Contents Frequency response: Nyquist and Bode plots

Contents1.1 Introduction…………………………..………....2

1.1.1 Time response……………………..2

1.1.2 Frequency response……………….2

1.2 Frequency response…………………………….3

1.3 Nyquist plot………………………….………….9

1.4 Bode plot..……………………………………...10

1.5 Check-in questions.…………………………...15

1.6 Summary…………..…………………………...16

Frequency response:

Nyquist and Bode plots

James E Pickering

[email protected]

Key learning points 1) You will be able to explain the effect a

sinusoidal input frequency has on the

magnitude and phase angle of a system output

through the use of Nyquist and Bode plots

2) You will be able to explain the bandwidth of a

Bode plot

mailto:[email protected]

1.1.1 Time response

▪ A system behaves subject to a specific input in the time-domain

1.1 Introduction

▪ Useful to understand the

frequency response

properties of a system

1.1.2 Frequency response

▪ System response due a sinusoidal input

▪ A range of frequencies used

▪ System behaviour is determined from the steady-state response to a sinusoidal input of the

following form:

𝑅 = 𝐴𝑠𝑖𝑛𝜔𝑡 (1-1)

Frequency response: Nyquist and bode plots

1.2 Frequency

response

▪ Why is this useful?

▪ Effectively, what range of

frequencies the system can

track well

▪ Undertaken on open-loop

and closed loop feedback

control systems

▪ Interesting practical

example, steps forming a

sine wave


▪ Steady state gain is frequency dependant and is given by:

▪ The phase shift 𝜙 is such that the response ‘lags’ behind the input or rather is shifted in phase

𝐺(𝑠)𝑌(𝑠)𝑈(𝑠)

𝐴𝑠𝑖𝑛𝜔𝑡 𝐵𝑠𝑖𝑛(𝜔𝑡 − 𝜙)

In this case, 𝐴 = 1 and 𝐵 = 0.8

3

𝑠2 + 2𝑠 + 4

𝑌(𝑠)𝑈(𝑠)


𝐵

𝐴

(1-2)

1.2 Frequency

response

▪ Sinusoidal signal applied

to a linear system

▪ Output will be a sinusoid

of the same frequency as

the input, but with

amplitude and phase

being dependent on the

frequency 𝜔 of the input

sinusoid


▪ The output is of the same frequency as the input, but with a phase shift

▪ The phase shift value is expressed as a negative quantity

One period = 360°

𝜙 Phase shift (angle in degrees)

3

𝑠2 + 2𝑠 + 4

𝑌(𝑠)𝑈(𝑠)


1.2 Frequency

response


▪ Consider the following continuous-time transfer function:

𝐺 𝑠 =5

2𝑠2 + 3𝑠 + 1=

2.5

𝑠2 + 1.5𝑠 + 0.5▪ Determine the poles:

−𝑏 ± 𝑏2 − 4𝑎𝑐

2𝑎=−1.5 ± (1.5)2−4(0.5)

2=−1.5 ± 2.25 − 2.0

2=−1.5 ± 0.5

2

Poles at -0.5 and -1.

▪ Equivalent 𝐺(𝑗𝜔) replacing 𝑠 by 𝑗𝜔, gives:

𝐺 𝑗𝜔 =2.5

(𝑗𝜔 + 1)(𝑗𝜔 + 0.5)

▪ For simplicity, when working in the frequency domain, rearrange to give all the linear

factors to be of the form (𝛽𝑗𝜔 + 1):

𝐺 𝑗𝜔 =2.5

𝑗𝜔 + 1 2𝑗𝜔 + 1 0.5=

5

𝑗𝜔 + 1 2𝑗𝜔 + 1

1.2 Frequency

response

▪ When exploring the

frequency domain, 𝐺 𝑠becomes 𝐺(𝑗𝜔), i.e.

replacing 𝑠 by 𝑗𝜔


▪ Each linear factor is dealt with separately

▪ Each will have a magnitude and phase which is a function of 𝜔

▪ In both the numerator and denominator, the magnitudes of each term multiple and the

phase angles are summed

▪ Then the overall magnitude is 𝑁

𝐷and the phase ∠𝑁 − ∠𝐷

▪ Hence, 𝑁 = 5 and ∠𝑁 = 0°

▪ Consider the general first order term 𝑗𝜔 + 1

▪ Magnitude is from Pythagoras, given by:

