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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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| + | <-- | ||
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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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| Defining $\tau = \frac{a_1}{a_0}$, | Defining $\tau = \frac{a_1}{a_0}$, | ||
| - | $$\frac{dy(t){dt} + \frac{1}{\tau} y(t) = \frac{1}{a_1} f(t) (Eq. 7.21)$$ | + | $$\frac{dy(t)}{dt} + \frac{1}{\tau} y(t) = \frac{1}{a_1} f(t) (Eq. 7.21)$$ |
| The initial condition on equation (7.21) is as before: | The initial condition on equation (7.21) is as before: | ||
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| $$y(t=t_0) = y_0 (Eq. 7.22)$$ | $$y(t=t_0) = y_0 (Eq. 7.22)$$ | ||
| - | The cases of the system | + | The cases of the system |
| **1. __Homogeneous response__** | **1. __Homogeneous response__** | ||
| - | For the homogenous | + | For the homogeneous |
| $y(t) = y_0e^\frac{-t}{\tau}$ for $t \geq 0 (Eq. 7.23)$ | $y(t) = y_0e^\frac{-t}{\tau}$ for $t \geq 0 (Eq. 7.23)$ | ||
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| **2. __Step Response__** | **2. __Step Response__** | ||
| - | For a step input of amplitude A, $f(t) = Au_0(t)$ where $u_0(t)$ is the unit step function defined in section 6.1. Substituting this input function into equation (7.21), | + | For a step input of amplitude A, $f(t) = Au_0(t)$ where $u_0(t)$ is the unit step function defined in section 6.1. Substituting this input function into equation (7.21): |
| + | |||
| + | $$\frac{dy(t)}{dt} + \frac{1}{\tau}y(t) = \frac{A}{a_1}, t> | ||
| + | |||
| + | Using the approach of case 2. of our previous mass-damper system example, we determine the system response to be: | ||
| + | |||
| + | $$y(t) = \frac{A}{a_0} \left[ 1-e^\frac{-1}{\tau} \right] | ||
| + | |||
| + | This response is shown graphically in Figure 7.7. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * A first order system is described by a first order differential equation. The order of the differential equation describing a system is the same as the number of independent energy storage elements in the system – a first order system has one independent energy storage element. (The number of energy dissipation elements is arbitrary, however.) | ||
| + | * The differential equation governing a first order system is of the form: | ||
| + | |||
| + | $$a_1 \frac{dy(t)}{dt} + a_0y(t) = f(t)$$ | ||
| + | |||
| + | Where $y(t)$ is the system output, $f(t)$ is the applied input to the system, and $a_0$ and $a_1$ are constants. | ||
| + | |||
| + | * The differential equation governing a first order system can also be written in the form: | ||
| + | |||
| + | $$\frac{dy(t)}{dt} + \frac{1}{\tau}y(t) = \frac{f(t)}{a_1}$$ | ||
| + | |||
| + | Where $y(t)$ is the system output, $f(t)$ is the applied input to the system, and $\tau$ is the system time constant. | ||
| + | * The system time constant is a primary parameter used to describe the response of first order systems. | ||
| + | * 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: | ||
| + | |||
| + | $$y(t) = y_0e^\frac{-t}{\tau}, | ||
| + | |||
| + | For the step input case, the so-called //step response// is: | ||
| + | |||
| + | $$y(t) = \frac{A}{a_0} \left[ 1-e^{-t}{\tau} \right], t>0$$ | ||
| + | |||
| + | * The system response consists of two parts: a // | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | 1. In Example 6.2, we examined a body which was subjected to external heating. The system is shown in the figure below. The mass of the body is m, the body material has a specific heat, $c_P$, and is at some temperature $T_B$. The surroundings are at an ambient temperature $T_0$. A heat input $q_{in}$ is applied to the body, the heat dissipation between the body and its surroundings is $q_{out}$. It is common to assume that the heat dissipation $q_{out}$ is proportional to the difference in temperature between the mass and its surrounds, so that $q_{out} = h(T_B - T_0)$. Incorporating this assumption into the governing equation for the system provided in Example 6.2, results in the following differential equation relating $q_{in}$ and $T_B$: | ||
| + | |||
| + | $$mc_p \frac{d(T_B-T_0)}{dt} + h(T_B-T_0) = q_{in}$$ | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | * If the body has some initial temperature $T_i$, and no heat is applied to the body (e.g. $q_{in} = 0$), | ||
| + | * What is the final temperature of the body? | ||
| + | * What is the time constant (in terms of $m$, $c_p$, and $h$)? | ||
| + | * If the body mass is doubled, what is the effect on the time constant? Does this agree with your expectations based on your intuition? | ||
| + | * Sketch the response of the body temperature ($T_B$) vs. time. Label the initial condition and time constant on the sketch. | ||
| + | * If the body is initially at a temperature $T_0$ (the same as the ambient surroundings) and a constant heat input $q_{in}$ is applied starting at $t=0$, | ||
| + | * What is the final temperature of the body (in terms of $m$, $c_p$, $q_{in}$, and $h$) | ||
| + | * What is the effect of doubling the heat input $q_{in}$ on the final temperature? | ||
| + | * What is the effect of doubling the mass on the final temperature. Does this agree with your intuition? | ||
| + | * What is the effect of doubling the mass on the time constant? Does this agree with your intuition? | ||
| + | * Sketch the response $T_B$ vs. time. Label the initial temperature, | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 7.2: Natural Response of RC Circuits ===== | ||
| + | In this section, we consider source-free circuits containing only resistors and a single capacitor – commonly referred to as RC circuits. Since these circuits contain only a single energy storage element – the capacitor – the governing equations for the circuits will be first order differential equations. Since the circuits are source-free, | ||
| + | |||
| + | To begin our investigation of the natural response of RC circuits, consider the simple resistor-capacitor combination shown in Figure 7.8. We assume that the capacitor is initially charged to some voltage, $V_0$, at time $t = 0$ (so that $v(0) = V_0$). We want to determine the capacitor voltage, $v(t)$, for $t>0$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Applying Kirchhoff’s current law at the positive terminal of the capacitor, as shown in Figure 7.9, results in: | ||
| + | |||
| + | $$i_C(t) + i_R(t) = 0 (Eq. 7.26)$$ | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Since $i_R(t) = \frac{v(t)}{R}$ and $i_C(t)=C \frac{dv(t)}{dt}$, | ||
| + | |||
| + | $$C \frac{dv(t)}{dt} + \frac{v(t)}{R}=0$$ | ||
| + | |||
| + | Separation of variables results in: | ||
| + | |||
| + | $$\frac{dv}{v} = - \frac{1}{RC}dt$$ | ||
| + | |||
| + | The integral of the above is: | ||
| + | |||
| + | $$\int_{V_0}^{v(t)} \frac{dv}{v} = - \frac{1}{RC} \int_0^t dt$$ | ||
| + | |||
| + | Which evaluates to: | ||
| + | |||
| + | $$ln \left[ v(t) \right] - lm \left[ V_0 \right] = - \frac{t}{RC}$$ | ||
| + | |||
| + | Or | ||
| + | |||
| + | $$ln \left[ \frac{v(t)}{V_0} \right] = - \frac{t}{RC}$$ | ||
| + | |||
| + | Since $e^{ln(x)} = x$, the above becomes: | ||
| + | |||
| + | $$v(t) = V_0e^\frac{-t}{RC}$$ | ||
| + | |||
| + | ---- | ||
| + | |||
| + | === Alternate Approach to Solving the Above Differential Equation === | ||
| + | Since $\frac{dv(t)}{dt} = - \frac{1}{RC}v(t)$ we see that the form of the voltage signal must not change as a result of differentiation. Thus, assume that the voltage signal is of the form: | ||
| + | |||
| + | $$v(t) = Ke^{-st}$$ | ||
| + | |||
| + | Where $K$ and $s$ are unknown constants. If we substitute this into the original differential equation, we obtain: | ||
