# Signal is always attenuated by passive bandpass filter

I'm trying to build a passive bandpass filter with the cutoff frequencies of 0.001 Hz and 10 Hz.

It seems that my signal is always attenuated. Why?

• 1) why are you using a triangle wave to simulate this? Realize that any waveform that is not a sinewave will contain harmonic frequencies which will always be attenuated. 2) You should do some calculations to see what attenuation you can expect. Expecting no attenuation at all, is that realistic? 3) have you studied the basics of filter design? The "just trying something" method might result in misunderstandings, dissapointsments and not getting what you want. While if you follow common filter design procedures, you know beforehand what you should get. Apr 19, 2021 at 11:46
• Since you are using a circuit simulation tool, see if the software allows you to plot the frequency response. particularly the node VF1 and the node to the left of R1 with respect to the input.
– AJN
Apr 19, 2021 at 12:58
• What is the procedure you followed to arrive at the values of the components ? Please use the edit link below the question to add more details to the question.
– AJN
Apr 19, 2021 at 13:00
• You cannot "solve this". Such a bandpass filter cannot have unity gain at the midfrequency.
– LvW
Apr 19, 2021 at 13:36
• i change the question and change the circuit. I also added bode plot. Apr 19, 2021 at 14:39

First, consider that with a passive bandpass filter, everything gets attenuated to some degree.

(With an ideal zero impedance source and infinite impedance load, a passive low-pass does not attenuate DC; a passive high-pass would not attenuate infinite frequency if it existed.)

I calculated the corner frequencies of your two filter stages. Neither matches the frequencies you mention in the question, and I note that the low-pass frequency as designed is lower than the high-pass frequency. Check your calculations for component values.

I can't tell what simulation software you're using. Does it have the capability to create a Bode plot (frequency-domain analyisis)? That might be more helpful to understanding.

• Yes, i noticed that the frequencies didnt match, so i changed the circuit and also added the bode plot. Apr 19, 2021 at 14:41
• @SaraMartins The updated image shows a CR high-pass with corner frequency of 0.6 Hz followed by a RC low-pass with a corner frequency of 0.001 Hz. Try moving the high-pass corner frequency lower than the low-pass corner frequency. Apr 19, 2021 at 14:52
• Thankss!!! I changed it and it works! Apr 19, 2021 at 15:13

You chose the wrong cut frequencies for the low and high pass functions.
Your low-pass corner freq is 1 mHz.
Your high-pass corner freq is 1 Hz.

If you make the low-pass cutoff 1 Hz and the high-pass 1 mHz, you'll get something closer to what you expect.

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following circuit:

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_1=\text{I}_2+\text{I}_3\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$\begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag2$$

Substitute $$\(2)\$$ into $$\(1)\$$, in order to get:

$$\begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag3$$

Now, we can solve for the transfer function:

$$\mathcal{H}:=\frac{\text{V}_2}{\text{V}_\text{i}}=\frac{\text{R}_2\text{R}_4}{\text{R}_2\left(\text{R}_3+\text{R}_4\right)+\text{R}_1\left(\text{R}_2+\text{R}_3+\text{R}_4\right)}\tag4$$

Where I used the following Mathematica-code:

In[1]:=Clear["Global*"];
FullSimplify[
Solve[{I1 == I2 + I3, I1 == (Vi - V1)/R1, I2 == V1/R2,
I3 == (V1 - V2)/R3, I3 == V2/R4}, {I1, I2, I3, V1, V2}]]

Out[1]={{I1 -> ((R2 + R3 + R4) Vi)/(R2 (R3 + R4) + R1 (R2 + R3 + R4)),
I2 -> ((R3 + R4) Vi)/(R2 (R3 + R4) + R1 (R2 + R3 + R4)),
I3 -> (R2 Vi)/(R2 (R3 + R4) + R1 (R2 + R3 + R4)),
V1 -> (R2 (R3 + R4) Vi)/(R2 (R3 + R4) + R1 (R2 + R3 + R4)),
V2 -> (R2 R4 Vi)/(R2 (R3 + R4) + R1 (R2 + R3 + R4))}}
`

When we want to apply the derivation from above to your circuit we need to use Laplace transform (I will use lower case function names for the functions that are in the (complex) s-domain, so $$\\text{y}\left(\text{s}\right)\$$ is the Laplace transform of the function $$\\text{Y}\left(t\right)\$$):

• $$\text{R}_1=\frac{1}{\text{sC}_1}\tag5$$
• $$\text{R}_4=\frac{1}{\text{sC}_2}\tag6$$

