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| learn:courses:real-analog-chapter-7:start [2017/01/09 19:00] – 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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| + | * Lecture Material: | ||
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| + | * Chapter 7 Videos: | ||
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| ====== 7. Introduction and Chapter Objectives ====== | ====== 7. Introduction and Chapter Objectives ====== | ||
| First order systems are, by definition, systems whose input-output relationship is a //first order// differential equation. A first order differential equation contains a //first order derivative// | First order systems are, by definition, systems whose input-output relationship is a //first order// differential equation. A first order differential equation contains a //first order derivative// | ||
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| $$v_h(t) = K_1e \frac{-bt}{m} | $$v_h(t) = K_1e \frac{-bt}{m} | ||
| - | Where $K_1# is (in this case) an unknown constant which will be determined from our initial conditions. | + | Where $K_1$ is (in this case) an unknown constant which will be determined from our initial conditions. |
| The particular solution is determined from: | The particular solution is determined from: | ||
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| * In this chapter, we considered two types of forcing functions: a zero-input case, in which $f(t) = 0$ and $y(t=0) = y_0$, and a step input case, in which $f(t) = Au_0(t)$ and $y(t=0) = 0$. For the zero-input case, the response is: | * In this chapter, we considered two types of forcing functions: a zero-input case, in which $f(t) = 0$ and $y(t=0) = y_0$, and a step input case, in which $f(t) = Au_0(t)$ and $y(t=0) = 0$. For the zero-input case, the response is: | ||
| - | $$y(t) = y_0e^\frac{-t}{\tau}$, t>0$$ | + | $$y(t) = y_0e^\frac{-t}{\tau}, |
| For the step input case, the so-called //step response// is: | For the step input case, the so-called //step response// is: | ||
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| $$y(t) = \frac{A}{a_0} \left[ 1-e^{-t}{\tau} \right], t>0$$ | $$y(t) = \frac{A}{a_0} \left[ 1-e^{-t}{\tau} \right], t>0$$ | ||
| - | * The system response consists of two parts: a // | + | * The system response consists of two parts: a // |
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| - | Since a capacitor cannot change its voltage instantaneously, | + | Since a capacitor cannot change its voltage instantaneously, |
| The system time constant can be determined from the capacitance and the equivalent resistance seen by the capacitor. The equivalent resistance can be determined by looking into the capacitor terminals with the switch at position B (recall that we are solving the differential equation for $t>0$). The appropriate circuit is shown below. The equivalent resistance consists of a parallel combination of the $12\Omega$ and $6\Omega$ resistors (note that the $4\Omega$ resistor and the voltage source are no longer relevant to the problem – they are isolated from the capacitor after the switch changes to position B). The equivalent resistance is thus: | The system time constant can be determined from the capacitance and the equivalent resistance seen by the capacitor. The equivalent resistance can be determined by looking into the capacitor terminals with the switch at position B (recall that we are solving the differential equation for $t>0$). The appropriate circuit is shown below. The equivalent resistance consists of a parallel combination of the $12\Omega$ and $6\Omega$ resistors (note that the $4\Omega$ resistor and the voltage source are no longer relevant to the problem – they are isolated from the capacitor after the switch changes to position B). The equivalent resistance is thus: | ||
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| The overall solution, then, from equations (7.48), (7.50), and (7.52) is: | The overall solution, then, from equations (7.48), (7.50), and (7.52) is: | ||
| - | $$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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