Hi dears,
I am studying the Bessel - Thompson frequency response and I have a doubt about the Sallen & Key realization.
The textbook I use is "Analog Filter Design" by M. E. Van Valkenburg and the Bessel response is covered in chapter 10 "Delay Filter".
There is a design example for a low pass active filter in this chapter (example 10.1), and I do not understand how to determine the cutoff frequency from the denormalization formula exposed:
\({D \over k_\text{r}k_\text{f}}\)
The filter is designed to provide a delay
\(D = 100\,\mu s\)
(while the cutoff frequency is not mentioned).
The order of the filter is \(n = 4\) and the complex conjugate poles pairs are:
\(
\begin{align}
&& p_p = a \pm \text{j}b\\
&& p_p1 = -2.1038 \pm \text{j}2.6574\\
&& p_p2 = -2.8962 \pm \text{j}0.8672\\
\end{align}
\)
And
\(
\begin{align}
&& \omega_0 = \sqrt(a^2 + b^2)\\
&& Q = {1 \over 2a}\omega_0\\
\end{align}
\)
For a Sallen & Key configuration:
The components for each cell are denormalizated from a scaling factor \(k_\text{r}\):
\(
\begin{align}
&& k_\text{r} = 10000\\
&& k_\text{f} = \omega_0\\
&& D = 100\,\mu s \\
\\
&& R_1 = R_2 = k_\text{r} \,(\Omega)\\
&& C_1 = 2Q \left ( {D \over k_\text{r}k_\text{f}} \right ) \,(\text{F})\\
&& C_2 = \left ( {1 \over 2Q}\right ) \left ({D \over k_\text{r}k_\text{f}} \right ) \,(\text{F})\\
\end{align}
\)
And the final circuits is:
Of course I tried to simulate it in LTspice:
And I can see through AC analysis that the cutoff frequency \(f_c\) at -3 dB is \(\approx 3,25\,\text{kHz}\)
As we can see from the attached Bode plot image.
At this point it is not clear to me how to relate the cutoff frequency \(f_c\) with the denormalization formula:
\({D \over k_\text{r}k_\text{f}}\)
Could please someone help me to understand it?
Thank you in advance.
simo_x
I am studying the Bessel - Thompson frequency response and I have a doubt about the Sallen & Key realization.
The textbook I use is "Analog Filter Design" by M. E. Van Valkenburg and the Bessel response is covered in chapter 10 "Delay Filter".
There is a design example for a low pass active filter in this chapter (example 10.1), and I do not understand how to determine the cutoff frequency from the denormalization formula exposed:
\({D \over k_\text{r}k_\text{f}}\)
The filter is designed to provide a delay
\(D = 100\,\mu s\)
(while the cutoff frequency is not mentioned).
The order of the filter is \(n = 4\) and the complex conjugate poles pairs are:
\(
\begin{align}
&& p_p = a \pm \text{j}b\\
&& p_p1 = -2.1038 \pm \text{j}2.6574\\
&& p_p2 = -2.8962 \pm \text{j}0.8672\\
\end{align}
\)
And
\(
\begin{align}
&& \omega_0 = \sqrt(a^2 + b^2)\\
&& Q = {1 \over 2a}\omega_0\\
\end{align}
\)
For a Sallen & Key configuration:
The components for each cell are denormalizated from a scaling factor \(k_\text{r}\):
\(
\begin{align}
&& k_\text{r} = 10000\\
&& k_\text{f} = \omega_0\\
