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Recent questions and answers in Digital Signal Processing
0
votes
1
answer
1
NIELIT 2017 July Scientist B (CS) - Section B: 31
At a room temperature of $300K$, calculate the thermal noise generated by two resistors of $10K\Omega$ and $20K\Omega$ when the bandwidth is $10KHz$. $1.2868\times10^{-6}V, 1.819\times10^{-6}V$ $6.08\times10^{-6}V, 15.77\times10^{-6}V$ $16.66\times10^{-6}V, 2.356\times10^{-6}V$ $1.66\times10^{-6}V, 0.23\times10^{-6}V$
At a room temperature of $300K$, calculate the thermal noise generated by two resistors of $10K\Omega$ and $20K\Omega$ when the bandwidth is $10KHz$. $1.2868\times10^{-6}V, 1.819\times10^{-6}V$ $6.08\times10^{-6}V, 15.77\times10^{-6}V$ $16.66\times10^{-6}V, 2.356\times10^{-6}V$ $1.66\times10^{-6}V, 0.23\times10^{-6}V$
answered
Oct 4, 2020
in
Digital Signal Processing
Madhav
194
views
nielit2017july-scientistb-cs
non-gate
0
votes
3
answers
2
NIELIT 2017 DEC Scientific Assistant A - Section B: 8
Microprocessors are used in which generation of computers? I – st Generation II – nd Generation III – rd Generation IV – th Generation
Microprocessors are used in which generation of computers? I – st Generation II – nd Generation III – rd Generation IV – th Generation
answered
Sep 20, 2020
in
Digital Signal Processing
Himanshu Kumar Gupta
213
views
nielit2017dec-assistanta
non-gate
microprocessors
0
votes
1
answer
3
NIELIT 2017 OCT Scientific Assistant A (CS) - Section D: 2
An ideal op-amp is an ideal voltage controlled current source voltage controlled voltage source current controlled current source current controlled voltage source
An ideal op-amp is an ideal voltage controlled current source voltage controlled voltage source current controlled current source current controlled voltage source
answered
Sep 20, 2020
in
Digital Signal Processing
Himanshu Kumar Gupta
83
views
nielit2017oct-assistanta-cs
non-gate
0
votes
0
answers
4
NIELIT 2017 OCT Scientific Assistant A (IT) - Section D: 1
Twelve $1\;\Omega$ resistances are used as edges to from a cube. The resistance between two diagonally opposite corners of the cube is $\frac{5}{6}\;\Omega$ $\frac{1}{6}\;\Omega$ $\frac{6}{5}\;\Omega$ $\frac{3}{2}\;\Omega$
Twelve $1\;\Omega$ resistances are used as edges to from a cube. The resistance between two diagonally opposite corners of the cube is $\frac{5}{6}\;\Omega$ $\frac{1}{6}\;\Omega$ $\frac{6}{5}\;\Omega$ $\frac{3}{2}\;\Omega$
asked
Aug 28, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
84
views
nielit2017oct-assistanta-it
non-gate
0
votes
0
answers
5
NIELIT 2017 OCT Scientific Assistant A (IT) - Section D: 11
In MOSFET fabrication, the channel; length is defined during the process of Isolation oxide growth Channel stop implantation Poly-silicon gate patterning Lithography step leading to the contact pad
In MOSFET fabrication, the channel; length is defined during the process of Isolation oxide growth Channel stop implantation Poly-silicon gate patterning Lithography step leading to the contact pad
asked
Aug 28, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
54
views
nielit2017oct-assistanta-it
non-gate
0
votes
0
answers
6
NIELIT 2017 OCT Scientific Assistant A (CS) - Section D: 1
Twelve $1\;\Omega$ resistances are used as edges to from a cube. The resistance between two diagonally opposite corners of the cube is $\frac{5}{6}\;\Omega$ $\frac{1}{6}\;\Omega$ $\frac{6}{5}\;\Omega$ $\frac{3}{2}\;\Omega$
