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| learn:courses:real-analog-chapter-7:start [2017/01/06 19:53] – [Section Summary] 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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| + | <-- | ||
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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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| {{ : | {{ : | ||
| - | Consistent with our approach of defining variables as voltages across capacitors and currents through inductors, we define the capacitor voltage as vc(t), as shown in the figure below. Also in the figure below, node A is defined | + | 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 |
| - | and the rules governing ideal op-amps are used to identify the node voltage | + | |
| + | {{ : | ||
| + | |||
| + | 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. | ||
| + | |||
| + | {{ : | ||
| + | [[{}/ | ||
| + | [[{}/ | ||