A Second onder contraining a
relay with dead line and hysteresis
shown below in figure obtain
the phase trajectory of
system for the
initial conditions
eco=.65, e(0)2 O. Does the
Sto system have a limit cycle? is
determine its amplitude and
period.
so
Answers
Answer:
\sf\large\underline\purple{Question:-}
Question:−
\sf{\implies 25^{n-1}+100=5^{2n-1}}⟹25
n−1
+100=5
2n−1
\sf\large\underline\purple{To\: Find:-}
ToFind:−
\sf{\implies The\: value\:of\:(n)=?}⟹Thevalueof(n)=?
\sf\large\underline\purple{Solution:-}
Solution:−
To calculate the value of (n) at first we have splitting the terms the simplify the given expression:-]
\tt{\implies 5^{2n-2}+100=5^{2n-1}}⟹5
2n−2
+100=5
2n−1
\tt{\implies 5^{2n-1}-5^{2n-2}=100}⟹5
2n−1
−5
2n−2
=100
\tt{\implies 5^{2n}*5^{-1}-5^{2n}*5^{2n}*5^{-2}=100}⟹5
2n
∗5
−1
−5
2n
∗5
2n
∗5
−2
=100
\tt{\implies 5^{2n}(5^{-1}-5^{-2})=100}⟹5
2n
(5
−1
−5
−2
)=100
\tt{\implies 5^{2n}\bigg(\dfrac{1}{5}-\dfrac{1}{25}\bigg)=100}⟹5
2n
(
5
1
−
25
1
)=100
\tt{\implies 5^{2n}\bigg(\dfrac{5-1}{25}\bigg)=100}⟹5
2n
(
25
5−1
)=100
\tt{\implies 5^{2n}*\dfrac{4}{25}=100}⟹5
2n
∗
25
4
=100
\tt{\implies 5^{2n}=25*25}⟹5
2n
=25∗25
\tt{\implies 5^{2n}=5^2*5^2}⟹5
2n
=5
2
∗5
2
\tt{\implies 5^{2n}=5^4}⟹5
2n
=5
4
\tt{\implies 2n=4\implies n=2}⟹2n=4⟹n=2
\sf\large{Hence,}Hence,
\sf{\implies The\: value\:of\:(n)=2}⟹Thevalueof(n)=2
\sf\large\underline\purple{Question:-}
Question:−
\tt{\implies \dfrac{(16*2^{n+1}-4*2^{n})}{(16*2^{2n+2}-2*2^{n+2})}
\sf\large\underline\purple{To\: Find:-}
ToFind:−
\sf{\implies To\: simplify\:the\: given\: expression:-}⟹Tosimplifythegivenexpression:−
\sf\large\underline\purple{Solution:-}
Solution:−
To calculate simplification, at first we have splitting the terms the simplify the given expression:-]
\tt{\implies \dfrac{16*2^{n}*2^{1}-4*2^{n}}{16*2^{2n}*2^{2}-2*2^{n}*2^{2}}
\tt{\implies \dfrac{2^{n}(16*2^{1}-4)}{2^n(16*2^{2}-2*2^{2}}
\tt{\implies \dfrac{16*2-4}{16*4-2*4}}⟹
16∗4−2∗4
16∗2−4
\tt{\implies \dfrac{32-4}{64-8}}⟹
64−8
32−4
\tt{\implies \dfrac{28}{56}=\dfrac{1}{2}}⟹
56
28
=
2
1