Class Wiki - User contributions [en]

Coupled Oscillator: Spring Pendulums

2009-12-14T00:56:28Z

Kendrick.mensink: /* Solution */

===Problem Statement===
----
Use State Space methods to write up the solution to a coupled pendulum problem. Describe the eigen-modes of the system.

===Solution===
----
By definition, the the state equation is stated as

<math>
\underline{\dot{x}} = \widehat{A} \, \underline{x} + \widehat{B} \, \underline{u}
</math>

Now, consider the motion equations described in the Solution section,

<math>m_1\ddot{x}_1+k_1x_1-k_2(x_2-x_1)=m_1\ddot{x}_1+k_1x_1-k_2x_2+k_2x_1=0</math>

<math>m_2\ddot{x}_2+k_2(x_2-x_1)-k_3x_2=m_2\ddot{x}_2+k_2x_2-k_2x_1-k_3x_2=0</math>

Solving for <math>\ddot{x}_1</math> and <math>\ddot{x}_2</math> yields,

:<math>\ddot{x}_1=-\dfrac{k_1}{m_1}x_1+\dfrac{k_2}{m_1}x_2-\dfrac{k_2}{m_1}x_1</math>

:<math>\ddot{x}_2=\dfrac{-k_2}{m_2}x_2+\dfrac{k_2}{m_2}x_1+\dfrac{k_3}{m_2}x_2</math>

Finally, we let <math>x_1 \frac{}{}</math>, <math>\dot{x}_1 \frac{}{}</math>, <math>x_2 \frac{}{}</math>, and <math>\dot{x_2} \frac{}{}</math> be the state variables. Thus,

<math>
\begin{bmatrix}
\dot{x}_1 \\
\ddot{x}_1 \\
\dot{x}_2 \\
\ddot{x}_2
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\frac{1}{m_1}(k_1+k_2) & 0 & \frac{k_2}{m_1} & 0 \\
0 & 0 & 0 & 1 \\
\frac{k_2}{m_2} & 0 & \frac{1}{m_2}(k_3-k_2) & 0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\end{bmatrix}
</math>

----
Created by Kendrick Mensink

Coupled Oscillator: Spring Pendulums

2009-12-14T00:53:25Z

Kendrick.mensink: /* Solution */

===Problem Statement===
----
Use State Space methods to write up the solution to a coupled pendulum problem. Describe the eigen-modes of the system.

===Solution===
----

----
Created by Kendrick Mensink

Coupled Oscillator: Spring Pendulums

2009-12-14T00:14:16Z

Kendrick.mensink: New page: ===Problem Statement=== ---- Use State Space methods to write up the solution to a coupled pendulum problem. Describe the eigen-modes of the system. ===Solution=== ----

===Problem Statement===
----
Use State Space methods to write up the solution to a coupled pendulum problem. Describe the eigen-modes of the system.

===Solution===
----

Coupled Double Pendulum Oscillator

2009-12-14T00:14:01Z

Kendrick.mensink: Removing all content from page

Coupled Double Pendulum Oscillator

2009-12-14T00:13:48Z

Kendrick.mensink: New page: ===Problem Statement=== ---- Use State Space methods to write up the solution to a coupled pendulum problem. Describe the eigen-modes of the system. ===Solution=== ----

===Problem Statement===
----
Use State Space methods to write up the solution to a coupled pendulum problem. Describe the eigen-modes of the system.

===Solution===
----

Fall 2009

2009-12-14T00:11:26Z

Kendrick.mensink: /* HW # 5 */

====HW # 5====
Put a link here to an example problem you made up and solved using Laplace Transforms, of the complete solution of a mechanical system or a circuit that is described by linear ordinary differential equations with constant coefficients. Make sure you start with the physical system, and end up with the time response of the system.

Use your wiki page to explain the problem and solution to one of your classmates. Have him or her certify that they have checked it for errors, by listing that on the page. Then go to [http://moodle.wallawalla.edu Moodle] and put a link to your page in the in-box for HW #5.

[[Laplace transforms:Series RLC circuit]]

[[Laplace transforms:Mass-Spring Oscillator]]

[[Laplace transforms:DC Motor circuit]]

[[Laplace transforms: Simple Electrical Network]]

[[Laplace transforms: R series with RC parallel circuit]]

[[Laplace transforms: Critically Damped Motion ]]

[[Laplace transforms: Under-damped Mass-Spring System on an Incline]]

[[Laplace transforms: Critically Damped Spring Mass system]]

[[Laplace Transforms: Vertical Motion of a Coupled Spring System]]

[[Laplace Transforms: Coupled Springs]]

====HW #12====
Coupled Oscillator Problem

[[Coupled Oscillator: Hellie]]

[[Coupled Oscillator: Coupled Mass-Spring System with Input]]

[[Coupled Oscillator: Coupled Mass-Spring System with Damping]]

[[Coupled Oscillator: Jonathan Schreven]]

[[Coupled Horizontal Spring Mass Oscillator]]

[[Coupled Oscillator: Double Pendulum]]

[[Coupled Oscillator: horizontal Mass-Spring]]

[[Coupled Oscillator: Spring Pendulums]]

==2009-2010 Contributors==

[[Ben Henry]]

[http://fweb.wallawalla.edu/class-wiki/index.php/Fall_2009/JonathanS Jonathan Schreven]

[http://fweb/class-wiki/index.php/Laplace_transforms:DC_Motor_circuit/ Kendrick Mensink]

[http://fweb/class-wiki/index.php/Laplace_transforms:_Critically_Damped_Motion Mark Bernet]

Fall 2009

2009-12-13T23:18:09Z

Kendrick.mensink: /* HW #12 */

====HW # 5====
Put a link here to an example problem you made up and solved using Laplace Transforms, of the complete solution of a mechanical system or a circuit that is described by linear ordinary differential equations with constant coefficients. Make sure you start with the physical system, and end up with the time response of the system.

Use your wiki page to explain the problem and solution to one of your classmates. Have him or her certify that they have checked it for errors, by listing that on the page. Then go to [http://moodle.wallawalla.edu Moodle] and put a link to your page in the in-box for HW #5.

[[Laplace transforms:Series RLC circuit]]

[[Laplace transforms:Mass-Spring Oscillator]]

[[Laplace transforms: DC Motor circuit]]

[[Laplace transforms: Simple Electrical Network]]

[[Laplace transforms: R series with RC parallel circuit]]

[[Laplace transforms: Critically Damped Motion ]]

[[Laplace transforms: Under-damped Mass-Spring System on an Incline]]

[[Laplace transforms: Critically Damped Spring Mass system]]

[[Laplace Transforms: Vertical Motion of a Coupled Spring System]]

[[Laplace Transforms: Coupled Springs]]

====HW #12====
Coupled Oscillator Problem

[[Coupled Oscillator: Hellie]]

[[Coupled Oscillator: Coupled Mass-Spring System with Input]]

[[Coupled Oscillator: Coupled Mass-Spring System with Damping]]

[[Coupled Oscillator: Jonathan Schreven]]

[[Coupled Horizontal Spring Mass Oscillator]]

[[Coupled Oscillator: Double Pendulum]]

[[Coupled Oscillator: horizontal Mass-Spring]]

[[Coupled Oscillator: Spring Pendulums]]

==2009-2010 Contributors==

[[Ben Henry]]

[http://fweb.wallawalla.edu/class-wiki/index.php/Fall_2009/JonathanS Jonathan Schreven]

[http://fweb/class-wiki/index.php/Laplace_transforms:DC_Motor_circuit/ Kendrick Mensink]

[http://fweb/class-wiki/index.php/Laplace_transforms:_Critically_Damped_Motion Mark Bernet]

Fall 2009

2009-12-13T23:16:59Z

Kendrick.mensink: /* HW # 5 */

====HW # 5====
Put a link here to an example problem you made up and solved using Laplace Transforms, of the complete solution of a mechanical system or a circuit that is described by linear ordinary differential equations with constant coefficients. Make sure you start with the physical system, and end up with the time response of the system.

Use your wiki page to explain the problem and solution to one of your classmates. Have him or her certify that they have checked it for errors, by listing that on the page. Then go to [http://moodle.wallawalla.edu Moodle] and put a link to your page in the in-box for HW #5.

