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barcode reader in asp.net codeproject = C = R1 R2 C1 C2 1 = 2 = Q R2 C 2 + R1 C 1 in Software
1 = C = R1 R2 C1 C2 1 = 2 = Q R2 C 2 + R1 C 1 QR Code 2d Barcode Reader In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. QR Code Printer In None Using Barcode drawer for Software Control to generate, create QR image in Software applications. = 1,000 rad/s R 1 C1 =1 R 2 C2
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Make QR Code In .NET Using Barcode maker for ASP.NET Control to generate, create Denso QR Bar Code image in ASP.NET applications. Making QR Code JIS X 0510 In VS .NET Using Barcode creation for .NET Control to generate, create QR Code ISO/IEC18004 image in .NET framework applications. 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: Generating QR In VB.NET Using Barcode creation for .NET framework Control to generate, create QR image in VS .NET applications. Code 128A Creation In None Using Barcode drawer for Software Control to generate, create Code 128 image in Software applications. H (s) = UPC Symbol Printer In None Using Barcode creation for Software Control to generate, create UPCA image in Software applications. Bar Code Printer In None Using Barcode creator for Software Control to generate, create bar code image in Software applications. 2 K C 1 = 2 + C s + 2 s 2 + 2s + 1 s C Q
Printing ECC200 In None Using Barcode generator for Software Control to generate, create Data Matrix image in Software applications. Bar Code Generator In None Using Barcode generation for Software Control to generate, create barcode image in Software applications. 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 Encode Code 93 Full ASCII In None Using Barcode generator for Software Control to generate, create Code 93 image in Software applications. Creating UPCA In Visual Basic .NET Using Barcode maker for .NET framework Control to generate, create UPC Code image in Visual Studio .NET applications. 15
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Bar Code Generator In ObjectiveC Using Barcode drawer for iPhone Control to generate, create bar code image in iPhone applications. Printing Data Matrix ECC200 In .NET Framework Using Barcode maker for VS .NET Control to generate, create Data Matrix ECC200 image in .NET framework applications. Thus, every secondorder 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 fourthorder Butterworth lter Making Bar Code In None Using Barcode maker for Font Control to generate, create barcode image in Font applications. Encoding Barcode In Java Using Barcode encoder for Android Control to generate, create bar code image in Android applications. EXAMPLE 155 Design of FourthOrder Butterworth Filter
Problem
Design a fourthorder lowpass 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 lowpass Sallen and Key lter prototype In the sinusoidal steady
state, s j
Analysis: Table 153 suggests that a Butterworth fourthorder 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 lowpass 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 lowfrequency 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

