ECE634S09_L17_BodePlots1

ECE634S09_L17_BodePlots1 - ECE634 Signals and Systems II,...

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ECE634 Signals and Systems II, Spring 2009 - Lecture 17, March 2 Daniel S. Brogan 1 4.8 Frequency Response of an LTIC System This section assumes that 0 sj j j σ ωω ω =+ =+ = , i.e. we are only concerned with the steady-state component of the response. Since the frequency response of an LTIC system can be defined as ( ) () ( ) jHj Hs Hj Hj e == the response to an input at some frequency defined as ( ) ( ) cos xt A t θ = + will produce the output ( ) ( ) ( ) ( ) cos y tx t H j A H j t H j + + Note that the magnitude and phase of the output may vary with frequency. Example 4.23 (Lathi p.424) Find the frequency response (amplitude and phase response) of a system whose transfer function is 0.1 5 s Hs s + = + Also, find the system response ( ) yt if the input ( ) x t is (a) cos 2 t (b) cos 10 50 t −° 0.1 5 j Hj j + = + 22 2 2 0.1 0.01 52 5 ++ 11 tan tan 0.1 5 −− ⎛⎞ ∠= ⎜⎟ ⎝⎠
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ECE634 Signals and Systems II, Spring 2009 - Lecture 17, March 2 Daniel S. Brogan 2 Sample values The figure below shows the frequency response of ( ) Hj ω . () ( ) 0 0.0200 0 0.01 0.0201 5.60 0.02 0.0204 11.1 0.05 0.0224 26.0 0.1 0.0283 43.9 0.2 0.0447 61.1 0.5 0.101 73.0 1 0.197 73.0 2 0.372 65.3 5 0.707 43.9 10 0.894 26.0 20 0.970 13.7 50 0.995 5.60 Inf 1.000 0
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ECE634 Signals and Systems II, Spring 2009 - Lecture 17, March 2 Daniel S. Brogan 3 The figure below shows the frequency response of ( ) Hj ω using frequency on a log scale. Since () ( ) ( ) () cos y tA H j t H j ωθ =+ + ( ) 2 cos 2 2 0.372cos 2 65.3 a yt Hj t Hj t + ° ( ) ( ) 10 cos 10 50 10 0.894cos 10 50 26.0 b t t =− ° + ° + ° 0.894cos 10 24.0 b yt t ≈− ° 4.9 Bode Plots Magnitude and phase of a factored transfer function 12 2 2 13 3 2 3 11 ss Ks a s a aa Ka a Hs bb ss b s bs b bs s b ⎛⎞ ++ ⎜⎟ ⎝⎠ == + +
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ECE634 Signals and Systems II, Spring 2009 - Lecture 17, March 2 Daniel S. Brogan 4 () 12 2 13 2 3 11 jj aa Ka a Hj bb b j b ωω ω ⎛⎞ ++ ⎜⎟ ⎝⎠ = +− + 2 2 133 Ka a b j bbb = + 2 2 1 2 1 3 3 Ka a b j j j a a b b b = + ∠+ + ∠− + Thus, a transfer function can be fully analyzed using real terms of not higher than the second order. While combining all of the phase terms required only addition and subtraction, combining the magnitude terms is much more complex. The standard simplification of this problems is to plot 10 20log rather than ( ) . This changes the magnitude equation to 10 10 10 10 1 2 2 2 10 10 10 3 20log 20log 20log 1 20log 1 20log 20log 1 20log 1 Ka a j j a a b j b =+ + + + −− + + The scaling factor consists of two portions, ‘2’ and ‘10’.
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This note was uploaded on 03/05/2009 for the course ECE 651 taught by Professor Messner during the Spring '09 term at New Hampshire.

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ECE634S09_L17_BodePlots1 - ECE634 Signals and Systems II,...

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