,differential equation for the 2nd order system

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I would like to know which differential equation I have to solve in order to get the answer to a unitary step like the one report in figure:

tell me if I'm wrong:
starting from :

$mfracd^2y_1dt+bfracdy_1dt+k y_1= 0$

I can compute the substitution $fracdydt = y_2 $

so the system of first order equation became :
begincases
y'_1 = y_2 \
y'_2 = -fracbmy_2 -frackm \
endcases
right ?
how can I set the constant ? which initial value should I use ?

EDIT
Maybe I'm not well explain! I don't wanna know the output but which value and boundary condition I have to use in the system that I wrote in order to get the set of response in function of the parameter $zeta$ (and also know what is $zeta$) I know the soultion but I wanna solve the differential equations and get this set of curve !! Consider my question like: which is the function to integrate .... with wich value or like :"I have to give an exercise: using a euler method find the solution of a unit step response for different value of $zeta$ and $omega_o$ "

Thanks for your precious support ... what is $x(t)$ ?

differential-equations dynamical-systems

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edited Jul 17 at 10:05

asked Jul 17 at 9:51

Drudox lebowsky

1165

add a comment | 

up vote
0
down vote

favorite

I would like to know which differential equation I have to solve in order to get the answer to a unitary step like the one report in figure:

tell me if I'm wrong:
starting from :

$mfracd^2y_1dt+bfracdy_1dt+k y_1= 0$

I can compute the substitution $fracdydt = y_2 $

so the system of first order equation became :
begincases
y'_1 = y_2 \
y'_2 = -fracbmy_2 -frackm \
endcases
right ?
how can I set the constant ? which initial value should I use ?

EDIT
Maybe I'm not well explain! I don't wanna know the output but which value and boundary condition I have to use in the system that I wrote in order to get the set of response in function of the parameter $zeta$ (and also know what is $zeta$) I know the soultion but I wanna solve the differential equations and get this set of curve !! Consider my question like: which is the function to integrate .... with wich value or like :"I have to give an exercise: using a euler method find the solution of a unit step response for different value of $zeta$ and $omega_o$ "

Thanks for your precious support ... what is $x(t)$ ?

differential-equations dynamical-systems

share|cite|improve this question

edited Jul 17 at 10:05

asked Jul 17 at 9:51

Drudox lebowsky

1165

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I would like to know which differential equation I have to solve in order to get the answer to a unitary step like the one report in figure:

tell me if I'm wrong:
starting from :

$mfracd^2y_1dt+bfracdy_1dt+k y_1= 0$

I can compute the substitution $fracdydt = y_2 $

so the system of first order equation became :
begincases
y'_1 = y_2 \
y'_2 = -fracbmy_2 -frackm \
endcases
right ?
how can I set the constant ? which initial value should I use ?

EDIT
Maybe I'm not well explain! I don't wanna know the output but which value and boundary condition I have to use in the system that I wrote in order to get the set of response in function of the parameter $zeta$ (and also know what is $zeta$) I know the soultion but I wanna solve the differential equations and get this set of curve !! Consider my question like: which is the function to integrate .... with wich value or like :"I have to give an exercise: using a euler method find the solution of a unit step response for different value of $zeta$ and $omega_o$ "

Thanks for your precious support ... what is $x(t)$ ?

differential-equations dynamical-systems

share|cite|improve this question

edited Jul 17 at 10:05

asked Jul 17 at 9:51

Drudox lebowsky

1165

I would like to know which differential equation I have to solve in order to get the answer to a unitary step like the one report in figure:

tell me if I'm wrong:
starting from :

$mfracd^2y_1dt+bfracdy_1dt+k y_1= 0$

I can compute the substitution $fracdydt = y_2 $

so the system of first order equation became :
begincases
y'_1 = y_2 \
y'_2 = -fracbmy_2 -frackm \
endcases
right ?
how can I set the constant ? which initial value should I use ?

