(t)]=\dfrac{2(s+1)}{s^2+4 s+7}$ then the initial and final values of $f(t)$ are respectively $0,2$ $2,0$ $0, 2 / 7$ $2 / 7,0$

If $F(s)=L[f(t)]=\dfrac{2(s+1)}{s^2+4 s+7}$ then the initial and final values of $f(t)$ are respectively$0,2$$2,0$$0, 2 / 7$$2 / 7,0$

24 views

asked Sep 3, 2022

1 votes

0 answers

1836

GATE ECE 2011 | Question: 46
In the circuit shown below, the current $\text{I}$ is equal to $1.4 \angle 0^{\circ} \; \mathrm{A}$ $2.0 \angle 0^{\circ} \; \mathrm{A}$ $2.8 \angle 0^{\circ} \; \mathrm{A}$ $3.2 \angle 0^{\circ} \; \mathrm{A}$

In the circuit shown below, the current $\text{I}$ is equal to$1.4 \angle 0^{\circ} \; \mathrm{A}$$2.0 \angle 0^{\circ} \; \mathrm{A}$$2.8 \angle 0^{\circ} \; \mathrm{A}$...

41 views

asked Sep 3, 2022

1 votes

0 answers

1837

GATE ECE 2011 | Question: 47
A numerical solution of the equation $f(x)=x+\sqrt{x}-3=0$ can be obtained using Newton-Raphson method. If the starting value is $x=2$ for the iteration, the value of $x$ that is to be used in the next step is $0.306$ $0.739$ $1.694$ $2.306$

A numerical solution of the equation $f(x)=x+\sqrt{x}-3=0$ can be obtained using Newton-Raphson method. If the starting value is $x=2$ for the iteration, the value of $x$...

43 views

asked Sep 3, 2022

1 votes

0 answers

1838

GATE ECE 2011 | Question: 48
The channel resistance of an $\text{N}$-channel $\text{JFET}$ shown in the figure below is $600 \; \Omega$ when the full channel thickness $\left(\mathrm{t}_{\mathrm{ch}}\right)$ of $10 \; \mu \mathrm{m}$ ... $480 \; \Omega$ $600 \; \Omega$ $750 \; \Omega$ $1000 \; \Omega$

The channel resistance of an $\text{N}$-channel $\text{JFET}$ shown in the figure below is $600 \; \Omega$ when the full channel thickness $\left(\mathrm{t}_{\mathrm{ch}}...

31 views

asked Sep 3, 2022

1 votes

0 answers

1839

GATE ECE 2011 | Question: 49
The channel resistance of an $\text{N}$-channel $\text{JFET}$ shown in the figure below is $600 \; \Omega$ when the full channel thickness $\left(\mathrm{t}_{\mathrm{ch}}\right)$ of $10 \; \mu \mathrm{m}$ is available for conduction. The built-in voltage of the gate ... $360 \; \Omega$ $917 \; \Omega$ $1000 \; \Omega$ $3000 \; \Omega$

The channel resistance of an $\text{N}$-channel $\text{JFET}$ shown in the figure below is $600 \; \Omega$ when the full channel thickness $\left(\mathrm{t}_{\mathrm{ch}}...

30 views

asked Sep 3, 2022

1 votes

0 answers

1840

GATE ECE 2011 | Question: 50
The input-output transfer function of a plant $H(s)=\frac{100}{s(s+10)^2}$. The plant is placed in a unity negative feedback configuration as shown in the figure below. The signal flow graph that DOES NOT model the plant $\operatorname{transfer~function~} H(s)$ is

The input-output transfer function of a plant $H(s)=\frac{100}{s(s+10)^2}$. The plant is placed in a unity negative feedback configuration as shown in the figure below.Th...

53 views

asked Sep 3, 2022