Simulating circuits - Revision history

Lynch: /* After a long time at v_{in} = 5 V */

2026-05-19T20:34:02Z

After a long time at v_{in} = 5 V

← Older revision		Revision as of 15:34, 19 May 2026
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	=== After a long time at <math>v_{in} = 5</math> V ===		=== After a long time at <math>v_{in} = 5</math> V ===

	(In the circuit above, 1 ms ~~classifies as~~ a "long time," as it is several times the time constant of decay.)		(In the circuit above, 1 ms is a "long time," as it is several times the time constant of decay.)

	The circuit is in steady state (no state variable is changing), so <math>i_L^' = v_C^' = 0</math>. Since <math>Cv_C^' = i_L</math>, <math>i_L = 0</math>. By KVL, <math>5 = v_L + v_R + v_C = Li_L^' + Ri_L + v_C = 0+0+v_C</math>, so <math>v_C = 5</math>. In summary,		The circuit is in steady state (no state variable is changing), so <math>i_L^' = v_C^' = 0</math>. Since <math>Cv_C^' = i_L</math>, <math>i_L = 0</math>. By KVL, <math>5 = v_L + v_R + v_C = Li_L^' + Ri_L + v_C = 0+0+v_C</math>, so <math>v_C = 5</math>. In summary,

Lynch: /* Immediately after the switch to v_{in} = 0 V */

2025-06-10T15:26:43Z

Immediately after the switch to v_{in} = 0 V

← Older revision		Revision as of 10:26, 10 June 2025
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	=== Immediately after the switch to <math>v_{in} = 0</math> V ===		=== Immediately after the switch to <math>v_{in} = 0</math> V ===

	The capacitor does not allow its voltage to change discontinuously and the inductor does not allow its current to change discontinuously. So <math>i_L = 0</math> and <math>v_C = 5</math>V. By KVL, <math>0 = v_L + v_R + v_C = L i_L^' + R i_L + 5</math> or <math>i_L^' = -5/L = -5/0.001 = -5000</math>A/s. Also $v_C^' = i_L/C =0$.		The capacitor does not allow its voltage to change discontinuously and the inductor does not allow its current to change discontinuously. So <math>i_L = 0</math> and <math>v_C = 5</math>V. By KVL, <math>0 = v_L + v_R + v_C = L i_L^' + R i_L + 5</math> or <math>i_L^' = -5/L = -5/0.001 = -5000</math>A/s. Also <math>v_C^' = i_L/C =0</math>.

	{\| class="wikitable"		{\| class="wikitable"

Lynch: /* Immediately after the switch to v_{in} = 5 V at t=0 */

2025-06-10T15:25:52Z

Immediately after the switch to v_{in} = 5 V at t=0

← Older revision		Revision as of 10:25, 10 June 2025
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	=== Immediately after the switch to <math>v_{in} = 5</math> V at t=0 ===		=== Immediately after the switch to <math>v_{in} = 5</math> V at t=0 ===

	The capacitor does not allow its voltage to change discontinuously and the inductor does not allow its current to change discontinuously. Otherwise there would be an infinite current or an infinite voltage, respectively. So <math>v_C = i_L = 0</math>. By KVL, <math>5 = v_L + v_R + v_C = L i_L^' + R i + 0 = L i_L^'</math> or <math>i_L^' = 5/L = 5/0.001 = 5000</math>A/s. Also $v_C^' = i_L/C =0$. (Note the '''second derivative''' of <math>v_C</math> is '''not''' zero.)		The capacitor does not allow its voltage to change discontinuously and the inductor does not allow its current to change discontinuously. Otherwise there would be an infinite current or an infinite voltage, respectively. So <math>v_C = i_L = 0</math>. By KVL, <math>5 = v_L + v_R + v_C = L i_L^' + R i + 0 = L i_L^'</math> or <math>i_L^' = 5/L = 5/0.001 = 5000</math>A/s. Also <math>v_C^' = i_L/C =0</math>. (Note the '''second derivative''' of <math>v_C</math> is '''not''' zero.)

	{\| class="wikitable"		{\| class="wikitable"

Lynch: /* Particular solution after the switch from v_{in} = 5V to v_{in} = 0 */

2025-06-08T16:45:03Z

Particular solution after the switch from v_{in} = 5V to v_{in} = 0

← Older revision		Revision as of 11:45, 8 June 2025
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	</math>		</math>

	To solve for the constants <math>A,B</math> for <math>i_L(t)<math> for the case where we switch the input voltage from 5V to 0, we use the initial conditions <math>i_L(0) = 0, i_L^'(0) = -5000</math>A/s. This gives us <math>A=0</math> and <math>B = -5000/23,848 = -0.21</math>A, for a response		To solve for the constants <math>A,B</math> for <math>i_L(t)</math> for the case where we switch the input voltage from 5V to 0, we use the initial conditions <math>i_L(0) = 0, i_L^'(0) = -5000</math>A/s. This gives us <math>A=0</math> and <math>B = -5000/23,848 = -0.21</math>A, for a response

