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| ====== Real Analog: Chapter 6 ====== | ====== Real Analog: Chapter 6 ====== | ||
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| ====== 6. Introduction and Chapter Objectives ====== | ====== 6. Introduction and Chapter Objectives ====== | ||
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| $${f} \left( t \right) = Ae^{-at}$$ | $${f} \left( t \right) = Ae^{-at}$$ | ||
| - | Where $a>0$. The function ${f}(t)$ is illustrated in Figure 6.18. The value of the function is //A// at $t = 0$ and decreases to zero as $t \rightarrow \infty$. As $t \rightarrow -infty$ the function increases without bound. The constant a dictates the rate at which the function decreases as time increases. | + | Where $a>0$. The function ${f}(t)$ is illustrated in Figure 6.18. The value of the function is //A// at $t = 0$ and decreases to zero as $t \rightarrow \infty$. As $t \rightarrow -\infty$ the function increases without bound. The constant a dictates the rate at which the function decreases as time increases. |
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| Or, using the unit step function to limit the function to positive values of time: | Or, using the unit step function to limit the function to positive values of time: | ||
| - | $${f} \left( t \right) = Ae\frac{1t}{\tau} \cdot u_o \left( t \right) | + | $${f} \left( t \right) = Ae\frac{-t}{\tau} \cdot u_o \left( t \right) |
| The function of equations (6.10) and (6.11) is illustrated in Figure 6.19. The time constant, $\tau$, is a positive number which dictates the rate at which the function will decay with time. When the time $t = \tau, {f} \left( t \right) Ae^{-1} = 0.368A$ and the function has decayed to 36.8% of its original value. In fact, __the function decreases by 36.8% every $\tau$ seconds__. Therefore, a signal with a small time constant decays more rapidly than a signal with a large time constant, as illustrated in Figure 6.20. | The function of equations (6.10) and (6.11) is illustrated in Figure 6.19. The time constant, $\tau$, is a positive number which dictates the rate at which the function will decay with time. When the time $t = \tau, {f} \left( t \right) Ae^{-1} = 0.368A$ and the function has decayed to 36.8% of its original value. In fact, __the function decreases by 36.8% every $\tau$ seconds__. Therefore, a signal with a small time constant decays more rapidly than a signal with a large time constant, as illustrated in Figure 6.20. | ||
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| * An exponential function, defined for $t>0$, is mathematically defined as: | * An exponential function, defined for $t>0$, is mathematically defined as: | ||
| - | $${f} \left( t \right) = AE^\frac{-t}{\tau} \cdot u_0 \left( t \right)$$ | + | $${f} \left( t \right) = Ae^\frac{-t}{\tau} \cdot u_0 \left( t \right)$$ |
| The function has an initial value, $A$, and a time constant, $\tau$. The time constant indicates how quickly the function decays; the value of the function decreases by 63.2% every $\tau$ seconds. Exponential functions are important to use because the solutions of linear, constant coefficient, | The function has an initial value, $A$, and a time constant, $\tau$. The time constant indicates how quickly the function decays; the value of the function decreases by 63.2% every $\tau$ seconds. Exponential functions are important to use because the solutions of linear, constant coefficient, | ||
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| - | We will denote the total magnetic flux created by the inductor by , as shown in Figure 6.28. For a linear inductor, the flux is proportional to the current passing through the wound wires. The constant of proportionality is the inductance, //L//: | + | We will denote the total magnetic flux created by the inductor by $\psi$, as shown in Figure 6.28. For a linear inductor, the flux is proportional to the current passing through the wound wires. The constant of proportionality is the inductance, //L//: |
| $$\psi(t) = Li(t) (Eq. 6.26)$$ | $$\psi(t) = Li(t) (Eq. 6.26)$$ | ||
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| * In the time period 0<t<1 msec, the voltage is constant and positive so the current will increase linearly. The total current change during this time period is the area under the voltage curve curve, divided by the inductance: $\frac{1}{0.01}(10V)(1x10^{-3} sec) = 1A$ | * In the time period 0<t<1 msec, the voltage is constant and positive so the current will increase linearly. The total current change during this time period is the area under the voltage curve curve, divided by the inductance: $\frac{1}{0.01}(10V)(1x10^{-3} sec) = 1A$ | ||
| * In the time period $1< | * In the time period $1< | ||
| - | * In the time period $2< | + | * In the time period $2< |
| * In the time period 3<t<4 seconds, the voltage is zero. The integral of zero over any time period is zero, so there is no change in current during this time range and the current remains constant at 0A. | * In the time period 3<t<4 seconds, the voltage is zero. The integral of zero over any time period is zero, so there is no change in current during this time range and the current remains constant at 0A. | ||
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