I came across this block in a larger circuit:


With a transfer function (in pole-zero form) of; $$ H(s) = \frac{s+30000}{s+5000} $$ which was found using KCL on the inverting terminal of the op-amp.

As I am not used to finding transfer functions using conventional circuit analysis and I wanted to find the "pure" transfer function (i.e. the general function not the pole-zero representation). However I found that substituting my transfer function in the place of the pole-zero, with a step response is not quite the same. (It also does not match simulation outputs compared to the pole-zero function)

My calculations are as follows:

$$ V^{+}=V_{in}=V^{-}=V_{out}\times\frac{R_{2}}{\frac{1}{sC}||R_{1}+R_{2}} $$ $$ \frac{V_{out}}{V_{in}}=\frac{\frac{\frac{R_{1}}{sC}}{\frac{1}{sC}+R_{1}}+R_{2}}{R_{2}}=1 + \frac{\frac{R_{1}}{sC}}{R_{2}(\frac{1}{sC}+R_{1})}=1+\frac{R_{1}}{R_{2}}\times\frac{{sC}}{{sC}}\times\frac{1}{1+R_{1}Cs} $$ From this I found the final transfer function to be: $$ H(s)=K\frac{1}{1+R_{1}Cs},\:\:\text{where}\:\:K=1+\frac{R_{1}}{R_{2}}\:\:\text{and}\:\:\omega_{p}=\frac{1}{R_{1}C} $$ The K value was derived from the DC gain of a non-inverting amplifier and the pole frequency equation was derived from the relationship shown here: https://en.wikipedia.org/wiki/Cutoff_frequency

For the transfer function of the larger circuit; this block's transfer function is multiplied with another (an RL filter is connected to the input of the op-amp); $$ T(s)=\frac{s}{s+\frac{R}{L}},\:\:R=80\Omega,\:\:L=10\text{mH},\:\:T(s)=\frac{s}{s+8000} $$ $$ H(s)\times T(s)=\frac{s(s+30000)}{(s+5000)(s+8000)} - \text{Using pole-zero function} $$ $$H(s)\times T(s)=\frac{30000s}{(s+5000)(s+8000)} - \text{Using transfer function} $$ The transfer function was then given a step function \$ \frac{0.6}{s} \$ and then simplified with partial fractions: $$ H(s)=\frac{5}{s+5000}-\frac{4.4}{s+8000} - \text{Using pole-zero function} $$ $$ H(s)=\frac{6}{s+5000}-\frac{6}{s+8000} - \text{Using transfer function} $$ Here is the graph of the two inverse laplace functions' step response: enter image description here

My question is this; is the "pure" transfer function I derived wrong or is there some error made afterwards that results in an incorrect final graph compared to the pole-zero one? The pole frequency equation is correct and the general form of the transfer function matches a low-pass so I am most confused.


3 Answers 3


The transfer function of the non-inverting circuit (which is no lowpass) is not correct. There is a simple math error.

The correct expression is:

Vout/Vin=1+ (R1/R2)*[1/(1+sR1C1)]

Hence, for s approaching infinity, the transfer function is approaching "1" (and not zero).

  • \$\begingroup\$ So instead of \$K=1+\frac{R_{1}}{R_{2}}\:\text{it should be}\:H(s)=1+K\frac{1}{1+sR_{1}C},\:\text{where}\:K=\frac{R_{1}}{R_{2}}\$? \$\endgroup\$
    – NBoss
    Sep 27, 2019 at 9:21
  • \$\begingroup\$ The Vout/Vin in your answer is the same as the Vin/Vout as OP!! \$\endgroup\$
    – Huisman
    Sep 27, 2019 at 10:47
  • \$\begingroup\$ The zero is missing in your expression LvW. \$\endgroup\$ Jun 25, 2020 at 11:25
  • \$\begingroup\$ @Verbal Kint...do you mean that the transfer function is not correct? I do not understand your comment. Of course, the zero is contained in the expression - but not visible without rewriting the formula. \$\endgroup\$
    – LvW
    Jun 25, 2020 at 15:03
  • \$\begingroup\$ False alarm, I missed the parenthesis after the +: entschuldigung : ) \$\endgroup\$ Jun 25, 2020 at 15:30

Your transfer function \$ \frac {V_{out} }{ V_{in} } \$ is correct.

