# Why is the transfer function of this non-inverting low-pass filter wrong?

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:

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.

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).

• So instead of $K=1+\frac{R_{1}}{R_{2}}\:\text{it should be}\:H(s)=1+K\frac{1}{1+sR_{1}C},\:\text{where}\:K=\frac{R_{1}}{R_{2}}$? Sep 27, 2019 at 9:21
• The Vout/Vin in your answer is the same as the Vin/Vout as OP!! Sep 27, 2019 at 10:47
• The zero is missing in your expression LvW. Jun 25, 2020 at 11:25
• @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.
– LvW
Jun 25, 2020 at 15:03
• False alarm, I missed the parenthesis after the +: entschuldigung : ) 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 $$s = \frac{ 1 }{ R_1 C }$$

EDIT
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.

• So what would the K value of the transfer function be in this case? Sep 27, 2019 at 11:32
• @NBoss What is K? The gain of the transfer function? Sep 27, 2019 at 11:52
• 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? Sep 27, 2019 at 11:55
• @Huisman....you are wrong. Please check the unit of your pole frequency expression. The pole frequency is as given by NBoss.
– LvW
Sep 27, 2019 at 12:10
• @LvW thanks, I see. Thanks for pointing out. 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:

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

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