| + | |||
| + | $$-Kse^{-st} = - \frac{K}{RC}e^{-st}$$ | ||
| + | |||
| + | This is satisfied if we choose $s= - \frac{1}{RC}$. Employing our initial condition, $v(0) = V_0$, gives: | ||
| + | |||
| + | $$Ke^\frac{-0}{RC} = K = V_0$$ | ||
| + | |||
| + | Results in: | ||
| + | |||
| + | $$v(t) = V_0e^\frac{-t}{RC}$$ | ||
| + | |||
| + | As before. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | The capacitor voltage response is shown graphically in Figure 7.10. The voltage response is a decaying exponential with a time constant: | ||
| + | |||
| + | $$\tau = RC (Eq. 7.27)$$ | ||
| + | |||
| + | Thus, if we increase the resistance without changing capacitance, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | We can also obtain the above result by writing the governing differential equation directly in terms of the time constant. Previously, applying KCL to the circuit of Figure 7.9, we obtained: | ||
| + | |||
| + | $$C \frac{dv(t)}{dt} + \frac{v(t)}{R} = 0$$ | ||
| + | |||
| + | Which can be re-written (by dividing the equation by the capacitance, | ||
| + | |||
| + | $$\frac{dv(t)}{dt} + \frac{v(t)}{RC}=0$$ | ||
| + | |||
| + | Since the time constant for this circuit is $\tau = RC$, the above can be re-written as: | ||
| + | |||
| + | $$\frac{dv(t)}{dt} + \frac{1}{\tau}v(t) = 0 (Eq. 7.28)$$ | ||
| + | |||
| + | With initial condition: | ||
| + | |||
| + | $$v(0) = V_0 (Eq. 7.29)$$ | ||
| + | |||
| + | The solution to equation (7.28), subject to the initial condition of equation (7.29) is (per our results above): | ||
| + | |||
| + | $v(t) = V_0e^{-t}{\tau}$, | ||
| + | |||
| + | Where the time constant is defined per equation (7.27). The above approach matches our previous result. We will use the problem description provided by equations (7.28) and (7.29) and the solution of the form (7.30) most commonly in the subsequent examples. It should be emphasized, however, that the results are not dependent upon the solution approach – either of the other two approaches presented above yield the same conclusions. | ||
| + | |||
| + | ==== Generalization to Multiple Resistors ==== | ||
| + | The resistance in the time constant of equation (7.27) can be more generally defined as the __equivalent overall resistance of the circuit as seen by the capacitor.__ Thus, if we remove the capacitor from the circuit and create a Thevenin equivalent resistance as seen by the capacitor, the time constant will be the product of the capacitance and this equivalent resistance. We illustrate this point with an example. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.1 ==== | ||
| + | Determine the voltage $v(t)$ for the circuit below if $v(0) = 5V$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | We will first solve the problem by writing the first order differential equation governing the system. To aid this process we re-draw the circuit as shown below, labeling the current through the capacitor, defining node “A”, and labeling the voltage at node “A”. For simplicity, we label the capacitor as having capacitance “C” in the figure below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The voltage at node A, $V_A = v + (3 \Omega) C \frac{dv}{dt}$ is obtained by applying KVL around the outer loop of the circuit. | ||
| + | |||
| + | Applying KCL at node A results in: | ||
| + | |||
| + | $$C \frac{dv}{dt} + \frac{V_A}{2\Omega} + \frac{V_A}{2\Omega} = C \frac{dv}{dt} + V_A = 0$$ | ||
| + | |||
| + | Which, when substituting $V_A = v+ 3C \frac{dv}{dt}$ results in: | ||
| + | |||
| + | $$4C \frac{dv}{dt} + V_A = 0$$ | ||
| + | |||
| + | This can be placed in the “standard form”, $\frac{dv(t)}{dt} + \frac{1}{\tau}v(t) = 0$, by dividing through by 4C: | ||
| + | |||
| + | $$\frac{dv}{dt} + \frac{1}{4C}V_A = 0$$ | ||
| + | |||
| + | Thus, the time constant $\tau = 4C = 8 x 10^{-6}$ seconds. | ||
| + | |||
| + | Removing the capacitor and using circuit reduction to determine an equivalent resistance results in the circuit shown to the left below. The parallel combination of two, $2\Omega$ resistors results in an equivalent $1\Omega$ resistor, as shown in the figure to the right below. The resulting series combinations simplifies to a single $4\Omega$ resistance, so $R_{eq} = 4\Omega$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The circuit time constant can now be determined from $R_{eq}$ and $C$. Thus, | ||
| + | |||
| + | $$\tau = R_{eq}C = (4\Omega)(2x10^{-6}F) = 8x10^{-6} seconds$$ | ||
| + | |||
| + | Since the initial capacitor voltage is given to be 5V, $v(t) = 5e^\frac{-t}{8x10^{-6}}=5e^{1.25x10^5t}V$. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Determining Initial Conditions ==== | ||
| + | The circuits we have considered so far in this chapter contain no sources, the circuits’ initial conditions are given. In general, we will need to determine the initial conditions from a given source and/or switching operation. For example, a conceptual circuit showing how the initial condition for the circuit of Figure 7.8 can be created is shown in Figure 7.11(a). The switch in Figure 7.11(a) has been closed for a long time; thus, just before the switch opens, the voltage across the capacitor is $V_0$. Opening the switch removes the source from the circuit of interest. Since the voltage across a capacitor cannot change suddenly, the capacitor still has voltage $V_0$ immediately after the switch opens. (Mathematically, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.2: Switched Circuit Natural Response ==== | ||
| + | Consider the circuit shown below. The switch is originally at position A; at time $t = 0$ seconds, the switch moves to position B in the circuit. We wish to determine the capacitor voltage, $v(t)$ for $t > 0$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Before time $t = 0$, we assume that the switch has been at position A for a long time – all transients have died off, and any voltages and currents in the circuit have become constant. Since the capacitor voltage-current relationship is $i=C\frac{dv}{dt}$, | ||
| + | |||
| + | $$v(0^-) = 12V \left[ \frac{12\Omega}{12\Omega + 4\Omega} \right] = 9V$$ | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Since a capacitor cannot change its voltage instantaneously, | ||
| + | |||
| + | The system time constant can be determined from the capacitance and the equivalent resistance seen by the capacitor. The equivalent resistance can be determined by looking into the capacitor terminals with the switch at position B (recall that we are solving the differential equation for $t>0$). The appropriate circuit is shown below. The equivalent resistance consists of a parallel combination of the $12\Omega$ and $6\Omega$ resistors (note that the $4\Omega$ resistor and the voltage source are no longer relevant to the problem – they are isolated from the capacitor after the switch changes to position B). The equivalent resistance is thus: | ||
| + | |||
| + | $$R_{eq} = \frac{(6\Omega)(12\Omega)}{6\Omega+12\Omega}=4\Omega$$ | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The system time constant is, therefore, $\tau = R_{eq}C = (4\Omega)(25x10^{-3}F) = 100x10^{-3}$ seconds or 0.1 seconds. Appropriate substitution of the initial condition and time constant into equation (7.30) gives: | ||
| + | |||
| + | $$v(t) = 9e^\frac{-t}{0.1} = 9e^{-10t}V$$ | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * The natural response of an RC circuit describes the capacitor voltage in a circuit consisting only of resistors and a single equivalent capacitance. The circuit is source-free; | ||
| + | * The differential equation for an unforced RC circuit is of the form: | ||
| + | |||
| + | $$\frac{dv(t)}{dt} + \frac{1}{R_{eq}C}v(t) = 0$$ | ||
| + | |||
| + | Where $R_{eq}$ is the equivalent resistance " | ||
