# Anti Aliasing filter Pass-band gain?

#### Nobby Nobbs

Joined Jul 31, 2014
11
Hi all
trying to figure this out, I have a circuit diagram that I have re produced in LTSpice, I have produced the frequency response graph. From this I need to work out Pass-band gain, break frequency, bandwidth and filter order.

here is the circuit from the book.

From this presuming I am not way off, I have got the break frequency to be just about 11KHz as this is a 3dB drop. The bandwidth is 11KHz as this is all to the left of the break frequency.
I think it would be classed as a 2nd order filter as the roll off is more than 20dB per decade and less than 40dB.

But I cant figure out how you get the Pass-band gain, I know how to work out the gain of an op amp, inverting and non inverting but not this and I cant seem to find a good explanation of how I do this.
The output on my graph is at 0dB, I often see this referred to as Unity gain but is this correct ? and if this is not correct how would you work out the gain for this circuit.

sorry if I have missed something glaringly obvious but I am very new to this.

Thanks for any help.

#### Attachments

• 1.2 KB Views: 35

#### MikeML

Joined Oct 2, 2009
5,444

On the plot pane, with the cursor over the Y-axis, left click the mouse. Change the plot axis to linear instead of decible (retain Bode). Now you can read volts out of the filter directlyin Volts instead of dB. The gain is simply V(output)/1 because the stimulus is 1VAC.

If you want, LTSpice will plot gain for you. Go to the plot pane, right click in the plot pane. Select ADD TRACE from the menu. Create this expression in the text box: V(output)/V(input)

You can either type the entire expression, or you can mouse over the list of nodes and branch currents that LTSpice keeps after a simulation run, and click on V(output), then type a "/", click on V(input), and finally type an "Enter". The resultant trace is dimensionless (volts/volts). Now change the stimulus to AC 2, and rerun the sim.

You carry plotting arbitrary expressions to the extreme. For examples, ask LTSpice to plot V(V1)/-I(V1). What do you get? What are the units of that expression? (You may have to set the plot y axis to linear Bode

Since your stimulus is AC 1 and your simulation directive is .AC, then you do not need the SINE(..) on the voltage source... The SINE would only be used if you are doing a time-domain .TRAN simulation.

Last edited:

#### Jony130

Joined Feb 17, 2009
5,528

#### Veracohr

Joined Jan 3, 2011
772
I think it would be classed as a 2nd order filter as the roll off is more than 20dB per decade and less than 40dB.
The order of the filter is determined by the number of reactive components (capacitors & inductors). Also, if you extend the upper frequency of the AC sweep, you might find the slope changes further up. You're looking at less than 1 decade above the cutoff frequency, the slope may continue to "bend down" into a steeper value at a higher frequency.

The output on my graph is at 0dB, I often see this referred to as Unity gain but is this correct ?
Yes. Unity = 1 = 0dB.

#### MrAl

Joined Jun 17, 2014
11,593
Hello there,

Strictly speaking, this isnt really a bandpass filter it's a low pass filter. It has a slight peaking that's all. Note that on the left side of the peak it never goes as low as 3db down but to the right it does, which means that calculating the 'center' frequency really isnt necessary the only point worth mentioning really is the -3db point which is to the right of the peak.

To do an analysis of this circuit or other circuit like this, you can follow the following steps:
1. Convert capacitors to impedances, zC=1/(s*C) where s is the complex frequency but you can look at it as just a variable for now.
2. Reduce any impedances that can be reduced to help make the equation forming a little easier. For example, we can combine R2 and C2 to make the impedance we'll call Z22:
Z22=R2+zC2=R2+1/(s*C2).
3. Power the circuit with Vin only while shorting out Vout. Note this puts C1 in parallel with Z22 so we can calculate Z122 as:
Z122=para(Z22,zC1)=para(Z22,1/(s*C1))
where the function para(a,b)=a*b/(a+b).
4. Calculate va, the voltage resulting from the application of Vin only with Vout shorted and this will be a function like Vin*Va(s).
5. Repeat the above procedure but this time short out Vin and allow Vout to preside, which results in a function like Vout*Vb(s).
6. Add the two and calculate the voltage at the junction of R2 and C2 and this results in a form like Vin*VA(s)+Vout*VB(s).
7. Because that is at the non inverting terminal of the op amp and it's a voltage follower (gain of +1) we can equate that implictely to Vout:
Vout=Vin*VA(s)+Vout*VB(s)
8. From that, calculate Vout explicitly:
Vout*(1-VB(s))=Vin*VA(s)
9. Calculate the transfer function:
Vout/Vin=VA(s)/(1-VB(s))=N/D
10. Replace every occurrence of 's' with "j*w".
11. Calculate the amplitude with:
Amplitude=|N|/|D|
12. Equate that to 1/sqrt(2):
|N|/|D|=1/sqrt(2)
13. Solve for w.
14. Divide w by 2*pi to get f the frequency of the -3db point.
Doing this results in f=21128.652 Hz
with values of R1=R2=10k, C1=1.8e-9, C2=490e-12.

