Effect of adding stages to a filter
Suppose we have a RC low pass filter of 1st order. If we add multiple stages every time we add a RC stage identical to our original does the slope of the curve in the bode plot db/Hz becomes wilder but the cutoff frequency remain the same?And is there a formula if we know the slope of the curve at the cutoff frequency of the 1st order filter to calculate the slope of the curve for a multistage filter?
2 answers
You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.
There are a number of misconceptions here.
bode plot db/Hz
First, it's "dB", not "db".
Second that's not a Bode plot. Those are dB/Log(Hz). In other words, a certain frequency ratio results in a fixed gain ratio as the slope approaches the asymptote. You were told in your previous question that dB/Hz doesn't make sense. If you're not going to listen answers, then there is no point in us writing them.
is there a formula if we know the slope of the curve at the cutoff frequency of the 1st order filter to calculate the slope of the curve for a multistage filter
For a first order low pass filter well past the rolloff frequency, gain is inversely proportional to frequency. For a second order, it's inversely proportional to the square of the frequency, and to the cube of the frequency for a third order, etc.
This is usually easier to express in dB. For first order, the voltage gain is -20 dB / Log(Hz), for a second order -40 dB / Log(Hz), etc.
Isn't the gain inversely proportional to frequency from the rolloff frequency of a filter? Why "well past"?
Because it approaches "gain inversely proportional to frequency" gradually. A single pole low pass filter isn't totally flat up to the rolloff frequency, then suddenly falls off proportional to frequency above that. There isn't a hard corner in the Bode plot. It's smoothed out. That's why I don't like the term "cutoff" frequency, since it incorrectly implies a hard transition.
If there was a hard transition, the gain at the rolloff frequency would still be 1 (0 dB). Instead, the gain has already started to fall off, and is actually -3 dB at the rolloff frequency. The further from the transition, the closer the gain becomes to the hard transition asymptotes.
Put another way, the real gain of the filter is off by 3 dB from the asymptotes at the rolloff frequency. Either side of that frequency, the gain approaches the asymptotes more closely. Therefore "well past" the rolloff frequency, the gain falls off inversely proportional with frequency. "Well below" the rolloff frequency, the gain is 1.
It depends on whther the stages are buffered, or not. If they are buffered then it's more simple, because an RC lowpass has the transfer function:
$$H_1(s)=\dfrac{\dfrac{1}{RC}}{s+\dfrac{1}{RC}}=\dfrac{1}{sRC+1} \tag{1}$$
And adding $N$ stages will mean $H(s)$ will simply be multiplied $N$ times, thus the overall result will be:
$$H(s)=\left(\dfrac{1}{sRC+1}\right)^N \tag{2}$$
Since the result of cascading is a simple exponentiation it results that the corner frequency will remain the same, but the attenuation will be $3N\space\text{dB}$ (for integer $N$).
If the stages are not buffered then you need to derive the transfer function for each final filter, since all the stages are influenced by the presence of both the previous, and the following ones. That's the sin of the passive filters. For example, for 2-, 3-, and 4-stages, once you solve for KCL/KVL/etc, you will get these transfer functions (considering $R=1,\space C=1$):
$$\begin{align} H_2(s)&=\dfrac{1}{s^2+3s+1} \\ H_3(s)&=\dfrac{1}{s^3+5s^2+6s+1} \\ H_4(s)&=\dfrac{1}{s^4+7s^3+15s^2+10s+1} \end{align}$$
Since forming the system of equations and then solving tend to take some time (even when automated), it's better to look for patterns. In this case, if you search on OEIS for the sequences in the denominators, you get to this page, which gives a nice, convenient formula for generating your transfer function:
$$H_n(s)=\dfrac{1}{\binom{2n-k}{k}s^{n-k}},\quad k=0 ... n,\quad n\ge 0$$
BTW, for the buffered stages, the pattern will be the regular $\binom{n}{k}$, since it's a simple expansion of $(x+1)^N$.
The Bode plots will be worse than the buffered stages, and they will be getting much worse with increasing $N$. Here is how the 1-, 2-, 3-, 4-stages look for buffered (red) vs unbuffered (blue):
The analytic values of the attenuations will be messier the higher the $N$, but you can still make your own test cases, mathematical or SPICE simulators, and you can see that the rate is approximately 6.3...6.5 dB/stage.
0 comment threads
0 comment threads