| + | |||
| + | * The RC circuit natural response is of the form: | ||
| + | |||
| + | $$v(t)=V_0e^\frac{-t}{\tau}, | ||
| + | |||
| + | Where $V_0$ is the initial voltage across the capacitor and $\tau$ is the circuit time constant. | ||
| + | |||
| + | * The time constant for __any__ first order system can be determined from the differential equation governing the system. If the governing differential equation is written in the form: | ||
| + | |||
| + | $$\frac{dv(t)}{dt} + \frac{1}{\tau}v(t) = 0$$ | ||
| + | |||
| + | The time constant $\tau$ can be determined by inspection. Thus, by comparison with the above differential equation for RC circuits, $\tau = R_{eq}C$. | ||
| + | |||
| + | * Alternatively, | ||
| + | |||
| + | $$\tau = R_{eq}C$$ | ||
| + | |||
| + | Where $R_{eq}$ is the equivalent resistance " | ||
| + | |||
| + | * The capacitor properties can be useful in determining initial conditions for an RC circuit: | ||
| + | * Capacitors behave like open-circuits when all circuit parameters are constant, and | ||
| + | * Capacitor voltages cannot change instantaneously | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | - The switch in the circuit below moves from position A to position B at time $t=0$. | ||
| + | - Write the differential equation governing $v(t), t>0$. | ||
| + | - Determine the time constant of the circuit from the differential equation of part 1. | ||
| + | - Use the capacitance and the equivalent resistance seen by the capacitor to check your answer to part 2. | ||
| + | - Determine the initial condition on the capacitor voltage, $v(t=0^+)$ | ||
| + | - Determine $v(t), t>0$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 7.3: Natural Response of RL Circuits ===== | ||
| + | In this section, we consider source-free circuits containing only resistors and a single inductor – commonly referred to as RL circuits. Like RC circuits, these circuits contain only a single energy storage element – the inductor – and the governing equations for the circuits will be first order differential equations. Since the circuits are source-free, | ||
| + | |||
| + | We will base our discussion of RL circuit natural responses on the series resistor-inductor circuit shown in Figure 7.12. We assume that the inductor has some initial current, $I_0$, flowing through it at time $t = 0$ (so that $i(0) = I_0$). We will determine the inductor current, $i(t)$, for $t>0$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Since the inductor' | ||
| + | |||
| + | $$L\frac{di(t)}{dt}+Ri(t)=0 | ||
| + | |||
| + | In this chapter, we will solve this differential equation using the “alternate approach” presented in section 7.2 for capacitors. This approach consists of assuming a form of the solution, based on the form of the differential equation being solved. The assumed solution will contain unknown constants; these constants will be determined by plugging the assumed solution into the original differential and forcing the solution to satisfy the original differential equation and initial conditions. Since the differential equation is linear and has constant coefficients, | ||
| + | |||
| + | The form of equation (7.31) indicates that i(t) must be a function which does not change form when it is differentiated ($L\frac{di(t)}{dt}$ must cancel out $Ri(t)$). The only function with this property is an exponential function. Thus, we assume that the current is of the form: | ||
| + | |||
| + | $$i(t)=Ke^{-st} | ||
| + | |||
| + | Where $K$ and $s$ are unknown constants. Substituting equation (7.32) into equation (7.31) results in: | ||
| + | |||
| + | $$L \left( -Kse^{-st} \right) + R \left(KE^{-st} \right) = 0$$ | ||
| + | |||
| + | The above simplifies to: | ||
| + | |||
| + | $$\left( R-Ls \right) Ke^{-st} = 0$$ | ||
| + | |||
| + | Which is satisfied if $s=\frac{R}{L}$ or $Ke^{-st}=0$. Choosing $Ke^{-st}=0$ results in the trivial solution $i(t)=0$, which will not, in general, satisfy the initial condition on the circuit. By the process of elimination, | ||
| + | |||
| + | $$i(t)=Ke^\frac{-tR}{L} | ||
| + | |||
| + | The unknown constant $K$ is determined by applying the initial condition, $i(0) = I_0$. Evaluating equation (7.33) at time $t = 0$, and equating the result to the given initial condition, we obtain: | ||
| + | |||
| + | $$i(0) = Ke^\frac{-0 \cdot R}{L} = K = I_0$$ | ||
| + | |||
| + | Thus, $K=I_0$, and the solution to the differential equation is: | ||
| + | |||
| + | $$i(t) = I_0e^\frac{-tR}{L} = I_0e^\frac{-t}{\tau} | ||
| + | |||
| + | And the circuit time constant is: | ||
| + | |||
| + | $$\tau = \frac{L}{R} | ||
| + | |||
| + | Equation (7.35) indicates that increasing L or decreasing R causes the time constant to increase. Conversely, decreasing L or increasing R decreases the time constant. A plot of the response of equation (7.34) is shown in Figure 7.13. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | ==== Generalization to Multiple Resistors ==== | ||
| + | As with the RC circuit, the resistance in the time constant of equation (7.35) can be more generally defined as the __equivalent overall resistance of the circuit as seen by the inductor__. Thus, if we remove the inductor from the circuit and create a Thevenin equivalent resistance as seen by the inductor, the time constant will be the quotient of the inductance and the equivalent resistance. | ||
| + | |||
| + | ==== Determining Initial Conditions ==== | ||
| + | Though the initial condition is given in the above example, in general we will need to determine the initial condition from the application of some source and/or switching condition. For example, the circuit of Figure 7.14 can be used to generate the initial condition of the example circuit above. In the circuit of Figure 7.14, we assume that the switch has been closed for a long time and all circuit voltages and currents have become constant. When all circuit operating conditions are constant, the inductor behaves like a short circuit and all the current applied by the current source goes through the inductor and the inductor current is $I_0$. Since the current through an inductor cannot change suddenly, the inductor still has current $I_0$ immediately after the switch opens. Mathematically, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | An example is provided below to illustrate the points made in this chapter. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.3: Switched RL Circuit Natural Response ==== | ||
| + | Consider the circuit shown below. The switch has been closed for a long time; at time $t = 0$ seconds, the switch suddenly opens. Determine the inductor current, $i(t)$ for $t > 0$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Since we are told that the switch has been closed for a long time, we assume that all voltages and currents in the circuit are constant before we open the switch. Therefore, before the switch is open the inductor behaves like a short circuit and the inductor current at time $t = 0^-$ can be determined by analyzing the circuit below: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | From the above circuit, the inductor current just before the switch is opened is, from Ohm’s law, $i(0^-) = \frac{50V}{25\Omega} = 2A$. Since an inductor cannot change its current suddenly, the current immediately after the circuit opens is $i(0^+)=i(0^-)=2A$. This provides our initial condition on the inductor current. | ||
| + | |||
| + | The time constant of the circuit is determined from the inductance and the equivalent resistance of the circuit seen by the inductor, after the switch opens. The equivalent resistance can be determined by analyzing the circuit shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | From the circuit above it can be seen that, to the inductor, the $25\Omega$ and $75\Omega$ resistors are in series. Thus, $R_{eq} = 25\Omega + 75\Omega = 100\Omega$. The system time constant is, therefore: | ||