[[Laplace transforms:Series RLC circuit]]

[[Laplace transforms:Mass-Spring Oscillator]]

[[Laplace transforms: DC Motor circuit]]

[[Laplace transforms: Simple Electrical Network]]

[[Laplace transforms: R series with RC parallel circuit]]

[[Laplace transforms: Critically Damped Motion ]]

[[Laplace transforms: Under-damped Mass-Spring System on an Incline]]

[[Laplace transforms: Critically Damped Spring Mass system]]

[[Laplace Transforms: Vertical Motion of a Coupled Spring System]]

[[Laplace Transforms: Coupled Springs]]

====HW #12====
Coupled Oscillator Problem

[[Coupled Oscillator: Hellie]]

[[Coupled Oscillator: Coupled Mass-Spring System with Input]]

[[Coupled Oscillator: Coupled Mass-Spring System with Damping]]

[[Coupled Oscillator: Jonathan Schreven]]

[[Coupled Horizontal Spring Mass Oscillator]]

[[Coupled Oscillator: Double Pendulum]]

[[Coupled Oscillator: horizontal Mass-Spring]]

[[Coupled Double Pendulum Oscillator]]

==2009-2010 Contributors==

[[Ben Henry]]

[http://fweb.wallawalla.edu/class-wiki/index.php/Fall_2009/JonathanS Jonathan Schreven]

[http://fweb/class-wiki/index.php/Laplace_transforms:DC_Motor_circuit/ Kendrick Mensink]

[http://fweb/class-wiki/index.php/Laplace_transforms:_Critically_Damped_Motion Mark Bernet]

Fall 2009

2009-12-13T23:10:05Z

Kendrick.mensink: /* HW #12 */

====HW # 5====
Put a link here to an example problem you made up and solved using Laplace Transforms, of the complete solution of a mechanical system or a circuit that is described by linear ordinary differential equations with constant coefficients. Make sure you start with the physical system, and end up with the time response of the system.

Use your wiki page to explain the problem and solution to one of your classmates. Have him or her certify that they have checked it for errors, by listing that on the page. Then go to [http://moodle.wallawalla.edu Moodle] and put a link to your page in the in-box for HW #5.

[[Laplace transforms:Series RLC circuit]]

[[Laplace transforms:Mass-Spring Oscillator]]

[[Laplace transforms:DC Motor circuit]]

[[Laplace transforms: Simple Electrical Network]]

[[Laplace transforms: R series with RC parallel circuit]]

[[Laplace transforms: Critically Damped Motion ]]

[[Laplace transforms: Under-damped Mass-Spring System on an Incline]]

[[Laplace transforms: Critically Damped Spring Mass system]]

[[Laplace Transforms: Vertical Motion of a Coupled Spring System]]

[[Laplace Transforms: Coupled Springs]]

====HW #12====
Coupled Oscillator Problem

[[Coupled Oscillator: Hellie]]

[[Coupled Oscillator: Coupled Mass-Spring System with Input]]

[[Coupled Oscillator: Coupled Mass-Spring System with Damping]]

[[Coupled Oscillator: Jonathan Schreven]]

[[Coupled Horizontal Spring Mass Oscillator]]

[[Coupled Oscillator: Double Pendulum]]

[[Coupled Oscillator: horizontal Mass-Spring]]

[[Coupled Double Pendulum Oscillator]]

==2009-2010 Contributors==

[[Ben Henry]]

[http://fweb.wallawalla.edu/class-wiki/index.php/Fall_2009/JonathanS Jonathan Schreven]

[http://fweb/class-wiki/index.php/Laplace_transforms:DC_Motor_circuit/ Kendrick Mensink]

[http://fweb/class-wiki/index.php/Laplace_transforms:_Critically_Damped_Motion Mark Bernet]

Laplace transforms:DC Motor circuit

2009-11-16T19:00:47Z

Kendrick.mensink: /* State Equations */

=== Problem ===
Find the steady state current ''i(t)'' through a DC motor represented by a series R-L-Motor circuit. The resistance (R) is from the armature winding. The inductance (L) is the equivalent inductance of the wire coil (which turns by current flowing through the coil in a permanent magnetic field). Jm is the combined moment of inertia of the armature and load. B is the coefficient of friction. The motor has input current ''i(t)'' and output angular velocity ω(t). Let v(t) = 110u(t)V, R = 20Ω, L = 50mH, k = 0.05N*m/A, Jm = 0.01 kg*m2, and B = e-4N*m*s.

[[Image:DCMotorPic.jpg]]

=== Solution ===
Since some of the elements of a motor are not typical circuit elements, we must solve this problem using Laplace transforms.

The torque is proportional to the armature current.

<math>T(t) = k i(t) \, </math>

Similarly, relating mechanical (''T(t)ω(t)'') and electrical (''vm(t)i(t)'') power, the conservation of energy requires the same proportionality between the voltage across the motor (''vm(t)'') and the angular velocity (''ω(t)'').

<math>v_m(t) = k \omega(t) \, </math>

We want to find the Laplace transfer function of the motor, and we define it as follows.

<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>

Summing the voltages around the series circuit gives us our differential equation.

<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>

Take the Laplace transform.

<math>V_s(s) = R I(s) + Ls( I(s) - i(0) ) + k\Omega(s) \, </math>

If we let i(0) equal zero, the transformed differential equation gives us Equation *1*.

<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s) *1* \, </math>

Repeat the process with the analogous mechanical differential equation. Here Jm is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.

<math>T(t) = J_m \frac{d\omega(t)}{dt} + B \omega(t)</math>

Transforming yields the following.

<math>T(s) = J_m s(\Omega(s) - \omega(0)) + B \Omega(s) \, </math>

Recall T(t) = k i(t), and so T(s) = k I(s). Again let ω(0) = 0. This gives us Equation *2*.

<math>k I(s) = (J_ms + B) \Omega(s) *2* \, </math>

Solve Equation *2* for I(s) and substitute that into Equation *1*.

<math>V_s(s) = (R + Ls) \frac{(J_ms + B)}{k} \Omega(s) + k\Omega(s)</math>

Solve this equation for Ω(s), some simplification steps are skipped here. Equation *3*

<math>\Omega(s) = \frac{\frac{k}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s) *3*</math>

===Final Value Theorem===

Now in order to apply the final value theorem we let Vs(s) = K/s. The s's in the limit definition and in K/s cancel. We can now see ω in steady state.

<math>\omega_{ss} = \lim_{s\rightarrow 0} s \Omega(s) = \frac{k}{RB + k^2}K = \omega(\infty)</math>

Take Equation *2* and Equation *3* and this time solve for I(s). This will get us a form of the final answer in the s domain.

<math>I(s) = \frac{(J_ms + B)}{k} \left [ \frac{\frac{k}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} \right ] K = \frac{\frac{s}{L} + \frac{B}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} K</math>

Finally was can apply the final value theorem to see what our steady state response will be for the armature current.

<math>i_{ss} = \lim_{s \to 0} sI(s) = \frac{B}{RB + k^2} K = i(\infty)</math>

=== Answers ===
Of interest are the ω(∞) and i(∞), found by plugging values into the steady state solutions above.

<math>\omega(\infty) = 14.91 radians/second</math>

<math>i(\infty) = 5.46 Amperes</math>

=== Bode Plot===

[[Image:Dcmotorbode.JPG]]

=== Break Points and Asymptotes===

The transfer function of the motor voltage over the voltage source.

<math>H(s) = \frac{V_m(s)}{V_s(s)} = \frac{\Omega(s)}{k} = \frac{\frac{1}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s)</math>

Now we need to plug in the given values to find our transfer function and characteristic equation.

<math>H(s) = \frac{2000}{s^2 + 401.8s + 737.6}V(s) = \frac{2000}{(s + 1.844)(s + 399.956)}V(s)</math>

At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.
At poles we see drops of 20 dB/decade, and we see 2 such drops at the appropriate values of 1.8 and 400 on the frequency graph.
You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log10(H(0)) of this value. In this case we'd have H(0)=2000/1.8*400=2.71. So our expected starting value should be 20log(2.71)=8.67, and it appears to be correct. If the graph was larger, it would be apparent that we have asymptotic behavior going down at 40 dB/decade to the right of 400 radians/second. This is because 400 is the largest root in the denominator of the transfer function and as s goes to infinity the bode plot will never go above that asymptote.

===Convolution===
For this we need the inverse Laplace transform of our H(s).

First use partial fraction expansion, or your fancy calculator, to expand the transfer function.

<math>\frac{2000}{(s + 1.844)(s + 399.956)} = \frac{A}{s + 1.844} + \frac{B}{s + 399.956}</math>

I found A = 5.0237 and B = -5.0237

Now we can take the inverse transform.

<math>h(t) = \mathcal{L}^{-1} \left [ \frac{5.0237}{s+1.844}-\frac{5.0237}{s+399.956} \right ]</math>

<math>h(t) = [5.0237e^{-1.844t} - 5.0237e^{-399.956t}]u(t) \,</math>

===State Equations===
First we need to write our state equations, which are rearranged forms of our original differential equations. The state variables will be current (i) and angular velocity (ω).

<math>\dfrac{di(t)}{dt}=-\frac{R}{L}i(t)+\frac{v_s(t)}{L}-\frac{v_m(t)}{L}</math>

<math>\dfrac{d\omega(t)}{dt}=\frac{k}{J_m}i(t)-\frac{B}{J_m}\omega(t)</math>

The following is an alternate way to represent the dc motor circuit as a matrix system of equations.

