So admittedly, it's been about 5 years since I've dealt with any kind of circuits and now I am going back and taking a controls course. On the very first problem I am doing, my answer is not agreeing with that in the back of the book. The controls book that I will be referencing is Control Systems Engineering, by Norman S. Nise 7th ed.

As far as I know I am doing everything exactly as described in the book (ch. 2). Anyway, here is the question (and I will show all of the work I have done as well).

Find the transfer function, enter image description here, for each network shown in Figure P2.3. [Section: 2.4].


  1. The first step is to write out all of the associated node equations, so I convert to the S-domain by:

Step #1

  1. My next step, is to represent everything as an admittance.

Step #2

  1. Now, I solve and get the following.

Step #3


In the book, they seem to be getting...


Am I actually going wrong somewhere, or is there an error with the book solution?


1 Answer 1


Consider the parallel RL network. The admittance is $$Y = \frac{1}{Z} = \frac{1}{s} + 0.5 \to Z = \frac{s}{0.5s + 1}$$

Now apply the voltage divider equation and multiply the numerator and denominator by 0.5s + 1 $$H = \frac{Z}{Z + 1} = \frac{s}{1.5s + 1} = \frac{2s}{3s + 2}$$

Your answer seems correct.

  • \$\begingroup\$ Okay, maybe it's not just me. It's just that it's been so long I don't quite have faith in my answer, and this is the very first problem on the very first homework (so what would be the chance of the book solution being wrong) and thus I kind of figured it was just me. \$\endgroup\$
    – Snoop
    May 28, 2016 at 0:10
  • \$\begingroup\$ Well, think about what is going on. The coil is a short circuit at DC. So at s=0 the output should be zero. As s approaches infinity, the coil's impedance will approach infinity reducing the system to a simple voltage divider. Can you see why their answer is wrong now? \$\endgroup\$
    – user110971
    May 28, 2016 at 0:15
  • \$\begingroup\$ Are you saying that because their answer is 1/(s + 2) that their transfer function will approach 0 as s approaches infinity? Does that have something to do with it? \$\endgroup\$
    – Snoop
    May 28, 2016 at 0:18
  • \$\begingroup\$ Yes. And when s=0 their answer also doesn't make sense. You get that kind of transfer function when you have an RC circuit. \$\endgroup\$
    – user110971
    May 28, 2016 at 0:20
  • \$\begingroup\$ Okay, that's very helpful. Most likely you are right, but just because I really don't know the material I will wait like a day and see what anyone else says (or doesn't) before accepting an answer. \$\endgroup\$
    – Snoop
    May 28, 2016 at 0:23

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