| + | |||
| + | $$\tau = \frac{L}{R_{eq}} = \frac{5x10^{-3}H}{100\Omega} = 5x10^{-5} seconds$$ | ||
| + | |||
| + | Substitution of the initial condition and time constant into equation (4) gives: | ||
| + | |||
| + | $$i(t) = 2e^\frac{-t}{5x10^{-5}}=2e^{-20, | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * The natural response of an RL circuit describes the inductor current in a circuit consisting only of resistors and a single equivalent inductance. The circuit is source-free; | ||
| + | * The differential equation for an unforced RL circuit is of the form: | ||
| + | |||
| + | $$\frac{di(t)}{dt}+\frac{R_{eq}}{L}i(t)=0$$ | ||
| + | |||
| + | Where $R_{eq}$ is the equivalent resistance " | ||
| + | |||
| + | * The RL circuit natural response is of the form: | ||
| + | |||
| + | $$i(t)=I_0e^\frac{-t}{\tau}$$, | ||
| + | |||
| + | Where $I_0$ is the initial voltage across the capacitor and $\tau$ is the circuit time constant. | ||
| + | |||
| + | * The time constant for a first order system can be determined from the differential equation governing the system. If the governing differential equation is written in the form: | ||
| + | |||
| + | $$\frac{di(t)}{dt}+\frac{1}{\tau}i(t)=0$$ | ||
| + | |||
| + | The time constant $\tau$ can be determined by inspection. Thus, by comparison with the above differential equation for RL circuits, $\tau = \frac{L}{R_{eq}}$. | ||
| + | |||
| + | * Alternatively, | ||
| + | |||
| + | $$\tau = \frac{L}{R_{eq}}$$ | ||
| + | |||
| + | Where $R_{eq}$ is the equivalent resistance “seen” by the inductor and L is the inductance in the circuit. (Notice, that to use this relation, we must accurately identify the circuit as a first order RL circuit before proceeding.) | ||
| + | |||
| + | * The inductor properties can be useful in determining initial conditions for an RL circuit: | ||
| + | * Inductors behave like short-circuits when all circuit parameters are constant, and | ||
| + | * Inductors currents cannot change instantaneously | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | - The switch in the circuit below moves from position A to position B at time $t=0$. | ||
| + | - Write the differential equation governing $i_L(t), t>0$. | ||
| + | - Determine the time constant of the circuit from the differential equation of part 1. | ||
| + | - Use the inductance and the equivalent resistance seen by the inductor to check your answer to part 2. | ||
| + | - Determine the initial condition on the inductor current, $i_L(t=0^+)$ | ||
| + | - Determine $I_L(t), t>0$ | ||
| + | - Determine the resistor current $i_R(t)$ just before and just after the switch moves. (e.g. determine $i_R(t=0^-)$ and $i_R(t=0^+)$.) Is the resistor current continuous with time? Is the current through the other resistor continuous with time? | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 7.4: Forced Response of First Order Circuits ===== | ||
| + | In sections 7.2 and 7.3, we were concerned with the natural response of electrical circuits containing a single energy storage element. For the natural response, any sources in the circuit were isolated from the circuit prior to determining the circuit response, so that the circuit being analyzed contained no sources. Thus, the circuit response of interest was entirely due to the energy initially stored in the circuit’s capacitors or inductors. In these cases, all voltages and currents in the circuit die out with time. | ||
| + | |||
| + | In this section, we consider the case in which voltage or current sources are present in the first order circuit being analyzed. In this case, we must concern ourselves not only with the initial conditions in the circuit, but also with any //driving// or //forcing// functions applied to the circuit. The response of a system in the presence of an external input such as a voltage or current source is commonly called the //forced response// of the system. A primary difference between the natural response and the forced response of a system is that, although the natural response of a system always decays to zero, the forced response has no such restriction. In fact, the forced response of the system will take the same form as the forcing function, as time goes to infinity. | ||
| + | |||
| + | The differential equations governing the forced response of first order circuits are still, as implied, first order – thus, the circuits presented here will contain only a single energy storage element. Figure 7.15 shows two examples of forced first order circuits; Figure 7.15(a) is a forced RC circuit and Figure 7.15(b) is a forced RL circuit. The voltage sources $v_s(t)$ in Figure 7.15 provide an arbitrary input voltage which can vary as a function of time. Without loss of generality, we will concern ourselves only with determining the forced response of the voltages across capacitors or the currents through inductors. In spite of the simplicity of the circuits shown in Figure 7.15, their analysis provides a framework within which to place the analysis of an arbitrary first order electrical circuit. These two circuits are, therefore, analyzed below in order to present the forced solution of an arbitrary first order differential equation. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | === Tip === | ||
| + | In all circuits we analyze, we will define as unknown variables the voltages across capacitors and the currents through inductors. We will then solve for these variables and determine any other necessary circuit parameters from these variables. | ||
| + | |||
| + | In any electrical circuit, we can determine //any// circuit parameter from the voltages across capacitors and the currents through inductors. The reason for this is that capacitor voltages describe the capacitor energy storage and inductor currents describe the inductor energy storage. If we know the energy stored in all energy storage elements, we have completely characterized the circuit’s operating parameters – mathematically, | ||
| + | |||
| + | Examples of this concept, in the context of Figures 7.15 include: | ||
| + | |||
| + | * The voltage drop across the resistor in Figure 7.15(a) is: $v_R(t)=v_s(t)-v_c(t)$ | ||
| + | * The current through the capacitor in Figure 7.15(a) is: $i_R(t)=\frac{v_s(t)-v_c(t)}{R}$ | ||
| + | * The voltage drop across the resistor in Figure 7.15(b) is: $v_R(t)=Ri_L(t)$ | ||
| + | |||
| + | KCL at the positive terminal of the capacitor of the circuit shown in Figure 7.15(a) provides: | ||
| + | |||
| + | $$\frac{v_s(t)-v_c(t)}{R}=C\frac{dv_c(t)}{dt} | ||
| + | |||
| + | Dividing through by the capacitance, | ||
| + | |||
| + | $$\frac{dv_c(t)}{dt}+\frac{1}{RC}v_c(t)=\frac{1}{RC}v_s(t) | ||
| + | |||
| + | KVL around the loop of the circuit of Figure 7.15(b) results in: | ||
| + | |||
| + | $$-v_s(t)+Ri_L(t)+L\frac{di_L(t)}{dt}=0 | ||
| + | |||
| + | Dividing through by the inductance, L, and rearranging results in the governing differential equation for the circuit of Figure 7.15(b): | ||
| + | |||
| + | $$\frac{di_L(t)}{dt}+\frac{R}{L}i_L(t) = \frac{1}{L}v_s(t) | ||
| + | |||
| + | We notice now that the $\frac{1}{RC}$ term in equation (7.37) corresponds to $\frac{1}{\tau}$, | ||
| + | |||
| + | $$\frac{dy(t)}{dt} + \frac{1}{\tau}y(t)=u(t) | ||
| + | |||
| + | Where $\tau$ is the time constant of the system, $u(t)$ is the input to the system, and $y(t)$ is the desired system parameter (a voltage across a capacitor or a current through an inductor, for example). Equation (7.40) can be solved, given knowledge of the initial conditions on $y(t)$, $y(0) = y_0$. A block diagram of the system described by equation (7.40) is shown in Figure 7.16. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The solution to any forced differential equation can be considered to be formed of two parts: the // | ||
