L.P. Butterworth filter gain at 3wc <= -50 dB

Thread Starter

J_Rod

Joined Nov 4, 2014
109
Good morning,
I have some questions about this problem about a L.P. (low pass) filter:

"Determine n, the order of a lowpass Butterworth filter, and the corresponding cutoff frequency wc, required to satisfy the following lowpass filter specifications. Find both the values of wc, the one that oversatisfies the passband specifications, and the one that oversatisfies the stopband specifications.

The gain at 3wc is required to be no greater than -50 dB."

For the lowpass filter, I think that for frequencies w < wc, the gain is 1 (no amplification or attenuation of the signal), and after that, the gain quickly drops to 0 (unity). I need to identify the minumum passband gain (at frequency wp) and maximum stopband gain (at frequency ws) to use the equations in my text for n and wc, but I'm not certain how those were given in the problem. Specifically, where is 3wc in relation to wp and ws? Intuitively, would 3wc = ws since the gain there is no greater than -50 dB? But then, what simplifying assumption(s) determine(s) where wp is?

Thank you for the help,
Sincerely,
J_Rod
 

crutschow

Joined Mar 14, 2008
23,306
The filter rolloff rate is determined by the filter order.
Each step in order increases the rolloff by an additional 6dB per octave or 20dB per decade, i.e. 1st order is -6dB/octave, 2nd order is 12dB/octave, etc.
This rolloff starts after the corner frequency (-3dB point).
So from that you should be able to determine the filter order required to get -50dB gain at 3wc.

Note that 1 is unity gain, not 0.
And the gain never reaches 0 at the output of a filter, it just gets arbitrarily small.
 

Papabravo

Joined Feb 24, 2006
12,389
You need to find the transfer function of a Butterworth lowpass filter. From the transfer function you should determine if the solution is unique, or might be satisfied bay a range of solutions. The gain at DC/lowfrequency should be close to 1, or 0 dB. As you approach the corner frequency the gain will drop to -3 dB. From \[\omega_c\] to \[3 \cdot \omega_c\] is the transition band and the gain can be in the range [-3 dB to -50 dB.]. For frequencies greater than than \[3 \cdot \omega_c\] the gain must be less than -50 dB.
 

Thread Starter

J_Rod

Joined Nov 4, 2014
109
The filter rolloff rate is determined by the filter order.
Each step in order increases the rolloff by an additional 6dB per octave or 20dB per decade, i.e. 1st order is -6dB/octave, 2nd order is 12dB/octave, etc.
This rolloff starts after the corner frequency (-3dB point).
So from that you should be able to determine the filter order required to get -50dB gain at 3wc.

Note that 1 is unity gain, not 0.
And the gain never reaches 0 at the output of a filter, it just gets arbitrarily small.
Would the filter have to be at least 3rd order since 3*-20 dB = -60 dB <= -50 dB? Is it possible for the filter to have a higher order, though, since that still satisfies the inequality?
 

Thread Starter

J_Rod

Joined Nov 4, 2014
109
You need to find the transfer function of a Butterworth lowpass filter. From the transfer function you should determine if the solution is unique, or might be satisfied bay a range of solutions. The gain at DC/lowfrequency should be close to 1, or 0 dB. As you approach the corner frequency the gain will drop to -3 dB. From \[\omega_c\] to \[3 \cdot \omega_c\] is the transition band and the gain can be in the range [-3 dB to -50 dB.]. For frequencies greater than than \[3 \cdot \omega_c\] the gain must be less than -50 dB.
Is wc the corner (or cutoff) frequency? It is defined as the point where the signal loses 1/2 power. Would this be the same frequency as the maximum passband frequency in this problem?
 

Papabravo

Joined Feb 24, 2006
12,389
Yes. The reason it is called the corner frequency is because it is the intersection of the passband gain of 0 dB and the asymptote of the rolloff which depends on the order.

Be careful.
The expression for voltage gain is

\[20 \cdot log(\frac{v_o}{v_i})\]

The expression for power gain is

\[10 \cdot log(\frac{p_o}{p_i})\]

Make sure you know which -3 dB point you are referring to.

For voltage:

\[10^{(-3/20)} \approx \frac{1}{\sqrt{2}}= 0.707\]

For power

\[10^{(-3/10)} \approx \frac{1}{2}\]
 
Last edited:

crutschow

Joined Mar 14, 2008
23,306
Be careful.
The expression for voltage gain is

\[20 \cdot log(\frac{v_o}{v_i})\]

The expression for power gain is

\[10 \cdot log(\frac{p_o}{p_i})\]

Make sure you know which -3 dB point you are referring to.

For voltage:

\[10^{(-3/20)} \approx \frac{1}{\sqrt{2}}= 0.707\]

For power

\[10^{(-3/10)} \approx \frac{1}{2}\]
But the -3dB point is always the half power point.
It's just that, due to the V^2 relation between power and voltage, the voltage is .707 down at the -3dB point.
 

Papabravo

Joined Feb 24, 2006
12,389
But the -3dB point is always the half power point.
It's just that, due to the V^2 relation between power and voltage, the voltage is .707 down at the -3dB point.
Yes and you need to make sure you don't get confused when working with this stuff. It can easily happen.
 

Veracohr

Joined Jan 3, 2011
688
Given how it's worded, I think yes ws=3wc for this question. As for wp I think it could be any frequency sufficiently far below wc that its amplitude equals the minimum passband gain, unaffected by the filter attenuation. Try wp=0.01wc.
 
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