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Linear Networks#

Superposition Principle#

Theorem: The Superposition Principle

Let \(\mathcal{N}\) be a linear network with

\[\boldsymbol{M}\boldsymbol{v} + \boldsymbol{N}\boldsymbol{i} = \boldsymbol{e}.\]

Each component of the branch voltage vector and each component of the branch current vector can be expressed as a linear combination of the components of \(\boldsymbol{e}\).

Proof

TODO

The superposition principle is powerful because it allows us to look at all ideal sources one by one and ignore the rest. The overall result is the sum of these partial results.

Algorithm: The Superposition Principle

  1. For each ideal source \(j\), calculate the branch voltage vector \(\boldsymbol{v}^{(j)}\) and the branch current vector \(\boldsymbol{i}^{(j)}\) by assuming that all other ideal sources are zero.
  1. The real branch voltage vector \(\boldsymbol{v}\) is the sum of the branch voltage vectors \(\boldsymbol{v}^{(j)}\) and the real branch current vector \(\boldsymbol{i}\) is the sum of the branch current vectors \(\boldsymbol{i}^{(j)}\).
\[\boldsymbol{v} = \sum_j \boldsymbol{v}^{(j)} \qquad \boldsymbol{i} = \sum_j \boldsymbol{i}^{(j)}\]

One-Port Equivalents#

Theorem: Hemholtz-Thévenin Equivalent

Every linear resistive network with only two accessible terminals which form a port is equivalent to an ideal voltage source and a linear resistor in series.

Hemholtz-Thévenin Equivalent

Definition: Internal Resistance

We call \(R_{\text{H-T}}\) the internal resistance of the network.

Proof

TODO

Theorem: Mayer-Norton Equivalent

Every linear resistive network with only two accessible terminals which form a port is equivalent to an ideal current source and a linear resistor in parallel.

Mayer-Norton Equivalent

Definition: Internal Conductance

We call \(G_{\text{M-N}}\) the internal conductance of the network.

Proof

TODO

Theorem:

The internal resistance and internal conductance of a linear network are related as follows:

\[R_{\text{H-T}} = - \frac{v_{\text{H-T}}(t)}{i_{\text{M-N}}(t)} \qquad G_{\text{M-N}} = - \frac{i_{\text{M-N}}(t)}{v_{\text{H-T}}(t)}\]
\[R_{\text{H-T}} = \frac{1}{G_{\text{M-N}}} \qquad G_{\text{M-N}} = \frac{1}{R_{\text{H-T}}}\]
Proof

TODO