I've seen some different forms for the equation of a transfer function and wonder which one is most correct and why? Does it depend on type of filter etc.?

Let's take an example and its transfer function: $$H(w)=\frac{V_o(w)}{V_i(w)}=\frac{R_2jwL}{R_1(R_2+jwL)+R_2jwL}$$                         Filter.

With some algebra you can get it on two different forms:

\$H_1(w)=\bigl(\frac{R_2}{R_1+R_2}\bigr)\frac{jw}{jw+\bigl(\frac{R_1R_2}{L(R_1+R_2}\bigl)}\$ where \$\frac{R_1R_2}{L(R_1+R_2)}\$ is the pole (and \$jw\$ is a zero?).

\$H_2(w)=\bigl(\frac{L}{R_1}\bigr)\frac{jw}{1+j\Biggl(\frac{w}{\bigl(\frac{R_1R_2}{L(R_1+R_2)}\bigl)}\Biggr)}\$ where \$\frac{R_1R_2}{L(R_1+R_2)}\$ is the pole (and \$jw\$ is a zero?).

\$H_1(0)=0\$, \$H_2(0)=0\$, \$H_1(\infty)=\frac{R_2}{R_1+R_2}\$ and \$H_2(\infty)=\frac{R_2}{R_1+R_2}\$.


They are clearly the same (if I've calculated it all right), but which form is most correct and standard to use? Does it differ from type of filter, i.e low vs. high pass etc.? It seems easier to use \$H_1\$ to read of the constant \$\frac{R_2}{R_1+R_2}\$ as \$w\$ approach \$\infty\$, although I've seen the form av \$H_2\$ more often and it seems standard two try to get a "\$1+...\$" in the denominator?

Thank you!


3 Answers 3


I'd be considering the gain factor as important and, for a high pass filter the gain is: -

\$\dfrac{R_2}{R_1+R_2}\$ at higher frequencies. I'd call it "K".

Clearly also for this type of filter, the effective series resistance is the parallel value of R1 and R2 hence I'd call that "R". Then I'd end up with a transfer function like this: -


This makes more sense because it uses gain (K) and the time constant (\$\tau\$) explicitly. If you are talking poles and zeros, it's probably better to use "s" rather than \$j\omega\$.

  • \$\begingroup\$ Okey, so the first form gives me the gain K as w goes to infinity (high frequencies). What's the way of thinking to get to the most appropriate formula? Is there always a "jw" (or "s") alone in the numerator or might it sometime just be a "1"? \$\endgroup\$
    – Defindun
    Commented Mar 3, 2018 at 10:03
  • \$\begingroup\$ In a high pass filter there will always be an s or jw in the numerator. In a low pass filter (say you replaced the inductor with a capacitor) then there wouldn't be an s or jw in the numerator. The most appropriate formula is the one that is most conveniently used or represents a meaning you wish to impart to the reader. My answer imparts knowledge about gain and time constant but if a formula is correct (but re-arranged) it may suit a different purpose. \$\endgroup\$
    – Andy aka
    Commented Mar 3, 2018 at 10:08
  • \$\begingroup\$ Okey, I think that explains it, thank you! And the "reason" there's always a "s" or "jw" in the numerator in high pass filters is because H(w) must go towards 0 when w goes towards 0? That would be the definition of a high pass filter I guess. So, the best approach would be to simplify the function to one of the above forms, analyze it as H(0) and H(infinity) and perhaps H(poles/zeros), i.e. see how the function behaves. And that should not depend on the "type of form". Then, if you want to, you can adjust the formula in order to present it in the most meaningful way for task at hand. \$\endgroup\$
    – Defindun
    Commented Mar 3, 2018 at 10:34
  • \$\begingroup\$ Yes, correct and correct. But there's not a really a correct form; just alternatives. For instance if you looked at 2nd order equations (and this is your task should you choose to take it LOL) there are plenty of ways of representing them but folk tend to adopt a "mechanical eng. approach" that factors out natural_resonant_frequency and damping ratio just because of the historical connection with flywheels and springs and mass. \$\endgroup\$
    – Andy aka
    Commented Mar 3, 2018 at 10:44
  • \$\begingroup\$ Ah a down vote. Now why should that happen? \$\endgroup\$
    – Andy aka
    Commented Mar 11, 2018 at 22:54

The first version in in the classical form $$H(s)=N(s)/D(s)$$ where both are written as $$(a_1+s)(a_2+s)\cdots (a_n+s)$$.

