Lecture19-FrequencyResponse

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Page : 1 EE406 Control Systems Lecture 19 : Frequency Response

UCSI University Faculty of Engineering Kuala Lumpur, Malaysia Department of Mechatronics

Lecture 19

Frequency Response

Mohd Sulhi bin Azman Lecturer

Department of Mechatronics UCSI University

[email protected]

1 August 2011

Contents

• Review

– Trigonometry and sinusoidal waves – Laws of logarithm

– Log-log and log-linear (semilog) scales • Frequency response technique • Bode plot

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Watch Out!

• Watch out if you are sleeping!

Review : Trigonometry

• General definition:

• Where:

– A=amplitude

– ω=angular frequency

– =phase shift – Period, T=1/f.

(

)

( ) sin

x t = A ω φt+

φ

2 f ω = π

(3)

Review : Logarithms

• Let us refresh the laws of logarithms:

( )

10

log log log log ln

log log log log log log log log

log10 1 ; log1 0

log log arg( ); where

n

x

e

n

y x y n

x x

xy x y

x

x y

y

x n x

z z j z z a jb

= ⇔ =

= =

= +

 

= −

   

=

= =

= + = +

Log-log and semilog scale

• Normally, we can choose a particular scale for our graph. The scale may be a 1-to-1 or 1-to-n scale. In another word, the scale may increase in ones or tens – it is up to you.

• But consider the following function:

• Let us create a table of values: 6

y

=

x

x 0 1 2 3 4 5 6 7 8 9 10

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Log-log and semilog scale

• The plot for this graph is given by:

-10≤x ≤10 0≤x ≤10

Log-log and semilog scale

• The graph may appear simpler, however, consider the variations in x and y.

• As x varies from 1 to 10, y varies from 1 to 1,000,000.

• Thus, several of these points would not be discernible on a graph so valuable information may be lost.

• This problem may be overcome by using a semilog scale which accommodates the large variation in y.

• The log y is plotted against x, rather than y against x.

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Log-log and semilog scale

• So, in the previous example, as x varies from 1 to 10, “log y” varies from 0 to 6.

• Let us tabulate the results:

• The plot of this graph is given in the next slide.

x 0 1 2 3 4 5 6 7 8 9 10

log y - 0 1.806 2.863 3.612 4.193 4.67 5.07 5.42 5.72 6

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Log-log and semilog scale

• Thus, in effect, the use of the log scale has compressed a large variation into one which is much smaller and easier to observe.

• Presentation of data on a logarithmic scale can be helpful when the data covers a large range of values – the logarithm reduces this to a more manageable range.

• A plot in which both scales are logarithmic is known as the log-log plot where log-log y is plotted against log-log x.

• We need to use log-log and semilog paper in producing these plots.

• Because each tickmark is a power of 10, they are referred to as a decade.

Log-log Graph Paper

1st_cycle ₂nd_cycle ₃rd_cycle

1

st

cy

cl

e

2

nd

cy

cl

e

3

rd

cy

cl

e

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Semilog Graph Paper

• The vertical axis has a log scale but a horizontal axis has a normal (linear and equally-spaced) scale, as shown in the following figure.

1

st

cy

cl

e

2

nd

cy

cl

e

3

rd

cy

cl

e

100

101

102

103

Frequency Response

• Engineers are often interested in how a circuit will respond to a sinusoidal input (i.e. current, voltage).

• Suppose that we have a linear system represented by the following diagram:

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Frequency Response

• Frequency response of a control system is defined as the steady-state response of the system when the sinusoidal input is applied at the input terminals.

• The sinusoidal input signal when applied to a linear system results in an output signal, which is sinusoidal in steady-state and differs from the input waveform only in amplitude and phase angle.

• Frequency response method determines experimentally the properties of complicated control systems without any difficulty as the sinusoidal test signals for various ranges of frequencies and amplitudes are easily available.

Frequency Response

• Frequency response function describing the sinusoidal steady-state behaviour of the system can be obtained by replacing s=jω in the transfer function G(s) of the system.

• The function G(jω) representing the sinusoidal steady-state behaviour of the system is a function of complex variable having magnitude and phase angle.

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Frequency Response

• We take note that any curve giving information regarding the amplitude (gain) or phase shift of the frequency function is known as the frequency response of the system.

