Upload
afonso-albuquerque
View
112
Download
2
Embed Size (px)
Citation preview
1
MEM 640 Lecture 2: Bode Plots
2
Bode Plots: Why Study Them?
Input: 0.1)( =tVi
Output (dots): )(tVo
• Output reaches 63.6% of steady-state at 0.47 ms
• Note: 1/0.00047 = 2127 radians/sec
Recall: RC Low-pass filter time response plot
3
Low Pass Filter Bode PlotDefinition: A Bode diagram consists of 2 plots. The first plots the output/input ratio[dB] versus frequency. The second plots the phase angle versus frequency.Typically a semi-log plot for frequency is used
Low Pass Filter Bode Plot Diagram:
-3 dB
2127 radians/sec
Thus the -3 dB point represents the frequency corresponding to about 1 time constant
4
Phase Shift Significance
Observe: Apply a sine voltage input (338 Hz) into a low-pass RC filter
Hence see that this is approximately the 45 degree lag shown on Bode plot
Bode phase plot on previous slide says 45-degree lag at 2127 radians/sec [338 Hz]
Period sec003.00019.00049.0 =−=T0.003 sec = 333.3 Hz
sec00042.000148.00019.0 =−=Δt
deg 502 =Δ
=Ttπφ
5
Bode Plot: Step-by-Step InstructionsProblem: Below is data collected by applying sine voltage inputs into a low pass RC filter. Sketch the Bode diagram on semi-log paper
6
Step 1: Label axis on semi-log paper. Choose units and be consistent!
Solution
7
Step 2: Plot your points
8
Step 3: Draw asymptotes, mark -3 dB point, cutoff frequency and label slopes
9
Step 4: On new semi-log paper, plot the phase angle versus frequencies
10
System ID Given the Bode PlotProblem: Given the following Bode Plot, calculate the transfer function
11
Step 1: Note that the magnitude is not zero at start and slope is -20 dB/dec
sKTF1=
Step 2: Slope decreases -20 dB/dec, hence have another pole. with time constant
52.0
1==τHence
15s1TF2+
=
Step 3: Slope increase +20 dB/dec, hence a zero has been added.
2.0=cωThe cutoff frequency
8.0=cωThe cutoff frequency The time constant is
25.18.0
1==τ
Hence
125.1TF3 += s
12
Step 4: Slope decreases another 20 dB/dec. This means a simple pole was added
The cutoff frequency is 10=cω The time constant
1.0101==τ
Hence
10.1s1TF4+
=
Step 5: Take product of all sub transfer functions
)11.0)(15(1)K(1.25s)(TF++
+==
ssssG
Step 6: Determine the value of K
Take decibel log of each side and replace s with ωj
( ) ( ) ( )11.0log22015log
220log2025.11log
220log20)(log20 222222 +−+−−++= ωωωωKsG
(1)
13
From Bode magnitude plot, see that at 1.0=ω have dB 60)(log20 =ωjG
Thus substituting this frequency into the (1)
4103.4969.0200673.0log20dB 60 −×−−++= K
90.40log20 =K
9.110=K
Hence
)11.0)(15()125.1(9.110)(
+++
=sss
ssG