The n-Extra Element Theorem.

This article is a continuation of the previous one.

When the extra element is a resistor...

Continuing where I left off from the previous article, let me address a question you might have... According to the EET, when a capacitor or inductor is removed, we can write the generalized (first order) transfer function as: H(\yellow s) = \frac{\blue{H_0} + \green{H^1}\red{\tau_1}\yellow s}{1 + \red{\tau_1}\yellow s}.

But what if a resistor is removed in the circuit? We can use this equivalent form of the EET (where Z = R): H(\yellow s) = \frac{\left.H\right\rvert_{\light Z=0} + \left.H\right\rvert_{\light Z\to\infty}\frac{\light Z}{\green{Z_d}}}{1 + \frac{\light Z}{\green{Z_d}}}. Where Z_d is the impedance of the network with the input zeroed (the Thevenin's impedance looking into the impedance Z). This formula also does not require calculation of the null impedance Z_n.

Now onto the nEET.

The 2-Extra Element Theorem.

While the EET is useful in a lot of cases, it falls short when there are multiple reactive elements (capacitors / inductors) in a circuit. The n-Extra Element Theorem can be applied in these cases. It is a generalization of the EET. We consider the cases when n = 2, then we can generalize it later.

The basic idea here is that we remove two elements and apply the Extra Element Theorem twice. This raises a question: How do we account for the effect of the second element while applying EET to the first?

There are four possible reference states when we have two elements, Z₁ & Z₂, to be removed: \begin{align} \yellow{Z_1} = 0,\ &\cyan{Z_2} = 0. \\ \yellow{Z_1} = 0,\ &\cyan{Z_2} \to \infty. \\ \yellow{Z_1} \to \infty,\ &\cyan{Z_2} = 0. \\ \yellow{Z_1} \to \infty,\ &\cyan{Z_2} \to \infty. \end{align} We can use any one of these reference states and write the final expression accordingly. However, like last time, it would be better if we could directly write a generalized second order transfer function that we can use to remove any combination of 2 inductors and capacitors and skip computing any Z_n by nulling the output.

Notation.

A small section on the notation used is required, since it can get pretty confusing.

• H₀ - DC gain of the circuit.
• Hⁱ - Gain of the circuit when element numbered 'i' is in its high frequency state (capacitors shorted, inductors opened), with the others in their DC state (capacitors opened and inductors shorted).
• H^ij - Gain of the circuit when elements numbered 'i' and 'j' are in their high frequency states with the other elements in their DC state. Note that H^ij = H^ji.
• τ_i - The time constant of element numbered 'i' with all other removed elements in their DC state.
• τ_i^j - The time constant of element numbered 'i' with element numbered 'j' in its high frequency state and all other removed elements in their DC state.

Generalized second order transfer function.

As with the generalized first order transfer function, we expect the generalized transfer function to be of the form: H(\yellow s) = \frac{\blue{\text{D.C. Gain}} + \green{\text{H.F. Gain}}\times\red{\text{Time Constant}}\times\yellow s + \green{\text{H.F. Gain}}\times\red{\text{Time Constant}^2}\times\yellow s^2}{1 + \red{\text{Time Constant}}\times\yellow s + \red{\text{Time Constant}^2}\times\yellow s^2}.

This is almost correct but this time we have to account for "interaction terms" between the two removed elements, since time constant of an element can change depending on the presence of other elements. Let us write our final transfer function, H(s) as: H(\yellow s) = \frac{N(\yellow s)}{D(\yellow s)}.

The denominator D(s) is easy to write: D(\yellow s) = 1 + \left(\red{\tau_1} + \red{\tau_2}\right)\yellow s + \red{\tau_1\tau_2^1}\yellow s^2. Note that \red{\tau_1\tau_2^1} = \red{\tau_2\tau_1^2}. This means you can choose to find τ₂¹ or τ₁² depending on which is easier to compute.