• Typical examples of frequency response graph are: – Polar plot (a.k.a. Nyquist plot)

– Bode plot

Polar & Bode Plot

• A polar plot is a plot of the magnitude and phase angle in polar coordinates for various values of frequencies ranging from zero to infinity.

• Bode plot on the other hand is a plot of magnitude and phase angle versus the frequencies in rectangular coordinates.

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Bode Plot

• In this lecture, I shall be focusing only the Bode plot. The polar plot is omitted and if you are interested, you can learn on your own.

• The Bode plot is generally:

– Plots of frequency response. Gain and phase are displayed in separate plots.

– Logarithmic plots.

– The horizontal axis is frequency - plotted on a log scale. It can be either f or ω.

– The vertical axis is gain, expressed in decibels - a logarithmic measure of gain.

– Sometimes, the vertical axis is simply a gain on a logarithmic scale.

Bode Plot

• And in general, a Bode plot consists of two components (subplot):

– The ratio of the amplitudes of the output signal and the input signal is plotted against frequency.

– The phase shift between the input and output signal is plotted against the frequency.

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Bode Plot

• Consider the following system:

• If the input signal is given as:

• Then, the output signal is:

input frequency amplitude

( ) sin

x t = X  ω t

 

output _phase amplitude _shift ( ) sin

y t Y ωt φ

 

 

= _ + _

 

 

Bode Plot

• Let us view this pictorially: given the following system:

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Bode Plot

• Note that the transfer function is given as:

• Therefore, the magnitude expression is:

• Note that we substitute s=jω because we are dealing with sinusoidal signal.

• The magnitude is plot in decibels (dB). ( ) ( )

( )

Y s G s

X s

=

[

] [

2

]

2

( ) ( )

( ) Re ( ) Im ( )

( )

Y s G s

X s Y j

G j G j G j

X j ω

ω ω ω

ω

=

= = +

( )

10

( ) 20 log

G s = _G jω _

A bit on decibel…

• A decibel (dB) is a ratio between two numbers on a logarithmic scale.

• A decibel is not itself a number, and cannot be treated as such in normal calculations.

• Widely used when dealing with sinusoidal function and waves.

• It confers a number of advantages, such as:

– the ability to conveniently represent very large or small numbers, a logarithmic scaling that roughly corresponds to the human perception of sound and light; and

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Bode Plot

• The phase shift is given as:

• Note that a positive angle is known as lead” and a negative angle is known as “phase-lag”

(

)

(

)

1 ( ) ( )

( ) ( ) ( )

( )

Im ( ) ( ) tan

Re ( ) Y s

G s

X s Y j G j

X j

G j G j

G j ω

ω

ω ω φ

ω

−

∠ = ∠

 

∠ = =  

 

Sketching Bode Plot

• There are two ways of sketching the Bode plot.

• The first method is the “tabulation method” where the table of values is generated. This is the easiest method.

• The second method is the analytical method. This is by far, the challenging method.

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How to Sketch a Bode Plot (1)

• The first method is the tabulation method.

• Before we generate the table of values, we need to find the magnitude and phase of the system.

• The magnitude is given by:

[

] [

2

]

2

( ) ( )

( ) Re ( ) Im ( )

( )

s j

Y s G s

X s Y j

G j G j G j

X j ω

ω ω

ω ω ω

ω

=

= = +

How to Sketch a Bode Plot (1)

• The magnitude, expressed in decibel is:

• To sketch the Bode plot, we will generate the table of values based from the equations presented in this and previous slide.

• The phase angle is given by the following relationship:

[

]

( ) 20 log ( )

M jω = G jω

( )

1 Im

( )

tan

Re G j G j

G j ω

φ ω

ω

−  

= ∠ = _ _

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Example 1

• Sketch the Bode plot for the following transfer function:

• Note : a transfer function of the form G(s)=1/s is said to be an integrator.

2 ( )

G s s

=

Solution to Example 1

• We take note that the transfer function is in the normal form. Therefore:

• Putting s=jω yield:

• And the magnitude is found by: 2 ( )

G s s

=

2 ( )

G j j ω

ω

=

2 2

( ) 0

G jω

ω ω

= =

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Solution to Example 1

• The magnitude in decibel is given by:

• We note that the gain (K) is K=6.02.