	<math>		<math>
	i_L(t) = -0.21 e^{-7500t} \sin 23,848 t.		i_L(t) = -0.21 e^{-7500t} \sin 23,848 t.
	</math>		</math>

	The time <math>t</math> for one-quarter of a cycle of oscillation satisfies <math>23,848 t = \pi/2</math>, or <math>t = 66 \mu</math>s. Plugging this time <math>t</math> into the response, we get <math>i_L = -128</math>mA, which agrees with the plot. We could do a similar calculation to get the particular solution for $v_C(t)$ using the initial conditions <math>v_C(0) = 5, v_C^'(0) = 0</math>.		The time <math>t</math> for one-quarter of a cycle of oscillation satisfies <math>23,848 t = \pi/2</math>, or <math>t = 66 \mu</math>s. Plugging this time <math>t</math> into the response, we get <math>i_L = -128</math>mA, which agrees with the plot. We could do a similar calculation to get the particular solution for <math>v_C(t)</math> using the initial conditions <math>v_C(0) = 5, v_C^'(0) = 0</math>.

Lynch: /* Particular solution after the switch from v_{in} = 5V to v_{in} = 0 */

2025-06-08T16:43:09Z

Particular solution after the switch from v_{in} = 5V to v_{in} = 0

← Older revision		Revision as of 11:43, 8 June 2025
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	<math>		<math>
	-7500e^{-7500t}(A \cos 23,848t + B \sin 23,848t) + e^{-7500t}(-23,848A \sin 23,848t + 23,848 B \cos 23,848t).		-7500e^{-7500t}(A \cos 23,848t + B \sin 23,848t) + e^{-7500t}(-23,848A \sin 23,848t + 23,848 B \cos 23,848t).
	<\math>		</math>

	To solve for the constants <math>A,B</math> for <math>i_L(t)<math> for the case where we switch the input voltage from 5V to 0, we use the initial conditions <math>i_L(0) = 0, i_L^'(0) = -5000</math>A/s. This gives us <math>A=0</math> and <math>B = -5000/23,848 = -0.21</math>A, for a response		To solve for the constants <math>A,B</math> for <math>i_L(t)<math> for the case where we switch the input voltage from 5V to 0, we use the initial conditions <math>i_L(0) = 0, i_L^'(0) = -5000</math>A/s. This gives us <math>A=0</math> and <math>B = -5000/23,848 = -0.21</math>A, for a response
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	i_L(t) = -0.21 e^{-7500t} \sin 23,848 t.		i_L(t) = -0.21 e^{-7500t} \sin 23,848 t.
	</math>		</math>
	The time <math>t</math> for one-quarter of a cycle of oscillation satisfies <math>23,848 t = \pi/2</math>, or <math>t = 66 \mu</math>s. Plugging this time <math>t</math> into the response, we get <math>i_L = -128</math>mA, which agrees with the plot. We could do a similar calculation to get the particular solution for $v_C(t)$ using the initial conditions $v_C(0) = 5, v_C^'(0) = 0$.		The time <math>t</math> for one-quarter of a cycle of oscillation satisfies <math>23,848 t = \pi/2</math>, or <math>t = 66 \mu</math>s. Plugging this time <math>t</math> into the response, we get <math>i_L = -128</math>mA, which agrees with the plot. We could do a similar calculation to get the particular solution for $v_C(t)$ using the initial conditions <math>v_C(0) = 5, v_C^'(0) = 0</math>.

Lynch: /* Analysis of the Example Circuit */

2025-06-08T16:42:28Z

Analysis of the Example Circuit

← Older revision		Revision as of 11:42, 8 June 2025
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	The decay time constant is <math>\tau = 1/7500 = 0.133</math>ms, and the response will decay to 2% after <math>4 \tau \approxeq 0.53</math>ms, which is why the 1ms time for holding the pulse in our example qualifies as "a long time." The period of oscillation is <math>T = 2\pi/\omega \approxeq 2\pi/23,848 = 0.26</math>ms, which is consistent with the plots above.		The decay time constant is <math>\tau = 1/7500 = 0.133</math>ms, and the response will decay to 2% after <math>4 \tau \approxeq 0.53</math>ms, which is why the 1ms time for holding the pulse in our example qualifies as "a long time." The period of oscillation is <math>T = 2\pi/\omega \approxeq 2\pi/23,848 = 0.26</math>ms, which is consistent with the plots above.