| + | |||
| + | $$y(t)=y_h(t)+y_p(t) | ||
| + | |||
| + | Where $y_h(t)$ is the homogeneous solution and $y_p(t)$ is the particular solution. | ||
| + | |||
| + | We will not attempt to analytically determine the solution of equation (7.40) for the general case of an arbitrary forcing function $u(t)$; instead, we will focus on specific types of inputs. Inputs of primary interest to us will consist of: | ||
| + | * Constant (step) input functions | ||
| + | * Sinusoidal input functions | ||
| + | |||
| + | The study of circuit responses to step functions is provided in Section 7.5. Sinusoidal input functions are discussed in later chapters. | ||
| + | |||
| + | ==== Generalization to Multiple Resistors ==== | ||
| + | As in the first order circuit natural response, the resistance in the time constant of equation (7.37) can be more generally defined as the __equivalent overall resistance of the circuit as seen by the energy storage element__. This conclusion follows directly from Thévenin’s theorem. The circuits of Figure 7.15 consist of energy storage elements (a capacitor and an inductor) which are connected to a circuit which can be considered to be the Thévenin equivalent circuit of a more complex circuit. Thus, the resistance R in the circuits of Figure 7.15 can be considered to be the equivalent (Thévenin) resistance of an arbitrary linear circuit to which an energy storage element is connected. Thus, the resistances in the governing differential equations (7.37) and (7.39) can be considered to be Thévenin equivalent resistances. These equations thus can be written as: | ||
| + | |||
| + | $$\frac{dv_c(t)}{dt}+ \frac{1}{R_{eq}C}v_c(t) = \frac{1}{R_{eq}C}v_s(t) | ||
| + | |||
| + | For an RC Circuit, and: | ||
| + | |||
| + | $$\frac{di_L(t)}{dt}+\frac{R_{eq}}{L}i_L(t) = \frac{1}{L}v_s(t) | ||
| + | |||
| + | For an RL circuit, where $R_{eq}$ is the equivalent (Thévenin) resistance seen by the energy storage element. This leads to the conclusion that the time constants for first order forced circuits can be written in terms of the Thevenin resistance seen by the energy storage element. The appropriate relationships are: | ||
| + | |||
| + | $$\tau = R_{eq}C | ||
| + | |||
| + | For RC circuits, and: | ||
| + | |||
| + | $$\tau = \frac{L}{R_{eq}} | ||
| + | |||
| + | for RL circuits. Note that this conclusion is consistent with our previous results for __unforced__ RC and RL circuits. We conclude this section with several examples in which we determine the differential equations governing first order electrical circuits. Note that we make no attempt to __solve__ these differential equations – in fact, we __cannot__ solve the differential equations, since we have not specified what the forcing functions are in the circuits below. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.4 ==== | ||
| + | Determine the differential equation relating $v_{in}(t)$ and $v_c(t)$ in the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | We will apply KCL at node “A”, as indicated in the figure below, to begin our analysis. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Thus, | ||
| + | |||
| + | $$i_1(t) = i_2(t) + i_3(t)$$ | ||
| + | |||
| + | Using the voltage-current relations to write these currents in terms of voltages results in: | ||
| + | |||
| + | $$\frac{v_{in}(t)-v_c(t)}{R}=\frac{v_c(t)}{R}+C\frac{dv_c(t)}{dt}$$ | ||
| + | |||
| + | After a little algebra, the above results can be written in our standard first order circuit form as: | ||
| + | |||
| + | $$\frac{dv_c(t)}{dt}+\frac{2}{RC}v_c(t)=\frac{1}{RC}v_{in}(t)$$ | ||
| + | |||
| + | And the time constant of the circuit is $\tau = \frac{RC}{2}$. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.5 ==== | ||
| + | Determine the differential equation relating $V_{in}(t)$ and $V_{out}(t)$ in the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Consistent with our approach of defining variables as voltages across capacitors and currents through inductors, we define the capacitor voltage as $v_c(t)$, as shown in the figure below. Also in the figure below, node A is defined and the rules governing ideal op-amps are used to identify the node voltage $V_A = 0V$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Applying KCL at node A in the circuit above gives: | ||
| + | |||
| + | $$i_1(t)=i_2(t)$$ | ||
| + | |||
| + | The currents can be written in terms of the voltages in the circuit to provide: | ||
| + | |||
| + | $$\frac{V_{in}(t)-0}{R} = C\frac{dv_c(t)}{dt}$$ | ||
| + | |||
| + | The capacitor voltage can be written in terms of $V_{OUT}$ (using KVL) as: | ||
| + | |||
| + | $$V_{OUT}(t)=-v_c(t)$$ | ||
| + | |||
| + | Thus, the governing differential equation for this circuit can be written as: | ||
| + | |||
| + | $$V_{OUT}=-\frac{1}{RC} \int V_{in}(t)dt$$ | ||
| + | |||
| + | Note that this circuit is performing an __integration__. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.6 ==== | ||
| + | Determine the differential equation relating $V_{in}(t)$ and $V_{out}(t)$ in the circuit below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | As in example 7.5, we define node A is defined and use the rules governing ideal op-amps to identify the node voltage $V_A = 0V$, as shown in the figure below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Writing KCL at node A directly in terms of the node voltages results in: | ||
| + | |||
| + | $$C\frac{dV_{in}(t)}{dt} = -\frac{V_{OUT}}{R}$$ | ||
| + | |||
| + | So that: | ||
| + | |||
| + | $$V_{OUT}=-RC\frac{dV_{in}(t)}{dt}$$ | ||
| + | |||
| + | And the output voltage is proportional to the __derivative__ of the input voltage. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Cross-Checking Results ==== | ||
| + | The above examples revolve entirely around determining the governing differential equation for the circuit. The actual circuit response depends upon the governing differential equation, the initial conditions, and the specific forcing function being applied to the circuit. In the above examples, the circuit time constants were inferred from the differential equation coefficients governing the forced response just as they were when we determined the natural response. | ||
| + | |||
| + | It is __always__ desirable((To the point of being absolutely essential!)) to check one’s results in as many ways as possible. With this in mind, we would like to check to see if the differential equation we have written for a given electrical circuit makes sense __before__ solving the equation for a specific forcing function. For first order circuits, at least, we can do this by determining a time constant __directly from the circuit itself__ and comparing this time constant with the time constant inferred from the governing differential equation. The time constant of any first order forced circuit can be obtained by calculating the Thévenin resistance seen by the energy storage element and using equations (7.44) and (7.45) to provide the time constant. We will now revisit example 7.4 using this approach to validate the differential equation we previously determined for this circuit. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.7 ==== | ||
| + | Check the time constant for the circuit of Example 7.4 by calculating the equivalent resistance seen by the capacitor. | ||
| + | |||
