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| learn:courses:real-analog-chapter-7:start [2017/04/28 22:05] – [Example 7.2: Switched Circuit Natural Response] Martha | learn:courses:real-analog-chapter-7:start [2022/10/04 11:38] (current) – external edit 127.0.0.1 | ||
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| ====== Real Analog: Chapter 7 ====== | ====== Real Analog: Chapter 7 ====== | ||
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| The overall solution, then, from equations (7.48), (7.50), and (7.52) is: | The overall solution, then, from equations (7.48), (7.50), and (7.52) is: | ||
| - | $$y(t)=K_e^\frac{-t}{\tau} + K_2 (Eq. 7.53)$$ | + | $$y(t)=K_1e^\frac{-t}{\tau} + K_2 (Eq. 7.53)$$ |
| The unknown constants $K_1$ and $K_2$ are typically determined from evaluating the circuit’s behavior for $t=0$ and $t \rightarrow \infty$ . | The unknown constants $K_1$ and $K_2$ are typically determined from evaluating the circuit’s behavior for $t=0$ and $t \rightarrow \infty$ . | ||
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| KCL at node “A” of the circuit shown above results in: | KCL at node “A” of the circuit shown above results in: | ||
| - | $$\frac{10-v_C)(100\Omega} = \frac{v_C}{200\Omega}+1x10^{-3} \frac{dv_C}{dt}$$ | + | $$\frac{10-v_C}{100 \Omega} = \frac{v_C}{200\Omega}+1x10^{-3} \frac{dv_C}{dt}$$ |
| Placing this in the form of equation (7.46) results in: | Placing this in the form of equation (7.46) results in: | ||
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