Note to solve for the peak, take the first derivative of |N|/|D| and set it equal to zero:
d(|N|/|D|)/dw=0
Solve for all the w that result from this and test to see which if any are the max. There will be at least one max.
That's the peak, but as previously mentioned it's not that interesting for this circuit because this is really a low pass filter.
Also note that we usually dont want to know the -3db point from the peak we want to know the -3db point from the passband, which is to the left of the peak and for a low pass like this the passband gain would be taken to be 0db even though it peaks a little higher.

So even if you try to solve for the "bandwidth" you wont find it to be possible in the same way as a bandpass filter is calculated, but you might consider the 'bandwidth' to be from 0Hz to 21.1kHz (approx).
The peak of this particular circuit is found to be very close to 11438.169 Hz.

Last edited:

#### JoeJester

Joined Apr 26, 2005
4,390
Since you used spice and more than likely used VOLTAGE, you will see from your graph that at approximately 22k, you are -6 dB. That is the half-power point when using voltage and that agrees with the calculations above by Mr Al.

There is no need to get rid of decibels if you know decibels.

#### MrAl

Joined Jun 17, 2014
11,593
Since you used spice and more than likely used VOLTAGE, you will see from your graph that at approximately 22k, you are -6 dB. That is the half-power point when using voltage and that agrees with the calculations above by Mr Al.

There is no need to get rid of decibels if you know decibels.
Hi Joe,

Actually 21.129kHz is the -3db point, and the -6db point occurs at 25.400kHz.
The response at 21.129kHz is 0.7071, and the response at 25.4kHz is 0.5000.
20*log(0.7071)=-3.01db (half power point)
20*log(0.5000)=-6.02db (half voltage point)

As a quick approximate calculation, we have:
w=218000.0-116000*A
where w=2*pi*f where f is the approximate frequency and
A is the voltage ratio to be solved for (such as 0.5 or 0.7071)
and this is valid from A=0.5 to A=1 on the right side of the peak. This is not as accurate as the above calculations but serves as a rough guide.

Last edited:

#### MrAl

Joined Jun 17, 2014
11,593
Hi Mike,

That's cheating The more software we use the less smart we become.

Seriously though, to understand the circuits like this one it is better to learn how to analyze using general circuit analysis. Taking a short cut will leave you short one day. Still good as a secondary method though. So at the beginning of your statement i would not say "Or" as much as i would say "And".

#### bertus

Joined Apr 5, 2008
22,291
Hello,

You could have a look at the attached PDF.

Bertus

#### Attachments

• 171.6 KB Views: 63

#### Nobby Nobbs

Joined Jul 31, 2014
11
Hi all and thanks for the great info as always, one other question regarding this circuit and the high pass one I have just built. In my text book they have refereed to the break frequency that I know is either a 3dB drop or 70%. but on this circuit they have used "Critical frequency" but with no explanation what this is, I have rightly or wrongly assumed this means the same thing as the break frequency in relation to the high pass filter is this correct or does Critical frequency mean something completely different.

Thanks again.

#### MrAl

Joined Jun 17, 2014
11,593
Hi,

Yes they sometimes quote "Critical frequency" when they mean "Cutoff frequency" although the phrase "Critical frequency" is a little more general and could mean different things in a different context. For a high pass or low pass however it would be taken to mean cutoff frequency which as you know is the 3db down frequency.

-3db is the same as a voltage gain of 0.7071 which is just 1/sqrt(2). That means that the response has decreased to about 71 percent at that frequency (down from 100 percent in the passband).

The bandpass filter has two of these "critical" frequencies, one on the left of the peak and one on the right, while the low pass and high pass only have one (unless you have a very high peak but then often the additional point is ignored).

If you look at bertus' pdf link it looks like it has some very good circuit examples with equations included. You might take a look at that and gain some insight and perhaps have some more questions you'd like to ask.