• To determine the vertical intercept, we simply let ω=1, and hence the second term is dropped out. Therefore, the graph’s vertical intercept at 20log(2).

2

( ) 20 log 20 log 2 20 log 6.02 20 log

G jω ω ω

ω

 

=  = − = −

 

• Let us generate the table of values for this function:

ω G(jω)

0.1 26.0206

1 6.0206

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Solution to Example 1

• Hence, our Bode plot of the magnitude is:

1

100 1000

log(jω) |G(jω)|

26.02

6.02

-14

-34

-54

10 0.1

• We can also verify our plot with MATLAB.

clc; num=[2]; den=[1 0]; sys=tf(num,den);

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Example 2

• Sketch the magnitude Bode plot for the following system:

• Note : this transfer function is said to a differentiator.

( )

G s =s

Solution to Example 2

• Since the function is in the standard Bode form, therefore, we have:

• Determining the magnitude gives:

• Putting in decibel yield: ( )

( )

G s s

G jω jω

= =

2 2

( ) 0

G jω = +ω =ω

( ) 20 log

(19)

• Let us generate the table of values for this function:

ω G(jω)

0.1 -20

1 0

10 20

100 40

1000 60

10000 80

Solution to Example 2

• Next, we plot this function:

1 100 1000

log(jω)

|G

(j

ω

)|

20

10 0.1

(20)

• MATLAB verification:

clc; num=[1 0];

den=[1]; sys=tf(num,den);

sgrid on; bode(sys)

How to Sketch a Bode Plot (2)

• Just like a root locus, there are few rules that you need to follow:

1. Determine the transfer function of the system.

2. Rewrite the transfer function into standard Bode form:

(

)

(

11

)

( ) ( )

( )

K s z Y s

G s

X s s s p

+

= =

+

1 1

1 ( )

1 s K z

z G s

s s p

p

 

+

 

 

=

 

+

 

 

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Pause for a moment.

• Exercise : Rewrite the following transfer function in standard Bode form.

• Answer:

(

)

(

)(

)

30 10 ( ) 3 50 s G s s s + = + + 2 1 10 ( ) 1 1 3 50 s G s s s   +     =     + +        

How to sketch a Bode plot?

• Continuing on:

3. Replace the term “s” with s=jω. Hence:

4. Next, find the magnitude for the above transfer function in decibels (dB).

1 1 1 1 1 ( ) 1 j K z z G s j j p p ω ω ω   +     =   +     1 1 1 ( ) 20 log

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How to sketch a Bode plot?

• Continuing on by expanding the previous expression:

• From the above equation, we should observe the significance of each of the above expression:

– Constant term, K.

– Poles and zeros at origin, jω.

– Poles and zeros not at the origin : or ( ) 1 1 1 1 expand expand 1 1 1 1

( ) 20 log 1 20 log 1

( ) 20 log log log 1 20 log log log 1

j j

G s K z j p

z p

j j

G s K z j p

z p

ω _ω ω

      =   + −   +                =  + +  + −  + +  +          1 log j 1

z ω   +     1 log j 1

p ω   +    

How to sketch a Bode plot?

• First, let us look at the effect of the constant term, K, first.

• We take note that the constant term contribute a straight horizontal line of magnitude 20log(K).

0.1 1 10 100

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• Second, let us look at the individual zeros and poles at the origin.

• Now, this is important:

– A zero at the origin occurs when there is an s or jω multiplying the numerator. Each occurrence of this causes a positively sloped line passing through ω = 1 with a rise of 20 db over a decade (steps of 10). In another word, the slope = 20dB/decade.

– A pole at the origin occurs when there are s or jω multiplying the denominator. Each occurrence of this causes a negatively sloped line passing through ω = 1 with a drop of 20 db over a decade. In another word, the slope = -20dB/decade

How to sketch a Bode plot?

• Let us study the first statement (zeros):

0.1 1 10 100

log(jω)

|G

(j

ω

)|

20dB

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• Let us now study the second statement (pole):

0.1 1 10 100

log(jω)

|G

(j

ω

)|

-20dB 20dB

• Third, let us look at the effect of zeros and poles not located at the origin.

