The system shown in the image is : * y[n] = Linear Non linear Both = None of the mentioned The system shown in the image is : * √x[n] y[n] = y[n − 1] + x[n] BIBO Stable BIBO Unstable Both None of the mer ed
The system shown in the image is : * y[n] = Linear Non linear Both = None of the mentioned The system shown in the image is : * √x[n] y[n] = y[n − 1] + x[n] BIBO Stable BIBO Unstable Both None of the mer ed
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter6: Power Flows
Section: Chapter Questions
Problem 6.66P
Related questions
Question
Solve all the questions correctly with explanation.
![The system shown in the image is : *
y[n] = √x[n]
Linear
Non linear
Both
None of the mentioned
The system shown in the image is : *
y[n] = y[n − 1] + x[n]
BIBO Stable
BIBO Unstable
Both
None of the mer
ed](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc7943fae-eca6-43c1-8ab1-3c716e641f7d%2F610aa574-015a-4729-9089-329fe7e04486%2Fyxlyz0g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The system shown in the image is : *
y[n] = √x[n]
Linear
Non linear
Both
None of the mentioned
The system shown in the image is : *
y[n] = y[n − 1] + x[n]
BIBO Stable
BIBO Unstable
Both
None of the mer
ed
![What is the fundamental
period of the
signal, x(n) = cos (n/18) + sin(n/18)
18
36
non-periodic
partly-periodic
The signal x(n) = 5 is a - *
Energy signal
Power signal
Periodic signal
None of the me,
ed](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc7943fae-eca6-43c1-8ab1-3c716e641f7d%2F610aa574-015a-4729-9089-329fe7e04486%2Fdfshpfh_processed.png&w=3840&q=75)
Transcribed Image Text:What is the fundamental
period of the
signal, x(n) = cos (n/18) + sin(n/18)
18
36
non-periodic
partly-periodic
The signal x(n) = 5 is a - *
Energy signal
Power signal
Periodic signal
None of the me,
ed
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Step 1: State the given data
VIEWStep 2: 1. Checking the linearity property of the system y[n]=sqrt(x[n]).
VIEWStep 3: 2. Checking the stability of the system y[n]=y[n-1]+x[n].
VIEWStep 4: 3. Finding the fundamental period of the signal, x[n].
VIEWStep 5: 4. Finding the type of the signal, x[n]=5 Whether it is energy o power or periodic signal.
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