Laplace transforms:DC Motor circuit - Revision history

Kendrick.mensink: /* State Equations */

2009-11-16T19:00:47Z

State Equations

← Older revision		Revision as of 12:00, 16 November 2009
Line 108:		Line 108:
	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).		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{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>		<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.		The following is an alternate way to represent the dc motor circuit as a matrix system of equations.

Kendrick.mensink: /* State Equations */

2009-11-16T18:59:50Z

State Equations

← Older revision		Revision as of 11:59, 16 November 2009
Line 106:		Line 106:

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

Kendrick.mensink: /* Convolution */

2009-11-16T18:48:36Z

Convolution

← Older revision		Revision as of 11:48, 16 November 2009
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	<math>h(t) = [5.0237e^{-1.844t} - 5.0237e^{-399.956t}]u(t) \,</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		Reviewed by Andrew Hellie

Kendrick.mensink: /* Convolution */

2009-10-30T05:03:07Z

Convolution

← Older revision		Revision as of 22:03, 29 October 2009
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	<math>h(t) = \mathcal{L}^{-1} \left [ \frac{5.0237}{s+1.844}-\frac{5.0237}{s+399.956} \right ]</math>		<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>		<math>h(t) = [5.0237e^{-1.844t} - 5.0237e^{-399.956t}]u(t) \,</math>

	----		----
	Reviewed by Andrew Hellie		Reviewed by Andrew Hellie

Kendrick.mensink: /* Convolution */

2009-10-30T05:01:23Z

Convolution

← Older revision		Revision as of 22:01, 29 October 2009
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	For this we need the inverse Laplace transform of our H(s).		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		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>		<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		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>

	----		----
	Reviewed by Andrew Hellie		Reviewed by Andrew Hellie

Kendrick.mensink: /* Convolution */

2009-10-30T03:57:43Z

Convolution

← Older revision		Revision as of 20:57, 29 October 2009
Line 91:		Line 91:

	===Convolution===		===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		Reviewed by Andrew Hellie

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

2009-10-30T03:23:28Z

Break Points and Asymptotes

← Older revision		Revision as of 20:23, 29 October 2009
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	The transfer function of the motor voltage over the voltage source.		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>		<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.		Now we need to plug in the given values to find our transfer function and characteristic equation.

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

2009-10-30T03:22:52Z

Break Points and Asymptotes

← Older revision		Revision as of 20:22, 29 October 2009
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	At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.		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.		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 20log<sub>10</sub> 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.		You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log<sub>10</sub>(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===		===Convolution===

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

2009-10-30T03:20:41Z

Break Points and Asymptotes

← Older revision		Revision as of 20:20, 29 October 2009
Line 88:		Line 88:
	At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.		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.		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 20log<sub>10</sub> 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		You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log<sub>10</sub> 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		Reviewed by Andrew Hellie

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

2009-10-30T03:18:51Z

Break Points and Asymptotes

← Older revision		Revision as of 20:18, 29 October 2009
Line 88:		Line 88:
	At zero's we would see an increase of 20 dB/decade, but the numerator has no zeros.		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.		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 20log<sub>10</sub> 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.		You can find the starting point on the dB scale by evaluating the transfer function when s=0 and taking 20log<sub>10</sub> 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		Reviewed by Andrew Hellie