• Note that the zeros and poles not located at the origin is given the following expression:

• The values z1and p1in each of these expression is called a critical

frequency (or break frequency or corner frequency).

• Below their critical frequency these terms do not contribute to the log magnitude of the overall plot. Above the critical frequency, they represent a ramp function of 20 db per decade.

• Zeros give a positive slope. • Poles produce a negative slope.

1 log j 1

z ω

 

+

 

 

1 log j 1

p ω

 

+

 

(25)

How to sketch a Bode plot?

• Here is what is going to happen (note that the slope is 20 dB per decade):

0.1 1 10 100

log(jω)

|G

(j

ω

)|

-20dB -20dB

zeros

poles

Corner/ critical/ break frequency p1or z1

• Lastly, to complete the sketch, then we need to superposition all the lines of the different

terms on the same plot.

• When we say superposition, it means that we need to ADD all the lines to provide the resultant lines.

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Example 3

• Sketch the Bode magnitude plot for the following transfer function:

• Note, we will use BOTH methods to plot the graph. You know, it doesn’t really matter which method you choose, because at the end of the day, you will get the same plot.

1 ( )

2 100

G s s

= +

Solution to Example 3 : Method (1)

• Step 1 : Repose the above equation in standard Bode form.

• If you notice, the standard Bode form for this particular transfer function is of the following form:

• And we note that K=0.01 and p₁=50.

(

)

1 1 1 1 1 /100 0.01

( )

2 100 2 50 2 ₁ 0.02 1

50 1

50 50

G s

s s

s s s

= = = ⋅ = =

+ +   ₊ +

+

 

 

1 ( )

1

K G s

s p

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Solution to Example 3 : Method (1)

• We note since we have a pole, then the graph will have a ramp going downwards at p₁=50.

• The magnitude of this transfer function is, after substituting s=jω:

( )

(

)

(

)

(

)

2

2 2

2

constant (K)

poles not at origin

0.01 1

( ) 20 log 20 log 0.01 log 1 0.0004

2 1 0.02

( ) 20 log 0.01 10 log 1 0.0004 ( ) 40 10 log 1 0.0004

G j

G j G j

ω ω

ω

ω ω

 

 

 

= _ _= _ − + _

+

 

= − +

= − − +

• Step 2 : Sketch!

– Here, we first sketch K at -40dB and corner frequency of the poles at p₁=50.

50

log(jω)

|G

(j

ω

)|

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Solution to Example 3 : Method (1)

• Step 3 : Lastly, use superposition to find the resultant plot. We add the red line and the orange line. The result would be:

• Now, this is Bode plot of the magnitude G(jω). 50 log(jω) |G (j ω )| -40dB

• Step 1 : Determine the magnitude of the equation.

( )

(

)

(

)

2 ₂ 2 2

2 2 2 2 2 2 1 ( ) 2 100 1 ( ) 2 100 1 1 ( ) 4 100 2 100 1

( ) 20 log 20 log1 10 log 4 100

4 100

( ) 10 log 4 100

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Solution to Example 3 : Method (2)

• Step 2 : Let us tabulate the results.

ω G(jω)

0.1 -40

1 -40.0004 10 -40.0432 100 -43.0103 1000 -60.0432 10000 -80.0004

• Let us now sketch the Bode plot:

1 100 1000

log(jω) |G(jω)|

-40dB

10 0.1

-60dB

-80dB

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• We can verify our solution with MATLAB.

num=[1]; den=[2 100]; sys=tf(num,den); sgrid on;

bode(sys)

Example 4

• Determine the Bode plot for the following system:

4 2

5 10 ( )

505 2500

s G s

s s

× =

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• First, we factorize the expression:

• Second, we write it in standard Bode form:

• And we take note of the gain (K)=20, zeros at the origin and poles at p₁=5 and p₂=500.