<math>\begin{bmatrix} \tfrac{di}{dt} \\ \tfrac{d\omega}{dt} \end{bmatrix}=\begin{bmatrix} -\frac{R}{L} & k \\ \frac{R}{J_m} & \frac{B}{J_m} \end{bmatrix} \begin{bmatrix} i \\ \omega \end{bmatrix} + \begin{bmatrix} \frac{1}{L} \\ 0 \end{bmatrix}V_s</math>
----
Reviewed by Andrew Hellie

Laplace transforms:DC Motor circuit

2009-11-16T18:59:50Z

Kendrick.mensink: /* State Equations */

=== Problem ===
Find the steady state current ''i(t)'' through a DC motor represented by a series R-L-Motor circuit. The resistance (R) is from the armature winding. The inductance (L) is the equivalent inductance of the wire coil (which turns by current flowing through the coil in a permanent magnetic field). Jm is the combined moment of inertia of the armature and load. B is the coefficient of friction. The motor has input current ''i(t)'' and output angular velocity ω(t). Let v(t) = 110u(t)V, R = 20Ω, L = 50mH, k = 0.05N*m/A, Jm = 0.01 kg*m2, and B = e-4N*m*s.

[[Image:DCMotorPic.jpg]]

=== Solution ===
Since some of the elements of a motor are not typical circuit elements, we must solve this problem using Laplace transforms.

The torque is proportional to the armature current.

<math>T(t) = k i(t) \, </math>

Similarly, relating mechanical (''T(t)ω(t)'') and electrical (''vm(t)i(t)'') power, the conservation of energy requires the same proportionality between the voltage across the motor (''vm(t)'') and the angular velocity (''ω(t)'').

<math>v_m(t) = k \omega(t) \, </math>

We want to find the Laplace transfer function of the motor, and we define it as follows.

<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>

Summing the voltages around the series circuit gives us our differential equation.

<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>

Take the Laplace transform.

<math>V_s(s) = R I(s) + Ls( I(s) - i(0) ) + k\Omega(s) \, </math>

If we let i(0) equal zero, the transformed differential equation gives us Equation *1*.

<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s) *1* \, </math>

Repeat the process with the analogous mechanical differential equation. Here Jm is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.

<math>T(t) = J_m \frac{d\omega(t)}{dt} + B \omega(t)</math>

Transforming yields the following.

<math>T(s) = J_m s(\Omega(s) - \omega(0)) + B \Omega(s) \, </math>

Recall T(t) = k i(t), and so T(s) = k I(s). Again let ω(0) = 0. This gives us Equation *2*.

<math>k I(s) = (J_ms + B) \Omega(s) *2* \, </math>

Solve Equation *2* for I(s) and substitute that into Equation *1*.

<math>V_s(s) = (R + Ls) \frac{(J_ms + B)}{k} \Omega(s) + k\Omega(s)</math>

Solve this equation for Ω(s), some simplification steps are skipped here. Equation *3*

<math>\Omega(s) = \frac{\frac{k}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s) *3*</math>

===Final Value Theorem===

Now in order to apply the final value theorem we let Vs(s) = K/s. The s's in the limit definition and in K/s cancel. We can now see ω in steady state.

<math>\omega_{ss} = \lim_{s\rightarrow 0} s \Omega(s) = \frac{k}{RB + k^2}K = \omega(\infty)</math>

Take Equation *2* and Equation *3* and this time solve for I(s). This will get us a form of the final answer in the s domain.

<math>I(s) = \frac{(J_ms + B)}{k} \left [ \frac{\frac{k}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} \right ] K = \frac{\frac{s}{L} + \frac{B}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} K</math>

Finally was can apply the final value theorem to see what our steady state response will be for the armature current.

<math>i_{ss} = \lim_{s \to 0} sI(s) = \frac{B}{RB + k^2} K = i(\infty)</math>

=== Answers ===
Of interest are the ω(∞) and i(∞), found by plugging values into the steady state solutions above.

<math>\omega(\infty) = 14.91 radians/second</math>

<math>i(\infty) = 5.46 Amperes</math>

=== Bode Plot===

[[Image:Dcmotorbode.JPG]]

=== Break Points and Asymptotes===

The transfer function of the motor voltage over the voltage source.

<math>H(s) = \frac{V_m(s)}{V_s(s)} = \frac{\Omega(s)}{k} = \frac{\frac{1}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s)</math>

Now we need to plug in the given values to find our transfer function and characteristic equation.

<math>H(s) = \frac{2000}{s^2 + 401.8s + 737.6}V(s) = \frac{2000}{(s + 1.844)(s + 399.956)}V(s)</math>

At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.
At poles we see drops of 20 dB/decade, and we see 2 such drops at the appropriate values of 1.8 and 400 on the frequency graph.
You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log10(H(0)) of this value. In this case we'd have H(0)=2000/1.8*400=2.71. So our expected starting value should be 20log(2.71)=8.67, and it appears to be correct. If the graph was larger, it would be apparent that we have asymptotic behavior going down at 40 dB/decade to the right of 400 radians/second. This is because 400 is the largest root in the denominator of the transfer function and as s goes to infinity the bode plot will never go above that asymptote.

===Convolution===
For this we need the inverse Laplace transform of our H(s).

First use partial fraction expansion, or your fancy calculator, to expand the transfer function.

<math>\frac{2000}{(s + 1.844)(s + 399.956)} = \frac{A}{s + 1.844} + \frac{B}{s + 399.956}</math>

I found A = 5.0237 and B = -5.0237

Now we can take the inverse transform.

<math>h(t) = \mathcal{L}^{-1} \left [ \frac{5.0237}{s+1.844}-\frac{5.0237}{s+399.956} \right ]</math>

<math>h(t) = [5.0237e^{-1.844t} - 5.0237e^{-399.956t}]u(t) \,</math>

===State Equations===
First we need to write our state equations, which are rearranged forms of our original differential equations. The state variables will be current (i) and angular velocity (ω).

<math>dfrac{di(t)}{dt}=-\frac{R}{L}i(t)+\frac{v_s(t)}{L}-\frac{v_m(t)}{L}</math>

<math>\dfrac{d\omega(t)}{dt}=\frac{k}{J_m}i(t)-\frac{B}{J_m}\omega(t)</math>
The following is an alternate way to represent the dc motor circuit as a matrix system of equations.

<math>\begin{bmatrix} \tfrac{di}{dt} \\ \tfrac{d\omega}{dt} \end{bmatrix}=\begin{bmatrix} -\frac{R}{L} & k \\ \frac{R}{J_m} & \frac{B}{J_m} \end{bmatrix} \begin{bmatrix} i \\ \omega \end{bmatrix} + \begin{bmatrix} \frac{1}{L} \\ 0 \end{bmatrix}V_s</math>
----
Reviewed by Andrew Hellie

Laplace transforms:DC Motor circuit

2009-11-16T18:48:36Z

Kendrick.mensink: /* Convolution */

=== Problem ===
Find the steady state current ''i(t)'' through a DC motor represented by a series R-L-Motor circuit. The resistance (R) is from the armature winding. The inductance (L) is the equivalent inductance of the wire coil (which turns by current flowing through the coil in a permanent magnetic field). Jm is the combined moment of inertia of the armature and load. B is the coefficient of friction. The motor has input current ''i(t)'' and output angular velocity ω(t). Let v(t) = 110u(t)V, R = 20Ω, L = 50mH, k = 0.05N*m/A, Jm = 0.01 kg*m2, and B = e-4N*m*s.

[[Image:DCMotorPic.jpg]]

=== Solution ===
Since some of the elements of a motor are not typical circuit elements, we must solve this problem using Laplace transforms.

The torque is proportional to the armature current.

<math>T(t) = k i(t) \, </math>

Similarly, relating mechanical (''T(t)ω(t)'') and electrical (''vm(t)i(t)'') power, the conservation of energy requires the same proportionality between the voltage across the motor (''vm(t)'') and the angular velocity (''ω(t)'').

<math>v_m(t) = k \omega(t) \, </math>

We want to find the Laplace transfer function of the motor, and we define it as follows.

<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>

Summing the voltages around the series circuit gives us our differential equation.

<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>

Take the Laplace transform.

<math>V_s(s) = R I(s) + Ls( I(s) - i(0) ) + k\Omega(s) \, </math>

If we let i(0) equal zero, the transformed differential equation gives us Equation *1*.

<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s) *1* \, </math>

Repeat the process with the analogous mechanical differential equation. Here Jm is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.

<math>T(t) = J_m \frac{d\omega(t)}{dt} + B \omega(t)</math>

Transforming yields the following.

<math>T(s) = J_m s(\Omega(s) - \omega(0)) + B \Omega(s) \, </math>

Recall T(t) = k i(t), and so T(s) = k I(s). Again let ω(0) = 0. This gives us Equation *2*.

<math>k I(s) = (J_ms + B) \Omega(s) *2* \, </math>

Solve Equation *2* for I(s) and substitute that into Equation *1*.