| + | The circuit of example 7.4 is shown below for reference. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | We can determine the equivalent resistance seen by the capacitor by replacing the voltage source with a short circuit and looking at the resistance seen across the capacitors terminals, as shown below: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The resistors are now in parallel, so that the equivalent resistance is $\frac{R}{2}$. The time constant is then $\tau = R_{eq}C = \frac{RC}{2}$, | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * The forced response of a first order circuit describes the response of the circuit in the presence of (in general) both a non-zero initial condition and an arbitrary time-varying input function. | ||
| + | * The differential equation describing the forced response of a first order circuit is of the form: | ||
| + | |||
| + | $$\frac{dy(t)}{dt}+\frac{1}{\tau}y(t)=f(t)$$ | ||
| + | |||
| + | Where $f(t)$ is the forcing function applied to the circuit. The time constant, | ||
| + | |||
| + | * The time constant for a first order forced system can also be determined directly from the circuit itself. The process is to determine the Thevenin resistance, $R_{eq}$, seen by the energy storage element and use that resistance in the appropriate time constant formula as introduced in sections 7.2 and 7.3. For an RL circuit, the time constant is $\tau= \frac{L}{R_{eq}}$, | ||
| + | * Although the time constant 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. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | 1. For the circuit below, write the differential equation governing $i_L(t)$. The input is the current source, $u(t)$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | 2. Determine the differential equation governing $i_c(t)$ in the circuit below: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | 3. Determine the differential equation governing $V_{out}(t)$ in the circuit below | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ===== 7.5: Step Response of First Order Circuits ===== | ||
| + | In section 7.4 we introduced the concept of the response of a first order circuit to an __arbitrary__ forcing function. We will not attempt to __solve__ this problem for an arbitrary forcing function; we will instead restrict our attention to specific forcing functions. In this section, we address the case in which the input consists of the sudden application of a constant voltage or current to a circuit; this type of input is typically modeled as a step function. The response of a system to this type of input, in the absence of any initial conditions, is called the //step response// of the system. | ||
| + | |||
| + | Figure 7.17 shows a conceptual circuit which applies a step input to an RC circuit, and an actual switched circuit which may be used to implement this forcing function. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | In section 7.3, we saw that the differential equation governing the forced response of a first order circuit is of the form: | ||
| + | |||
| + | $$\frac{dy(t)}{dt}+ \frac{1}{\tau}y(t) = u(t) (Eq. 7.46)$$ | ||
| + | |||
| + | Where $\tau$ is the time constant of the system, $u(t)$ is the input to the system, and $y(t)$ is the desired system parameter (for example, the voltage across a capacitor or the current through an inductor). In the case of a step input to the system, the input $u(t)$ to the system is a constant, | ||
| + | |||
| + | $$u(t)=Au_0(t) | ||
| + | |||
| + | Where $u_0(t)$ is the unit step function. | ||
| + | |||
| + | We also saw, in section 7.3, that the solution of equation (7.46) can be written as the superposition of a homogeneous solution and the particular solution, | ||
| + | |||
| + | $$y(t)=y_h(t)+y_p(t) | ||
| + | |||
| + | The // | ||
| + | |||
| + | The homogeneous response is the solution to the homogeneous differential equation: | ||
| + | |||
| + | $$\frac{dy(t)}{dt} + \frac{1}{\tau}y(t)=0 | ||
| + | |||
| + | In Section 7.1, we showed that the form of the solution of the homogeneous response is: | ||
| + | |||
| + | $$y_h(t)=K_1e^\frac{-t}{\tau} | ||
| + | |||
| + | However, now we cannot determine $K_1$ directly at this point. Any conditions we can generate from the circuit with which to determine unknown coefficients apply to the entire forced solution, not the homogeneous or particular solution individually. | ||
| + | |||
| + | Our next step, therefore, is to determine the particular solution by substituting the input of equation (7.47) into the differential equation (7.46) and solving for $y_p(t)$. The appropriate differential equation is, therefore: | ||
| + | |||
| + | $$\frac{dy_p(t)}{dt}+\frac{1}{\tau}y_p(t)=Au_0(t) | ||
| + | |||
| + | The particular solution is appropriate after the homogeneous solution dies out. Thus, we evaluate equation (7.51) as $t \rightarrow \infty$. | ||
| + | $$Au_0(t) = | ||
| + | \begin{cases} | ||
| + | 0, t < 0\\ | ||
| + | A, t > 0\\ | ||
| + | \end{cases} | ||
| + | $$ | ||
| + | |||
| + | If the right-hand side of equation (7.51) is a constant, then the left-hand side of equation (6) must be a constant and the individual terms in the left-hand side of equation (7.51) must be constants. It follows, then, that $y_p(t)$ is a constant and that $\frac{dy_p}{dt}=0$. Therefore, as $t \rightarrow \infty$, equation (7.51) becomes: | ||
| + | |||
| + | $$\frac{1}{\tau}y_p(t)=A \Rightarrow y_p(t) = K_2 (Eq. 7.52)$$ | ||
| + | |||
| + | The overall solution, then, from equations (7.48), (7.50), and (7.52) is: | ||
| + | |||
| + | $$y(t)=K_1e^\frac{-t}{\tau} + K_2 (Eq. 7.53)$$ | ||
| + | |||
| + | The unknown constants $K_1$ and $K_2$ are typically determined from evaluating the circuit’s behavior for $t=0$ and $t \rightarrow \infty$ . | ||
| + | |||
| + | We illustrate the overall solution process with three examples below. The first two examples are of passive first order circuit responses; the third example is of an active first order circuit. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.8 ==== | ||
| + | Determine the voltage across the capacitor, $v_C(t)$ for $t>0$ in the circuit below. The switch has been open for a long time, and the initial voltage across the capacitor is zero. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | When the switch is closed, the circuit is as shown below. The capacitor current has been labeled for later convenience. Note that the direction of the capacitor current is set to agree with the polarity of the capacitor voltage, according to the passive sign convention. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | KCL at node “A” of the circuit shown above results in: | ||
| + | |||
| + | $$\frac{10-v_C}{100 \Omega} = \frac{v_C}{200\Omega}+1x10^{-3} \frac{dv_C}{dt}$$ | ||
| + | |||
| + | Placing this in the form of equation (7.46) results in: | ||
| + | |||
| + | $$\frac{dv_C}{dt}+15v_c = 100$$ | ||
| + | |||
| + | Thus, the time constant $\tau = \frac{1}{15}$ sec. From equation (7.53) above, the form of the solution is: | ||
| + | |||
| + | $$v_C(t)=K_1e^\frac{-t}{\tau}+K_2$$ | ||
| + | |||
| + | We now apply the given initial condition, $v_C(0)=0$ to get: | ||
| + | |||
| + | $$0=K_1e^\frac{-t}{\tau}+K_2 \Rightarrow K_1 + K_2 = 0 (*)$$ | ||
| + | |||
| + | Another condition for determining the unknown constants is the capacitor voltage as $t \rightarrow \infty$. As $t \rightarrow \infty$, for a constant input, the capacitor becomes an open circuit. Thus, the circuit above becomes: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | And the capacitor voltage can be determined from voltage division to be: | ||
| + | |||
| + | $$v_C(t \rightarrow \infty) = 10V \frac{200\Omega}{100\Omega} + 200\Omega = 6.67V$$ | ||
| + | |||
| + | Substituting this result into the expression for the capacitor voltage results in: | ||
| + | |||
| + | $$6.67V = K_1e^\frac{-\infty}{\tau} + K_2 \Rightarrow K_2 = 6.67V (* *)$$ | ||
| + | |||
| + | Equations (*) and (* *) provide two equations in two unknowns. Solving these results in: | ||
| + | |||
| + | $$K_1=-6.67V$$ | ||
| + | |||
| + | $$K_2=6.67V$$ | ||
