please have step by step solution and explain. thank you very much x(t) —e(t)G(s)y(t)1/s Consider a feedback-based causal LTI system with input x(t) shown in Figure6.41, where G(s) is the transfer function of a "loop filter" and 1/s is an integrator. The idea isto use the error e(t) to drive these, so as to make y(t) track x(t).(a) Find the transfer functions H(s) =Y(s)X(s)and He(s)=E(s)X(s)*(b) Now set G(s) = 10. This is termed "proportional feedback,” where the feedback is propor-tional to the error.(i) Find y(t) and e(t) for x(t) = sin 10t.(ii) How do your answers change (provide a qualitative discussion) for x(t) = sint and x(t) =sin 100t?=(iii) For x(t) = u(t) (unit step), find and sketch y(t) and e(t), and specify their asymptotic valuesas t∞.(iv) For x(t) = tu(t) (ramp starting at time zero), find the asymptotic value of the error e(t) ast → ∞.Hint: You can simply use the final value theorem in (iii).(c) Redo (b)(iv) for G(s) = 10 + ½ (“proportional plus integral" feedback).S

A negative feedback closed loop system has the following blocks: G(s), and an integrator 1s.

Answered: Consider a feedback-based causal LTI…

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please have step by step solution and explain. thank you very much

Consider a feedback-based causal LTI system with input x(t) shown in Figure
6.41, where G(s) is the transfer function of a "loop filter" and 1/s is an integrator. The idea is
to use the error e(t) to drive these, so as to make y(t) track x(t).
(a) Find the transfer functions H(s) =
Y(s)
X(s)
and He(s)
=
E(s)
X(s)*
(b) Now set G(s) = 10. This is termed "proportional feedback,” where the feedback is propor-
tional to the error.
(i) Find y(t) and e(t) for x(t) = sin 10t.
(ii) How do your answers change (provide a qualitative discussion) for x(t) = sint and x(t) =
sin 100t?
=
(iii) For x(t) = u(t) (unit step), find and sketch y(t) and e(t), and specify their asymptotic values
as t∞.
(iv) For x(t) = tu(t) (ramp starting at time zero), find the asymptotic value of the error e(t) as
t → ∞.
Hint: You can simply use the final value theorem in (iii).
(c) Redo (b)(iv) for G(s) = 10 + ½ (“proportional plus integral" feedback).
S

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for your answer for part b ii what happened to x(j10)? sin10t should be equal to =10/s^2+100, since its the laplace transform. i need more explaination on the y(t) answer too. how did the angle become sin? i am so confused on your answer

The input signal is given by,
x(t)=sin(10t).
Its angular frequency is w = 10 rad/s.
The corresponding output and error signals in frequency domain are
Y(j10)=H(10) X(10)
1
=
Z-tan
1(100) × 1
×120°,
102+100
=
·Z-45°;
√2
E(j10) = H¸(j10)·X(j10)
1
√10²+100
1
= - 45°.
√2
In time-domain,
[90°-tan-1 (100)] × 120°,
1
y(t)=
sin(10t―45º),
√2
e(t) =
sin(10t+45°).
√2

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