The numerator N(s) is similar to the denominator but you will notice that the time constants have the high frequency gains multiplied with them: N(\yellow s) = \blue{H_0} + \left(\green{H^1}\red{\tau_1} + \green{H^2}\red{\tau_2}\right)\yellow s + \green{H^{12}}\red{\tau_1\tau_2^1}\yellow s^2. This means that you don't have to recompute these time constants again. You just need the high frequency gain terms.

Thus we have: H(\yellow s) = \frac{\blue{H_0} + \left(\green{H^1}\red{\tau_1} + \green{H^2}\red{\tau_2}\right)\yellow s + \green{H^{12}}\red{\tau_1\tau_2^1}\yellow s^2}{1 + \left(\red{\tau_1} + \red{\tau_2}\right)\yellow s + \red{\tau_1\tau_2^1}\yellow s^2}.

Example - Second order RC low-pass filter.

Let us do one example problem to really understand how to use this formula. Given below is a standard passive second order RC low-pass filter. Let us find the transfer function of this network using 2EET. We shall take capacitors C₁ and C₂ as the extra elements.

Let us find H₀. Both the capacitors are set to their DC state.

Clearly we have V_o = V_i. Which means: \blue{H_0} = 1. Let us find each of the time constants. Starting with τ₁, we short the voltage source and set C₂ to its DC state.

From the above diagram, we get: \red{\tau_1} = R_1C_1. Similarily for τ₂...

we get: \red{\tau_2} = (R_1 + R_2)C_2. Now we should find either τ₁² or τ₂¹. For completeness sake let us find both. But in actual circuits you should find only one of them (whichever is easier to compute). Starting with τ₂¹, we keep C₂ as it is, and put C₁ in its high frequency state.

Notice that in the above diagram, R₁ plays no role (both of the terminals of R₁ are at same potential). Hence, \red{\tau_2^1} = R_2C_2. Similarily for τ₁², we keep C₁ but set C₂ in its high frequency state.

This time, R₁ and R₂ are parallel: \red{\tau_1^2} = \left(R_1\parallel R_2\right)C_1.

Now let us compute all the high freqency gains, H¹, H² and H¹².

Notice how in all three diagrams (H¹, H², H¹² computations respectively) the output is shorted. Which means V_o = 0. Hence: \green{H^1} = \green{H^2} = \green{H^{12}} = 0.

Substituting the values in our generalized transfer function: H(\yellow s) = \frac{\blue{H_0} + \left(\green{H^1}\red{\tau_1} + \green{H^2}\red{\tau_2}\right)\yellow s + \green{H^{12}}\red{\tau_1\tau_2^1}\yellow s^2}{1 + \left(\red{\tau_1} + \red{\tau_2}\right)\yellow s + \red{\tau_1\tau_2^1}\yellow s^2}. We get: H(\yellow s) = \frac{\blue{1} + \left(\green{0}\times\red{R_1C_1} + \green{0}\times\red{\left(R_1+R_2\right)C_2}\right)\yellow s + \green{0}\times\red{R_1C_1\times R_2C_2}\yellow s^2}{1 + \left(\red{R_1C_1} + \red{\left(R_1 + R_2\right)C_2}\right)\yellow s + \red{R_1C_1\times R_2C_2}\yellow s^2}. H(\yellow s) = \frac{1}{1 + \left(R_1C_1 + \left(R_1 + R_2\right)C_2\right)\yellow s + R_1R_2C_1C_2\yellow s^2}.

The n-Extra Element Theorem.

The n-Extra Element Theorem is now very similar to the 2EET case we showed. Let us discuss how to find the transfer function for any order network. We again let: H(\yellow s) = \frac{N(\yellow s)}{D(\yellow s)}. This time, n elements are removed from the network. Let us discuss how N(s) and D(s) are written.

The denominator, D(s).

The denominator of an n-order network will be an n-order polynomial in s. D(\yellow s) = 1 + b_1\yellow s^1 + b_2\yellow s^2 + \cdots + b_n\yellow s^n. Note that the dimensions of b_i are [Tⁱ].