Twelve $1\;\Omega$ resistances are used as edges to from a cube. The resistance between two diagonally opposite corners of the cube is $\frac{5}{6}\;\Omega$ $\frac{1}{6}\;\Omega$ $\frac{6}{5}\;\Omega$ $\frac{3}{2}\;\Omega$
asked
Aug 28, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
66
views
nielit2017oct-assistanta-cs
non-gate
0
votes
0
answers
7
NIELIT 2017 OCT Scientific Assistant A (CS) - Section D: 3
The final value theorem is used to find the Steady state value of the system output Initial value of the system output Transient behavior of the system output None of these
The final value theorem is used to find the Steady state value of the system output Initial value of the system output Transient behavior of the system output None of these
asked
Aug 28, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
68
views
nielit2017oct-assistanta-cs
non-gate
0
votes
0
answers
8
NIELIT 2017 OCT Scientific Assistant A (CS) - Section D: 4
For the discrete signal $x[n] = a^{n}u[n],a>0$ the $z$-transform is $\frac{(z+a)}{z}$ $\frac{(z-a)}{z}$ $\frac{z}{(z-a)}$ $\frac{z}{(z+a)}$
For the discrete signal $x[n] = a^{n}u[n],a>0$ the $z$-transform is $\frac{(z+a)}{z}$ $\frac{(z-a)}{z}$ $\frac{z}{(z-a)}$ $\frac{z}{(z+a)}$
asked
Aug 28, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
71
views
nielit2017oct-assistanta-cs
non-gate
0
votes
0
answers
9
NIELIT 2017 OCT Scientific Assistant A (CS) - Section D: 5
For a periodic signal $v(t) = 30\sin 100t + 10\cos 300t + 6\sin(500t + \frac{\pi}{4}),$ the fundamental frequency in rad/s $100$ $300$ $500$ None of these
For a periodic signal $v(t) = 30\sin 100t + 10\cos 300t + 6\sin(500t + \frac{\pi}{4}),$ the fundamental frequency in rad/s $100$ $300$ $500$ None of these
asked
Aug 28, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
70
views
nielit2017oct-assistanta-cs
non-gate
0
votes
0
answers
10
NIELIT 2017 OCT Scientific Assistant A (CS) - Section D: 7
If the number of bits per sample in a PCM system is increased from a $n$ to $n+1,$ the improvement in signal to quantization nose ratio will be $3\;dB$ $6\;dB$ $2n\;dB$ $n\;dB$
If the number of bits per sample in a PCM system is increased from a $n$ to $n+1,$ the improvement in signal to quantization nose ratio will be $3\;dB$ $6\;dB$ $2n\;dB$ $n\;dB$
asked
Aug 28, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
59
views
nielit2017oct-assistanta-cs
non-gate
0
votes
0
answers
11
NIELIT 2017 OCT Scientific Assistant A (CS) - Section D: 8
A carrier $Ac\cos(\omega c)t$ is frequency modulated by a signal $Em\cos(\omega m)t.$ The modulation index is $mf.$ The expression for the resulting FM signal is $Ac\cos [\omega ct + mf\sin(\omega m)t]$ ... $Ac\cos [\omega ct + 2\pi mf Em \cos(\omega m)t/\omega m]$
A carrier $Ac\cos(\omega c)t$ is frequency modulated by a signal $Em\cos(\omega m)t.$ The modulation index is $mf.$ The expression for the resulting FM signal is $Ac\cos [\omega ct + mf\sin(\omega m)t]$ $Ac\cos [\omega ct + mf\cos(\omega m)t]$ $Ac\cos [\omega ct + \pi mf\sin \omega m t]$ $Ac\cos [\omega ct + 2\pi mf Em \cos(\omega m)t/\omega m]$
asked
Aug 28, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
57
views
nielit2017oct-assistanta-cs
non-gate
0
votes
0
answers
12
NIELIT 2017 OCT Scientific Assistant A (CS) - Section D: 11
In MOSFET fabrication, the channel; length is defined during the process of Isolation oxide growth Channel stop implantation Poly-silicon gate patterning Lithography step leading to the contact pad
In MOSFET fabrication, the channel; length is defined during the process of Isolation oxide growth Channel stop implantation Poly-silicon gate patterning Lithography step leading to the contact pad