&& D = 100\,\mu s \\
\\
&& R_1 = R_2 = k_\text{r} \,(\Omega)\\
&& C_1 = 2Q \left ( {D \over k_\text{r}k_\text{f}} \right ) \,(\text{F})\\
&& C_2 = \left ( {1 \over 2Q}\right ) \left ({D \over k_\text{r}k_\text{f}} \right ) \,(\text{F})\\
\end{align}
\)
And the final circuits is:
Of course I tried to simulate it in LTspice:
Rich (BB code):
Version 4
SHEET 1 1444 928
WIRE 128 336 64 336
WIRE 320 336 192 336
WIRE 512 352 448 352
WIRE 704 352 576 352
WIRE 64 416 64 336
WIRE 64 416 48 416
WIRE 80 416 64 416
WIRE 176 416 160 416
WIRE 240 416 176 416
WIRE 320 432 320 336
WIRE 320 432 304 432
WIRE 352 432 320 432
WIRE 448 432 448 352
WIRE 448 432 432 432
WIRE 464 432 448 432
WIRE 560 432 544 432
WIRE 624 432 560 432
WIRE 240 448 224 448
WIRE 704 448 704 352
WIRE 704 448 688 448
WIRE 736 448 704 448
WIRE 624 464 608 464
WIRE 224 496 224 448
WIRE 320 496 320 432
WIRE 320 496 224 496
WIRE 608 512 608 464
WIRE 704 512 704 448
WIRE 704 512 608 512
FLAG -240 368 0
FLAG -384 528 vdd
FLAG -384 608 0
FLAG 272 400 0
FLAG 176 480 0
FLAG 272 464 vdd
FLAG 656 416 0
FLAG 560 496 0
FLAG 656 480 vdd
FLAG 736 448 Out
IOPIN 736 448 Out
FLAG -32 416 Vin
IOPIN -32 416 In
FLAG -240 288 Vin
SYMBOL voltage -240 272 M0
WINDOW 123 24 124 Left 2
WINDOW 39 24 152 Left 2
SYMATTR Value2 AC 1 0
SYMATTR SpiceLine Rser=0 Cpar=0
SYMATTR InstName V1
SYMATTR Value 1
SYMBOL Opamps\\UniversalOpamp2 272 432 M180
SYMATTR InstName U1
SYMBOL cap 192 352 M270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName C1
SYMATTR Value 3.45n
SYMBOL cap 160 416 R0
SYMATTR InstName C2
SYMATTR Value 3.17n
SYMBOL Opamps\\UniversalOpamp2 656 448 M180
SYMATTR InstName U2
SYMBOL cap 576 368 M270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName C3
SYMATTR Value 4.76n
SYMBOL cap 544 432 R0
SYMATTR InstName C4
SYMATTR Value 1.83n
SYMBOL res 64 432 M270
WINDOW 0 32 56 VTop 2
WINDOW 3 0 56 VBottom 2
SYMATTR InstName R1
SYMATTR Value 1k
SYMATTR SpiceLine tol=1 pwr=0.25
SYMBOL voltage -384 512 R0
WINDOW 123 0 0 Left 2
WINDOW 39 24 124 Left 2
SYMATTR SpiceLine Rser=1
SYMATTR InstName V2
SYMATTR Value 10
SYMBOL res 176 432 M270
WINDOW 0 32 56 VTop 2
WINDOW 3 0 56 VBottom 2
SYMATTR InstName R2
SYMATTR Value 1k
SYMATTR SpiceLine tol=1 pwr=0.25
SYMBOL res 448 448 M270
WINDOW 0 32 56 VTop 2
WINDOW 3 0 56 VBottom 2
SYMATTR InstName R3
SYMATTR Value 1k
SYMATTR SpiceLine tol=1 pwr=0.25
SYMBOL res 560 448 M270
WINDOW 0 32 56 VTop 2
WINDOW 3 0 56 VBottom 2
SYMATTR InstName R4
SYMATTR Value 1k
SYMATTR SpiceLine tol=1 pwr=0.25
TEXT -360 696 Left 2 !.ac oct 100000 1 300kAs we can see from the attached Bode plot image.
At this point it is not clear to me how to relate the cutoff frequency \(f_c\) with the denormalization formula:
\({D \over k_\text{r}k_\text{f}}\)
Could please someone help me to understand it?
Thank you in advance.
simo_x
Attachments
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