<math>V_s(s) = (R + Ls) \frac{(J_ms + B)}{k} \Omega(s) + k\Omega(s)</math>

Solve this equation for Ω(s), some simplification steps are skipped here. Equation *3*

<math>\Omega(s) = \frac{\frac{k}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s) *3*</math>

===Final Value Theorem===

Now in order to apply the final value theorem we let Vs(s) = K/s. The s's in the limit definition and in K/s cancel. We can now see ω in steady state.

<math>\omega_{ss} = \lim_{s\rightarrow 0} s \Omega(s) = \frac{k}{RB + k^2}K = \omega(\infty)</math>

Take Equation *2* and Equation *3* and this time solve for I(s). This will get us a form of the final answer in the s domain.

<math>I(s) = \frac{(J_ms + B)}{k} \left [ \frac{\frac{k}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} \right ] K = \frac{\frac{s}{L} + \frac{B}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} K</math>

Finally was can apply the final value theorem to see what our steady state response will be for the armature current.

<math>i_{ss} = \lim_{s \to 0} sI(s) = \frac{B}{RB + k^2} K = i(\infty)</math>

=== Answers ===
Of interest are the ω(∞) and i(∞), found by plugging values into the steady state solutions above.

<math>\omega(\infty) = 14.91 radians/second</math>

<math>i(\infty) = 5.46 Amperes</math>

=== Bode Plot===

[[Image:Dcmotorbode.JPG]]

=== Break Points and Asymptotes===

The transfer function of the motor voltage over the voltage source.

<math>H(s) = \frac{V_m(s)}{V_s(s)} = \frac{\Omega(s)}{k} = \frac{\frac{1}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s)</math>

Now we need to plug in the given values to find our transfer function and characteristic equation.

<math>H(s) = \frac{2000}{s^2 + 401.8s + 737.6}V(s) = \frac{2000}{(s + 1.844)(s + 399.956)}V(s)</math>

At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.
At poles we see drops of 20 dB/decade, and we see 2 such drops at the appropriate values of 1.8 and 400 on the frequency graph.
You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log10(H(0)) of this value. In this case we'd have H(0)=2000/1.8*400=2.71. So our expected starting value should be 20log(2.71)=8.67, and it appears to be correct. If the graph was larger, it would be apparent that we have asymptotic behavior going down at 40 dB/decade to the right of 400 radians/second. This is because 400 is the largest root in the denominator of the transfer function and as s goes to infinity the bode plot will never go above that asymptote.

===Convolution===
For this we need the inverse Laplace transform of our H(s).

First use partial fraction expansion, or your fancy calculator, to expand the transfer function.

<math>\frac{2000}{(s + 1.844)(s + 399.956)} = \frac{A}{s + 1.844} + \frac{B}{s + 399.956}</math>

I found A = 5.0237 and B = -5.0237

Now we can take the inverse transform.

<math>h(t) = \mathcal{L}^{-1} \left [ \frac{5.0237}{s+1.844}-\frac{5.0237}{s+399.956} \right ]</math>

<math>h(t) = [5.0237e^{-1.844t} - 5.0237e^{-399.956t}]u(t) \,</math>

===State Equations===
First we need to write our state equations
The following is an alternate way to represent the dc motor circuit as a matrix system of equations.

<math>\begin{bmatrix} \tfrac{di}{dt} \\ \tfrac{d\omega}{dt} \end{bmatrix}=\begin{bmatrix} -\frac{R}{L} & k \\ \frac{R}{J_m} & \frac{B}{J_m} \end{bmatrix} \begin{bmatrix} i \\ \omega \end{bmatrix} + \begin{bmatrix} \frac{1}{L} \\ 0 \end{bmatrix}V_s</math>
----
Reviewed by Andrew Hellie

Laplace transforms:DC Motor circuit

2009-10-30T05:03:07Z

Kendrick.mensink: /* Convolution */

Laplace transforms:DC Motor circuit

2009-10-30T05:01:23Z

Kendrick.mensink: /* Convolution */

Laplace transforms:DC Motor circuit

2009-10-30T03:57:43Z

Kendrick.mensink: /* Convolution */

=== Problem ===
Find the steady state current ''i(t)'' through a DC motor represented by a series R-L-Motor circuit. The resistance (R) is from the armature winding. The inductance (L) is the equivalent inductance of the wire coil (which turns by current flowing through the coil in a permanent magnetic field). Jm is the combined moment of inertia of the armature and load. B is the coefficient of friction. The motor has input current ''i(t)'' and output angular velocity ω(t). Let v(t) = 110u(t)V, R = 20Ω, L = 50mH, k = 0.05N*m/A, Jm = 0.01 kg*m2, and B = e-4N*m*s.

[[Image:DCMotorPic.jpg]]

=== Solution ===
Since some of the elements of a motor are not typical circuit elements, we must solve this problem using Laplace transforms.

The torque is proportional to the armature current.

<math>T(t) = k i(t) \, </math>

Similarly, relating mechanical (''T(t)ω(t)'') and electrical (''vm(t)i(t)'') power, the conservation of energy requires the same proportionality between the voltage across the motor (''vm(t)'') and the angular velocity (''ω(t)'').

<math>v_m(t) = k \omega(t) \, </math>

We want to find the Laplace transfer function of the motor, and we define it as follows.

<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>

Summing the voltages around the series circuit gives us our differential equation.

<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>

Take the Laplace transform.

<math>V_s(s) = R I(s) + Ls( I(s) - i(0) ) + k\Omega(s) \, </math>

If we let i(0) equal zero, the transformed differential equation gives us Equation *1*.

<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s) *1* \, </math>

Repeat the process with the analogous mechanical differential equation. Here Jm is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.

<math>T(t) = J_m \frac{d\omega(t)}{dt} + B \omega(t)</math>

Transforming yields the following.

<math>T(s) = J_m s(\Omega(s) - \omega(0)) + B \Omega(s) \, </math>

Recall T(t) = k i(t), and so T(s) = k I(s). Again let ω(0) = 0. This gives us Equation *2*.

<math>k I(s) = (J_ms + B) \Omega(s) *2* \, </math>

Solve Equation *2* for I(s) and substitute that into Equation *1*.

<math>V_s(s) = (R + Ls) \frac{(J_ms + B)}{k} \Omega(s) + k\Omega(s)</math>

Solve this equation for Ω(s), some simplification steps are skipped here. Equation *3*

<math>\Omega(s) = \frac{\frac{k}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s) *3*</math>

===Final Value Theorem===

Now in order to apply the final value theorem we let Vs(s) = K/s. The s's in the limit definition and in K/s cancel. We can now see ω in steady state.

<math>\omega_{ss} = \lim_{s\rightarrow 0} s \Omega(s) = \frac{k}{RB + k^2}K = \omega(\infty)</math>

Take Equation *2* and Equation *3* and this time solve for I(s). This will get us a form of the final answer in the s domain.

<math>I(s) = \frac{(J_ms + B)}{k} \left [ \frac{\frac{k}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} \right ] K = \frac{\frac{s}{L} + \frac{B}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} K</math>

Finally was can apply the final value theorem to see what our steady state response will be for the armature current.

<math>i_{ss} = \lim_{s \to 0} sI(s) = \frac{B}{RB + k^2} K = i(\infty)</math>

=== Answers ===
Of interest are the ω(∞) and i(∞), found by plugging values into the steady state solutions above.

<math>\omega(\infty) = 14.91 radians/second</math>

<math>i(\infty) = 5.46 Amperes</math>

=== Bode Plot===

[[Image:Dcmotorbode.JPG]]

=== Break Points and Asymptotes===

The transfer function of the motor voltage over the voltage source.

<math>H(s) = \frac{V_m(s)}{V_s(s)} = \frac{\Omega(s)}{k} = \frac{\frac{1}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s)</math>

Now we need to plug in the given values to find our transfer function and characteristic equation.

<math>H(s) = \frac{2000}{s^2 + 401.8s + 737.6}V(s) = \frac{2000}{(s + 1.844)(s + 399.956)}V(s)</math>

At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.
At poles we see drops of 20 dB/decade, and we see 2 such drops at the appropriate values of 1.8 and 400 on the frequency graph.
You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log10(H(0)) of this value. In this case we'd have H(0)=2000/1.8*400=2.71. So our expected starting value should be 20log(2.71)=8.67, and it appears to be correct. If the graph was larger, it would be apparent that we have asymptotic behavior going down at 40 dB/decade to the right of 400 radians/second. This is because 400 is the largest root in the denominator of the transfer function and as s goes to infinity the bode plot will never go above that asymptote.

===Convolution===
For this we need the inverse Laplace transform of our H(s).

