1
\$\begingroup\$

I am studying the book: Control system engineering for Norman S Nise.

I left the college few years ago, and I forgot lots of thing about Kirchhof's laws (Voltage and current).

I tried to solve the following exercise:

PROBLEM: Given the network of Figure 2.6(a), find the transfer function, h(s)/V(s).

Figure:

enter image description here

The author said in a previous part that:

[Sum of impedances around Mesh 1] I1 - [Sum of impedances common between Meshes] I2 = Sum of applied voltges around mesh 1

But as I remember, the both currents I1 and I2 are in the same directions, so it should + instead of - ?

And at the end he said that the transfer function is G(s) = I2/V(s). And as I remembered the transfer function should be Vc(s)/V(s).

And when we should use the Nodal analysis instead of Mesh analysis ? When there is a lots of calculations ?

Can anyone explain it to me ?

After transforming the components

enter image description here

\$\endgroup\$
2
\$\begingroup\$

But as I remember, the both currents I1 and I2 are in the same directions, so it should + instead of - ?

If you look at the total current through the inductor, then you see that \$I_1(s)\$ goes from top to bottom, while \$I_2(s)\$ goes from bottom to top. They go in opposite directions, hence the minus sign.

And at the end he said that the transfer function is G(s) = I2/V(s). And as I remembered the transfer function should be Vc(s)/V(s).

Transfer functions can be from either voltage or current, to a voltage or current. Any combination is possible, and it just depends on what you want to treat as the input and output. In this case, the author is calculating the transconductance, ie. the output current \$I_2(s)\$ for a given input voltage \$V(S)\$.

And when we should use the Nodal analysis instead of Mesh analysis ? When there is a lots of calculations?

Both Nodal analysis and Mesh analysis can be used to solve any circuit. But you usually can save yourself some work if you choose the method appropriately.

Generally:

  1. Mesh analysis uses the KVL equations, ie. it works with voltages. It is easier to use when there are ideal voltage sources in the circuit.

  2. Nodal analysis uses the KCL equations, meaning currents. If you have ideal current sources in your circuit, then this is probably easier.

  3. If you have both ideal voltage and current sources, then you will need to apply Thevenin/Norton equivalents, or you will need to use source-shifting. Or you can also use Modified Nodal analysis which is the method used in simulators.

  4. If you have many components/nodes, but only few loops, mesh analysis will lead to fewer equations, which is easier to solve.

  5. If you have a highly branching circuit (ie. lots of loops and not as many nodes), then Nodal analysis is preferred as it will lead to less equations.


Appendix

The Thevenin and Norton equivalents are shown in this figure:

Equivalents

They are exactly equivalent if the following relationship holds:

$$V_{Th} = R_N\cdot I_{N}$$

$$R_N = R_{Th}$$

Just make sure you can draw a circle around the structure, and that there are only two connections leaving/entering that circle. If that is not the case, then you can't replace them by one another.

So you can use this to turn voltage sources into current sources and vice-versa.

The equivalents can be used more generally, but this is how you'd use it for Nodal analysis or Mesh current analysis.

\$\endgroup\$
  • \$\begingroup\$ Well explained. And now I remembered that we should follow the current to check the sign. Anyway, can you edit your answer and explain to me how to use Thevenin/Northon ? \$\endgroup\$ – alim1990 Jun 28 '18 at 11:20
2
\$\begingroup\$

To determine transfer functions in a swift and efficient manner, the fast analytical circuits techniques or FACTs are an invaluable tool to learn. The basic principle is to determine the time constants of a circuit in certain conditions (excitation reduced to 0 and with a nulled response) through separate small sketches. With some skill, you can even determine the whole transfer function in your head, without writing a line of algebra.

For this circuit, the below drawing illustrates the steps. You start with \$s=0\$ for which the inductor is replaced by a short circuit and the capacitor is open-circuited. The gain in this mode is simply 0. Then, you reduce the excitation (\$V_{in}\$) to 0 V (a short circuit) and "look" through both energy-storing elements connections to determine the resistance.

enter image description here

The resistance you have is now combined with the energy-storing element you were observing to form the time constant: \$\tau=RC\$ or \$\tau=\frac{L}{R}\$. During the observation on one energy-storing element, the second is put in its dc state (shorted inductor and open-circuited capacitor). Because you have an inductor and a capacitor, this is a second-order circuit. The denominator of your transfer function is thus: \$D(s)=1+sb_1+s^2b_2\$ which can be reorganized as \$D(s)=1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2\$. You have \$b_1=\tau_1+\tau_2\$ and \$b_2=\tau_1\tau_{12}\$. The term \$\tau_{12}\$ implies that element 1 is set in its high-frequency state while you determine the resistance driving element 2.

The numerator \$N(s)\$ is defined as \$N(s)=H_0+s(H^1\tau_1+H^2\tau_2)+s^2H^{12}\tau_1\tau_{12}\$. You simply reuse the time constants you have already determined in \$D(s)\$ and you associate them with simple gains: \$H^1\$ is the gain when element 1 is set in its high-frequency state (the second is dc) and \$H^{12}\$ implies that both elements are in high-frequency state. In this example, some of the gain are 1, some 0 so its easy.

If you do the sketches ok, then the below Mathcad sheet shows the frequency response you obtain:

enter image description here

As you can see, no KVL or KCL, just inspection of simple drawings. If you made a mistake, it is easy to correct the error as you fix an intermediate step rather than the whole process in case of brute-force approach. Also, it is easy to factor the final expression and make it fit a low-entropy format in which the mid-band gain clearly appears (see \$H_6\$ in the sheet). Reorganizing the expression at the end is extremely important as the result will contain design goals such as the mid-band gain for instance.

These FACTs are truly an excellent tool to learn if you are determining transfer functions of any kind with passive or active circuits. This document will give you a smooth introduction to the subject. Once you'll taste them, you won't go back to the classical approach : )

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.