| + | |||
| + | And the capacitor voltage becomes: | ||
| + | |||
| + | $$v_C(t) = 6.67 \left( 1-e^{-15t} \right)$$ | ||
| + | |||
| + | The value for $\tau$ can be checked by determining the equivalent resistance seen by the capacitor. To do this, we kill the sources and look into the capacitor terminals. The appropriate circuit is shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The equivalent resistance is $R_{eq} = \frac{(100\Omega)(200\Omega)}{100\Omega+200\Omega} = 66.67\Omega$ and $\tau = R_{eq}C=(66.67\Omega)(1mH)= \frac{1}{15}$ seconds, which checks our previous result. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.9 ==== | ||
| + | Determine the current through the inductor, $i_L(t)$, in the circuit below. No energy is stored in the circuit prior to $t = 0$ seconds. The applied current input consists of a $12A$ step input applied at $t = 0$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | The circuit is shown below for $t>0$, with the inductor current labeled for ease of reference. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | KVL around the rightmost loop in the circuit results in the following voltage across the $3\Omega$ resistor: | ||
| + | |||
| + | $$v_{3\Omega} = 4 \frac{di_L}{dt} + 6i_L$$ | ||
| + | |||
| + | Employing this result and applying KCL at node A results in: | ||
| + | |||
| + | $$-12=i_L + \frac{1}{3\Omega} \left( 4 \frac{di_L}{dt} + 6i_L \right)$$ | ||
| + | |||
| + | Placing the above equation in the form of equation (7.46) results in: | ||
| + | |||
| + | $$\frac{di_L}{dt} + \frac{0}{4}i_L=9$$ | ||
| + | |||
| + | From this, we see that the circuit time constant is $\tau = \frac{4}{9}$ and the form of $i_L(t)$ is, from equation (7.53): | ||
| + | |||
| + | $$i_L(t)=K_1e^\frac{-4t}{9} + K_2$$ | ||
| + | |||
| + | From the given initial condition: | ||
| + | |||
| + | $$i_L(0)=K_1+K_2=0$$ | ||
| + | |||
| + | As $t \rightarrow \infty$, the inductor becomes a short circuit and the above circuit becomes: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Current division allows us to determine that $i_L(t \rightarrow \infty) = 12A \frac{3\Omega}{3\Omega+6\Omega}=4A$. Substituting this into equation governing $i_L(t)$, we obtain: | ||
| + | |||
| + | $$i_L(t \rightarrow \infty) = 4A = K_1e^{-\infty} + K_2 \Rightarrow K_2 = 4A$$ | ||
| + | |||
| + | Thus, the current through the inductor is: | ||
| + | |||
| + | $$i_L(t)=4(1-e^\frac{-4t}{9})$$ | ||
| + | |||
| + | Note that, once again, we can check our value for the time constant by killing any sources and determining the equivalent resistance seen by the inductor. The appropriate circuit is shown below: | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | From the above circuit, the equivalent resistance is simply a series combination of the $3\Omega$ and $6\Omega$ resistors. Thus, $R_{eq}=9\Omega$. For an RL circuit, the time constant $\tau = \frac{L}{R}=\frac{4}{9}$ seconds. This agrees with the previous result we obtained from examining the form of the governing differential equation. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.10 ==== | ||
| + | The switch in the circuit below closes at time $t = 0$. Find $V_{OUT}(t)$, | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | For time $t>0$, the switch is closed and the circuit is as shown on the figure below. Labeling node A on this circuit as shown, we can determine from the ideal operational amplifier rules that $V_A = 0V$. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Applying KCL at node A results in: | ||
| + | |||
| + | $$\frac{V_{IN}-0}{20\Omega}= \frac{v_C(t)}{20\Omega} + 2\frac{dv_C(t)}{dt}$$ | ||
| + | |||
| + | Since $v_C(t)=0-V_{OUT}(t)$, | ||
| + | |||
| + | $$\frac{V_{IN}}{20\Omega}= \frac{0-V_{OUT}(t)}{20\Omega} - 2 \frac{dV_{OUT}(t)}{dt}$$ | ||
| + | |||
| + | Placing this in the form of equation (7.46) results in: | ||
| + | |||
| + | $$\frac{dV_{OUT}(t)}{dt}+40V_{OUT}(t)=-V_{IN}$$ | ||
| + | |||
| + | Thus, the time constant $\tau=40$ sec. From equation (7.53) above, the form of the solution is: | ||
| + | |||
| + | $$V_{OUT}(t)=K_1e^\frac{-t}{\tau}+K_2$$ | ||
| + | |||
| + | We now apply the given initial condition, $v_C(0)=0$ to get: | ||
| + | |||
| + | $$0=K_1e^\frac{-0}{\tau}+K_2 \Rightarrow K_1+K_2 = 0 (*)$$ | ||
| + | |||
| + | The output voltage as $t \rightarrow \infty$ can be determined from by open-circuiting the capacitor and analyzing the resulting circuit. The circuit with the capacitor open-circuited is simply an inverting voltage amplifier with a gain of one, so: | ||
| + | |||
| + | $$V_{OUT}(t \rightarrow \infty) = -V_{IN}$$ | ||
| + | |||
| + | Substituting this result into the expression for the output voltage results in: | ||
| + | |||
| + | $$-V_{IN}=K_1e^\frac{-\infty}{\tau}+K_2 \Rightarrow K_2 = -V_{IN} | ||
| + | |||
| + | Equations (*) and (* *) provide two equations in two unknowns. Solving these results in: | ||
| + | |||
| + | $$K_1=V_{IN}$$ | ||
| + | $$K_2= - V_{IN}$$ | ||
| + | |||
| + | And the output voltage is: | ||
| + | |||
| + | $$V_{OUT}(t)= - V_{IN} \left( 1-e^\frac{-t}{40} \right)$$ | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Cross-Checking Results ==== | ||
| + | As in section 7.4, the examples above emphasize writing the governing differential equation for the circuit and determining the time constant from this differential equation. We also noted in section 7.4 that cross-checking results is crucial to producing reliable analyses, and that the time constant can also be determined directly from the circuit by calculating an equivalent resistance seen by the energy storage element. This cross-check was performed in exercises 7.8 and 7.9 above, as it should be. | ||
| + | |||
| + | For the case of a constant (or step) forcing function, we can use the final value of the solution as an additional cross-check of our results. In the above examples, we determined the final value of the response, $y(t \rightarrow \infty$), based on the circuit behavior with capacitors replaced by open circuits and inductors replaced by short circuits. We can also determine the final value directly from the differential equation itself by examining the response of the differential equation as $r \rightarrow \infty$. This value is called the // | ||
| + | |||
| + | The general differential equation governing the step response of a first order circuit is given by equation (7.51) above. If we examine the solution to this differential equation as $t \rightarrow \infty$, we obtain only the particular solution, $y_p(t)$. (The homogeneous solution must go to zero as $t \rightarrow \infty$, leaving only the particular solution.) Thus, as $t \rightarrow \infty$ equation (7.51) becomes: | ||
| + | |||
| + | $$\frac{dy_p(t)}{dt}+\frac{1}{\tau}y_p(t)=A | ||
| + | |||
| + | Since the particular solution has the same form as the forcing function, and the forcing function is a constant, the derivative of the particular solution with respect to time is zero, and equation (7.54) becomes: | ||
| + | |||
| + | $$\frac{1}{\tau}y_p=A$$ | ||
| + | |||
| + | And the steady state response, $y_{ss}$, is $y_{ss}=A \cdot \tau$. This value must agree with the final value obtained by short- circuiting inductors, open circuiting capacitors, and determining the final value from the circuit itself. | ||
| + | |||
| + | We apply this cross-check to the circuits of examples 7.8 and 7.9 below. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.11 ==== | ||
| + | In example 7.8, we determined (directly from the circuit behavior) that the final value of the capacitor voltage was: | ||
| + | |||
| + | $$v_C(t \rightarrow \infty) = 6.67V$$ | ||
| + | |||
| + | And that the governing differential equation for the circuit was: | ||
| + | |||
| + | $$\frac{dv_C}{dt}+15v_C=100$$ | ||