First use partial fraction expansion, or your fancy calculator, to expand the transfer function
<math>\frac{2000}{(s + 1.844)(s + 399.956)} = \frac{A}{s + 1.844} + \frac{B}{s + 399.956}</math>

I found A = 5.0237 and B = -5.0237
----
Reviewed by Andrew Hellie

Laplace transforms:DC Motor circuit

2009-10-30T03:23:28Z

Kendrick.mensink: /* Break Points and Asymptotes */

Laplace transforms:DC Motor circuit

2009-10-30T03:22:52Z

Kendrick.mensink: /* Break Points and Asymptotes */

=== Problem ===
Find the steady state current ''i(t)'' through a DC motor represented by a series R-L-Motor circuit. The resistance (R) is from the armature winding. The inductance (L) is the equivalent inductance of the wire coil (which turns by current flowing through the coil in a permanent magnetic field). Jm is the combined moment of inertia of the armature and load. B is the coefficient of friction. The motor has input current ''i(t)'' and output angular velocity ω(t). Let v(t) = 110u(t)V, R = 20Ω, L = 50mH, k = 0.05N*m/A, Jm = 0.01 kg*m2, and B = e-4N*m*s.

[[Image:DCMotorPic.jpg]]

=== Solution ===
Since some of the elements of a motor are not typical circuit elements, we must solve this problem using Laplace transforms.

The torque is proportional to the armature current.

<math>T(t) = k i(t) \, </math>

Similarly, relating mechanical (''T(t)ω(t)'') and electrical (''vm(t)i(t)'') power, the conservation of energy requires the same proportionality between the voltage across the motor (''vm(t)'') and the angular velocity (''ω(t)'').

<math>v_m(t) = k \omega(t) \, </math>

We want to find the Laplace transfer function of the motor, and we define it as follows.

<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>

Summing the voltages around the series circuit gives us our differential equation.

<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>

Take the Laplace transform.

<math>V_s(s) = R I(s) + Ls( I(s) - i(0) ) + k\Omega(s) \, </math>

If we let i(0) equal zero, the transformed differential equation gives us Equation *1*.

<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s) *1* \, </math>

Repeat the process with the analogous mechanical differential equation. Here Jm is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.

<math>T(t) = J_m \frac{d\omega(t)}{dt} + B \omega(t)</math>

Transforming yields the following.

<math>T(s) = J_m s(\Omega(s) - \omega(0)) + B \Omega(s) \, </math>

Recall T(t) = k i(t), and so T(s) = k I(s). Again let ω(0) = 0. This gives us Equation *2*.

<math>k I(s) = (J_ms + B) \Omega(s) *2* \, </math>

Solve Equation *2* for I(s) and substitute that into Equation *1*.

<math>V_s(s) = (R + Ls) \frac{(J_ms + B)}{k} \Omega(s) + k\Omega(s)</math>

Solve this equation for Ω(s), some simplification steps are skipped here. Equation *3*

<math>\Omega(s) = \frac{\frac{k}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s) *3*</math>

===Final Value Theorem===

Now in order to apply the final value theorem we let Vs(s) = K/s. The s's in the limit definition and in K/s cancel. We can now see ω in steady state.

<math>\omega_{ss} = \lim_{s\rightarrow 0} s \Omega(s) = \frac{k}{RB + k^2}K = \omega(\infty)</math>

Take Equation *2* and Equation *3* and this time solve for I(s). This will get us a form of the final answer in the s domain.

<math>I(s) = \frac{(J_ms + B)}{k} \left [ \frac{\frac{k}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} \right ] K = \frac{\frac{s}{L} + \frac{B}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} K</math>

Finally was can apply the final value theorem to see what our steady state response will be for the armature current.

<math>i_{ss} = \lim_{s \to 0} sI(s) = \frac{B}{RB + k^2} K = i(\infty)</math>

=== Answers ===
Of interest are the ω(∞) and i(∞), found by plugging values into the steady state solutions above.

<math>\omega(\infty) = 14.91 radians/second</math>

<math>i(\infty) = 5.46 Amperes</math>

=== Bode Plot===

[[Image:Dcmotorbode.JPG]]

=== Break Points and Asymptotes===

The transfer function of the motor voltage over the voltage source.

<math>H(s) = \frac{V_m(s)}{V_s(s)} = \frac{\Omega(s)}{k} = \Omega(s) = \frac{\frac{1}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s)</math>

Now we need to plug in the given values to find our transfer function and characteristic equation.

<math>H(s) = \frac{2000}{s^2 + 401.8s + 737.6}V(s) = \frac{2000}{(s + 1.844)(s + 399.956)}V(s)</math>

At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.
At poles we see drops of 20 dB/decade, and we see 2 such drops at the appropriate values of 1.8 and 400 on the frequency graph.
You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log10(H(0)) of this value. In this case we'd have H(0)=2000/1.8*400=2.71. So our expected starting value should be 20log(2.71)=8.67, and it appears to be correct. If the graph was larger, it would be apparent that we have asymptotic behavior going down at 40 dB/decade to the right of 400 radians/second. This is because 400 is the largest root in the denominator of the transfer function and as s goes to infinity the bode plot will never go above that asymptote.

===Convolution===

----
Reviewed by Andrew Hellie

Laplace transforms:DC Motor circuit

2009-10-30T03:20:41Z

Kendrick.mensink: /* Break Points and Asymptotes */

=== Problem ===
Find the steady state current ''i(t)'' through a DC motor represented by a series R-L-Motor circuit. The resistance (R) is from the armature winding. The inductance (L) is the equivalent inductance of the wire coil (which turns by current flowing through the coil in a permanent magnetic field). Jm is the combined moment of inertia of the armature and load. B is the coefficient of friction. The motor has input current ''i(t)'' and output angular velocity ω(t). Let v(t) = 110u(t)V, R = 20Ω, L = 50mH, k = 0.05N*m/A, Jm = 0.01 kg*m2, and B = e-4N*m*s.

[[Image:DCMotorPic.jpg]]

=== Solution ===
Since some of the elements of a motor are not typical circuit elements, we must solve this problem using Laplace transforms.

The torque is proportional to the armature current.

<math>T(t) = k i(t) \, </math>

Similarly, relating mechanical (''T(t)ω(t)'') and electrical (''vm(t)i(t)'') power, the conservation of energy requires the same proportionality between the voltage across the motor (''vm(t)'') and the angular velocity (''ω(t)'').

<math>v_m(t) = k \omega(t) \, </math>

We want to find the Laplace transfer function of the motor, and we define it as follows.

<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>

Summing the voltages around the series circuit gives us our differential equation.

<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>

Take the Laplace transform.

<math>V_s(s) = R I(s) + Ls( I(s) - i(0) ) + k\Omega(s) \, </math>

If we let i(0) equal zero, the transformed differential equation gives us Equation *1*.

<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s) *1* \, </math>

Repeat the process with the analogous mechanical differential equation. Here Jm is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.

<math>T(t) = J_m \frac{d\omega(t)}{dt} + B \omega(t)</math>

Transforming yields the following.

<math>T(s) = J_m s(\Omega(s) - \omega(0)) + B \Omega(s) \, </math>

Recall T(t) = k i(t), and so T(s) = k I(s). Again let ω(0) = 0. This gives us Equation *2*.

<math>k I(s) = (J_ms + B) \Omega(s) *2* \, </math>

Solve Equation *2* for I(s) and substitute that into Equation *1*.

<math>V_s(s) = (R + Ls) \frac{(J_ms + B)}{k} \Omega(s) + k\Omega(s)</math>

Solve this equation for Ω(s), some simplification steps are skipped here. Equation *3*

<math>\Omega(s) = \frac{\frac{k}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s) *3*</math>

===Final Value Theorem===

Now in order to apply the final value theorem we let Vs(s) = K/s. The s's in the limit definition and in K/s cancel. We can now see ω in steady state.

<math>\omega_{ss} = \lim_{s\rightarrow 0} s \Omega(s) = \frac{k}{RB + k^2}K = \omega(\infty)</math>

Take Equation *2* and Equation *3* and this time solve for I(s). This will get us a form of the final answer in the s domain.

<math>I(s) = \frac{(J_ms + B)}{k} \left [ \frac{\frac{k}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} \right ] K = \frac{\frac{s}{L} + \frac{B}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} K</math>

Finally was can apply the final value theorem to see what our steady state response will be for the armature current.

<math>i_{ss} = \lim_{s \to 0} sI(s) = \frac{B}{RB + k^2} K = i(\infty)</math>

=== Answers ===
Of interest are the ω(∞) and i(∞), found by plugging values into the steady state solutions above.

<math>\omega(\infty) = 14.91 radians/second</math>

<math>i(\infty) = 5.46 Amperes</math>

=== Bode Plot===

[[Image:Dcmotorbode.JPG]]

=== Break Points and Asymptotes===

The transfer function of the motor voltage over the voltage source.

<math>H(s) = \frac{V_m(s)}{V_s(s)} = \frac{\Omega(s)}{k} = \Omega(s) = \frac{\frac{1}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s)</math>

Now we need to plug in the given values to find our transfer function and characteristic equation.

