Bode Plotting Band Pass Filter

Thread Starter

Mechatronical

Joined Feb 15, 2014
30
When you read my contributions carefully, you will understand that THIS function applies for Q<0.5 only.
Pleaase, read all the contributions in this thread again and try to UNDERSTAND!

Comment: Regarding comparison of the various terms I repeat here part of my response#15:
If you have the actual transfer function (derived from the circuit) in the same form you can compare the various terms and derive the expressions for ωp and Q
I've posted the result which I got from scilab:
http://s10.postimg.org/d9hjzlpeh/Scilabbb.png
It wasn't a flat line, but it appears that the response time is very fast.
It's the only sense I make out of it. So should I assume that the settling time is 0.003 seconds?

The transfer function which I derived from the circuit shown in the link below;
http://www.electronics-tutorials.ws/filter/fil51.gif
which has values: R2=R1=10000Ω,C1=10nF,C2=1nF, is;
H(jw) = R2C1jw / [(R2C2jw+1)(R1C1jw+1)]

Low frequency cut off is = 1/2∏R1C1 = 1591 Hz = ωl
High frequency cut off is = 1/2∏R1C1 = 15915 Hz = ωh
Bandwidth B = 14324 Hz
Center Frequency ω0= sqrt(ωl*ωh) = 5032
Quality Factor = ω0/B = 0.5032
Or maybe it's ω0/B = 2∏ω0/B = 2∏*5032/10000=3.16

As you see, the quality factor is above 0.5

I have the actual transfer function (derived from the circuit) H(jw). So can't you compare the various terms and derive the expressions for ωp and Q for me?
 
Last edited:

Thread Starter

Mechatronical

Joined Feb 15, 2014
30
When you read my contributions carefully, you will understand that THIS function applies for Q<0.5 only.
Pleaase, read all the contributions in this thread again and try to UNDERSTAND!

Comment: Regarding comparison of the various terms I repeat here part of my response#15:
If you have the actual transfer function (derived from the circuit) in the same form you can compare the various terms and derive the expressions for ωp and Q
I came across, stumbled upon, and fell over something interesting, with a little dirt in my face:

http://s27.postimg.org/4to49f3hf/Low_Q_bandpass_filter.png

It's stated that ω0 = 1000 Hz = 2∏ω0 and Q = 4.
From Q = ω0/B --> B = ω0/Q = 2∏ω0/4 = 1570
So B is measured in Hz, while ω0 should be in 2∏ω0?

If so, my bandpass filter's Q factor is Q/B = 2∏ω0/B = 2∏5031/10000 = 3.16

SO... Perhaps I should be adding the H(s) with Q and ω0 into the plant in Scilab, and not the H(s) where jw is replaced by s?

Why on earth can I do it with all other plants and transfer functions in regulation engineering, but not the H(s) with jw replaced with s?

It's a bit handicap, if I can't add resistors and capacitors values into a plant, and depend on the Q factor and ω0 in order to see the response time.
 

LvW

Joined Jun 13, 2013
1,752
Mechatronical - I really don`t know what to do or how to answer.

Nevertheless, I will try it:

(Quote): "Quality Factor = ω0/B = 0.5032
Or maybe it's ω0/B = 2∏ω0/B = 2∏*5032/10000=3.16
As you see, the quality factor is above 0.5"

That means:
If you are not satisfied with a result you simply multiply it with 2*Pi. Fine.
Who gave you such an advice?

* Why don`t you believe me that the pole Q for your circuit always has a maximum of Q=0.5 ?

* Do you need a proof? OK - here it comes:

If you would follow my advice to develop the product of the two brackets in the denominator of your transfer function (for comparison with the general transfer function expressed in terms of Q and wo) you would find that for your circuit

* the pole frequency (center frequency) is wo=1/SQRT(R1R2C1C2) and
* Qp=Q=SQRT(R1R2C1C2)/(R1C1+R2C2).

You can try all R and C values and you will see that Q=0.5 for R1=R2 and C1=C2 only.
For ALL OTHER values always Q<0.5.

