Question 1 |
A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles
an AND gate | |
an OR gate | |
an XOR gate | |
a NAND gate |
Question 1 Explanation:
Truth table of XOR gate
\begin{array}{cc|c} A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}
So, from the XOR gate truth table it is clear that the bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch.
\begin{array}{cc|c} A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}
So, from the XOR gate truth table it is clear that the bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch.
Question 2 |
Consider a vector field {\overrightarrow{A}}( \overrightarrow{r}). The closed loop line integral \int \overrightarrow{A}\cdot \overrightarrow{di} can be expressed as
\int \int ( \overline{V}\times \overrightarrow{A} )\cdot \overrightarrow{ds} over the closed surface bounded by the loop | |
\int \int \int ( \overline{V}\cdot \overrightarrow{A} )dv over the closed volume bounded by the loop
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\int \int \int ( \overline{V}\cdot \overrightarrow{A} )dv over the open volume bounded by the loop
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\int \int ( \overline{V}\times \overrightarrow{A} )\cdot \overrightarrow{ds} over the open surface bounded by the loop |
Question 2 Explanation:
According to Stoke's theorem
\oint_{c} \vec{A} \cdot \vec{d} i=\iint_{s}(\nabla \times \vec{A}) \cdot \overrightarrow{d s}
\oint_{c} \vec{A} \cdot \vec{d} i=\iint_{s}(\nabla \times \vec{A}) \cdot \overrightarrow{d s}
Question 3 |
Two systems with impulse responses h_{1}\left ( t \right ) and h_{2}\left ( t \right ) are connected in cascade. Then the overall impulse response of the cascaded system is given by
product of h_{1}\left ( t \right ) and h_{2}\left ( t \right )
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sum of h_{1}\left ( t \right ) and h_{2}\left ( t \right )
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convolution of h_{1}\left ( t \right ) and h_{2}\left ( t \right )
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subtraction of h_{2}\left ( t \right ) from h_{1}\left ( t \right )
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Question 3 Explanation:
The overall impulse response h(t) of the cascade
system is given by:
h(t)=h_{1}(t) * h_{2}(t)
h(t)=h_{1}(t) * h_{2}(t)
Question 4 |
In a forward biased pn junction diode, the sequence of events that best describes the mechanism of current flow is
injection, and subsequent diffusion and recombination of minority carriers | |
injection, and subsequent drift and generation of minority carriers | |
extraction, and subsequent diffusion and generation of minority carriers | |
extraction, and subsequent drift and recombination of minority carriers |
Question 4 Explanation:
In a forward biased pn-junction diode, the current
flow is due to diffusion of majority carriers and
recombination of minority carriers.
Question 5 |
In IC technology, dry oxidation (using dry oxygen) as compared to wet oxidation (using steam or water vapor) produces
superior quality oxide with a higher growth rate | |
inferior quality oxide with a higher growth rate | |
inferior quality oxide with a lower growth rate | |
superior quality oxide with a lower growth rate |
Question 5 Explanation:
Dry oxidation has better quality over wet oxidation.
Dry oxidation is slower over wet oxidation.
Dry oxidation is slower over wet oxidation.
Question 6 |
The maximum value of \theta until which the approximation \sin \theta \approx \theta holds to within 10% error is
10^{\circ}
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18^{\circ} | |
50^{\circ}
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90^{\circ}
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Question 6 Explanation:
10^{\circ}=\frac{10 \pi}{180}=0.1745
\sin 10^{\circ}=0.1736
So, for 10^{\circ} \rightarrow \sin \cong \theta holds within 10 % error
\begin{aligned} 18^{\circ} &=\frac{18 \times \pi}{180}=0.3142 \\ \sin 18^{\circ} &=0.3090 \end{aligned}
So, for 18^{\circ} \rightarrow \sin \theta \equiv \theta holds within 10% error
50^{\circ}=\frac{50 \times \pi}{180}=0.8727
\sin 50^{\circ}=0.766
So, for 50^{\circ} \rightarrow \sin \theta \cong \theta does not hold within 10 % error
90^{\circ}=\frac{90 \times \pi}{180}=1.571
\sin 90^{\circ}=1
So, for 90^{\circ} \rightarrow \sin \theta \cong \theta does not hold within 10 % error.