	=== Particular solution ~~for~~ ~~our~~ ~~example~~ ===		=== Particular solution after the switch from <math>v_{in} = 5</math>V to <math>v_{in} = 0</math> ===

			The time derivative of the response is

			<math>
			-7500e^{-7500t}(A \cos 23,848t + B \sin 23,848t) + e^{-7500t}(-23,848A \sin 23,848t + 23,848 B \cos 23,848t).
			<\math>

			To solve for the constants <math>A,B</math> for <math>i_L(t)<math> for the case where we switch the input voltage from 5V to 0, we use the initial conditions <math>i_L(0) = 0, i_L^'(0) = -5000</math>A/s. This gives us <math>A=0</math> and <math>B = -5000/23,848 = -0.21</math>A, for a response

			<math>
			i_L(t) = -0.21 e^{-7500t} \sin 23,848 t.
			</math>
			The time <math>t</math> for one-quarter of a cycle of oscillation satisfies <math>23,848 t = \pi/2</math>, or <math>t = 66 \mu</math>s. Plugging this time <math>t</math> into the response, we get <math>i_L = -128</math>mA, which agrees with the plot. We could do a similar calculation to get the particular solution for $v_C(t)$ using the initial conditions $v_C(0) = 5, v_C^'(0) = 0$.

Lynch: /* General solution for v_{in} = 0 */

2025-06-08T16:22:11Z

General solution for v_{in} = 0

← Older revision		Revision as of 11:22, 8 June 2025
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	</math>		</math>

	The decay time constant is $\tau = 1/7500 = 0.133$ms, and the response will decay to 2% after <math>4 \tau \approxeq 0.53</math>ms, which is why the 1ms time for holding the pulse in our example qualifies as "a long time." The period of oscillation is <math>T = 2\pi/\omega = 2\pi/23,848 ~~\approxeq~~ 0.26$ms, which is consistent with the plots above.		The decay time constant is <math>\tau = 1/7500 = 0.133</math>ms, and the response will decay to 2% after <math>4 \tau \approxeq 0.53</math>ms, which is why the 1ms time for holding the pulse in our example qualifies as "a long time." The period of oscillation is <math>T = 2\pi/\omega \approxeq 2\pi/23,848 = 0.26</math>ms, which is consistent with the plots above.

	=== Particular solution for our example ===		=== Particular solution for our example ===

Lynch: /* General solution for v_{in} = 0 */

2025-06-08T16:20:23Z

General solution for v_{in} = 0

← Older revision		Revision as of 11:20, 8 June 2025
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	<math>		<math>
	v_C(t) \; or \; i_L(t) = e^{-~~7500~~}t (A \cos 23,848t + B \sin 23,848t).		v_C(t) \; or \; i_L(t) = e^{-7500t} (A \cos 23,848t + B \sin 23,848t).
	</math>		</math>

			The decay time constant is $\tau = 1/7500 = 0.133$ms, and the response will decay to 2% after <math>4 \tau \approxeq 0.53</math>ms, which is why the 1ms time for holding the pulse in our example qualifies as "a long time." The period of oscillation is <math>T = 2\pi/\omega = 2\pi/23,848 \approxeq 0.26$ms, which is consistent with the plots above.

	=== Particular solution for our example ===		=== Particular solution for our example ===

Lynch at 16:15, 8 June 2025

2025-06-08T16:15:39Z

← Older revision		Revision as of 11:15, 8 June 2025
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	The previous differential equations have the same form and the same characteristic equation, <math>s^2 + (R/L) s + (1/LC) = 0</math>. Their solutions differ only by their different initial conditions.		The previous differential equations have the same form and the same characteristic equation, <math>s^2 + (R/L) s + (1/LC) = 0</math>. Their solutions differ only by their different initial conditions.

	Plugging in the numerical values for R, L, and C, the characteristic equation is <math>s^2 + 15,000s + 6.25\times10^8 = 0</math>.		Plugging in the numerical values for R, L, and C, the characteristic equation is <math>s^2 + 15,000s + 6.25\times10^8 = 0</math>, which has roots at <math>-7500 \pm 23,848j</math>. The roots are complex conjugates, so the solution is underdamped with the form

			<math>
			v_C(t) \; or \; i_L(t) = e^{-7500}t (A \cos 23,848t + B \sin 23,848t).
			</math>

	=== Particular solution for our example ===		=== Particular solution for our example ===

Lynch: /* General solution for v_{in} = 0 */

2025-06-08T16:11:03Z

General solution for v_{in} = 0

← Older revision		Revision as of 11:11, 8 June 2025
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	=== General solution for <math>v_{in} = 0</math> ===		=== General solution for <math>v_{in} = 0</math> ===

			The previous differential equations have the same form and the same characteristic equation, <math>s^2 + (R/L) s + (1/LC) = 0</math>. Their solutions differ only by their different initial conditions.

			Plugging in the numerical values for R, L, and C, the characteristic equation is <math>s^2 + 15,000s + 6.25\times10^8 = 0</math>.

	=== Particular solution for our example ===		=== Particular solution for our example ===