Last comment:
For a bandpass realization you always have the choice between many different bandpass topologies. However, there are some simple circuit alternatives which allow only Q values below 0.5.
And your circuit belongs to this class - as I have mentioned several times.
Didn´t you get it?
In case you want a Q value above 0.5 you are required to use another circuit topology - Sallen-Key or MFB (multi-feedack) or GIC-realizations or integrator based,...).

By the way: I really don`t know what you have simulated in Scilab. It looks VERY strange!
Why not using one of the classic circuit simulation packages?
 

t_n_k

Joined Mar 6, 2009
5,455
Of course a band pass filter has a response time. It doesn't matter if it's AC or DC. What matters is, how quickly does the filter react, when it receives a signal, at all. The input is an AC signal.
You missed the point.

I agree that there must be a transient response, but of what form?

Showing the generalized step response from the link you posted is confusing. That response shown in your link will clearly not be the form of the response for a second order BPF.

Again, you persist in insisting that you can have a BPF Q exceeding unity when the roots of the second order function are both real. This isn't possible.

The latest link you posted is relevant to the discussion.....

http://s27.postimg.org/4to49f3hf/Low...ass_filter.png

It is readily found that the roots of the second order function in this case are complex.

Plugging in the values yields a second order form ..

\(25132.741 + s + 0.0006366*s^2\)

for which the complex roots are ...

\(-785.39816 \pm 6233.9047i\)

I will post a method for extracting the transient response for this system, using Scilab.
 

t_n_k

Joined Mar 6, 2009
5,455
Create an '*.sce' file using the following code and load into Scilab using the SciNotes application:

Rich (BB code):
// define/declare the variable s
s=poly(0,'s');
// Enter the system parameters
Q=4
K=6.7
fo=1000
wo=2*%pi*fo
// Create the system Transfer Function
h=K*s/((Q/wo)*s^2+s+Q*wo)
// Define the frequency domain continuous linear system equivalent
H=syslin('c',h)
//Plot the Bode response for interest
subplot(2,1,1)
// Set the frequency range as 100Hz to 10KHz
frq=[100:0.1:10000]
bode(H,frq)
// Specify the time response interval as 0 to 10msec
t=[0:1e-5:10e-3]
// Create the time step response
resp=csim('step',t,H)
// Plot the time response
subplot(2,1,2)
plot(t,resp)
You will obtain an output like this image - only clearer.
 

Attachments

Thread Starter

Mechatronical

Joined Feb 15, 2014
30
Create an '*.sce' file using the following code and load into Scilab using the SciNotes application:

Rich (BB code):
// define/declare the variable s
s=poly(0,'s');
// Enter the system parameters
Q=4
K=6.7
fo=1000
wo=2*%pi*fo
// Create the system Transfer Function
h=K*s/((Q/wo)*s^2+s+Q*wo)
// Define the frequency domain continuous linear system equivalent
H=syslin('c',h)
//Plot the Bode response for interest
subplot(2,1,1)
// Set the frequency range as 100Hz to 10KHz
frq=[100:0.1:10000]
bode(H,frq)
// Specify the time response interval as 0 to 10msec
t=[0:1e-5:10e-3]
// Create the time step response
resp=csim('step',t,H)
// Plot the time response
subplot(2,1,2)
plot(t,resp)
You will obtain an output like this image - only clearer.
Thank you so very much. God bless you.
 

Thread Starter

Mechatronical

Joined Feb 15, 2014
30
Mechatronical - I really don`t know what to do or how to answer.

Nevertheless, I will try it:

(Quote): "Quality Factor = ω0/B = 0.5032
Or maybe it's ω0/B = 2∏ω0/B = 2∏*5032/10000=3.16
As you see, the quality factor is above 0.5"

That means:
If you are not satisfied with a result you simply multiply it with 2*Pi. Fine.
Who gave you such an advice?

* Why don`t you believe me that the pole Q for your circuit always has a maximum of Q=0.5 ?