| + | |||
| + | We wish to check our final value of capacitor voltage relative to the differential equation behavior. | ||
| + | |||
| + | We can determine the final value of capacitor voltage by assuming that the voltage in the differential equation is constant, and setting its derivative to zero. Thus, the steady-state capacitor voltage can be determined from: | ||
| + | |||
| + | $$15v_{ss}=100$$ | ||
| + | |||
| + | So that $v_{ss}=\frac{100}{15}=6.67V$ | ||
| + | |||
| + | and agrees with the final value obtained by replacing the capacitor with an open circuit, as was done in example 7.8. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.12 ==== | ||
| + | In example 7.9, we determined (directly from the circuit behavior) that the final value of the inductor current was: | ||
| + | |||
| + | $$i_L(t \rightarrow \infty) = 4A$$ | ||
| + | |||
| + | and that the governing differential equation for the circuit was: | ||
| + | |||
| + | $$\frac{di_L}{dt}+\frac{9}{4}i_L=9$$ | ||
| + | |||
| + | We wish to check our final value of inductor current relative to the differential equation behavior. | ||
| + | |||
| + | We can determine the final value of inductor current by assuming that the current in the differential equation is constant, and setting its derivative to zero. Thus, the steady-state capacitor voltage can be determined from: | ||
| + | |||
| + | $$\frac{9}{4}i_{ss}=9$$ | ||
| + | |||
| + | So that $i_{ss}=9 \cdot \frac{4}{9} = 4A$ | ||
| + | |||
| + | which agrees with the final value obtained by replacing the inductor with a short circuit, as was done in example 7.9. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== DC Gain ==== | ||
| + | The steady-state response of a circuit to a step input provides an important parameter which is often used to characterize the circuit’s behavior. This parameter is called the DC gain, and is essentially the steady state response, normalized by the magnitude of the input step function. Mathematically, | ||
| + | |||
| + | DC gain $=\frac{y_{ss}}{A} | ||
| + | |||
| + | So the DC gain is simply the ratio of the output amplitude to the input amplitude, as $t \rightarrow \infty$. This parameter is of comparable importance to the characterization of first order circuits as the time constant. If we know the time constant and the DC gain of the circuit, we can immediately sketch the response of the circuit to __any__ step input. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Example 7.13 ==== | ||
| + | Determine the DC gain for the circuit of example 7.8. | ||
| + | |||
| + | In example 7.8, the input voltage amplitude was 10V. The steady state output, the capacitor voltage, had an amplitude of: | ||
| + | |||
| + | $$v_C(t \rightarrow \infty)=6.67V$$ | ||
| + | |||
| + | Thus, the DC gain is simply the ratio of the input magnitude to the (steady-state) output magnitude: | ||
| + | |||
| + | DC gain $= \frac{6.67V}{10V}=0.67$ | ||
| + | |||
| + | The DC gain can also be determined from the governing differential equation. This is probably easiest to do if we replace the original 10V source with an arbitrary voltage, $V_{in}$, as shown below. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | Re-deriving the governing differential equation, as was done in example 7.8, results in: | ||
| + | |||
| + | $$\frac{dv_C}{dt}+15v_C=10V_{in}$$ | ||
| + | |||
| + | If we are only concerned with the steady-state response, the derivative term can be set to zero and: | ||
| + | |||
| + | $$15v_{ss}=10V_{in}$$ | ||
| + | |||
| + | So that $\frac{v_{ss}}{V_{in}}=\frac{10}{15}=0.667$ | ||
| + | |||
| + | Which agrees with the DC gain determined by calculating the steady state response to a specific input voltage and taking the same ratio. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Section Summary ==== | ||
| + | * The step response of a first order circuit describes the response of a first order circuit to an applied step function. Typically, the term “step response” implies that all initial conditions in the circuit are zero, but this is not a requirement for application of any of the concepts presented in this section | ||
| + | * The differential equation describing the forced response of a first order circuit is of the form: | ||
| + | |||
| + | $$\frac{dy(t)}{dt}+\frac{1}{\tau}y(t)=Au_0(t)$$ | ||
| + | |||
| + | Where $u_0(t)$ is the unit step function defined in section 6.2. The time constant, $\tau$, of the circuit is readily obtained from the differential equation when it is written in the above form. | ||
| + | |||
| + | * The form of the step response of a first order system is: | ||
| + | |||
| + | $$y(t)=K_1e^\frac{-1}{\tau}+K_2$$ | ||
| + | |||
| + | * The time constant in the solution above can be determined from either the governing differential equation or directly from the circuit itself. | ||
| + | * The unknown constants in the response are determined from initial, $y(0^+)$, and final, $y(t\rightarrow \infty)$, conditions. The initial conditions must be determined from the circuit itself. The final conditions can be determined from either the circuit itself or from the governing differential equation. | ||
| + | * Although both the time constant and the final value of the response can be determined from either the governing differential equation or the circuit itself, it is strongly recommended that both approaches be used and the results compared with one another to provide a cross-check of your analysis. | ||
| + | * The DC gain of a circuit provides the ratio of the output amplitude to the input amplitude, as $t \rightarrow \infty$, if the input is a constant value. The DC gain and the time constant are often used to characterize the response of a first order circuit. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | ==== Exercises ==== | ||
| + | - The initial current in the circuit below is zero. (e.g. $i_L(t=0^-) = 0A$.) | ||
| + | - Write the differential equation governing $i_L(t)$. | ||
| + | - From your result in part 1, determine the time constant of the circuit. | ||
| + | - Write the form of the current $i_L(t)$. | ||
| + | - Use conditions at $t = 0$ and $t \rightarrow \infty$ to determine the unknown constants in the expression for $i_L(t)$ in part 3. | ||
| + | - Determine the equivalent resistance seen by the inductor to check your answer from part 2. | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | 2. An input voltage input, $v(t)$, is applied to a first order electrical circuit. The differential equation governing the resulting current, $i(t)$, through an inductor is determined to be: | ||
| + | |||
| + | $$2\frac{di(t)}{dt}+3i(t)=5v(t)$$ | ||
| + | |||
| + | What is the DC gain of the circuit? What are the units of DC gain for this circuit? What is the time constant of the circuit? | ||
| + | |||
| + | 3. What is the DC gain of the circuit below? $u(t)$ is the voltage input to the circuit and $v(t)$ is the response. What are the units of the DC gain for this circuit? | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | 4. The differential equation governing the circuit shown below is determined to be: | ||
| + | |||
| + | $$\frac{dv(t)}{dt}+\frac{1}{9}v(t)=i(t)$$ | ||
| + | |||
| + | Where the current $i(t)$ is the input to the circuit and the voltage $v(t)$ is the circuit response. __Without re- deriving the differential equation governing the circuit__, do you feel that the given differential equation above accurately describes the circuit response? | ||
| + | |||
| + | {{ : | ||
| + | |||
| + | 5. The differential equation governing the circuit shown below is determined to be: | ||
| + | |||
| + | $$3\frac{dv_C(t)}{dt}+v_C(t)=v(t)$$ | ||
| + | |||
| + | 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. | ||
| + | |||
| + | {{ : | ||
| + | [[{}/ | ||
| + | [[{}/ | ||