Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part60

Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part60

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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 11 29 Copyright © Orchard Publications Low-Pass Analog Filter Prototypes This can be shown to be true by (11.43), that is, or Therefore, and (11.54) Substitution of (11.53) and (11.54) into (11.51) yields (11.55) or or (11.56) We have seen that when , there is a maximum at . At this frequency, (11.49) reduces to (11.57) and for a unity gain, when . However, for unity gain when , we must have . This is because at , we must have in accordance with(11.45). Then, the relation reduces to or C k x () kcos 1 x x 1 cos = C k x 1 for 1 x 1 ≤≤ C k 2 ωω C max 1 = A min 2 α 1 ε 2 + ------------- = r dB 10log 10 A 2 A 2 ------------ 10log 10 α α 1 ε 2 + -------------------------- 10log 10 1 ε 2 + == = log 10 1 ε 2 + r 10 ------- = 1 ε 2 + 10 r 10 = ε 2 10 r 10 1 = ko d d = ω 0 = A 2 0 α = α 1 = d d = ke v e n = α 1 ε 2 + = ω 0 = C k 0 1 = A 2 ω α 1 ε 2 C k 2 C + ------------------------------------------ = A 2 0 α 1 ε 2 C k 2 0 + ----------------------------- α 1 ε 2 + 1 = α 1 ε 2 + =
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Chapter 11 Analog and Digital Filters 11 30 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications For this choice of , the magnitude response at maxima, corresponds to and this will be maximum when resulting in or Example 11.7 Derive the transfer function for the , Chebyshev Type I function that has pass band ripple , unity gain, and normalized cutoff frequency at . Solution: From (11.49), (11.58) and since , for unity gain, we must have . Then, (11.58) becomes For and Also, from (11.56), Then, α A 2 ω max () 1 ε 2 + 1 ε 2 C k 2 ω ω C + ----------------------------------------------------- = C k 2 ω ω C 0 = A 2 ω 1 ε 2 + 1 ------------- 1 ε 2 + == A ω 1 ε 2 + = Gs k2 = r dB 1 dB = DC ω C 1 rad s = A 2 ω α 1 ε 2 C k 2 ωω C + ------------------------------------------ = ke v e n = α 1 ε 2 + = A 2 ω 1 ε 2 + 1 ε 2 C k 2 C + ----------------------------------------- = = C 2 x 2x 2 1 = C k 2 C C k 2 ω 2 ω 2 1 2 4 ω 4 4 ω 1 + = ε 2 10 r 10 1 10 11 0 1 1.259 1 0.259 = = A 2 ω 10 . 2 5 9 + . 2 5 9 4 ω 4 4 ω 1 + + ------------------------------------------------------------ 1.259 1.036 ω 4 1.036 ω 2 1.259 + ------------------------------------------------------------------
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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 11 31 Copyright © Orchard Publications Low-Pass Analog Filter Prototypes and with , We find the poles from the roots of the denominator using the MATLAB script below.
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Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part60

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