▪ Phase is given by:

𝑗𝜔 𝜔

∠

1

where 𝜔 = 𝑟𝑎𝑑/𝑠

1.2 Frequency

response

▪ Recall from the previous

slide:

𝐺 𝑗𝜔

=2.5

𝑗𝜔 + 1 2𝑗𝜔 + 1 0.5

=5

𝑗𝜔 + 1 2𝑗𝜔 + 1

= 𝜔 2 + 1 (1-3)

∠ = tan−1𝑜𝑝𝑝

𝑎𝑑𝑗, i.e. tan−1

𝜔

1or tan−1𝜔

(1-4)


▪ In the general case 𝛽𝑗𝜔 + 1

▪ Magnitude is given by:

▪ Phase is given by:

Frequency (𝐑𝒂𝒅/𝒔)

Numerator Denominator Overall 𝑮(𝒋𝝎)

𝒋𝝎 + 𝟏 𝟐𝒋𝝎 + 𝟏

∠ ∠ ∠ ∠

0.01 5 0 1.000 0.5729° 1.0002 1.1458° 4.9988 -1.7187°

0.05 5 0 1.0012 2.8624° 1.0050 5.7106° 4.9690 -8.5730°

0.1 5 0 1.0050 5.7106° 1.0198 11.3099° 4.8786 -17.0205°

0.2 5 0 1.0198 11.3099° 1.0770 21.8014° 4.5522 -33.1113°

0.5 5 0 1.1180 26.5651° 1.4142 45° 3.1623 -71.5651°

1 5 0 1.414 45° 2.236 63.434° 1.5811 -108.4349°

2 5 0 2.2361 63.4349° 4.1231 75.9638° 0.5423 -139.3987°

5 5 0 5.0990 78.6901° 10.0499 84.2894° 0.0976 -162.9795°

10 5 0 10.0499 84.2894° 20.0250 87.1376° 0.0248 -171.4270°

e.g. 𝝎 = 𝟏𝒓𝒂𝒅/𝒔

Consider:

= (𝛽𝜔) 2 + 1∠ = tan−1 𝛽𝜔

Denominator - 𝒋𝝎 + 𝟏 :

= 2 = 1.414∠ = tan−1 1 = 45°

Denominator - 𝟐𝒋𝝎 + 𝟏 :

= 5 = 2.236∠ = tan−1 2 = 63.434°

Overall - 𝑮(𝒋𝝎):= 5/(1.414)(2.236)= 1.581

∠ = 0 − 108.434= −108.434°

1.2 Frequency

response

= (𝛽𝜔) 2+1 (1-5)

∠ = tan−1𝛽𝜔 (1-6)


▪ Polar plot on Argand diagram of open loop frequency locus

✓ Amplitude ratio is radial coordinate

✓ Phase angle is the angular coordinate

Frequency (𝑹𝒂𝒅/𝒔)

∠

0.01 4.9988 -1.7187

0.05 4.9690 -8.5730

0.1 4.8786 -17.0205

0.2 4.5522 -33.1113

0.5 3.1623 -71.5651

1 1.5811 -108.4349

2 0.5423 -139.3987

5 0.0976 -162.9795

10 0.0248 -171.4270

1.3 Nyquist plot


▪ Two logarithmic plots

✓ Amplitude ratio versus frequency

✓ Phase angle versus frequency

▪ Amplitude measured in dB, converted as follows:

• Phase measured in degrees

Frequency (𝑅𝑎𝑑/𝑠) (dB)

