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# Lecture4

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• 1. Asymptotic Approximations: Bode PlotsThe log-magnitude and phase frequency re-sponse curves as functions of log ω are calledBode plots or Bode diagrams.Consider the following transfer function: K(s + z1)(s + z2)...(s + zk ) G(s) = m s (s + p1)(s + p2)...(s + pn)The magnitude frequency response K|(jω + z1)||(jω + z2 )|...|(jω + zk )| |G(jω)| = |(jω)m||(jω + p1 )||(jω + p2)|...|(jω + pn)|Converting into dB 20 log |G(jω)| = 20 log K + 20 log |(jω + z1)| +20 log |(jω + z2)| + ... + 20 log |(jω + zk )| −20 log |(jω)m| − 20 log |(jω + p1)| −20 log |(jω + p2)|... − 20 log |(jω + pn)| 1
• 2. Thus if we know the magnitude response ofeach term, we can ﬁnd the total magnituderesponse by adding zero terms’ magnitude re-sponses and subtracting pole terms’ magnituderesponses.Similarly if we know the phase response of eachterm, we can ﬁnd the total phase response byadding zero terms’ phase responses and sub-tracting pole terms’ phase responses.Sketching Bode plots can be simpliﬁed becausethey can be approximated as a sequence ofstraight lines. 2
• 3. Bode plots for G(s) = (s + a)Let s = jω, we have ω G(jω) = (jω + a) = a(j + 1) aAt low frequencies, when ω → 0, G(jω) ≈ aThe magnitude response in dB is 20 log M = 20 log awhere M = |G(jω)| and is a constant.At high frequencies, where ω ≫ a ω G(jω) ≈ a(j ) = ω 90◦ aThe magnitude response in dB is 20 log M = 20 log ωIf we plot 20 log M (y) against log ω (x), thisbecomes a straight line y = 20x, i.e. the linehas a slope of 20. 3
• 4. We call the straight-line approximation asymp-totes. The low frequency approximation iscalled the low-frequency asymptote, and thehigh-frequency approximation is called the high-frequency asymptote. The frequency, a, iscalled the break away frequency.Let us turn to the phase response  0◦  ω→0 ω  G(jω) = arctan( ) = 45◦ ω=a a  90◦  ω≫a 4
• 5. Figure above: Bode plots of (s + a);(a) Mag-nitude plot; (b) Phase plot. 5
• 6. It is often convenient to normalize the magni-tude and scale the frequency so that the log-magnitude plot will be 0 dB at a break fre-quency of unity.To normalize (s+a) we factor out the quantitya and form a[(s/a) + 1]. By deﬁning a new fre-quency variable s1 = s/a, the normalized scaledfunction is s1 + 1. To obtain the original fre-quency response, the magnitude and frequencyare multiplied by the quantity a.The actual magnitude curve is never greaterthan 3.01 dB from the asymptote. The maxi-mum diﬀerence occurs at break away frequency.The maximum diﬀerence for the phase curve is5.71◦, which occurs at the decades above andbelow the break frequency. 6
• 7. 7
• 8. 1Bode plots for G(s) = s+aG(s) = s+a = a( s1 1 a +1)The function has a low-frequency asymptoteof 20 log(1/a) which is found by letting s → 0.The Bode plot is constant unit the break fre-quency, a is reached. The plot is then approxi-mated by the high frequency asymptote foundby letting s → ∞. Thus at high frequencies 1 1G(jω) ≈ a( s ) |s→jω = ω − 90◦ aOr, in dB, 20 log M = −20 log ωThe Bode log-magnitude will decrease at arate of 20dB/decade after the break frequency.The phase plot is the negative of that of G(s) =s + a. It begins at 0◦ and reaches −90◦ at highfrequency, going through −45◦ at the breakfrequency. 8
• 9. Bode plots for G(s) = sG(s) = s has only a high-frequency asymp-tote. The magnitude plot is straight line with20dB/decade slope passing 0 dB when ω = 1.The phase plot has a constant 90◦.Bode plots for G(s) = 1 sThe magnitude plot for G(s) = 1 is a straight sline with -20dB/decade slope passing 0 dBwhen ω = 1.The phase plot has a constant −90◦. 9
• 10. Figure above; Normalized and scaled Bode plotsfor (a) G(s) = s; (b) G(s) = 1 ; (c) G(s) = s 1 .s + a; and (d) G(s) = s+a 10