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| learn:courses:real-analog-chapter-4:start [2017/04/21 22:51] – [Norton’s Theorem] Martha | learn:courses:real-analog-chapter-4:start [2022/10/04 11:38] (current) – external edit 127.0.0.1 | ||
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| ====== Real Analog: Chapter 4 ====== | ====== Real Analog: Chapter 4 ====== | ||
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| - | We previously wrote mesh equations for this circuit (for a specific value of $V_{in}$) in [[https:// | + | We previously wrote mesh equations for this circuit (for a specific value of $V_{in}$) in [[/ |
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| $$V= 2 \left( i_1 - 3V_x \right)$$ | $$V= 2 \left( i_1 - 3V_x \right)$$ | ||
| - | The above four equations provide an inpuot-output description of the circuit. If desired, they can be combined to eliminate all variables except $V_{in}$ and $V$ and re-written in the form $V = f \{ V_{in} \}$ per equation (1). __Note that all information about the original system, except the relationship between the input and output signals, is lost once we do this__. | + | The above four equations provide an inppot-output description of the circuit. If desired, they can be combined to eliminate all variables except $V_{in}$ and $V$ and re-written in the form $V = f \{ V_{in} \}$ per equation (1). __Note that all information about the original system, except the relationship between the input and output signals, is lost once we do this__. |
| The system-level block diagram for the circuit might then be drawn as: | The system-level block diagram for the circuit might then be drawn as: | ||
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| - | Applying KCL at $v_0$, we obtain: $-2A + \frac{v_0 - 6V}{6\Omega} + \frac{v_0}{3|Omega} = 0 \Rightarrow v_0 = v_{OC} = 6V$. Thus, the Thévenin equivalent circuit is on the left below. Re-introducing the load resistance, as shown on the right below, allows us to easily analyze the overall circuit. | + | Applying KCL at $v_0$, we obtain: $-2A + \frac{v_0 - 6V}{6\Omega} + \frac{v_0}{3\Omega} = 0 \Rightarrow v_0 = v_{OC} = 6V$. Thus, the Thévenin equivalent circuit is on the left below. Re-introducing the load resistance, as shown on the right below, allows us to easily analyze the overall circuit. |
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