<math>H(s) = \frac{2000}{s^2 + 401.8s + 737.6}V(s) = \frac{2000}{(s + 1.844)(s + 399.956)}V(s)</math>

At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.
At poles we see drops of 20 dB/decade, and we see 2 such drops at the appropriate values of 1.8 and 400 on the frequency graph.
You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log10 of this value. In this case we'd have a value of 2000/1.8*400=2.71. So our expected starting value should be 20log(2.71)=8.67, and it appears to be correct. If the graph was larger, it would be apparent that we have asymptotic behavior going down at 40 dB/decade to the right of 400 radians/second. This is because this is the largest root in the denominator of the transfer function and as s goes to infinity the bode plot will never go above that asymptote.

===Convolution===

----
Reviewed by Andrew Hellie

Laplace transforms:DC Motor circuit

2009-10-30T03:18:51Z

Kendrick.mensink: /* Break Points and Asymptotes */

=== Problem ===
Find the steady state current ''i(t)'' through a DC motor represented by a series R-L-Motor circuit. The resistance (R) is from the armature winding. The inductance (L) is the equivalent inductance of the wire coil (which turns by current flowing through the coil in a permanent magnetic field). Jm is the combined moment of inertia of the armature and load. B is the coefficient of friction. The motor has input current ''i(t)'' and output angular velocity ω(t). Let v(t) = 110u(t)V, R = 20Ω, L = 50mH, k = 0.05N*m/A, Jm = 0.01 kg*m2, and B = e-4N*m*s.

[[Image:DCMotorPic.jpg]]

=== Solution ===
Since some of the elements of a motor are not typical circuit elements, we must solve this problem using Laplace transforms.

The torque is proportional to the armature current.

<math>T(t) = k i(t) \, </math>

Similarly, relating mechanical (''T(t)ω(t)'') and electrical (''vm(t)i(t)'') power, the conservation of energy requires the same proportionality between the voltage across the motor (''vm(t)'') and the angular velocity (''ω(t)'').

<math>v_m(t) = k \omega(t) \, </math>

We want to find the Laplace transfer function of the motor, and we define it as follows.

<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>

Summing the voltages around the series circuit gives us our differential equation.

<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>

Take the Laplace transform.

<math>V_s(s) = R I(s) + Ls( I(s) - i(0) ) + k\Omega(s) \, </math>

If we let i(0) equal zero, the transformed differential equation gives us Equation *1*.

<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s) *1* \, </math>

Repeat the process with the analogous mechanical differential equation. Here Jm is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.

<math>T(t) = J_m \frac{d\omega(t)}{dt} + B \omega(t)</math>

Transforming yields the following.

<math>T(s) = J_m s(\Omega(s) - \omega(0)) + B \Omega(s) \, </math>

Recall T(t) = k i(t), and so T(s) = k I(s). Again let ω(0) = 0. This gives us Equation *2*.

<math>k I(s) = (J_ms + B) \Omega(s) *2* \, </math>

Solve Equation *2* for I(s) and substitute that into Equation *1*.

<math>V_s(s) = (R + Ls) \frac{(J_ms + B)}{k} \Omega(s) + k\Omega(s)</math>

Solve this equation for Ω(s), some simplification steps are skipped here. Equation *3*

<math>\Omega(s) = \frac{\frac{k}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s) *3*</math>

===Final Value Theorem===

Now in order to apply the final value theorem we let Vs(s) = K/s. The s's in the limit definition and in K/s cancel. We can now see ω in steady state.

<math>\omega_{ss} = \lim_{s\rightarrow 0} s \Omega(s) = \frac{k}{RB + k^2}K = \omega(\infty)</math>

Take Equation *2* and Equation *3* and this time solve for I(s). This will get us a form of the final answer in the s domain.

<math>I(s) = \frac{(J_ms + B)}{k} \left [ \frac{\frac{k}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} \right ] K = \frac{\frac{s}{L} + \frac{B}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} K</math>

Finally was can apply the final value theorem to see what our steady state response will be for the armature current.

<math>i_{ss} = \lim_{s \to 0} sI(s) = \frac{B}{RB + k^2} K = i(\infty)</math>

=== Answers ===
Of interest are the ω(∞) and i(∞), found by plugging values into the steady state solutions above.

<math>\omega(\infty) = 14.91 radians/second</math>

<math>i(\infty) = 5.46 Amperes</math>

=== Bode Plot===

[[Image:Dcmotorbode.JPG]]

=== Break Points and Asymptotes===

The transfer function of the motor voltage over the voltage source.

<math>H(s) = \frac{V_m(s)}{V_s(s)} = \frac{\Omega(s)}{k} = \Omega(s) = \frac{\frac{1}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s)</math>

Now we need to plug in the given values to find our transfer function and characteristic equation.

<math>H(s) = \frac{2000}{s^2 + 401.8s + 737.6}V(s) = \frac{2000}{(s + 1.844)(s + 399.956)}V(s)</math>

At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.
At poles we see drops of 20 dB/decade, and we see 2 such drops at the appropriate values of 1.8 and 400 on the frequency graph.
You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log10 of this value. In this case we'd have a value of 2000/1.8*400=2.71. So our expected starting value should be 20log(2.71)=8.67, and it appears to be correct. If the graph was larger, it would be apparent that we have asymptotic behavior going down at 40 dB/decade to the right of
----
Reviewed by Andrew Hellie

Laplace transforms:DC Motor circuit

2009-10-30T03:15:23Z

Kendrick.mensink: /* Solution */

=== Problem ===
Find the steady state current ''i(t)'' through a DC motor represented by a series R-L-Motor circuit. The resistance (R) is from the armature winding. The inductance (L) is the equivalent inductance of the wire coil (which turns by current flowing through the coil in a permanent magnetic field). Jm is the combined moment of inertia of the armature and load. B is the coefficient of friction. The motor has input current ''i(t)'' and output angular velocity ω(t). Let v(t) = 110u(t)V, R = 20Ω, L = 50mH, k = 0.05N*m/A, Jm = 0.01 kg*m2, and B = e-4N*m*s.

[[Image:DCMotorPic.jpg]]

=== Solution ===
Since some of the elements of a motor are not typical circuit elements, we must solve this problem using Laplace transforms.

The torque is proportional to the armature current.

<math>T(t) = k i(t) \, </math>

Similarly, relating mechanical (''T(t)ω(t)'') and electrical (''vm(t)i(t)'') power, the conservation of energy requires the same proportionality between the voltage across the motor (''vm(t)'') and the angular velocity (''ω(t)'').

<math>v_m(t) = k \omega(t) \, </math>

We want to find the Laplace transfer function of the motor, and we define it as follows.

<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>

Summing the voltages around the series circuit gives us our differential equation.

<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>

Take the Laplace transform.

<math>V_s(s) = R I(s) + Ls( I(s) - i(0) ) + k\Omega(s) \, </math>

If we let i(0) equal zero, the transformed differential equation gives us Equation *1*.

<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s) *1* \, </math>

Repeat the process with the analogous mechanical differential equation. Here Jm is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.

<math>T(t) = J_m \frac{d\omega(t)}{dt} + B \omega(t)</math>

Transforming yields the following.

<math>T(s) = J_m s(\Omega(s) - \omega(0)) + B \Omega(s) \, </math>

Recall T(t) = k i(t), and so T(s) = k I(s). Again let ω(0) = 0. This gives us Equation *2*.

<math>k I(s) = (J_ms + B) \Omega(s) *2* \, </math>

Solve Equation *2* for I(s) and substitute that into Equation *1*.

<math>V_s(s) = (R + Ls) \frac{(J_ms + B)}{k} \Omega(s) + k\Omega(s)</math>

Solve this equation for Ω(s), some simplification steps are skipped here. Equation *3*

<math>\Omega(s) = \frac{\frac{k}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s) *3*</math>

===Final Value Theorem===

Now in order to apply the final value theorem we let Vs(s) = K/s. The s's in the limit definition and in K/s cancel. We can now see ω in steady state.

<math>\omega_{ss} = \lim_{s\rightarrow 0} s \Omega(s) = \frac{k}{RB + k^2}K = \omega(\infty)</math>

Take Equation *2* and Equation *3* and this time solve for I(s). This will get us a form of the final answer in the s domain.

<math>I(s) = \frac{(J_ms + B)}{k} \left [ \frac{\frac{k}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} \right ] K = \frac{\frac{s}{L} + \frac{B}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} K</math>

Finally was can apply the final value theorem to see what our steady state response will be for the armature current.

<math>i_{ss} = \lim_{s \to 0} sI(s) = \frac{B}{RB + k^2} K = i(\infty)</math>

=== Answers ===
Of interest are the ω(∞) and i(∞), found by plugging values into the steady state solutions above.

<math>\omega(\infty) = 14.91 radians/second</math>

<math>i(\infty) = 5.46 Amperes</math>

=== Bode Plot===

[[Image:Dcmotorbode.JPG]]

=== Break Points and Asymptotes===

The transfer function of the motor voltage over the voltage source.

<math>H(s) = \frac{V_m(s)}{V_s(s)} = \frac{\Omega(s)}{k} = \Omega(s) = \frac{\frac{1}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s)</math>

Now we need to plug in the given values to find our transfer function and characteristic equation.