EDIT
Maybe I'm not well explain! I don't wanna know the output but which value and boundary condition I have to use in the system that I wrote in order to get the set of response in function of the parameter $zeta$ (and also know what is $zeta$) I know the soultion but I wanna solve the differential equations and get this set of curve !! Consider my question like: which is the function to integrate .... with wich value or like :"I have to give an exercise: using a euler method find the solution of a unit step response for different value of $zeta$ and $omega_o$ "

Thanks for your precious support ... what is $x(t)$ ?

differential-equations dynamical-systems

share|cite|improve this question

edited Jul 17 at 10:05

asked Jul 17 at 9:51

Drudox lebowsky

1165

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 10:05

edited Jul 17 at 10:05

edited Jul 17 at 10:05

asked Jul 17 at 9:51

Drudox lebowsky

1165

asked Jul 17 at 9:51

Drudox lebowsky

1165

asked Jul 17 at 9:51

Drudox lebowsky

1165

1165

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
1
down vote



accepted

From the image you can read of a period of about $6$ which confirms the label of a frequency $ω_0=1$. Thus the basis is a harmonic oscillator $y''+(y-1)=0$. In addition there is a friction term that leads to the represented exponential decay, where one can guess that the friction coefficient is $2zeta$ (as per the normalized form in the answer of Mostafa Ayaz). In total this makes an equation
$$y''+2zeta y'+y=x.$$
As per usual in such situations of exploring system responses it is assumed that all functions are zero for $t<0$, so that the initial conditions are $y(0)=0$ and $y'(0)=0$. The unit step function is $x(t)=1$ for $tge 0$, and of course $x(t)=0$ for $t<0$. Then solve
$$y''+2zeta y'+y=1$$
for $tge 0$.

t = np.linspace(0,30,1501)
for zeta in [0.1,0.2,0.5,1,2,4]:
 sol = odeint(lambda y,t:[y[1], 1-y[0]-2*zeta*y[1]],[0,0], t)
 plt.plot(t,sol[:,0], label="$zeta=$%.2f"%zeta);
plt.grid(); plt.legend(); plt.show()

to get

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edited Jul 17 at 10:39

answered Jul 17 at 10:18

LutzL

49.8k31849

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up vote
1
down vote

The differential equation you propose, when disturbed, will oscillate around y=0. The differential equation you should consider is $$mfracd^2ydt^2+bfracdydt+ky=F(t)$$ where F(t) is a simple step function, namely, F=0 for t<0 and F=1/k for t>0. This will make the displacement oscillate around 1 as above. The solution to this equation can be achieved by several standard methods such as the Laplace transform listed @Mostafa Ayaz but in this case, the Laplace transform of the forcing function, F(t), must also be included.

share|cite|improve this answer

answered Jul 17 at 10:22

Dr Peter McGowan

4637

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up vote
1
down vote

The output has such a Laplace transform$$Y(s)=dfrac1s^2+2zetaomega_0s+omega_0^2$$which is the output of the following ODE$$y''(t)+2zetaomega_0y'(t)+omega_0^2y(t)=x(t)$$with initial rest

share|cite|improve this answer

edited Jul 17 at 10:36

answered Jul 17 at 9:58

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you it was a typo
  – Mostafa Ayaz
  Jul 17 at 10:36
  

  
  
  
  

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

From the image you can read of a period of about $6$ which confirms the label of a frequency $ω_0=1$. Thus the basis is a harmonic oscillator $y''+(y-1)=0$. In addition there is a friction term that leads to the represented exponential decay, where one can guess that the friction coefficient is $2zeta$ (as per the normalized form in the answer of Mostafa Ayaz). In total this makes an equation
$$y''+2zeta y'+y=x.$$
As per usual in such situations of exploring system responses it is assumed that all functions are zero for $t<0$, so that the initial conditions are $y(0)=0$ and $y'(0)=0$. The unit step function is $x(t)=1$ for $tge 0$, and of course $x(t)=0$ for $t<0$. Then solve
$$y''+2zeta y'+y=1$$
for $tge 0$.

t = np.linspace(0,30,1501)
for zeta in [0.1,0.2,0.5,1,2,4]:
 sol = odeint(lambda y,t:[y[1], 1-y[0]-2*zeta*y[1]],[0,0], t)
 plt.plot(t,sol[:,0], label="$zeta=$%.2f"%zeta);
plt.grid(); plt.legend(); plt.show()

to get

share|cite|improve this answer

edited Jul 17 at 10:39

answered Jul 17 at 10:18

LutzL

49.8k31849

add a comment | 

up vote
1
down vote



accepted

From the image you can read of a period of about $6$ which confirms the label of a frequency $ω_0=1$. Thus the basis is a harmonic oscillator $y''+(y-1)=0$. In addition there is a friction term that leads to the represented exponential decay, where one can guess that the friction coefficient is $2zeta$ (as per the normalized form in the answer of Mostafa Ayaz). In total this makes an equation
$$y''+2zeta y'+y=x.$$
As per usual in such situations of exploring system responses it is assumed that all functions are zero for $t<0$, so that the initial conditions are $y(0)=0$ and $y'(0)=0$. The unit step function is $x(t)=1$ for $tge 0$, and of course $x(t)=0$ for $t<0$. Then solve
$$y''+2zeta y'+y=1$$
for $tge 0$.

