Bode Plots

bode plots - 10.1

10. BODE PLOTS

Topics: • Bode plots Objectives:  To be able to describe the response of a system using Bode plots

10.1 INTRODUCTION

When a phasor transform is applied to a transfer function the result can be expressed as a magnitude and angle that are functions of frequency. The magnitude is the gain, and the angle is the phase shift. In the previous chapter these values were calculated for a single frequency and then multiplied by the input values to get an output value. At different frequencies the transfer function value will change. The transfer function gain and phase angle can be plotted as a function of frequency to give an overall picture of system response.

bode plots - 10.2

 Aside: Consider a ’graphic equalizer’ commonly found on home stereo equipment. The spectrum can be adjusted so that high or low tones are emphasized or muted. The  position of the sliders adjusts the envelope that the audio signal is filtered through. The sliders trace out a Bode gain plot. In theoretical terms the equalizer can be described with a transfer function. As the slides are moved the transfer function is changed, and the bode plot shifts. In the example below the slides are positioned to  pass more of the lower frequencies. The high frequencies would not be passed passe d clearly, and might sound somewhat muffled. 40-120

120-360

360-1K

1K-3.2K

3.2K-9.7K

9.7K-20K  

gain

 f(Hz)

bode plots - 10.2

 Aside: Consider a ’graphic equalizer’ commonly found on home stereo equipment. The spectrum can be adjusted so that high or low tones are emphasized or muted. The  position of the sliders adjusts the envelope that the audio signal is filtered through. The sliders trace out a Bode gain plot. In theoretical terms the equalizer can be described with a transfer function. As the slides are moved the transfer function is changed, and the bode plot shifts. In the example below the slides are positioned to  pass more of the lower frequencies. The high frequencies would not be passed passe d clearly, and might sound somewhat muffled. 40-120

120-360

360-1K

1K-3.2K

3.2K-9.7K

9.7K-20K  

gain

 f(Hz)

bode plots - 10.3

 x ( ω ) 1 ------------ = ----------------------------------------------------------------------2 F ( ω )  j ( 3000 ω ) + ( 2000 – 1000 ω )

1  x ( ω ) ------------ = ----------------------------------------------------------------------2 F ( ω )  j ( 3000 ω ) + ( 2000 – 1000 ω )

2

( – j ) ( 3000 ω ) + ( 2000 – 1000 ω )

------------------------------------------------------------------------------2 ( – j ) ( 3000 ω ) + ( 2000 – 1000 ω )

2

 x ( ω ) ( 2000 – 1000 ω ) – j ( 3000 ω ) ------------ = -------------------------------------------------------------------------2 2 2 F ( ω ) ( 2000 – 1000 ω ) + ( 3000 ω ) 2 2

( 2000 – 1000 ω ) + ( 3000 ω )

2

1  x ( ω ) = ------------------------------------------------------------------------------ = ---------------------------------------------------------------------------------------2 2 2 2 F ( ω ) 2 2 ( 2000 – 1000 ω ) + ( 3000 ω ) ( 2000 – 1000 ω ) + ( 3000 ω ) – 3000 ω –3 ω   = atan  ------------θ = atan  ----------------------------------------2 2  for  ( ω ≤ 2 )    – 1000 ω + 2000 2–ω –3 ω θ = atan   -------------2 – π    2–ω

Figure 10.2

 for  ( ω > 2 )

A phasor transform example

The results in Figure 10.2 are normally left in variable form so that they may be analyzed for a range of frequencies. An example of this type of analysis is done in Figure 10.3. 10.3. A set of frequencies is used for calculations. These need to be converted from Hz to

bode plots - 10.4

 f(Hz)

(rad/sec)

0 0.001 0.01 0.1 1 10 100 1000

0 0.006283 0.06283 0.6283 6.283 62.83 628.3 6283

Gain

Gain (dB)

 Note: the gain values will cover many magnitudes of  values. To help keep the graphs rational we will "compress" the values by converting them to dB (decibels) using the following formula. gain db = 20 log ( gain )

θ (rad.)

θ (deg.)

 Note: negative phase angles mean that the mass motion lags the  force.

 Note: The frequencies chosen should be chosen to cover the points with the greatest  amount of change.