The second form explicitely displays the pole and zero frequency and the DC gain: $$A_0 \frac{1}{(1+s/\omega_0)(1+s/\omega_1)}$$ etc.

As you said, both forms are identical and independent of filter. They are just two (equally valid) representations and equally correct.

In my opinion, the first one is more frequently used in control theory whereas the second more in circuits.

I personally prefer the second because you immediately see the poles and zeros and the DC gain.

If you go to higher order systems, you realize that for the first representation it is not as straight forward to "see" the DC gain at first glance. You have to multiply them out:

$$H(s)=\frac{(b_1+s)(b_2+s)}{(a_1+s)(a_2+s)} = \frac{b_1 b_2}{a_1 a_2} \frac{(1+\frac{s}{b_1})(1+\frac{s}{b_2})}{(1+\frac{s}{a_1})(1+\frac{s}{a_2})} $$

In your specific example, there is no DC gain since it has inductive character. However, L/R1 is the "differentiator gain".

In my opinion, the following example makes it more clear: Suppose you have an ideal integrator followed by a doublet:

$$ H(s) = \frac{1}{s} K \frac{s + z_1}{s + p_1} $$

Assume the doublet occurs after the integration frequency (e.g. a parasitic pole and a zero due to finite output resistance of an opamp). Can you read off from this immediately how much "V per s" the integrator integrates from the formula above?

However, if I factor z1 and p1:

$$ H(s) = \frac{1}{s} \frac{K z_1}{p_1} \frac{1 + \frac{s}{z_1}}{1 + \frac{s}{p_1}} $$

Note that for $$s \gg \{z_1,p_1\}$$:

$$ \frac{1 + \frac{s}{z_1}}{1 + \frac{s}{p_1}} \approx 1 $$

Now you can quickly see that the integration constant is K z1/p1.

  • \$\begingroup\$ Okey, so in the second form A_0 is the DC gain and is L/R_1, i.e. the value at w close to 0. Whereas in the first form the factor R_2/(R_1+R_2) gives the gain as w goes to infinity (high frequencies)? And can't I see the poles and zeros in the first form as well? It's the same in the denominator? R_1R_2/(L(R_1+R_2)). \$\endgroup\$
    – Defindun
    Commented Mar 3, 2018 at 9:57
  • \$\begingroup\$ In your case you do not have a "DC gain" but you see the "differentiation constant" which you would not immediately see from your first from. I updated my answer to make it a bit more clear. \$\endgroup\$
    – divB
    Commented Mar 3, 2018 at 11:01

Using the fast analytical circuits techniques or FACTs is the best way to form the right low-entropy formula in the smallest amount of time. It can be shown that any first-order circuit can be put under the following form: \$H(s)=\frac{H_0+sH^1\tau_1}{1+s\tau_1}\$ in which \$H_0\$ is the gain of the circuit at dc (\$s = 0\$), \$H^1\$ is the gain when the energy-storing element (here the inductor) is set in its high-frequency state (an open circuit for the inductor) and finally, \$\tau_1\$ is the time constant determined when the excitation source is set to 0 V.

The below drawing shows the 3 steps to determine the transfer function without writing a single line of algebra:

enter image description here

In dc, the inductor is a short circuit thus \$H_0 = 0\$. Now, set \$V_{in}\$ to 0 V and replace it by a short circuit to determine the resistance "seen" between the inductor's terminals: you "see" \$R=R_1||R_2\$ immediately leading to the circuit time constant \$\tau_1=\frac{L_1}{R_1||R_2}\$. We know that the pole is the inverse of the time constant in a 1st-order circuit, therefore \$\omega_p=\frac{R_1||R_2}{L_1}\$. Now, set \$L_1\$ in high-frequency state and solve the transfer function by immediate inspection: \$H^1=\frac{R_2}{R_1+R_2}\$. This is it, we have everything we need to determine the transfer function:


However, this is not the correct way of writing this transfer function. Factor \$s\tau_1\$ in the numerator and the denominator and you end up with the following expression which is truly a low-entropy formula:


This is the correct way of writing this transfer function, showing the high-frequency plateau gain (your design goal) and the inverted pole in the denominator. This is the most compact form you can think of which gives you the insight you need for your design goals. The below Mathcad sheet shows the resulting plots in various forms:

enter image description here

enter image description here

The fast analytical techniques are truly well adapted to solving these linear circuits (passive or active) up to the order \$n\$ as they always deliver this low-entropy format in the least amount of time. Even better, you often get there without writing a line of algebra. A tutorial is here for those who would like to further dig the subject (highly recommended for students and engineers).


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