DeCarlo Ch21 Solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

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SOLUTIONS CHAPTER 21 PROBLEMS S OLUTION TO 21.1. (a) Low pass (b) High pass S OLUTION TO 21.2. S OLUTION TO 21.3. (a) »n = 0.65378; »d = [1 0.80381643 0.82306043]; »w = 0:0.005:2; »h = freqs(n,d,w); »plot(w, 20*log10(abs(h))) »grid »xlabel('Frequency rads/s') »ylabel('dB Gain') »
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0 0.5 1 1.5 2 -15 -10 -5 0 Frequency rads/s dB Gain TextEnd (b) »poles = roots(d) poles = -4.0191e-01 + 8.1335e-01i -4.0191e-01 - 8.1335e-01i (c) »% Poles of new transfer function »wp = 2*pi*750' wp = 4.7124e+03 »wp = 2*pi*750; »polesnew = poles*wp polesnew = -1.8939e+03 + 3.8328e+03i -1.8939e+03 - 3.8328e+03i »% All zeros remain at infinity.
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Further H ( s ) = H NLP ( s p ) = ( p ) 2 s 2 + 0.80381643 p s + 0.82306043( p ) 2 = 2.2207 × 10 7 s 2 + 3.7879 × 10 3 s + 1.8277 × 10 7 S OLUTION TO 21.4. (a) The 2 nd order normalized LP transfer function is H NLP ( s ) = 1 s 2 + 2 s + 1 . This must be frequency scaled by K f = 1000 π . Hence, H ( s ) = H NLP ( s K f ) = ( K f ) 2 s 2 + K f 2 s + ( K f ) 2 = 9.8696 × 10 6 s 2 + 4.4429 × 10 3 s + 9.8696 × 10 6 (b) Using MATLAB, »n = (1000*pi)^2; »d = [1 sqrt(2)*pi*1e3 (1000*pi)^2]; »w = 0:1:2*pi*1500; »h = freqs(n,d,w); »plot(w/(2*pi),abs(h)) »grid »xlabel('Frequency in Hz') »ylabel('Magnitude') »plot(w/(2*pi),20*log10(abs(h))) »grid »xlabel('Frequency in Hz') »ylabel('Magnitude in dB')
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0 500 1000 1500 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Frequency in Hz Magnitude TextEnd 0 500 1000 1500 -20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0 Frequency in Hz Magnitude in dB TextEnd (c) »n n = 9.8696e+06 »d d = 1.0000e+00 4.4429e+03 9.8696e+06 »w = j*2000*pi; »mag = abs(n/(w^2 + d(2)*w + d(3))) mag = 2.4254e-01
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S OLUTION TO 21.5. (a) max is that value of e that places the magnitude response curve through A max at ϖ = ϖ p . Therefore A max = 10log 10 H ( j p ) 2 = 10log 10 1 + max 2 p p 2 n = 10log 10 1 + max 2 ( 29 Therefore max 2 = 10 0.1 A max - 1 which upon a square root yields the final answer. (b) Similarly, min puts the magnitude response curve through the A min spec. Hence A min = 10log 10 H ( j s ) 2 = 10log 10 1 + min 2 s p 2 n Therefore min 2 = 10 0.1 A min - 1 s p 2 n which is equivalent to the required formula. S OLUTION TO 21.6. The relationship of ε and ϖ c is given by the formula: c = p ( ) 1 n . Further, max in putting the magnitude response curve through the A max spec produces c min , and min in putting the magnitude response curve through the A min spec produces c max . Hence, from the solution to problem 5, c min = p ( max ) 1 n = p 2 n 10 0.1 A max - 1 ϖ c max = p ( min ) 1 n = s 2 n 10 0.1 A min - 1 S OLUTION TO 21.7. (a) From above material, the second order Butterworth NLP transfer function is H NLP 2 ( s ) = 1 s 2 + 2 s + 1 and from tables, the third order is
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H NLP 3 ( s ) = 1 s 3 + 2 s 2 + 2 s + 1 (b) n1 = 1; d1 = [1 sqrt(2)
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DeCarlo Ch21 Solutions - SOLUTIONS CHAPTER 21 PROBLEMS...

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