For b₁, you would sum each time constant: b_1 = \red{\tau_1} + \red{\tau_2} + \cdots + \red{\tau_n} = \sum_{i}\red{\tau_i}.

Now for b₂, you would have to consider the sum of combinations of products of two time constants: b_2 = \red{\tau_1\tau_2^1} + \red{\tau_1\tau_3^1} + \red{\tau_2\tau_3^2} + \red{\tau_2\tau_4^2} + \cdots = \sum_{i<j}\red{\tau_i\tau_j^i}. Note that due to the symmetry τ_iτ_jⁱ = τ_jτ_i^j.

For b₃, you have to consider the sum of combinations of products of three time constants: b_3 = \red{\tau_1\tau_2^1\tau_3^{12}} + \red{\tau_2\tau_3^2\tau_4^{23}} + \cdots = \sum_{i<j<k}\red{\tau_i\tau_j^i\tau_k^{ij}}. Here, τ_i^jk = τ_i^kj, since it doesn't matter if elements numbered 'j' and 'k' are set to their high frequency state or elements numbered 'k' and 'j' are set to their high frequency state with the others being in their DC state. Obviously order doesn't matter here. However when constructing each term, we follow the order τ_iτ_jⁱτ_k^ij. Using these two facts we can write: \begin{align} \red{\tau_k^{ij}} &= \red{\tau_k^{ji}}, \\ \implies \red{\tau_i\tau_j^i\tau_k^{ij}} &= \red{\tau_j\tau_i^j\tau_k^{ji}}. \end{align} Which means you can choose the term which is easier to compute, and you can get equivalent terms by shuffling the indicies.

You can probably see the pattern now. This pattern continues till b_n, which turns out to be: b_n = \red{\tau_1\tau_2^1\tau_3^{12}\cdots\tau_n^{1,2,3,\cdots,(n-1)}}.

The numerator, N(s).

The numerator has a similar structure to the denominator, but each term containing a time constant is multiplied by a corresponding high frequency gain term. N(\yellow s) = \blue{H_0} + a_1\yellow s^1 + a_2\yellow s^2 + \cdots + a_n\yellow s^n.

H₀ is the usual DC gain (each element is set to its DC state).

a₁ is the sum of each time constant term multiplied by the corresponding high frequency gain term. a_1 = \red{\tau_1}\green{H^1} + \red{\tau_2}\green{H^2} + \cdots + \red{\tau_n}\green{H^n} = \sum_{i}\red{\tau_i}\green{H^i}.

a₂ is the sum of products of two time constants along with the corresponding high frequency gain term. a_2 = \red{\tau_1\tau_2^1}\green{H^{12}} + \red{\tau_1\tau_3^1}\green{H^{13}} + \red{\tau_2\tau_3^2}\green{H^{23}} + \red{\tau_2\tau_4^2}\green{H^{24}} + \cdots = \sum_{i<j}\red{\tau_i\tau_j^i}\green{H^{ij}}.

a₃ is the sum of products of three time constants along with the corresponding high frequency gain term. a_3 = \red{\tau_1\tau_2^1\tau_3^{12}}\green{H^{123}} + \red{\tau_2\tau_3^2\tau_4^{23}}\green{H^{234}} + \cdots = \sum_{i<j<k}\red{\tau_i\tau_j^i\tau_k^{ij}}\green{H^{ijk}}.

You get the pattern... Finally a_n will be: a_n = \red{\tau_1\tau_2^1\tau_3^{12}\cdots\tau_n^{1,2,3,\cdots,(n-1)}}\green{H^{1,2,3,\cdots,(n-1)}}.

Resultant transfer functions of various orders.

Here are the final generalized transfer functions till the fourth order for reference.

First order.

H(\yellow s) = \frac{\blue{H_0} + \green{H^1}\red{\tau_1}\yellow s}{1 + \red{\tau_1}\yellow s}.

Second order.