* Do you need a proof? OK - here it comes:

If you would follow my advice to develop the product of the two brackets in the denominator of your transfer function (for comparison with the general transfer function expressed in terms of Q and wo) you would find that for your circuit

* the pole frequency (center frequency) is wo=1/SQRT(R1R2C1C2) and
* Qp=Q=SQRT(R1R2C1C2)/(R1C1+R2C2).

You can try all R and C values and you will see that Q=0.5 for R1=R2 and C1=C2 only.
For ALL OTHER values always Q<0.5.

Last comment:
For a bandpass realization you always have the choice between many different bandpass topologies. However, there are some simple circuit alternatives which allow only Q values below 0.5.
And your circuit belongs to this class - as I have mentioned several times.
Didn´t you get it?
In case you want a Q value above 0.5 you are required to use another circuit topology - Sallen-Key or MFB (multi-feedack) or GIC-realizations or integrator based,...).

By the way: I really don`t know what you have simulated in Scilab. It looks VERY strange!
Why not using one of the classic circuit simulation packages?
I'll try using your method:
Q=SQRT(R1R2C1C2)/(R1C1+R2C2)
For R1=R2=2 and C1=C2=3

Q=SQRT(2*2*3*3)/(2*3+2*3) = 1.73
I thought it should be less than 0.5?

Look at the equations in the link below.
http://s2.postimg.org/kvi2i7n15/Center_Frequency.png
 

Thread Starter

Mechatronical

Joined Feb 15, 2014
30
You missed the point.

I agree that there must be a transient response, but of what form?

Showing the generalized step response from the link you posted is confusing. That response shown in your link will clearly not be the form of the response for a second order BPF.

Again, you persist in insisting that you can have a BPF Q exceeding unity when the roots of the second order function are both real. This isn't possible.

The latest link you posted is relevant to the discussion.....

http://s27.postimg.org/4to49f3hf/Low...ass_filter.png

It is readily found that the roots of the second order function in this case are complex.

Plugging in the values yields a second order form ..

\(25132.741 + s + 0.0006366*s^2\)

for which the complex roots are ...

\(-785.39816 \pm 6233.9047i\)

I will post a method for extracting the transient response for this system, using Scilab.
I think you're most qualified to answer this. Please look at the link below:
http://s2.postimg.org/kvi2i7n15/Center_Frequency.png
In addition the frequencies are:
ωl = 1/2∏R1C1 = 1591 Hz
ωl = 1/2∏R2C2 = 15915 Hz
ω0 = sqrt(1591*15915)=5032 Hz
B = 14324 Hz
Q is given by ω0/B where ω0 = 2∏ω0;
Q = 2∏ω0/B = 3.16

Your feedback please?
 

t_n_k

Joined Mar 6, 2009
5,455
I think you're most qualified to answer this. Please look at the link below:
http://s2.postimg.org/kvi2i7n15/Center_Frequency.png
In addition the frequencies are:
ωl = 1/2∏R1C1 = 1591 Hz
ωl = 1/2∏R2C2 = 15915 Hz
ω0 = sqrt(1591*15915)=5032 Hz
B = 14324 Hz
Q is given by ω0/B where ω0 = 2∏ω0;
Q = 2∏ω0/B = 3.16

Your feedback please?
There's not much I can add. Your resulting values are simply those obtained after plugging your stated filter cut-off frequency values into the general terms defined in the link. Nothing particularly significant in that.

I think both LvW & I are puzzled as to where you are heading with this. If you intend to become an expert filter designer, then you may need to set down a clear plan of action to get you to your destination. That may require your initial securing of a much stronger & coherent mathematical foundation. I'm not sure either of us can help you any further at the moment.
 

LvW

Joined Jun 13, 2013
1,752
Something I could add is the following:

The angular frequency is defined as ω=2π*f (rad/s) and f in Hz=1/s.

In mechatronical`s last contribution we can read: ωl = 1/2∏R1C1 = 1591 Hz.
That would mean: ω=1/2π*τ or 2π*ω=1/τ, which obviously is wrong.

Again my recommendation: Read a textbook with some basics on electronics.
(At least you should learn that resistances and capacitances must carry a unit, see end of your post#27)
 
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