(

)(

)

4 4

2

5 10 5 10

( )

505 2500 5 500

s s

G s

s s s s

× ×

= =

+ + + +

4

5 10 20

( )

5 500 1 1 1 1

5 500 5 500

s s

G s

s s s s

×

= =

      

⋅  +   +   +  + 

      

Solution to Example 4

• Next, we calculate the magnitude in decibel:

constant (K) _zero 20 ( ) 20 log

1 1

5 500

( ) 20 log 20 log log 1 log 1

5 500

( ) 20 log 20 20 log 20 log 1 20 log 1

5 500

( ) 26.02 20 log

j G j

j j

G j j

j j

G j j

ω

ω _ω _ω

ω ω ω ω ω ω ω ω ω ω       =      ₊ ₊     _ _{ } _        =  + −  + −  +            = + −  + −  +      = + at origin

20 log 1 20 log 1

5 500

jω jω

   

−  + −  + 

   

(32)

• And we plot individually:

1

log(jω)

|G

(j

ω

)|

26dB

5 ₅₀₀

-20dB

• And we superpose all the lines (add them):

• You may verify your answer by using the “tabulation method” – do it yourself!

log(jω)

|G

(j

ω

)|

(33)

Solution to Example 4

clc; num=[50000 1]; den=[1 505 2500];

sys=tf(num,den); sgrid on; bode(sys)

Example 5

• Sketch the Bode plot of the magnitude G(s) for the following system:

(

5

)

( )

1

G s

s s

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• Step 1 : Rewrite the transfer function in standard Bode form:

• We take note that the above form is already in standard Bode form.

• The gain is K=5. And there will be a ramp with negative slope at ω=1 (due to the term 1/s – integrator).

(

5

)

( )

1

G s

s s

= +

Solution to Example 5

• Next, calculate the magnitude in decibel after substituting s=jω.

• The vertical intercept is found by approximating ω=0.1.

(

)

(

)

(

)

(

)

1 pole at origin pole not at origin i.e. at =1 i.e. but at p =1

5 ( ) 20 log ( ) 20 log

1

( ) 20 log 5 log log 1

( ) 14 20 log 20 log 1

G j G j

j j

G j j j

ω

ω ω

ω ω ω

 

= =  

+

 

 

= _ − + + _

= − − +

(

)

( ) 14 20 log 20 log 1 33.17 33

(35)

• However, making approximation such as ω=0.1 is not accurately correct. The correct method is:

( )

(

)

(

)

(

)

2 2

2 4 2

2 2

4 2

5 5 5

( )

1

5 5

( )

5 ( ) 20 log

( ) 20 log 5 10 log

( ) 14 10 log

( ) 14 10 log 0.1 0.1 34

G j

j j j j

G j G j G j G j G j ω

ω ω ω ω ω ω

ω

ω ω

ω

ω ω

ω ω ω

ω = = = + − − + = = + − +   =   +   = − + = − + = − + ≃

Express the magnitude in decibels.

Substitute s=jω

Calculate the magnitude.

Substitute s=0.1.

Solution to Example 5

• The approximate magnitude Bode plot is:

(36)

• Superposing all the lines gives the following the final plot:

1

log(jω)

|G

(j

ω

)|

0.1

34dB

10

clc; num=[5]; den=[1 1 0]; sys=tf(num,den);

(37)

Bode Phase Plot

• Now, remember that I told you earlier that Bode plot consists of two plots, right?

• The first plot is the magnitude Bode plot. The second plot is the phase plot dealing with phase shifts.

• The second plot is really a plot of phase angle versus the input frequency.

• Let us now learn how to calculate the phase Bode plot.

Review : Complex Phasors

• Suppose that we have the following complex numbers:

• In polar form, we have:

• Take careful note of the highlighted section above.

( )( )

1 2 1 1 2 2 A

Z =z z = a + jb a + jb

( )

3 4 3

3 4 4 4 4

B

r r

Z

r r

θ _{θ θ}

θ ∠

= = ∠ −

∠

3 3 3 4 4 4 B

z a jb Z

z a jb

+ = =

+ and

and

( )( ) ( )

1 2 1 1 2 2 1 2 1 2 A

(38)

Bode Phase Plot

• We consider the following transfer function:

• Now, writing it in standard Bode form, we have:

(

)

(

11

)

( ) ( )

( )

K s z Y s

G s

X s s s p

+ = = + 1 1 1 1 1 ( ) 1 j K z z G s j j p p ω ω ω   +     =   +    

Bode Phase Plot

• Next, the cumulative phase angle associated with this function is given by:

• Let us now take a look at the effect of each terms in Bode phase plot.