However, the error is in the following:

$$H(s)=1+\frac{R_{1}}{R_{2}} \cdot \frac{1}{1+R_{1}Cs} \neq \bigg(1+\frac{R_{1}}{R_{2}} \bigg)\frac{1}{1+R_{1}Cs}$$

You are also right that the DC gain is

$$H(0)=1+\frac{R_{1}}{R_{2}} \cdot \frac{1}{1+R_{1}Cs} = 1+\frac{R_{1}}{R_{2}}$$

There is a pole1) at \$ \omega_p = \frac{ 1 }{ R_1C } \$. You can not simply conclude that from the transfer function above. Better to rewrite the transfer function as a \$ \text{gain factor} \times \frac{ \text{polynomial with zeros} }{ \text{polynomial with poles} }\$:

$$H(s) = 1 + \frac{R_1}{R_2} \cdot \frac{ 1 }{ 1+R_{1}Cs }$$

$$H(s) = 1 + \frac{ R_1 }{ R_2+R_1 R_2 Cs }$$

$$H(s) = \frac{ R_2+R_1 R_2 Cs }{ R_2+R_1 R_2 Cs } + \frac{ R_1 }{ R_2+R_1 R_2 Cs }$$

$$H(s) = \frac{ 1}{ R_2} \frac{ (R_1 + R_2) + R_1 R_2 Cs }{ 1 + R_1 Cs }$$

The pole lies indeed at $$ s = \frac{ 1 }{ R_1 C } $$

1) As LvW correctly mentions the cut-off frequency \$ \omega_c\$ is NOT identical to the pole frequency \$ \omega_p=\frac{ 1 }{ R_1 C } \$. This would be true for a real lowpass only.

  • \$\begingroup\$ So what would the K value of the transfer function be in this case? \$\endgroup\$
    – NBoss
    Sep 27, 2019 at 11:32
  • \$\begingroup\$ @NBoss What is K? The gain of the transfer function? \$\endgroup\$
    – Huisman
    Sep 27, 2019 at 11:52
  • \$\begingroup\$ Yes, for an inverting amplifier transfer function the K is usually \$-\frac{R_{2}}{R_{1}}\$ is there such as K value in this case? \$\endgroup\$
    – NBoss
    Sep 27, 2019 at 11:55
  • 1
    \$\begingroup\$ @Huisman....you are wrong. Please check the unit of your pole frequency expression. The pole frequency is as given by NBoss. \$\endgroup\$
    – LvW
    Sep 27, 2019 at 12:10
  • \$\begingroup\$ @LvW thanks, I see. Thanks for pointing out. \$\endgroup\$
    – Huisman
    Sep 27, 2019 at 12:18

This simple circuit features a pole and a zero. It is important to express the transfer function in a low-entropy form so that the pole and the zero clearly appear as well as the dc gain. The way to write it in this way would be: \$H(s)=H_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\$.

There are many ways to determine this expression but I personally favor the fast analytical circuits techniques or FACTs as described in my book: determine the dc gain \$H_0\$ when the capacitor is open-circuited (\$s=0\$). Then set the excitation to 0 V (replace the source by a short circuit) and determine the resistance \$R\$ "seen" from the capacitor terminals in this mode. This will give you the natural time constant \$RC_1\$ and the pole is the inverse of the time constant in a 1st-order system. For the zero, simply replace the cap. by a short circuit and determine the gain \$H^1\$ in this mode:

enter image description here

Once you have these values on hand, capture them in a solver like Mathcad, rearrange to fit the aforementioned format and voilà:

enter image description here

I added the cutoff frequency calculation as well as the phase minimum.


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