SYSTEM DYNAMICS>LOOSELEAF<
SYSTEM DYNAMICS>LOOSELEAF<
3rd Edition
ISBN: 9781260163087
Author: Palm
Publisher: MCG
Question
Chapter 9, Problem 9.1P
To determine

(a)

The steady state response.

Expert Solution
Check Mark

Answer to Problem 9.1P

The steady state response is,

yss=13.5582sin(1.5t0.8622).

Explanation of Solution

Given Information:

Transfer function is given by,

T(s)=Y(s)F(s)=2514s+18.

Input signal is, f(t)=15sin1.5t.

Calculation:

From the input given, f(t)=15sin1.5t, the angular frequency is,

ω=1.5rad/s.

Transfer function is,

T(s)=2514s+18T(jω)=2514(jω)+18T(jω)=2514(j1.5)+18T(jω)=2518+j21

So, the magnitude of transfer function is,

M=25182+212M=0.90388

Phase angle,

ϕ=0tan1(2118)ϕ=0.8622rad

Thus, the steady state response,

yss=15Msin(1.5t+ϕ)yss=13.5582sin(1.5t0.8622).

Conclusion:

The steady state response is,

yss=13.5582sin(1.5t0.8622).

To determine

(b)

The steady state response.

Expert Solution
Check Mark

Answer to Problem 9.1P

The steady state response is,

yss=20.8013sin(1.5t+0.588).

Explanation of Solution

Given Information:

Transfer function is given by,

T(s)=Y(s)F(s)=15s3s+4.

Input signal is, f(t)=5sin2t.

Calculation:

From the input given, f(t)=5sin2t, the angular frequency is,

ω=2rad/s.

Transfer function is,

T(s)=15s3s+4T(jω)=15(jω)3(jω)+4T(jω)=15(j2)3(j2)+4T(jω)=j304+j6

So, the magnitude of transfer function is,

M=3042+62M=4.16025

Phase angle,

ϕ=π2tan1(64)ϕ=0.588rad

Thus, the steady state response,

yss=5Msin(2t+ϕ)yss=20.8013sin(1.5t+0.588).

Conclusion:

The steady state response is,

yss=20.8013sin(1.5t+0.588).

To determine

(c)

The steady state response.

Expert Solution
Check Mark

Answer to Problem 9.1P

The steady state response is,

yss=1.8605sin(100t+0.5191).

Explanation of Solution

Given Information:

Transfer function is given by,

T(s)=Y(s)F(s)=s+50s+150.

Input signal is, f(t)=3sin100t.

Calculation:

From the input given,, the angular frequency is,

ω=100rad/s.

Transfer function is,

T(s)=s+50s+150T(jω)=(jω)+50(jω)+150T(jω)=50+j100150+j100

So, the magnitude of transfer function is,

M=502+10021502+1002M=0.62017

Phase angle,

ϕ=tan1(10050)tan1(100150)ϕ=0.5191rad

Thus, the steady state response,

yss=3Msin(100t+ϕ)yss=1.8605sin(100t+0.5191).

Conclusion:

The steady state response is,

yss=1.8605sin(100t+0.5191).

To determine

(d)

The steady state response.

Expert Solution
Check Mark

Answer to Problem 9.1P

The steady state response is,

yss=2.4634sin(50t0.5238).

Explanation of Solution

Given Information:

Transfer function is given by,

T(s)=Y(s)F(s)=33200s+100s+33.

Input signal is, f(t)=8sin50t.

Calculation:

From the input given,, the angular frequency is,

ω=50rad/s.

Transfer function is,

T(s)=33200(s+100s+33)T(jω)=33200((jω)+100(jω)+33)T(jω)=33200(100+j5033+j50)

So, the magnitude of transfer function is,

M=33200×1002+502332+502M=0.30793

Phase angle,

ϕ=tan1(50100)tan1(5033)ϕ=0.5238rad

Thus, the steady state response,

yss=8Msin(50t+ϕ)yss=2.4634sin(50t0.5238).

Conclusion:

The steady state response is,

yss=2.4634sin(50t0.5238).

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