barcode reader in asp.net codeproject = C = R1 R2 C1 C2 1 = 2 = Q R2 C 2 + R1 C 1 in Software

Drawer QR in Software = C = R1 R2 C1 C2 1 = 2 = Q R2 C 2 + R1 C 1

1 = C = R1 R2 C1 C2 1 = 2 = Q R2 C 2 + R1 C 1
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= 1,000 rad/s R 1 C1 =1 R 2 C2
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R1 C2 + (1 K) R2 C1
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Comments: What type of response does the lter analyzed above have We can compare the lter response to that of a quadratic Butterworth lter (or other lter family) by determining the Q of the lter Once the gain and cutoff frequency have been de ned, Q is the parameter that distinguishes, say, a Butterworth from a Chebyshev lter The Butterworth polynomial of order 2 is given in Table 153 as (s 2 + 2s + 1) If we compare this expression to the denominator of equation 1519, we obtain:
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H (s) =
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2 K C 1 = 2 + C s + 2 s 2 + 2s + 1 s C Q
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Since the expressions for the quadratic polynomials of Table 153 are normalized to unity gain and cutoff frequency, we know that K = 1 and C = 1, and therefore we can solve for the value of Q in a Butterworth lter by setting 1 1 = 2 or Q = = 0707 Q 2
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Thus, every second-order Butterworth lter is characterized by a Q of 0707; this corresponds to a damping ratio = 05Q 1 = 0707, that is, to a lightly underdamped response The next example considers the characteristics of a fourth-order Butterworth lter
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EXAMPLE 155 Design of Fourth-Order Butterworth Filter
Problem
Design a fourth-order low-pass Butterworth lter using two quadratic Sallen and Key sections
Solution
Known Quantities: Filter response; desired gain and cutoff frequency Find: Component values R1 , R2 , C1 , C2 , RA , RB , for each lter section Schematics, Diagrams, Circuits, and Given Data: Gain = 100; cutoff frequency =
400 rad/s
Assumptions: Use low-pass Sallen and Key lter prototype In the sinusoidal steady
state, s j
Analysis: Table 153 suggests that a Butterworth fourth-order lter is composed of the product of two quadratic responses Our rst objective is to determine the Q of each of these two quadratic responses, so that we can design each of the two Sallen and Key quadratic sections Comparing the standard quadratic low-pass lter response to the rst of the two Butterworth polynomials for a normalized lter with K = 1 and C = 1, we have:
H (s) =
2 1 K C = 2 s + 07654s + 1 2 + C s + 2 s C Q
and, for a normalized lter with K = 1 and C = 1, we can solve for the value of Q1 , the Q of the rst section: 1 = 07654 Q1 or Q1 = 1 = 13065 07654
Repeating the procedure for the second section we obtain: 1 = 18478 Q2 or Q2 = 1 = 05412 18478
Having determined these values, we can now proceed to design two separate quadratic sections with the values of Q computed above, and each with gain K = 10 (so that the product of the two sections yields a low-frequency gain of 100, as speci ed), and cutoff frequency C = 400 rad/s The responses for the two sections are: H (s) = =
2 16 106 K C = 2 C s + 2 s + 30616s + 16 105 s 2 + Q1 C
10 625 10 6 s 2 + 1914 10 3 s + 1
Part II
Electronics
and H (s) = =
2 16 106 K C = 2 C s + 2 s + 73912s + 16 105 s 2 + Q2 C
10 625 10 6 s 2 + 462 10 3 s + 1
One of the important features of the Sallen and Key lter prototype is that we can choose the values for the resistors that set the circuit gain independently of those of the resistors that set the cutoff frequency (the converse is not true) Thus, we can separately select K = 10 for both stages by requiring that 1 + RA /RB = 10, for example, RA = 100 k , RA = 111 k Next, we compute the component values using equations 1520 Since we have only two equations and four unknowns, two values will have to be selected arbitrarily We can write: C = and 1 = Q R2 R1 C2 + C1 R1 R2 C2 R1 + (1 K) C1 R2 C1 C2 1 1 C C1 C2 = R 1 R2 C1 C2 R1 R 2 R1 R 2 = 1 C C1 C2
Rearrange the last equation as: 1 = Q R2 R1 C2 + C1 R1 R2 C2 + (1 K) C1 C1 C2
and operate a change of variables to obtain: x= R2 R1 c x c= C2 + (1 K) C1 C1 C2 a= C2 C1 b= 1 Q
b = ax + or
ax 2 bx + c = 0 There is always a positive root in the preceding equation, as a and b are both positive One can easily show that there is another positive root if 1 +K 1< 4Q2 C2 < K 1 C1
and in that case there are two solutions for R1 , R2 ; also, there are real solutions only if C2 < (K 1)C1 Solving the preceding equations gives: x= b+ R2 = R1 b2 4 ac 2a
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