So, we can rewrite the transfer function as:

$$\mathscr{H}\left(\text{s}\right)=\frac{\text{R}_2\cdot\frac{1}{\text{sC}_2}}{\text{R}_2\left(\text{R}_3+\frac{1}{\text{sC}_2}\right)+\frac{1}{\text{sC}_1}\left(\text{R}_2+\text{R}_3+\frac{1}{\text{sC}_2}\right)}=$$ $$\frac{\text{C}_1\text{R}_2\text{s}}{\text{C}_1\text{C}_2\text{R}_2\text{R}_3\text{s}^2+\left(\text{C}_1\text{R}_2+\text{C}_2\left(\text{R}_2+\text{R}_3\right)\right)\text{s}+1}=$$ $$\frac{\underbrace{\text{C}_1\text{R}_2}_{=\space\epsilon_1}\cdot\text{s}}{\underbrace{\text{C}_1\text{C}_2\text{R}_2\text{R}_3}_{=\space\epsilon_2}\cdot\text{s}^2+\left(\underbrace{\text{C}_1\text{R}_2+\text{C}_2\left(\text{R}_2+\text{R}_3\right)}_{=\space\epsilon_3}\right)\cdot\text{s}+1}=\frac{\epsilon_1\text{s}}{\epsilon_2\text{s}^2+\epsilon_3\text{s}+1}\tag7$$

Now, when working with sinusoidal signals we can use $$\\text{s}:=\text{j}\omega\$$ (where $$\\text{j}^2=-1\$$ and $$\\omega=2\pi\text{f}\$$ with $$\\text{f}\$$ is the frequency of the input signal in Hertz). So we get:

$$\underline{\mathscr{H}}\left(\text{j}\omega\right)=\frac{\epsilon_1\text{j}\omega}{\epsilon_2\left(\text{j}\omega\right)^2+\epsilon_3\text{j}\omega+1}=\frac{\epsilon_1\omega\text{j}}{1-\epsilon_2\omega^2+\epsilon_3\omega\text{j}}\tag8$$

So, the absolute value of the transfer function is given by:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\epsilon_1\omega}{\sqrt{\left(1-\epsilon_2\omega^2\right)^2+\left(\epsilon_3\omega\right)^2}}\tag9$$

Now, we can solve for the maximum:

$$\frac{\partial\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|}{\partial\omega}=0\space\Longrightarrow\space\omega_0:=\omega=\frac{1}{\sqrt{\epsilon_2}}\tag{10}$$

Now, we can solve for the cut-off frequencies:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{1}{\sqrt{2}}\cdot\left|\underline{\mathscr{H}}\left(\text{j}\omega_0\right)\right|=\frac{1}{\sqrt{2}}\cdot\frac{\epsilon_1}{\epsilon_3}\space\Longrightarrow\space\omega_\pm:=\omega=\frac{\sqrt{\epsilon_3^2+4\epsilon_2}\pm\epsilon_3}{2\epsilon_2}\tag{11}$$

We can conclude that $$\\omega_+>\omega_-\$$.

In your circuit we know that $$\\omega_+=2\pi\cdot10=20\pi\space\text{rad/sec}\$$ and $$\\omega_-=2\pi\cdot\frac{1}{1000}=\frac{\pi}{500}\space\text{rad/sec}\$$. Using the statement from above, we can see that $$\20\pi>\frac{\pi}{500}\$$. So we need to solve:

$$\begin{cases} \frac{\sqrt{\epsilon_3^2+4\epsilon_2}+\epsilon_3}{2\epsilon_2}=20\pi\\ \\ \frac{\sqrt{\epsilon_3^2+4\epsilon_2}-\epsilon_3}{2\epsilon_2}=\frac{\pi}{500} \end{cases}\tag{12}$$

Solving $$\(12)\$$ gives:

$$\epsilon_2=\frac{25}{\pi^2}\approx2.53303\space\left[\text{second}^2\right]\space\space\space\wedge\space\space\space\epsilon_3=\frac{9999}{20\pi}\approx159.139\space\left[\text{second}\right]\tag{13}$$

This implies that:

$$\omega_0=\frac{1}{\sqrt{\epsilon_2}}=\frac{1}{\sqrt{\frac{25}{\pi^2}}}=\sqrt{\frac{\pi^2}{25}}=\frac{\sqrt{\pi^2}}{\sqrt{25}}=\frac{\pi}{5}\approx0.628319\space\text{rad/sec}\tag{14}$$

So we need to solve:

$$\begin{cases} \text{C}_1\text{C}_2\text{R}_2\text{R}_3=\frac{25}{\pi^2}\\ \\ \text{C}_1\text{R}_2+\text{C}_2\left(\text{R}_2+\text{R}_3\right)=\frac{9999}{20\pi} \end{cases}\tag{15}$$

Now, I let you work out the exact values for the components using $$\(15)\$$.

• I think you forgot to read the OP's question: "It seems that my signal is always attenuated. I want to know why." They already have a circuit. They don't need or want to analyze it. May 2, 2021 at 11:42
• @ElliotAlderson thx for the downvote. May 2, 2021 at 11:44
• I think I made it clear why I downvoted. The answer is "not useful". It does not come anywhere close to addressing the OP's question. It provides a rigorous mathematical analysis that does provide the intuitive answer that the OP is looking for. May 2, 2021 at 11:49
• @ElliotAlderson as I said: thx for the downvote. May 2, 2021 at 11:51
• @ElliotAlderson By the way, the component values they've chosen do NOT give the right cutoff frequencies they want. With those components they get $\text{f}_+\approx0.633617\space\text{Hz}$ and $\text{f}_-\approx0.000976008\space\text{Hz}$. So what you stated is maybe right, but they have chosen the wrong values, that is why I wrote my answer. May 2, 2021 at 12:29