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
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
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
Frequency response: Nyquist and bode plots
▪ 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)
𝐴𝑠𝑖𝑛𝜔𝑡 𝐵𝑠𝑖𝑛(𝜔𝑡 − 𝜙)
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
Frequency response: Nyquist and bode plots
▪ 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)
𝐴𝑠𝑖𝑛𝜔𝑡 𝐵𝑠𝑖𝑛(𝜔𝑡 − 𝜙)
1.2 Frequency
response
Frequency response: Nyquist and bode plots
▪ 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
Frequency response: Nyquist and bode plots
▪ 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)
Frequency response: Nyquist and bode plots
▪ 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)
Frequency response: Nyquist and bode plots
▪ 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
Frequency response: Nyquist and bode plots
▪ 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)
Frequency response: Nyquist and bode plots
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
Frequency response: Nyquist and bode plots
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
Frequency response: Nyquist and bode plots
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
Frequency response: Nyquist and bode plots
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.
Frequency response: Nyquist and bode plots
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
Frequency response: Nyquist and bode plots
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
Frequency response: Nyquist and bode plots