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| learn:courses:real-analog-chapter-6:start [2017/04/27 20:47] – [Exponential Functions] Martha | learn:courses:real-analog-chapter-6:start [2022/10/04 11:38] (current) – external edit 127.0.0.1 | ||
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| ====== Real Analog: Chapter 6 ====== | ====== Real Analog: Chapter 6 ====== | ||
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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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