So, the maximum value of \theta for the approximation
\sin \theta \cong \theta \text{ holds to within } 10 \% \text{ error is } 18^{\circ}
\sin 10^{\circ}=0.1736
So, for 10^{\circ} \rightarrow \sin \cong \theta holds within 10 % error
\begin{aligned} 18^{\circ} &=\frac{18 \times \pi}{180}=0.3142 \\ \sin 18^{\circ} &=0.3090 \end{aligned}
So, for 18^{\circ} \rightarrow \sin \theta \equiv \theta holds within 10% error
50^{\circ}=\frac{50 \times \pi}{180}=0.8727
\sin 50^{\circ}=0.766
So, for 50^{\circ} \rightarrow \sin \theta \cong \theta does not hold within 10 % error
90^{\circ}=\frac{90 \times \pi}{180}=1.571
\sin 90^{\circ}=1
So, for 90^{\circ} \rightarrow \sin \theta \cong \theta does not hold within 10 % error.
So, the maximum value of \theta for the approximation
\sin \theta \cong \theta \text{ holds to within } 10 \% \text{ error is } 18^{\circ}
Question 7 |
The divergence of the vector field \overrightarrow{A}=x\hat{a_{x}}+y\hat{a_{y}}+z\hat{a_{x}} is
0
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1/3
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1
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3
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Question 7 Explanation:
\begin{array}{l} \nabla \cdot \vec{A}=\frac{\partial A_{x}}{\partial x}+\frac{\partial A_{y}}{\partial y}+\frac{\partial A_{z}}{\partial z} \\ \nabla \cdot \vec{A}=\frac{\partial}{\partial x}(x)+\frac{\partial}{\partial y}(y)+\frac{\partial}{\partial z}(z)=1+1+1 \\ \nabla \cdot \vec{A}=3 \end{array}
Question 8 |
The impulse response of a system is h\left ( t \right )= tu\left ( t \right ). For an input u(t-1), the output is
\frac{t^{2}}{2}u\left ( t \right )
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\frac{t(t-1)}{2}u(t-1)
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\frac{\left ( t-1 \right )^{2}}{2}u\left ( t-1 \right )
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\frac{t^{2}-1}{2}u\left ( t-1 \right )
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Question 8 Explanation:
\begin{aligned} h(t)&=t u(t) \\ \text{Taking }&\text{Laplace transform}\\ H(s)&=\frac{1}{s^{2}} \\ x(t) &=u(t-1) \\ \text { Taking }&\text{Laplace transform } \\ X(s)&=\frac{e^{-s}}{s} \\ \frac{Y(s)}{X(s)}&=H(s) \\ Y(s)&=H(s) X(s) \\ Y(s)&=\frac{1}{s^{2}} \cdot \frac{e^{-s}}{s}=\frac{e^{-s}}{s^{3}} \\ \text{Taking } &\text{the inverse Laplace transform } \\ y(t)&=\frac{(t-1)^{2}}{2} u(t-1) \\ \end{aligned}
Question 9 |
The Bode plot of a transfer function G(s) is shown in the figure below.
The gain \left ( 20\log\left | G\left ( s \right ) \right | \right ) is 32 dB and -8 dB at 1 rad/s and 10 rad/s respectively. The phase is negative for all \omega. Then G(s) is
The gain \left ( 20\log\left | G\left ( s \right ) \right | \right ) is 32 dB and -8 dB at 1 rad/s and 10 rad/s respectively. The phase is negative for all \omega. Then G(s) is
\frac{39.8}{s}
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\frac{39.8}{s^{2}}
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\frac{32}{s}
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\frac{32}{s^{2}}
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Question 9 Explanation:
10 \mathrm{rad} / \mathrm{s} to 1 \mathrm{rad} / \mathrm{s} is 1 decade
32-(-8)=40 d B
So, the slope is 40dB/decade it means there are two poles at origin, it means either option (B) or option (D) is correct put \omega=1rad/sec in both the options.