asked
Aug 28, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
54
views
nielit2017oct-assistanta-cs
non-gate
0
votes
0
answers
13
NIELIT 2017 OCT Scientific Assistant A (CS) - Section D: 12
The open-loop transfer function of a feedback control system is $G(s)\cdot H(s) = 1/(s+1)^{3}.$ The gain margin of the system is $2$ $4$ $8$ $16$
The open-loop transfer function of a feedback control system is $G(s)\cdot H(s) = 1/(s+1)^{3}.$ The gain margin of the system is $2$ $4$ $8$ $16$
asked
Aug 28, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
66
views
nielit2017oct-assistanta-cs
non-gate
0
votes
1
answer
14
NIELIT 2016 DEC Scientist B (CS) - Section B: 60
The sequence of operation in which PCM is done is Sampling, quantizing, encoding Quantizing, sampling, encoding Quantizing, encoding, sampling None of the above
The sequence of operation in which PCM is done is Sampling, quantizing, encoding Quantizing, sampling, encoding Quantizing, encoding, sampling None of the above
answered
Jun 10, 2020
in
Digital Signal Processing
Ollie
986
views
nielit2016dec-scientistb-cs
non-gate
0
votes
1
answer
15
NIELIT 2016 DEC Scientist B (CS) - Section B: 41
The IETF standard documents are called: RFC RCF ID none of the above
The IETF standard documents are called: RFC RCF ID none of the above
answered
Jun 10, 2020
in
Digital Signal Processing
Ollie
99
views
nielit2016dec-scientistb-cs
non-gate
0
votes
1
answer
16
NIELIT 2016 DEC Scientist B (CS) - Section B: 17
$T1$ carrier system is used: For delta modulation Industrial noise For frequency modulated signals None of the above
$T1$ carrier system is used: For delta modulation Industrial noise For frequency modulated signals None of the above
answered
May 25, 2020
in
Digital Signal Processing
Mohit Kumar 6
172
views
nielit2016dec-scientistb-cs
non-gate
0
votes
1
answer
17
NIELIT 2016 MAR Scientist C - Section C: 48
In a microprocessor, WAIT states are used to make the processor wait during a DMA operation make the processor wait during a power interrupt processing make the processor wait during a power shutdown interface slow peripherals to the processor
In a microprocessor, WAIT states are used to make the processor wait during a DMA operation make the processor wait during a power interrupt processing make the processor wait during a power shutdown interface slow peripherals to the processor
answered
May 19, 2020
in
Digital Signal Processing
Mohit Kumar 6
138
views
nielit2016mar-scientistc
non-gate
microprocessors
0
votes
1
answer
18
NIELIT 2016 DEC Scientist B (CS) - Section B: 25
A low pass filter is Passes the frequencies lower than the specified cut off frequency Used to recover signal from sampled signal Rejects higher frequencies All of the above
A low pass filter is Passes the frequencies lower than the specified cut off frequency Used to recover signal from sampled signal Rejects higher frequencies All of the above
answered
Apr 25, 2020
in
Digital Signal Processing
Sabirazain
242
views
nielit2016dec-scientistb-cs
non-gate
0
votes
1
answer
19
NIELIT 2016 MAR Scientist C - Section C: 61
Microprogramming is a technique for writing small programs effectively programming output/input routines programming the microprocessors programming the control steps of a computer
Microprogramming is a technique for writing small programs effectively programming output/input routines programming the microprocessors programming the control steps of a computer