H(\yellow s) = \frac{\blue{H_0} + \left(\green{H^1}\red{\tau_1} + \green{H^2}\red{\tau_2}\right)\yellow s + \green{H^{12}}\red{\tau_1\tau_2^1}\yellow s^2}{1 + \left(\red{\tau_1} + \red{\tau_2}\right)\yellow s + \red{\tau_1\tau_2^1}\yellow s^2}.

Third order.

H(\yellow s) = \frac{\blue{H_0} + \left(\green{H^1}\red{\tau_1} + \green{H^2}\red{\tau_2} + \green{H^3}\red{\tau_3}\right)\yellow s + \left(\green{H^{12}}\red{\tau_1\tau_2^1} + \green{H^{13}}\red{\tau_1\tau_3^1} + \green{H^{23}}\red{\tau_2\tau_3^2}\right)\yellow s^2 + \green{H^{123}}\red{\tau_1\tau_2^1\tau_3^{12}}\yellow s^3}{1 + \left(\red{\tau_1} + \red{\tau_2} + \red{\tau_3}\right)\yellow s + \left(\red{\tau_1\tau_2^1} + \red{\tau_1\tau_3^1} + \red{\tau_2\tau_3^2}\right)\yellow s^2 + \red{\tau_1\tau_2^1\tau_3^{12}}\yellow s^3}

Fourth order.

For the fourth order, we define H(s) = N(s) / D(s), where:

\begin{align} N(\yellow s) = \blue{H_0} &+ \left(\green{H^1}\red{\tau_1} + \green{H^2}\red{\tau_2} + \green{H^3}\red{\tau_3} + \green{H^4}\red{\tau_4}\right)\yellow s \\ &+ \left(\green{H^{12}}\red{\tau_1\tau_2^1} + \green{H^{13}}\red{\tau_1\tau_3^1} + \green{H^{14}}\red{\tau_1\tau_4^1} + \green{H^{23}}\red{\tau_2\tau_3^2} + \green{H^{24}}\red{\tau_2\tau_4^2} + \green{H^{34}}\red{\tau_3\tau_4^3}\right)\yellow s^2 \\ &+ \left(\green{H^{123}}\red{\tau_1\tau_2^1\tau_3^{12}} + \green{H^{124}}\red{\tau_1\tau_2^1\tau_4^{12}} + \green{H^{134}}\red{\tau_1\tau_3^1\tau_4^{13}} + \green{H^{234}}\red{\tau_2\tau_3^2\tau_4^{23}}\right)\yellow s^3 \\ &+ \green{H^{1234}}\red{\tau_1\tau_2^1\tau_3^{12}\tau_4^{123}}\yellow s^4 \end{align}

\begin{align} D(\yellow s) = 1 &+ \left(\red{\tau_1} + \red{\tau_2} + \red{\tau_3} + \red{\tau_4}\right)\yellow s \\ &+ \left(\red{\tau_1\tau_2^1} + \red{\tau_1\tau_3^1} + \red{\tau_1\tau_4^1} + \red{\tau_2\tau_3^2} + \red{\tau_2\tau_4^2} + \red{\tau_3\tau_4^3}\right)\yellow s^2 \\ &+ \left(\red{\tau_1\tau_2^1\tau_3^{12}} + \red{\tau_1\tau_2^1\tau_4^{12}} + \red{\tau_1\tau_3^1\tau_4^{13}} + \red{\tau_2\tau_3^2\tau_4^{23}}\right)\yellow s^3 \\ &+ \red{\tau_1\tau_2^1\tau_3^{12}\tau_4^{123}}\yellow s^4 \end{align}

Now what?

This concludes the article. We did not solve a 3rd order network as the article is already getting long. But you can always do that.

Why do we care about these coefficients (a_i, b_i)? Actually these coefficients play an important role in determining many factors of the system. nEET gives a direct relation between individual components of the system and these coefficients which can be very helpful for desiging circuits.

Finally, an excerise for the reader: what if both a capacitor and a resistor are removed from a network? (Hint: Check the beginning of the article for inspiration.)