20 \log \left[\frac{39.8}{(1)^{2}}\right]=32 \mathrm{dB}
20 \log \left[\frac{32}{(1)^{2}}\right]=30.1 \mathrm{dB}
So, option (B) is correct option =\frac{39.8}{s^{2}}
32-(-8)=40 d B
So, the slope is 40dB/decade it means there are two poles at origin, it means either option (B) or option (D) is correct put \omega=1rad/sec in both the options.
20 \log \left[\frac{39.8}{(1)^{2}}\right]=32 \mathrm{dB}
20 \log \left[\frac{32}{(1)^{2}}\right]=30.1 \mathrm{dB}
So, option (B) is correct option =\frac{39.8}{s^{2}}
Question 10 |
In the circuit shown below what is the output voltage \left ( V_{out} \right ) if a silicon transistor Q and an ideal op-amp are used?
-15V
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-0.7V
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+0.7V
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+15V
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Question 10 Explanation:
Due to virtual short
V_{c} = V_{B} = 0V
(collector voltage of transistor 0)
and given that base voltage of transistor 0
V_{B} = O_{V}
So, V_{C} = O_{V}
It means collector to base of transistor Oare short circuited.
If any junction of transistor is short circuited then the junction acts as reverse bias. So, the C-B junction is reverse bias. The given op-amp is an inverting configuration which have positive voltage as input. So the output voltage of op-amp will be negative voltage.
V_{\text{out}} = -ve voltage
Emitter voltage of transistor
V_{E} = -ve voltage
given transistor is n-p-ntransistor and
V_{B} = OV
V_{E} = -ve voltage
So, the E-8 junction will be forward bias. Thus, the transistor is in active region and will behave as closed switch.
So,
\begin{aligned} V_{B E} &=0.7 \mathrm{V} \quad \text { (for silicon transistor) } \\ \mathrm{V}_{\mathrm{B}}-\mathrm{V}_{E} &=0.7 \mathrm{V} \\ \mathrm{V}_{\mathrm{E}} &=\mathrm{V}_{\mathrm{B}}-0.7 \\ \mathrm{V}_{\mathrm{E}} &=0-0.7 \\ \mathrm{V}_{\mathrm{E}} &=-0.7 \mathrm{V} \end{aligned}
V_{c} = V_{B} = 0V
(collector voltage of transistor 0)
and given that base voltage of transistor 0
V_{B} = O_{V}
So, V_{C} = O_{V}
It means collector to base of transistor Oare short circuited.
If any junction of transistor is short circuited then the junction acts as reverse bias. So, the C-B junction is reverse bias. The given op-amp is an inverting configuration which have positive voltage as input. So the output voltage of op-amp will be negative voltage.
V_{\text{out}} = -ve voltage
Emitter voltage of transistor
V_{E} = -ve voltage
given transistor is n-p-ntransistor and
V_{B} = OV
V_{E} = -ve voltage
So, the E-8 junction will be forward bias. Thus, the transistor is in active region and will behave as closed switch.
So,
\begin{aligned} V_{B E} &=0.7 \mathrm{V} \quad \text { (for silicon transistor) } \\ \mathrm{V}_{\mathrm{B}}-\mathrm{V}_{E} &=0.7 \mathrm{V} \\ \mathrm{V}_{\mathrm{E}} &=\mathrm{V}_{\mathrm{B}}-0.7 \\ \mathrm{V}_{\mathrm{E}} &=0-0.7 \\ \mathrm{V}_{\mathrm{E}} &=-0.7 \mathrm{V} \end{aligned}
There are 10 questions to complete.