Figure 10.3

A phasor transform example (continued)

bode plots - 10.5

 Assuming there is excitation from two sinusoidal sources in addition to the static load, as defined by the equation, F ( t ) = 1000 + 20 sin ( 20 t ) + 10 sin ( 0.5 t ) ( N )

The result for each component can be evaluated separately

· 2π  x ( t ) = 1000( _________ ) sin  0 t + ______  ---------    360  2π + 20( _________ ) sin  20 t + _______  ---------    360  2π + 10( __________ ) sin  0.5 t – ________  ---------    360  gain from calculations or Bode plot 

 phase angle from calculations or Bode plot 

 x ( t ) = 0.5 + ______ sin ( 20 t + _____ ) + _______ sin ( 0.5 t + ______ )

Figure 10.4

A phasor transform example (continued)

1

   6  .    0    1     s    t   o    l   p   e    d   o    b

2

3

4

5

6

789 1

2

3

4

5

6

789 1

2

3

4

5

6

789 1

2

3

4

5

6

789

bode plots - 10.7

Figure 10.5

4 cycle semi-log graph paper

Plot the points from Figure 10.3 in the semi-log graph paper in Figure 10.5. The general layout is pictured below. gain (dB) -40dB/dec

 phase (deg)

0.1

1

10

100

1000

 f(Hz)

Figure 10.6 

Drill problem: Plot the points from Figure 10.3 on graph paper

bode plots - 10.8

steps_per_dec = 6; decades = 6; start_freq = 0.1;  // the transfer function function foo=G(w) D = %i * w; foo = (D + 5) / (D^2 + 100*D + 10000); endfunction  // this section writes the values to a datafile that may be graphed in a spreadsheet fd = mopen("data.txt", "w"); for step = 0:(steps_per_dec * decades), f = start_freq * 10 ^ (step / steps_per_dec); // calculate the next frequency w = f * 2 * %pi); // convert the frequency to radians [gain, phase] = polar(G(w));// find the gain and convert it to mag and angle gaindb = 20 * log10(gain);// convert magnitude to dB phasedeg = 180 * phase / %pi;// convert to degrees mfprint(fd, "%f, %f, %f \n", f, gaindb, phasedeg); end mclose(fd);  // to graph it directly the following is used D = poly(0, ’D’); h = syslin(’c’, (D + 5) / (D^2 + 100*D + 10000) ) ;

bode plots - 10.9

 Draw the Bode plot for the transfer function by hand or  with computer.

 D + 3 -----------------------------------------------------2  D + 10000 D + 10000

bode plots - 10.10

An approximate technique for constructing a gain Bode plot is shown in Figure 10.10. This method involves looking at the transfer function and reducing it to roots in the numerator and denominator. Once in that form, a straight line approximation for each term can be drawn on the graph. An initial gain is also calculated to shift the results up or down. When done, the straight line segments are added to produce a more complex straight line curve. A smooth curve is then drawn over top of this curve.

 Bode plots for transfer functions can be approximated with the following steps. 1. Plot the straight line pieces. a) The gain at 0rad/sec is calculated and used to find an initial offset. For example this transfer function starts at 10(D+1)/  (D+1000)=10(0+1)/(0+1000)=0.01=-40dB. b) Put the transfer function in root form to identify corner frequencies. For example (D+1)/(D+1000) will have corner frequencies at 1 and 1000 rad/sec. c) Curves that turn up or down are drawn for each corner frequency. At each corner frequency a numerator term causes the graph to turn up, each term in the dominator causes the graph to turn down. The slope up or down is generally +/- 20dB/decade  for each term. Also note that squared (second-order) terms would  have a slope of +/-40dB/decade. 2. The effect of each term is added up to give the resulting straight line approximation. 3. When the smooth curve is drawn, there should commonly be a 3dB dif ference at the corner frequencies. In second-order systems the damping

bode plots - 10.11

 Note: Some of the straight line approximation issues are discussed below. Why is there 3dB between a first order corner and the smooth plot, and the phase angle is 45 degrees of the way to +/- 90 degrees.

1 G ( j ω ) = -----------------ω c + j ω the initial gain is 1 1 G ( 0 ) = ----------------- = -----ω c + j 0 ωc at the corner frequency ωc = ω 1 1 1 π G ( j ω ) = --------------------- = ---------------------- = ------------- ∠– --ω c + j ω c π ωc 2 4 ω c 2 ∠--4 Therefore the difference is G(jω) π π π 1 1 diff  = --------------- = ------- ∠– --- = 20 log  ------- ∠– --- = – 3.01 dB ∠– -- 2 4 G(0) 4 4 2 Why does a first order pole go down at 20dB/dec?

1 G ( j ω ) = -----------------ωc + j ω

ω c > j ω

before the corner frequency, G( jω)

1

–1

20 log ( ω )

20 log ( ω )

bode plots - 10.12

Why does a pole make the phase angle move by -90deg after the corner frequency?

