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| learn:courses:real-analog-chapter-7:start [2017/01/06 23:50] – [7.5: Step Response of First Order Circuits] 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}$$ |
| Placing this in the form of equation (7.46) results in: | Placing this in the form of equation (7.46) results in: | ||
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| ==== Section Summary ==== | ==== Section Summary ==== | ||
| + | * The step response of a first order circuit describes the response of a first order circuit to an applied step function. Typically, the term “step response” implies that all initial conditions in the circuit are zero, but this is not a requirement for application of any of the concepts presented in this section | ||
| + | * The differential equation describing the forced response of a first order circuit is of the form: | ||
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| + | $$\frac{dy(t)}{dt}+\frac{1}{\tau}y(t)=Au_0(t)$$ | ||
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| + | Where $u_0(t)$ is the unit step function defined in section 6.2. The time constant, $\tau$, of the circuit is readily obtained from the differential equation when it is written in the above form. | ||
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| + | * The form of the step response of a first order system is: | ||
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| + | $$y(t)=K_1e^\frac{-1}{\tau}+K_2$$ | ||
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| + | * The time constant in the solution above can be determined from either the governing differential equation or directly from the circuit itself. | ||
| + | * The unknown constants in the response are determined from initial, $y(0^+)$, and final, $y(t\rightarrow \infty)$, conditions. The initial conditions must be determined from the circuit itself. The final conditions can be determined from either the circuit itself or from the governing differential equation. | ||
| + | * Although both the time constant and the final value of the response can be determined from either the governing differential equation or the circuit itself, it is strongly recommended that both approaches be used and the results compared with one another to provide a cross-check of your analysis. | ||
| + | * The DC gain of a circuit provides the ratio of the output amplitude to the input amplitude, as $t \rightarrow \infty$, if the input is a constant value. The DC gain and the time constant are often used to characterize the response of a first order circuit. | ||
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| + | ==== Exercises ==== | ||
| + | - The initial current in the circuit below is zero. (e.g. $i_L(t=0^-) = 0A$.) | ||
| + | - Write the differential equation governing $i_L(t)$. | ||
| + | - From your result in part 1, determine the time constant of the circuit. | ||
| + | - Write the form of the current $i_L(t)$. | ||
| + | - Use conditions at $t = 0$ and $t \rightarrow \infty$ to determine the unknown constants in the expression for $i_L(t)$ in part 3. | ||
| + | - Determine the equivalent resistance seen by the inductor to check your answer from part 2. | ||
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| + | 2. An input voltage input, $v(t)$, is applied to a first order electrical circuit. The differential equation governing the resulting current, $i(t)$, through an inductor is determined to be: | ||
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| + | $$2\frac{di(t)}{dt}+3i(t)=5v(t)$$ | ||
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| + | What is the DC gain of the circuit? What are the units of DC gain for this circuit? What is the time constant of the circuit? | ||
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| + | 3. What is the DC gain of the circuit below? $u(t)$ is the voltage input to the circuit and $v(t)$ is the response. What are the units of the DC gain for this circuit? | ||
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| + | 4. The differential equation governing the circuit shown below is determined to be: | ||
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| + | $$\frac{dv(t)}{dt}+\frac{1}{9}v(t)=i(t)$$ | ||
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| + | Where the current $i(t)$ is the input to the circuit and the voltage $v(t)$ is the circuit response. __Without re- deriving the differential equation governing the circuit__, do you feel that the given differential equation above accurately describes the circuit response? | ||
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| + | 5. The differential equation governing the circuit shown below is determined to be: | ||
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| + | $$3\frac{dv_C(t)}{dt}+v_C(t)=v(t)$$ | ||
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| + | Where the voltage $v(t)$ is the input to the circuit and the capacitor voltage $v_C(t)$ is the circuit response. __Without re- deriving the differential equation governing the circuit__, do you feel that the given differential equation above accurately describes the circuit response? Justify your answer. | ||
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