t = np.linspace(0,30,1501)
for zeta in [0.1,0.2,0.5,1,2,4]:
 sol = odeint(lambda y,t:[y[1], 1-y[0]-2*zeta*y[1]],[0,0], t)
 plt.plot(t,sol[:,0], label="$zeta=$%.2f"%zeta);
plt.grid(); plt.legend(); plt.show()

to get

share|cite|improve this answer

edited Jul 17 at 10:39

answered Jul 17 at 10:18

LutzL

49.8k31849

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

From the image you can read of a period of about $6$ which confirms the label of a frequency $ω_0=1$. Thus the basis is a harmonic oscillator $y''+(y-1)=0$. In addition there is a friction term that leads to the represented exponential decay, where one can guess that the friction coefficient is $2zeta$ (as per the normalized form in the answer of Mostafa Ayaz). In total this makes an equation
$$y''+2zeta y'+y=x.$$
As per usual in such situations of exploring system responses it is assumed that all functions are zero for $t<0$, so that the initial conditions are $y(0)=0$ and $y'(0)=0$. The unit step function is $x(t)=1$ for $tge 0$, and of course $x(t)=0$ for $t<0$. Then solve
$$y''+2zeta y'+y=1$$
for $tge 0$.

t = np.linspace(0,30,1501)
for zeta in [0.1,0.2,0.5,1,2,4]:
 sol = odeint(lambda y,t:[y[1], 1-y[0]-2*zeta*y[1]],[0,0], t)
 plt.plot(t,sol[:,0], label="$zeta=$%.2f"%zeta);
plt.grid(); plt.legend(); plt.show()

to get

share|cite|improve this answer

edited Jul 17 at 10:39

answered Jul 17 at 10:18

LutzL

49.8k31849

From the image you can read of a period of about $6$ which confirms the label of a frequency $ω_0=1$. Thus the basis is a harmonic oscillator $y''+(y-1)=0$. In addition there is a friction term that leads to the represented exponential decay, where one can guess that the friction coefficient is $2zeta$ (as per the normalized form in the answer of Mostafa Ayaz). In total this makes an equation
$$y''+2zeta y'+y=x.$$
As per usual in such situations of exploring system responses it is assumed that all functions are zero for $t<0$, so that the initial conditions are $y(0)=0$ and $y'(0)=0$. The unit step function is $x(t)=1$ for $tge 0$, and of course $x(t)=0$ for $t<0$. Then solve
$$y''+2zeta y'+y=1$$
for $tge 0$.

t = np.linspace(0,30,1501)
for zeta in [0.1,0.2,0.5,1,2,4]:
 sol = odeint(lambda y,t:[y[1], 1-y[0]-2*zeta*y[1]],[0,0], t)
 plt.plot(t,sol[:,0], label="$zeta=$%.2f"%zeta);
plt.grid(); plt.legend(); plt.show()

to get

share|cite|improve this answer

edited Jul 17 at 10:39

answered Jul 17 at 10:18

LutzL

49.8k31849

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 10:39

edited Jul 17 at 10:39

edited Jul 17 at 10:39

answered Jul 17 at 10:18

LutzL

49.8k31849

answered Jul 17 at 10:18

LutzL

49.8k31849

answered Jul 17 at 10:18

LutzL

49.8k31849

49.8k31849

add a comment | 

add a comment | 

up vote
1
down vote

The differential equation you propose, when disturbed, will oscillate around y=0. The differential equation you should consider is $$mfracd^2ydt^2+bfracdydt+ky=F(t)$$ where F(t) is a simple step function, namely, F=0 for t<0 and F=1/k for t>0. This will make the displacement oscillate around 1 as above. The solution to this equation can be achieved by several standard methods such as the Laplace transform listed @Mostafa Ayaz but in this case, the Laplace transform of the forcing function, F(t), must also be included.