ω 1 1 ∠0 1 G ( j ω ) = ------------------ = --------------------------------------------------- = ----------------------- ∠– atan  ------  ω c ω c + j ω ω 2 2 2 2 ωc + ω ω c + ω ∠ atan  ------

 ω c

ω c > j ω

before the corner frequency,

ω 0 angle ( G ( j ω ) ) = – atan  ------ = – atan  ------ = 0 after the corner frequency,

 ω c

ω c = 0 seconds considering the effects of gravity. 7. The following differential equation is supplied, with initial conditions. ·· · ·  y + y + 7 y = F   y ( 0 ) = 1  y ( 0 ) = 0 F ( t ) = 10

t > 0

a) Write the equation in state variable form. b) Solve the differential equation numerically. c) Solve the differential equation using calculus techniques. d) Find the frequency response (gain and phase) for the transfer function using the phasor transform. Sketch the bode plots. 8. You are given the following differential equation for a spring damper pair. d  5 x +  ----- x = F   dt 

F ( t ) = 10 sin ( 100 t )

a) Write the transfer function for the differential equation if the input is F. b) Apply the phasor transform to the transfer function to find magnitude and phase as functions of frequency. c) Draw a Bode plot for the system using either approximate or exact techniques. d) Use the Bode plot to find the response to; F ( t ) = 10 sin ( 100 t ) e) Put the differential equation in state variable form and use a calculator to find values in time for the given input. F  = 10 sin ( 100 t )

t

x

bode plots - 10.25

10.6 PRACTICE PROBLEM SOLUTIONS

1. 20dB 0dB 0dB

-20dB

1 -----2π

100 --------2π

1000 -----------2π

-40dB/dec

5 ------2π

90deg

0deg

-45deg

-180deg

bode plots - 10.26

2.  y ( D ) ( D + 10 ) ( D + 5 ) ( D + 10 ) ------------ = ---------------------------------------- = --------------------2  x ( D ) ( D + 5 ) ( D + 5 )

6dB 0dB 0.8

1.6

0.8

1.6

90

-90  x ( t ) = 20 cos ( 9 t + 0.3 )

Aside: the numbers should be obtained from the graphs, but I have calculated them  y ( D + 10 ) 9 j + 10 5 – 9 j 50 + 81 – 45 j -- = --------------------- =  -----------------   -------------- = -------------------------------- = 1.236 – 0.425 j  9 j + 5   5 – 9 j  x ( D + 5 ) 25 + 81  y -- =  x

– 0.425 2 2 1.236 + 0.425 ∠ atan  ----------------  = 1.307 ∠–0.3312 ra d   1.236 

 y -- = 2.33 dB ∠– 9.49 °

bode plots - 10.27

(cont’d Aside: This can also be done entirely with phasors in polar notation  y ( D + 10 ) 9 j + 10 13.45 ∠0.733 13.45 -- = --------------------- =  -----------------  = ------------------------------- = ------------- ∠0.733 – 1.064 = 1.31 ∠– 0.331  9 j + 5   x ( D + 5 ) 10.30 ∠1.064 10.30  x = 20 ∠0.3  y ------------------ = 1.31 ∠– 0.331 20 ∠0.3  y = 1.31 ( 20 ) ∠( – 0.331 + 0.3 ) = 26.2 ∠– 0.031  y ( t ) = 26.2 sin ( 9 t – 0.031 )

bode plots - 10.28

3. for the numerator, (zero) for the denominator, (poles)

ra d  1000 --------- = 159 Hz s 2

2

2

 D + 2 ω n ζ D + ωn =  D + 5 D + 100 2

ω n = 100 ra d  s

∴ω n = 10 ---------

 f n = 1.59 Hz

2 ωn ζ = 5 5

∴ζ = -------------- = 0.25 2 ( 10 )  f d  =  f n 1 – ζ

for the initial gain

2

(underdamped)

= 1.54 Hz

0 + 1000 G ( 0 ) = -------------------------------------- = 10 = 20 dB 2 0 + 5 ( 0 ) + 100

Gain (dB) 20 159

0 0.1

1

1.54

10

100 -40dB/dec

f(Hz) 1000

bode plots - 10.29

4. 5

a)

V o ( t ) = 0.050 sin ( 10 t – 1.561)

b)

16Hz

159Hz

1.6KHz -20dB/dec

0 deg -90 deg

5. a) b) c)

–3

8.89 sin ( 624 t + 1.571 ) + 0.141 ×10 sin ( 6.2 t + 2.356 )

bode plots - 10.30

6. 0.0667 m a)  x --- = ----------------------- ---F   D 2 + 33.3 N  b)

ω = 1

∴ D = 1 j

 x m 0.0667 –3 --- = ----------------------------- = 2.07 × 10 ∠0 ra d ---F  ( 1 j ) 2 + 33.3  N  F ( t ) = 10 cos ( t + 1 ) N 