1 1 1 1 1 1 1 1 1

( ) 1 1

1

j

K z

z j j

G s K z j p

z p

j

j p

p

ω

ω _ω ω

(39)

Bode Phase Plot

• Effect of positive constant, K.

• If K is positive, start your graph (with zero slope) at 0 degrees.

• If K is negative, start your graph with zero slope at 180 degrees (or -180 degrees, they are the same thing).

Bode Phase Plot

• Let us now look at the effect of zeros and poles at the origin.

• Zeros at the origin cause a constant +90 degree shift for each zero.

• Poles at the origin cause a constant -90 degree shift for each pole.

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Bode Phase Plot

• Effect of zeros at origin:

• Effect of poles at origin: 1

log(jω)

0.1 ₁₀

Ph

as

e

90°

1

log(jω)

0.1 ₁₀

Ph

as

e

-90°

Bode Phase Plot

• Let us now look at the effect of zeros not at origin.

• The rule says that for every zero, slope the line up at 45 degrees per decade when ω=z_n/10, that is, 1 decade before the break frequency.

• Multiple zeros means that the slope is steeper.

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Bode Phase Plot

• Effect of zeros not at origin:

• From our transfer function in standard Bode form, the phase line of 45° for the zeros not at origin will slope-up starting at z₁/10 until z₁*10.

log(jω) 0.1

Ph

as

e

45°

10 1 _decade decade 90°

Bode Phase Plot

• Let us now look at the effect of poles not at origin.

• The rule says that for every pole, slope the line down at 45 degrees per decade when ω=p_n/10 that is, 1 decade before the break frequency.

• Multiple poles means that the slope is steeper.

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Bode Phase Plot

• Effect of poles not at origin:

• From our transfer function in standard Bode form, the phase line of 45° for the poles not at origin will slope-down starting at p₁/10 until p₁*10.

log(jω) 0.1

Ph

as

e

-45°

10 1 _decade decade

-90°

Bode Phase Plot

• Important note:

– When drawing the phase angle shift for not-at-the-origin zeros and poles, first locate the critical/corner/braking frequency of the zero or pole.

– Then start the transition 1 decade before, following a slope of ±45° per decade.

– Continue the transition until reaching the frequency one decade past the critical frequency.

• Do not forget that you still need to do superposition for lines drawn for each term.

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Example 6

• Given the following transfer function:

• Sketch the Bode phase plot. Then, verify your result with MATLAB.

2 ( )

G s s

=

• Now, the transfer function is in standard Bode form.

• Finding the magnitude in decibel gives:

• Now, our Bode phase plot is (in next slide): 2

( )

G s s

=

constant, K

poles at origin 2

( ) 20 log 20 log 2 20 log 6.02 20 log

G j j j

j

ω ω ω

ω

 

=  = − = −

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Solution to Example 6

• The constant is positive, so it has not much of an effect in terms of the phase.

• We have a pole at origin, then, there will be a phase shift of -90°. And since there are no poles located at the origin, the our Bode phase plot is just a simple straight line.

Ph

as

e

log(jω)

0.1 1 10

-90°

Solution to Example 6

clc; num=[2]; den=[1 0]; sys=tf(num,den);

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Example 7

• Consider the transfer function discussed in Example 5 having the following transfer function.

• Let us now determine the Bode phase plot.

(

5

)

( )

1

G s

s s

= +

• First, write the transfer function in standard Bode form. We see that the transfer function is already in standard Bode form:

• The gain, K is 5. It has no effect on the Bode phase plot. But we have two poles – one pole is at the origin (ω=1) and the other pole not located at the origin is also at (ω=1).

• It simply means that the pole at the origin will shift the graph -90° and for the other pole, at ω=1, there will be a slope going down at 45° per decade. The starting point for the slope to go down is at p1/10=1/10=0.1 and the ending point for the slope is at p1*10.

(

5

)

( )

1

G s

s s

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Solution to Example 7

• Let us now sketch the Bode phase plot for individual term:

log(jω)

0.1 1 10

-90°

Ph

as

e

-135° 45°

-180° 45°

0.01

101

100

10-1

10-2

clc; num=[5]; den=[1 1 0]; sys=tf(num,den);

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Next Step

• Textbook reference : Chapter 10.

• Homework 14 has been posted on the course website. Attempt them. You do not have to submit Homework 14 as it will not be graded.