answered
Apr 4, 2020
in
Digital Signal Processing
Vipin Tiwari
105
views
nielit2016mar-scientistc
co-and-architecture
microprogramming
0
votes
0
answers
20
NIELIT 2016 MAR Scientist C - Section C: 42
A stable multivibrator are used as comparator circuit squaring circuit frequency to voltage converter voltage to frequency converter
A stable multivibrator are used as comparator circuit squaring circuit frequency to voltage converter voltage to frequency converter
asked
Apr 2, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
69
views
nielit2016mar-scientistc
non-gate
0
votes
0
answers
21
NIELIT 2016 MAR Scientist C - Section C: 43
The astable multivibrator has two quasi stable states two stable states one stable and one quasi-stable state none of these
The astable multivibrator has two quasi stable states two stable states one stable and one quasi-stable state none of these
asked
Apr 2, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
60
views
nielit2016mar-scientistc
non-gate
0
votes
0
answers
22
NIELIT 2017 OCT Scientific Assistant A (CS) - Section B: 27
At $100\%$ modulation, the power in each sideband is _______ of that of carrier. $50\%$ $40\%$ $60\%$ $25\%$
At $100\%$ modulation, the power in each sideband is _______ of that of carrier. $50\%$ $40\%$ $60\%$ $25\%$
asked
Apr 1, 2020
in
Digital Signal Processing
Lakshman Patel RJIT
70
views
nielit2017oct-assistanta-cs
non-gate
0
votes
0
answers
23
KPGCET-CSE-2019-49
The communication network comprises number of base stations. The scope of each base station covers the circular area. To avoid the overlapping and gaping of areas, the scope of each base station is accounted by Rectangle Pentagon Hexagon Octagon
The communication network comprises number of base stations. The scope of each base station covers the circular area. To avoid the overlapping and gaping of areas, the scope of each base station is accounted by Rectangle Pentagon Hexagon Octagon
asked
Aug 4, 2019
in
Digital Signal Processing
gatecse
34
views
kpgcet-cse-2019
non-gate
data-communication
0
votes
0
answers
24
KPGCET-CSE-2019-51
In a communicative network, though the scope of base station is circular area, the network scope is represented by______ shape, as it is nearest to the circular area. triangular rectangular pentagonal hexagonal.
In a communicative network, though the scope of base station is circular area, the network scope is represented by______ shape, as it is nearest to the circular area. triangular rectangular pentagonal hexagonal.
asked
Aug 4, 2019
in
Digital Signal Processing
gatecse
23
views
kpgcet-cse-2019
non-gate
data-communication
2
votes
3
answers
25
UGCNET-June2016-III: 58
Consider a discrete memoryless channel and assume that H(x) is the amount of information per symbol at the input of the channel; H(y) is the amount of information per symbol at the output of the channel. H(x $\mid$ y) is the amount of uncertainty remaining on x knowing y; and I(x;y)is the ... (y $\mid$ x)]; p(x) max [H(x)-H(x $\mid$ y)]; p(x) max H(x $\mid$y); p(x)
Consider a discrete memoryless channel and assume that H(x) is the amount of information per symbol at the input of the channel; H(y) is the amount of information per symbol at the output of the channel. H(x $\mid$ y) is the amount of uncertainty remaining on x knowing y; and I(x;y)is the information transmission. ... -H(y $\mid$ x)]; p(x) max [H(x)-H(x $\mid$ y)]; p(x) max H(x $\mid$y); p(x)
answered
May 3, 2019
in
Digital Signal Processing
Adnan Ashraf
967
views