∴F ( ω ) = ( 10 ∠1 ra d ) N   x m –3 -------------------------------- = 2.07 × 10 ∠0 ra d --- N  ( 10 ∠1 rad ) N  m  N  –3 ∴ x ( ω ) = ( 10 ) 2.07 × 10 ∠( 0 ra d + 1 ra d ) m –3

∴ x ( ω ) = 2.07 × 10 ∠0 ra d ---- ( 10 ∠1 ra d ) N  ∴ x ( ω ) = 0.0207 ∠1 radm ∴ x ( t ) = 0.0207 cos ( t + 1 ) m c)

2

F   D + 33.3 N  --- = ----------------------- --- x 0.0667 m

bode plots - 10.31

d)

2

2

2

 D + 2 ω n ζ D + ω n =  D + 33.3

for the denominator, (poles)

rad  s

ω n = 5.77 --------ζ = 0

 f n = 0.918 Hz

(undamped)

 f d  =  f n 1 – ζ

2

= 0.918 Hz

0.0667 G ( 0 ) = ---------------- = 2 = –54 dB 33.3

for the initial gain Gain (dB)

∞

-54 f(Hz)

0 0.1 -94

-134 phase (deg)

1

10

100 -40dB/dec

-20dB/dec

bode plots - 10.32

7. a)

·  y = v · v = – v – 7 y + F 

b)  y i + 1

=

vi + 1

 y i

0

+h

vi

given

1  y i

0

+

– 7 –1 v i

 y 0

F 

1

=

F  = 10

0

v0

using h=0.001s  y 1 v1

=

1

+ 0.001

0

etc.. until

 y 100 v 100

c)

 y ( t ) = ( – 0.437 e

– 0.5 t 

0

1

1 + 0 –7 –1 0 10 1.428 m

=

m 0.000 ---s

cos ( 2.598 t – 0.19 ) + 1.429 ) m

bode plots - 10.33

d)

 y 1 1 1 --- = -------------------------- = ----------------------------------- = ------------------------------------2 2 F   D2 + D + 7 ( j ω ) + j ω + 7 ( 7 – ω ) + j ( ω )

1  y = -----------------------------------------F  2 2 2 (7 – ω ) + ω

∠0 2 2 ∠θ = --------------------------------------------2--- = ∠( 0 – angle ( ω, 7 – ω ) ) = ∠– angle ( ω, 7 – ω ) ∠angle ( ω, 7 – ω )

-20dB -40dB 0.01Hz

1Hz

100Hz

0.01Hz

1Hz

100Hz

0deg

bode plots - 10.34

8. a)

1  x --- = ------------F  5 + D

b)

1 1 1 ∠0 1  x ω --- = ------------- = --------------- = ----------------------------------------------- = ----------------------∠– atan  ----  5 F  5 + D 5 + j ω 2 2 ω 2 2 5 +ω 5 + ω ∠ atan  ----  5

c)

gain(dB)

1 initial gain = 20 log  ------------  = –14 dB  5 + 0 5 corner freq. = ------ = 0.8 Hz 2π -20dB/dec

-14dB

phase(deg) 0° – 45 ° – 90 ° 0.08Hz 0.8Hz d)

8Hz

100  f  = --------- = 16 Hz 2π From the Bode plot,

gain = – 40 dB = 0.01

bode plots - 10.35

e)

0.186

0.2

0

Xi

− 0.099 0.2

0

0.5

1

0

i ⋅h

0.999

– 5 t 

f)

1 e  x = --- – --------5 5

t

x

0.0

0

0.002

9.98e-4

0.004

5.92e-3

0.006

0.015

0.008

0.026

0.010

0.041

10.7 ASSIGNMENT PROBLEMS

1. For the transfer functions below, draw the bode plots using computer software.

( D + 2 π ) D ---------------------------------------------------------( D + 200 π ) ( D + 0.02 π )

( D + 2 π ) ------------------------------------------------------2 2  D + 50 π D + 10000 π

bode plots - 10.36

b) Apply a phasor (Fourier) transform to the differential equation and develop equations for the system gain and phase shift as a function of input frequency. c) Draw a Bode plot using the equations found in part b) on the attached log paper. d) Draw a straight line approximation of the system transfer function on the attached log paper.

10.8 LOG SCALE GRAPH PAPER

Please notice that there are a few sheets of 2 and 4 cycle log paper attached, make additional copies if required, and if more cycles are required, sheets can be cut and pasted together. Also note that better semi-log paper can be purchased at technical bookstores, as well at most large office supply stores.

bode plots - 10.37

1

2

3

4

5

6

7 8 9 1

2

3

4

5

6

7 8 9