∠

0.01 4.9988 13.98 -1.7187

0.05 4.9690 13.93 -8.5730

0.1 4.8786 13.77 -17.0205

0.2 4.5522 13.16 -33.1113

0.5 3.1623 10.00 -71.5651

1 1.5811 3.98 -108.4349

2 0.5423 -5.32 -139.3987

5 0.0976 -20.21 -162.9795

10 0.0248 -32.11 -171.4270

1.4 Bode plot𝐴𝑅 𝑑𝐵 = 20log(𝐴𝑅) (1-7)


Bandwidth considerations

▪ The bandwidth is the maximum frequency that an input sinusoidal signal will be tracked well by a

control system

▪ High bandwidth value corresponds to a fast response in the time-domain

▪ Low bandwidth value corresponds to a slow response in the time-domain

1.4 Bode plot

▪ Closed loop tracking control

is typically low pass in

nature with unity gain at low

frequencies

▪ Consideration to the

application and sampled data

control system Bandwidth


1.4 Bode plot

▪ The cut-off/corner

frequency is defined as

the boundary in a

system’s frequency

response at which energy

in the system is reduced

to 70.79%

Bandwidth

At low frequencies, output equals input

At high frequencies, output is not equal to input

At high frequencies, phase goes towards -180°

At low frequencies, phase is near 0 °

-3dB

Cut-off frequency

~0.50𝑟𝑎𝑑/𝑠~ 0.08 𝐻𝑧


Bandwidth considerations: 1st order

▪ Consider the following closed loop transfer function:

𝐺𝐶𝐿 𝑠 =𝜔0

𝜏𝑠 + 𝜔0=

1

𝑠 + 1

▪ Note that 𝜔𝐵𝑊 = 𝜔0, with the following given:

𝜏 =1

𝜔0

▪ Higher bandwidth corresponds to faster response

1.4 Bode plot

𝑦(𝑡) = 1 − 𝑒𝑡


1.4 Bode plot

▪ Consider the general form of a

second order transfer function:

𝐺 𝑠 =𝜔𝑛2

𝑠2 + 2𝜁𝜔𝑛𝑠 + 𝜔𝑛2

where 𝜔𝐵𝑊 = 𝜔𝑛

▪ Considering 𝜁=0.7, the rise

time in time-domain is given by

(Franklin, G.F., Powell, J.D.,

Emami-Naeini, 2015):

≅1.8

𝜔𝑛

▪ Higher bandwidth corresponds

to faster response

Bode Plots Overview. Available at: https://lpsa.swarthmore.edu/Bode/Bode.html [Assessed 01/10/2020]

Franklin, G.F., Powell, J.D., Emami-Naeini, A. and Sanjay, H.S., 2015. Feedback control of dynamic

systems. London: Pearson.


https://lpsa.swarthmore.edu/Bode/Bode.html

Exercise 1

Considering the following continuous-time transfer function:

𝐺 𝑠 =2

(𝑠 + 1)(𝑠 + 0.25)

Determine the magnitude and phase when subject to the following sinusoidal inputs:

i. 0.1𝑟𝑎𝑑/𝑠ii. 1𝑟𝑎𝑑/𝑠iii. 5𝑟𝑎𝑑/𝑠

Exercise 2

Considering the following continuous-time transfer function:

𝐺 𝑠 =3

𝑠2 + 1.6𝑠 + 0.6

Determine the magnitude and phase when subject to the following sinusoidal inputs:

i. 0.1𝑟𝑎𝑑/𝑠ii. 1𝑟𝑎𝑑/𝑠iii. 5𝑟𝑎𝑑/𝑠

1.5 Check-in

questions


1.6 Summary

▪ The frequency response has been explored and its practical

application

▪ The relationship between the sinusoidal input frequency and system

output magnitude and phase angle has been discovered

▪ The bandwidth of a bode plot has been explained and it’s importance

Contents1.1 Introduction…………………………..………....2

1.1.1 Time response……………………..2

1.1.2 Frequency response……………….2

1.2 Frequency response…………………………….3

1.3 Nyquist plot………………………….………….9

1.4 Bode plot..……………………………………...10

1.5 Check-in questions.…………………………...15

1.6 Summary…………..…………………………...16

Key learning points 1) You will be able to explain the effect a

sinusoidal input frequency has on the

magnitude and phase angle of a system output

through the use of Nyquist and Bode plots

2) You will be able to explain the bandwidth of a

Bode plot


Documents

Contents Frequency response: Nyquist and Bode plots