ugcnetjune2016iii
digital-image-processing
discrete-memoryless-channel
2
votes
1
answer
26
UGCNET-Dec2014-III: 66
A Butterworth lowpass filter of order $n$, with a cutoff frequency at distance $D_{0}$ from the origin, has the transfer function $H(u, v)$ given by $\frac{1}{1+\left[\frac{D(u, v)}{D_{0}}\right]^{2n}}$ $\frac{1}{1+\left[\frac{D(u, v)}{D_{0}}\right]^{n}}$ $\frac{1}{1+\left[\frac{D_{0}}{D(u, v)}\right]^{2n}}$ $\frac{1}{1+\left[\frac{D_{0}}{D(u, v)}\right]^{n}}$
A Butterworth lowpass filter of order $n$, with a cutoff frequency at distance $D_{0}$ from the origin, has the transfer function $H(u, v)$ given by $\frac{1}{1+\left[\frac{D(u, v)}{D_{0}}\right]^{2n}}$ $\frac{1}{1+\left[\frac{D(u, v)}{D_{0}}\right]^{n}}$ $\frac{1}{1+\left[\frac{D_{0}}{D(u, v)}\right]^{2n}}$ $\frac{1}{1+\left[\frac{D_{0}}{D(u, v)}\right]^{n}}$
answered
Jan 5, 2017
in
Digital Signal Processing
Sanjay Sharma
661
views
ugcnetdec2014iii
digital-image-processing
butterworth-lowpass-filter
3
votes
1
answer
27
UGCNET-Dec2014-III: 64
Given two spatial masks $S_{1}= \begin{bmatrix} 0&1&0 \\ 1&-4&0 \\ 0&1&0 \end{bmatrix}$ $S_{2} = \begin{bmatrix} 1&1&1 \\ 1&-8&1 \\ 1&1&1 \end{bmatrix}$ The Laplacian of an image at all points $(x, y)$ can be ... of the following can be used as the spatial mask ? only $S_{1}$ only $S_{2}$ Both $S_{1}$ and $S_{2}$ None of these
Given two spatial masks $S_{1}= \begin{bmatrix} 0&1&0 \\ 1&-4&0 \\ 0&1&0 \end{bmatrix}$ $S_{2} = \begin{bmatrix} 1&1&1 \\ 1&-8&1 \\ 1&1&1 \end{bmatrix}$ The Laplacian of an image at all points $(x, y)$ can be implemented by convolving the image with spatial mask. Which of the following can be used as the spatial mask ? only $S_{1}$ only $S_{2}$ Both $S_{1}$ and $S_{2}$ None of these
answered
Dec 23, 2016
in
Digital Signal Processing
Sanjay Sharma
1.3k
views
ugcnetdec2014iii
digital-image-processing
spatial-mask
–2
votes
0
answers
28
3sin(200rt) is amplitude modulated with 600
3sin(200rt) is amplitude modulated with 600s(6000rt) Find the spectral response and expression of the AM signal generated
3sin(200rt) is amplitude modulated with 600s(6000rt) Find the spectral response and expression of the AM signal generated
asked
Nov 28, 2016
in
Digital Signal Processing
rahuldb
176
views
non-gate
1
vote
1
answer
29
UGCNET-Dec2015-III: 51
Blind image disconvolution is Combination of blur identification and image restoration Combination of segmentation and classification Combination of blur and non-blur image None of the above
Blind image disconvolution is Combination of blur identification and image restoration Combination of segmentation and classification Combination of blur and non-blur image None of the above
answered
Aug 11, 2016
in
Digital Signal Processing
Sanjay Sharma
828
views
ugcnetdec2015iii
digital-image-processing
discovolution
3
votes
1
answer
30
UGCNET-June2012-II: 18
Page Shift Keying (PSK) Method is used to modulate digital signal at 9600 bps using 16 level. Find the line signals and speed (i.e., modulation rate) 2400 bauds 1200 bauds 4800 bauds 9600 bauds
Page Shift Keying (PSK) Method is used to modulate digital signal at 9600 bps using 16 level. Find the line signals and speed (i.e., modulation rate) 2400 bauds 1200 bauds 4800 bauds 9600 bauds
answered
Jul 5, 2016
in
Digital Signal Processing
Sanjay Sharma
2.3k
views
ugcnetjune2012ii
non-gate
digital-signal-processing
0
votes
1
answer
31
net12
answered
May 6, 2016
in
Digital Signal Processing
rude
98
views
signal-processing
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