<math>H(s) = \frac{2000}{s^2 + 401.8s + 737.6}V(s) = \frac{2000}{(s + 1.844)(s + 399.956)}V(s)</math>

At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.
At poles we see drops of 20 dB/decade, and we see 2 such drops at the appropriate values of 1.8 and 400 on the frequency graph.
You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log10 of this value. In this case we'd have a value of 2000/1.8*400=2.71. So our expected starting value should be 20log(2.71)=8.67, and it appears to be correct.
----
Reviewed by Andrew Hellie

Laplace transforms:DC Motor circuit

2009-10-30T03:13:23Z

Kendrick.mensink: /* Break Points and Asymptotes */

=== Problem ===
Find the steady state current ''i(t)'' through a DC motor represented by a series R-L-Motor circuit. The resistance (R) is from the armature winding. The inductance (L) is the equivalent inductance of the wire coil (which turns by current flowing through the coil in a permanent magnetic field). Jm is the combined moment of inertia of the armature and load. B is the coefficient of friction. The motor has input current ''i(t)'' and output angular velocity ω(t). Let v(t) = 110u(t)V, R = 20Ω, L = 50mH, k = 0.05N*m/A, Jm = 0.01 kg*m2, and B = e-4N*m*s.

[[Image:DCMotorPic.jpg]]

=== Solution ===
Since some of the elements of a motor are not typical circuit elements, we must solve this problem using Laplace transforms.

The torque is proportional to the armature current.

<math>T(t) = k i(t) \, </math>

Similarly, relating mechanical (''T(t)ω(t)'') and electrical (''vm(t)i(t)'') power, the conservation of energy requires the same proportionality between the voltage across the motor (''vm(t)'') and the angular velocity (''ω(t)'').

<math>v_m(t) = k \omega(t) \, </math>

We want to find the Laplace transfer function of the motor, and we define it as follows.

<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>

Summing the voltages around the series circuit gives us our differential equation.

<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>

Take the Laplace transform.

<math>V_s(s) = R I(s) + Ls( I(s) - i(0) ) + k\Omega(s) \, </math>

If we let i(0) equal zero, the transformed differential equation gives us Equation *1*.

<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s) *1* \, </math>

Repeat the process with the analogous mechanical differential equation. Here Jm is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.

<math>T(t) = J_m \frac{d\omega(t)}{dt} + B \omega(t)</math>

Transforming yields the following.

<math>T(s) = J_m s(\Omega(s) - \omega(0)) + B \Omega(s) \, </math>

Recall T(t) = k i(t), and so T(s) = k I(s). Again let ω(0) = 0. This gives us Equation *2*.

<math>k I(s) = (J_ms + B) \Omega(s) *2* \, </math>

Solve Equation *2* for I(s) and substitute that into Equation *1*.

<math>V_s(s) = (R + Ls) \frac{(J_ms + B)}{k} \Omega(s) + k\Omega(s)</math>

Solve this equation for Ω(s), some simplification steps are skipped here. Equation *3*

<math>\Omega(s) = \frac{\frac{k}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s) *3*</math>

Now in order to apply the final value theorem we let Vs(s) = K/s. The s's in the limit definition and in K/s cancel. We can now see ω in steady state.

<math>\omega_{ss} = \lim_{s\rightarrow 0} s \Omega(s) = \frac{k}{RB + k^2}K = \omega(\infty)</math>

Take Equation *2* and Equation *3* and this time solve for I(s). This will get us a form of the final answer in the s domain.

<math>I(s) = \frac{(J_ms + B)}{k} \left [ \frac{\frac{k}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} \right ] K = \frac{\frac{s}{L} + \frac{B}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} K</math>

Finally was can apply the final value theorem to see what our steady state response will be for the armature current.

<math>i_{ss} = \lim_{s \to 0} sI(s) = \frac{B}{RB + k^2} K = i(\infty)</math>

=== Answers ===
Of interest are the ω(∞) and i(∞), found by plugging values into the steady state solutions above.

<math>\omega(\infty) = 14.91 radians/second</math>

<math>i(\infty) = 5.46 Amperes</math>

=== Bode Plot===

[[Image:Dcmotorbode.JPG]]

=== Break Points and Asymptotes===

The transfer function of the motor voltage over the voltage source.

<math>H(s) = \frac{V_m(s)}{V_s(s)} = \frac{\Omega(s)}{k} = \Omega(s) = \frac{\frac{1}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s)</math>

Now we need to plug in the given values to find our transfer function and characteristic equation.

<math>H(s) = \frac{2000}{s^2 + 401.8s + 737.6}V(s) = \frac{2000}{(s + 1.844)(s + 399.956)}V(s)</math>

At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.
At poles we see drops of 20 dB/decade, and we see 2 such drops at the appropriate values of 1.8 and 400 on the frequency graph.
You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log10 of this value. In this case we'd have a value of 2000/1.8*400=2.71. So our expected starting value should be 20log(2.71)=8.67, and it appears to be correct.
----
Reviewed by Andrew Hellie

Laplace transforms:DC Motor circuit

2009-10-30T03:12:38Z

Kendrick.mensink: /* Break Points and Asymptotes */

=== Problem ===
Find the steady state current ''i(t)'' through a DC motor represented by a series R-L-Motor circuit. The resistance (R) is from the armature winding. The inductance (L) is the equivalent inductance of the wire coil (which turns by current flowing through the coil in a permanent magnetic field). Jm is the combined moment of inertia of the armature and load. B is the coefficient of friction. The motor has input current ''i(t)'' and output angular velocity ω(t). Let v(t) = 110u(t)V, R = 20Ω, L = 50mH, k = 0.05N*m/A, Jm = 0.01 kg*m2, and B = e-4N*m*s.

[[Image:DCMotorPic.jpg]]

=== Solution ===
Since some of the elements of a motor are not typical circuit elements, we must solve this problem using Laplace transforms.

The torque is proportional to the armature current.

<math>T(t) = k i(t) \, </math>

Similarly, relating mechanical (''T(t)ω(t)'') and electrical (''vm(t)i(t)'') power, the conservation of energy requires the same proportionality between the voltage across the motor (''vm(t)'') and the angular velocity (''ω(t)'').

<math>v_m(t) = k \omega(t) \, </math>

We want to find the Laplace transfer function of the motor, and we define it as follows.

<math>\Omega(s) = \mathcal{L}[\omega(t)]/v_s(s)</math>

Summing the voltages around the series circuit gives us our differential equation.

<math>v_s(t) = R i(t) + L \frac{di(t)}{dt} + k \omega(t)</math>

Take the Laplace transform.

<math>V_s(s) = R I(s) + Ls( I(s) - i(0) ) + k\Omega(s) \, </math>

If we let i(0) equal zero, the transformed differential equation gives us Equation *1*.

<math>V_s(s) = R I(s) + LsI(s) + k\Omega(s) *1* \, </math>

Repeat the process with the analogous mechanical differential equation. Here Jm is the moment of inertia of the combined armature, shaft, and load. B is the coefficient of friction.

<math>T(t) = J_m \frac{d\omega(t)}{dt} + B \omega(t)</math>

Transforming yields the following.

<math>T(s) = J_m s(\Omega(s) - \omega(0)) + B \Omega(s) \, </math>

Recall T(t) = k i(t), and so T(s) = k I(s). Again let ω(0) = 0. This gives us Equation *2*.

<math>k I(s) = (J_ms + B) \Omega(s) *2* \, </math>

Solve Equation *2* for I(s) and substitute that into Equation *1*.

<math>V_s(s) = (R + Ls) \frac{(J_ms + B)}{k} \Omega(s) + k\Omega(s)</math>

Solve this equation for Ω(s), some simplification steps are skipped here. Equation *3*

<math>\Omega(s) = \frac{\frac{k}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s) *3*</math>

Now in order to apply the final value theorem we let Vs(s) = K/s. The s's in the limit definition and in K/s cancel. We can now see ω in steady state.

<math>\omega_{ss} = \lim_{s\rightarrow 0} s \Omega(s) = \frac{k}{RB + k^2}K = \omega(\infty)</math>

Take Equation *2* and Equation *3* and this time solve for I(s). This will get us a form of the final answer in the s domain.

<math>I(s) = \frac{(J_ms + B)}{k} \left [ \frac{\frac{k}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} \right ] K = \frac{\frac{s}{L} + \frac{B}{J_mL}}{s(s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL})} K</math>

Finally was can apply the final value theorem to see what our steady state response will be for the armature current.

<math>i_{ss} = \lim_{s \to 0} sI(s) = \frac{B}{RB + k^2} K = i(\infty)</math>

=== Answers ===
Of interest are the ω(∞) and i(∞), found by plugging values into the steady state solutions above.

<math>\omega(\infty) = 14.91 radians/second</math>

<math>i(\infty) = 5.46 Amperes</math>

=== Bode Plot===

[[Image:Dcmotorbode.JPG]]

=== Break Points and Asymptotes===

The transfer function of the motor voltage over the voltage source.

