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 \light{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 \green{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 \red{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, \yellow{Z_1} & \cyan{Z_2}, 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 \red{Z_n}
by nulling the output.
Notation.
A small section on the notation used is required, since it can get pretty confusing.
\blue{H_0} - DC gain of the circuit.
\green{H^i} - 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).
\green{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 \green{H^{ij}} = \green{H^{ji}}.
\red{\tau_i} - The time constant of element numbered 'i' with all other removed elements in their DC state.
\red{\tau_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(\yellow{s}) as:
H(\yellow s) = \frac{N(\yellow s)}{D(\yellow s)}.
The denominator D(\yellow{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 \red{\tau_2^1} or \red{\tau_1^2} depending on which is easier to compute.
The numerator N(\yellow{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 C_1 and C_2 as the extra elements.
Let us find \blue{H_0}. Both the capacitors are set to their DC state.
Clearly we have \magenta{V_o} = \blue{V_i}. Which means:
\blue{H_0} = 1.
Let us find each of the time constants. Starting with \red{\tau_1}, we short the voltage source and set C_2 to its DC state.
From the above diagram, we get:
\red{\tau_1} = R_1C_1.
Similarily for \red{\tau_2}...
we get:
\red{\tau_2} = (R_1 + R_2)C_2.
Now we should find either \red{\tau_2^1} or \red{\tau_1^2}. 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 \red{\tau_2^1}, we keep C_2 as it is, and put C_1 in its
high frequency state.
Notice that in the above diagram, R_1 plays no role (both of the terminals of R_1 are at same potential). Hence,
\red{\tau_2^1} = R_2C_2.
Similarily for \red{\tau_1^2}, we keep C_1 but set C_2 in its high frequency state.
This time, R_1 and R_2 are parallel:
\red{\tau_1^2} = \left(R_1\parallel R_2\right)C_1.
Now let us compute all the high freqency gains, \green{H^1}, \green{H^2} and \green{H^{12}}.
Notice how in all three diagrams (\green{H^1}, \green{H^2}, \green{H^{12}} computations respectively) the output
is shorted. Which means \magenta{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(\yellow{s}) and D(\yellow{s}) are written.
The denominator, D(\yellow{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 [T^i].
For b_1, 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 b_2, 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\lt j}\red{\tau_i\tau_j^i}.
Note that due to the symmetry \red{\tau_i\tau_j^i} = \red{\tau_j\tau_i^j}.
For b_3, 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\lt j\lt k}\red{\tau_i\tau_j^i\tau_k^{ij}}.
Here, \red{\tau_i^{jk}} = \red{\tau_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 \red{\tau_i\tau_j^i\tau_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(\yellow{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.
\blue{H_0} is the usual DC gain (each element is set to its DC state).
a_1 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}.
a_2 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\lt j}\red{\tau_i\tau_j^i}\green{H^{ij}}.
a_3 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\lt j\lt 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.
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.)