share|cite|improve this answer

answered Jul 17 at 10:22

Dr Peter McGowan

4637

add a comment | 

up vote
1
down vote

The differential equation you propose, when disturbed, will oscillate around y=0. The differential equation you should consider is $$mfracd^2ydt^2+bfracdydt+ky=F(t)$$ where F(t) is a simple step function, namely, F=0 for t<0 and F=1/k for t>0. This will make the displacement oscillate around 1 as above. The solution to this equation can be achieved by several standard methods such as the Laplace transform listed @Mostafa Ayaz but in this case, the Laplace transform of the forcing function, F(t), must also be included.

share|cite|improve this answer

answered Jul 17 at 10:22

Dr Peter McGowan

4637

add a comment | 

up vote
1
down vote

up vote
1
down vote

The differential equation you propose, when disturbed, will oscillate around y=0. The differential equation you should consider is $$mfracd^2ydt^2+bfracdydt+ky=F(t)$$ where F(t) is a simple step function, namely, F=0 for t<0 and F=1/k for t>0. This will make the displacement oscillate around 1 as above. The solution to this equation can be achieved by several standard methods such as the Laplace transform listed @Mostafa Ayaz but in this case, the Laplace transform of the forcing function, F(t), must also be included.

share|cite|improve this answer

answered Jul 17 at 10:22

Dr Peter McGowan

4637

The differential equation you propose, when disturbed, will oscillate around y=0. The differential equation you should consider is $$mfracd^2ydt^2+bfracdydt+ky=F(t)$$ where F(t) is a simple step function, namely, F=0 for t<0 and F=1/k for t>0. This will make the displacement oscillate around 1 as above. The solution to this equation can be achieved by several standard methods such as the Laplace transform listed @Mostafa Ayaz but in this case, the Laplace transform of the forcing function, F(t), must also be included.

share|cite|improve this answer

answered Jul 17 at 10:22

Dr Peter McGowan

4637

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 10:22

Dr Peter McGowan

4637

answered Jul 17 at 10:22

Dr Peter McGowan

4637

answered Jul 17 at 10:22

Dr Peter McGowan

4637

4637

add a comment | 

add a comment | 

up vote
1
down vote

The output has such a Laplace transform$$Y(s)=dfrac1s^2+2zetaomega_0s+omega_0^2$$which is the output of the following ODE$$y''(t)+2zetaomega_0y'(t)+omega_0^2y(t)=x(t)$$with initial rest

share|cite|improve this answer

edited Jul 17 at 10:36

answered Jul 17 at 9:58

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you it was a typo
  – Mostafa Ayaz
  Jul 17 at 10:36
  

  
  
  
  

add a comment | 

up vote
1
down vote

The output has such a Laplace transform$$Y(s)=dfrac1s^2+2zetaomega_0s+omega_0^2$$which is the output of the following ODE$$y''(t)+2zetaomega_0y'(t)+omega_0^2y(t)=x(t)$$with initial rest

share|cite|improve this answer

edited Jul 17 at 10:36

answered Jul 17 at 9:58

Mostafa Ayaz

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you it was a typo
  – Mostafa Ayaz
  Jul 17 at 10:36
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

The output has such a Laplace transform$$Y(s)=dfrac1s^2+2zetaomega_0s+omega_0^2$$which is the output of the following ODE$$y''(t)+2zetaomega_0y'(t)+omega_0^2y(t)=x(t)$$with initial rest

share|cite|improve this answer

edited Jul 17 at 10:36

answered Jul 17 at 9:58

Mostafa Ayaz

8,6023630

The output has such a Laplace transform$$Y(s)=dfrac1s^2+2zetaomega_0s+omega_0^2$$which is the output of the following ODE$$y''(t)+2zetaomega_0y'(t)+omega_0^2y(t)=x(t)$$with initial rest

share|cite|improve this answer

edited Jul 17 at 10:36

answered Jul 17 at 9:58

Mostafa Ayaz

8,6023630

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 10:36

edited Jul 17 at 10:36

edited Jul 17 at 10:36

answered Jul 17 at 9:58

Mostafa Ayaz

8,6023630

answered Jul 17 at 9:58

Mostafa Ayaz

8,6023630

answered Jul 17 at 9:58

Mostafa Ayaz

8,6023630

8,6023630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you it was a typo
  – Mostafa Ayaz
  Jul 17 at 10:36
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you it was a typo
  – Mostafa Ayaz
  Jul 17 at 10:36
  

  
  
  
  

Thank you it was a typo
– Mostafa Ayaz
Jul 17 at 10:36

Thank you it was a typo
– Mostafa Ayaz
Jul 17 at 10:36

add a comment | 

 

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