<math>H(s) = \frac{V_m(s)}{V_s(s)} = \frac{\Omega(s)}{k} = \Omega(s) = \frac{\frac{1}{J_mL}}{s^2 + (\frac{B}{J_m} + \frac{R}{L}) s + \frac{RB + k^2}{J_mL}}V(s)</math>

Now we need to plug in the given values to find our transfer function and characteristic equation.

<math>H(s) = \frac{2000}{s^2 + 401.8s + 737.6}V(s) = \frac{2000}{(s + 1.844)(s + 399.956)}V(s)</math>

At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.
At poles we see drops of 20 dB/decade, and we see 2 such drops at the appropriate values of 1.8 and 400 on the frequency graph.
You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log10 of this value. In this case we'd have a value of 2000/1.8*400=2.71. So 20log(2.71)=8.67
----
Reviewed by Andrew Hellie

Laplace transforms:DC Motor circuit

2009-10-29T23:33:06Z

Kendrick.mensink: /* Break Points and Asymptotes */

Laplace transforms:DC Motor circuit

2009-10-29T23:30:40Z

Kendrick.mensink: /* Break Points and Asymptotes */

Laplace transforms:DC Motor circuit

2009-10-29T23:30:16Z

Kendrick.mensink: /* Break Points and Asymptotes */

Laplace transforms:DC Motor circuit

2009-10-29T23:29:54Z

Kendrick.mensink: /* Break Points and Asymptotes */

Laplace transforms: R series with RC parallel circuit

2009-10-29T18:36:55Z

Kendrick.mensink: /* Use Convolution to find the output of the system */

===Problem Statement===

:Find the Voltage across the capacitor for t>=0:

:Voltage across capacitor at t({0-})=0

[[Image:lna_hw_5.jpg|400px|thumb|left|Fig (1)]]

: <math>v(t0-)=0 Volts\,</math>

: <math>v(t)=10 Volts\,</math>

: <math>R_1=20 \boldsymbol{\Omega}\,</math>

: <math>R_2=30 \boldsymbol{\Omega}\,</math>

: <math>C=.1 Farad\,</math>

Use Loop Equations to solve for the currents in <math>i_1\,</math> and <math>i_2\,</math>

:Loop 1 (Resistor Branch)

:<math>v(t)=R1(i_1+i_2)+R2(i_1)\,</math>

:<math>10=20(i_1+i_2)+30(i_1)\,</math>

:<math>i_1=(10-20i_2)/50\,</math>___________________________________equation (1)

:Loop 2 (Capacitor Branch)

:<math>v(t)=R1(i_1+i_2)+\dfrac{1}{C}\int{i_2 dt}\,</math>

:<math>10=20(i_1+i_2)+\dfrac{1}{.1}\int{i_2 dt}\,</math>_______________________equation (2)

Solve equations (1) and (2) simultaneously

:Substituting equation (1) into equation (2) gives...

:<math>20((10-20i_2)/50+i_2)+\dfrac{1}{.1}\int{i_2 dt}=10\,</math>

:simplifies to...

:<math>12i_2+10\int{i_2 dt}=6\,</math>

Take the Laplace Transform to move to the S-domain

:<math>\mathcal{L}\left\{\int_{0-}^\infty{(i_2) dt}\right\}=I_2/s\,</math>

:<math>\mathcal{L}\left\{(1)\right\}=1/s\,</math>

:<math>12I_2+10I_2/S=6/S\,</math>

:<math>I_2(12+10/S)=6/S\,</math>

:<math>I_2=6/(12S+10)\,</math>

:<math>I_2=(1/2)(1/(S+(5/6))\,</math>

Take the inverse Laplace transform to move back into the t-domain

:<math>i_2=(1/2)(e^{-(5/6)t})\,</math>

:substitute this equation back into equation (1)

:<math>i_1=(10-20(.5e^{-(5t/6)}))/50\,</math>

:<math>i_1=(1/5)(1-e^{-(5t/6)})\,</math>

Voltage on Capacitor

:<math>v_{capacitor}=10/20(i_1+i_2)\,</math>

:<math>v_{capacitor}=10-20((1/5)(1-e^{-(5t/6)})+(1/2)(e^{-(5t/6)}))\,</math>

:<math>v_{capacitor}=10-4+4e^{-(5t/6)}-10e{-(5t/6)}\,</math>

===Answer===

::<math>v_{capacitor}=6-6e^{-(5t/6)}\,</math> Volts

----

===Apply the Initial and Final Value Theorems to find the initial and final values===

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>

:<math>V(S)=6/s-6(1/(s+(5/6))\,</math>

:Initial Value:

::<math>\lim_{s\rightarrow \infty} sV(s)=6s/s-6s(1/(s+(5/6))\,</math>

::<math>\lim_{s\rightarrow \infty} sV(s)=0\,</math>

:::Initial Value = 0 Volts

:Final Value:

::<math>\lim_{s\rightarrow 0} sV(s)=6s/s-6s(1/(s+(5/6))\,</math>

::<math>\lim_{s\rightarrow 0} sV(s)=6\,</math>

:::Final Value = 6 Volts

:<math>v(t0)=0\,</math> Volts

:<math>v(t{\infty})=6\,</math> Volts

===Bode Plot===

:<math>V_{in}(t)=10\,</math>

:<math>V_{out}(t)=6-6*e^-((5/6)t)\,</math>

:<math>H(S)=V(s)_{in}/V(s)_{out}\,</math>

:<math>H(S)=1/(6s)-1/(6(s-5))\,</math>

:simplified to...

:<math>H(S)=-30/(36s(s-5))\,</math>

[[Image:Bode_plot_2.jpg‎|700px|thumb|left|Fig (1)]]

----

===How to use break points and asymptotes to obtain the magnitude frequency response of the system...===

The break points are the values of s in H(s) that make the numerator and or the denominator 0.

The location of the break points is the magnitude frequency response of the system.

Zeros are where the numerator is equal to zero.

Poles are when the denominator is equal to zero.

===Use Convolution to find the output of the system===

----
Written by: Andrew Hellie

Checked by: Kendrick Mensink

Laplace transforms:DC Motor circuit

2009-10-28T17:59:12Z

Kendrick.mensink: /* Break Points and Asymptotes */

Laplace transforms:DC Motor circuit

2009-10-28T17:58:46Z

Kendrick.mensink: /* Break Points and Asymptotes */

Laplace transforms:DC Motor circuit

2009-10-28T17:58:35Z

Kendrick.mensink: /* Break Points and Asymptotes */

Laplace transforms:DC Motor circuit

2009-10-28T17:58:08Z

Kendrick.mensink: /* Bode Plot */

Laplace transforms:DC Motor circuit

2009-10-28T17:43:16Z

Kendrick.mensink: /* Bode Plot */

Laplace transforms:DC Motor circuit

2009-10-28T17:43:06Z

Kendrick.mensink: /* Bode Plot */

Laplace transforms:DC Motor circuit

2009-10-28T17:41:52Z

Kendrick.mensink: /* Bode Plot */

Laplace transforms:DC Motor circuit

2009-10-28T17:41:34Z

Kendrick.mensink: /* Bode Plot */

Laplace transforms:DC Motor circuit

2009-10-28T17:40:10Z

Kendrick.mensink: /* Bode Plot */

Laplace transforms:DC Motor circuit

2009-10-28T17:39:44Z

Kendrick.mensink: /* Bode Plot */

File:Dcmotorbode.JPG

2009-10-28T17:38:57Z

Kendrick.mensink:

Laplace transforms:DC Motor circuit

2009-10-28T17:37:12Z

Kendrick.mensink: /* Answers */

Laplace transforms:DC Motor circuit

2009-10-26T18:14:16Z

Kendrick.mensink: /* Solution */

Laplace transforms:DC Motor circuit

2009-10-26T18:09:13Z

Kendrick.mensink:

Laplace transforms:DC Motor circuit

2009-10-26T17:42:24Z

Kendrick.mensink: /* Problem */

Laplace transforms:DC Motor circuit

2009-10-26T17:42:11Z

Kendrick.mensink: /* Problem */

Laplace transforms:DC Motor circuit

2009-10-26T17:38:37Z

Kendrick.mensink: /* Solution */

Laplace transforms:DC Motor circuit

2009-10-26T17:35:46Z

Kendrick.mensink: /* Problem */

File:DCMotorPic.jpg

2009-10-26T17:34:49Z

Kendrick.mensink:

Fall 2009

2009-10-22T18:07:54Z

Kendrick.mensink: /* 2009-2010 Contributors */

Laplace transforms:DC Motor circuit

2009-10-22T14:32:43Z

Kendrick.mensink: /* Answers */

Laplace transforms:DC Motor circuit

2009-10-22T14:14:33Z

Kendrick.mensink: /* Solution */

Laplace transforms:DC Motor circuit

2009-10-22T14:04:09Z

Kendrick.mensink: /* Problem */