Frequency Response and Bode Plots

Supplementary notes for ME2143 (optional)

Peter Chen, Ph.D.

Associate Professor

Department of Mechanical Engineering

Faculty of Engineering

National University of Singapore

http://guppy.mpe.nus.edu.sg/peter chen/

c�2000-2014

1

2 Frequency Response and Bode Plots (For use by ME2143 only)

1 Introduction

We normally analyze the stability and performance of a control system by exam-

ining its response (i.e., output) in two domains. One is the so-called time domain,

in which the output of a system is analyzed with respect to time. The other is the

so-called frequency domain, in which the output of a system is analyzed with re-

spect to its characteristics of oscillation. In the frequency domain, the independent

variable is the frequency of a signal (normally denoted by the symbol ω).

In these notes, we review some basic concepts of representing sinusoids as

complex numbers, and analyze the response of a linear system when subjected

to a sinusoidal input. We then discuss a method of representing such responses

graphically using diagrams called Bode plots.

2 Sinusoids and their complex function representa-

tion

We refer to a sinusoid as a time function of the form

y = A sin(ωt + φ) , (1)

where A is called the (peak) amplitude and ω is called the frequency of oscillation

(whose unit is rad/sec). For example, y = 5 sin(0.5t + π4) is a sinusoid. Another

example of a sinusoid is y = sin(ωt + π2) = cos(ωt). We use sin (instead of cos)

in the definition of a sinusoid as a matter of convention.

A complex number can be expressed as a complex sinusoid. A complex num-

ber s = σ + jω can be graphically identified as a point in the s-plane as shown in

Figure 1, where

r = |s| =�σ 2 + ω2 , (2)

and

θ = tan−1�ωσ

�. (3)

From Figure 1, we see that

σ = r cos θ , (4)

and

ω = r sin θ. (5)

Therefore, a complex number s = σ + jω can be expressed as

s = σ + jω = r cos θ + jr sin θ = r(cos θ + j sin θ) . (6)

Peter Chen 3

Figure 1: Polar coordinates of a complex number.

Note that complex sinusoids can be expressed in terms of complex exponen-

tials by Euler’s identity (also called Euler’s formula). Euler’s identity has the form

e jθ = cos θ + j sin θ , (7)

or

e− jθ = cos θ − j sin θ . (8)

Hence, we can express a complex number s compactly as

s = r(cos θ + j sin θ) = re jθ . (9)

3 The concept of frequency response

We define the frequency response of a system as the steady-state response of the

system to a sinusoidal input. Frequency response analysis is based on the fact that,

when a system is subject to a sinusoidal input of frequency ω, the response of the

system is also sinusoidal at the same frequency. The output sinusoid differs from

the input sinusoid only in amplitude and phase angle. An explanation of this fact

can be found in textbooks, and so we will not discuss the general proof here. We

will, however, present a simple example to illustrate this fact.

Example 3.1 Determine the response of the first-order system subject to a sinu-

soidal input, as shown in Figure 2.

Solution: The input to the system, i.e., R(s) = Aωs2+ω2 , is the Laplace transform of

r(t) = A sin(ωt), which is a sinusoid. The amplitude of the sinusoid is A and the

frequency is ω. The output of the system is

C(s) = G(s)R(s) =�

1

s + 1

��Aω

s2 + ω2

�. (10)

Since

(s2 + ω2) = (s + jω)(s − jω) , (11)

4 Frequency Response and Bode Plots (For use by ME2143 only)

Figure 2: A first-order system subject to a sinusoidal input.

we can write equation (10) as a partial-fraction expansion, i.e.,

C(s) = Aω

(s + 1)(s + jω)(s − jω)= a

s + 1+ b

s + jω+ c

s − jω. (12)

The coefficients a, b, and c can be calculated as follows.

a =�

Aω

(s + jω)(s − jω)

�

s=−1= Aω

1+ ω2, (13)

b =�

Aω

(s + 1)(s − jω)

�

s=− jω

= − A

2 j (1− jω), (14)

c =�

Aω

(s + 1)(s + jω)

�

s= jω

= A

2 j (1+ jω). (15)

Thus,

C(s) = Aω

1+ ω2

1

s + 1+ A

2 j

�1

1+ jω

1

s − jω− 1

1− jω

1

s + jω

�, (16)

and the response in the time domain is

c(t) = Aω

1+ ω2e−t + A

2 j

�1

1+ jωe jωt − 1

1− jωe− jωt

�. (17)

Let A and φ1 denote the magnitude and the phase angle of 11+ jω

respectively, i.e.,

1

1+ jω= A � φ1 = Ae jφ1 . (18)

Then

A =����

1

1+ jω

���� =

|1||1+ jω| =

1√1+ ω2

, (19)

and

φ1 = ��

1

1+ jω

�= � (1+ j0)− � (1+ jω) = − tan−1(ω) . (20)

Similarly, we can write

1

1− jω=����

1

1− jω

���� ��

1

1− jω

�= Ae− jφ1, (21)

Peter Chen 5

where A and φ1 are as defined in equations (19) and (20).

Substituting equations (18) and (21) into equation (17) yields

c(t) = Aω

1+ ω2e−t + AA

2 j

e jφ1e jωt − e− jφ1e− jωt

= Aω

1+ ω2e−t + AA

2 j

�e j (ωt+φ1) − e− j (ωt+φ1)

�. (22)

From Euler’s identity, i.e., equations (7) and (8), we have

e j (ωt+φ1) − e− j (ωt+φ1)

= cos(ωt + φ1) + j sin(ωt + φ1)−�cos(ωt + φ1) − j sin(ωt + φ1)

�

= 2 j sin(ωt + φ1) . (23)

So equation (22) becomes

c(t) = Aω

1+ ω2e−t + AA

2 j

�2 j sin(ωt + φ1)

�

= Aω

1+ ω2e−t + AA sin(ωt + φ1) . (24)

Recall that the frequency response of a system is the steady-state response of

the system to a sinusoidal signal. We can now obtain the frequency response of

the first-order system (shown in Figure 2) by determining c(t) as t goes to infinity,

i.e.,

css = limt→∞

c(t) = limt→∞

�Aω

1+ ω2e−t + AA sin(ωt + φ1)

�. (25)

Since limt→∞ e−t → 0, equation (25) becomes

css = A sin(ωt + φ1) , (26)

where

A = AA . (27)

From equation (26), we see that the steady-state response of the first-order

system to a sinusoidal input is also sinusoidal and at the same frequency as the

input. As a specific case, let the input be

r(t) = 2 sin(5t) . (28)

Then we have

A = 2 ,

ω = 5 ,

A = 1√1+ ω2

= 1√1+ 52

= 0.196 ,

φ = − tan−1(5) = −78.69◦ = −1.373 rad ,

A = AA = 2× 0.196 = 0.392 .

6 Frequency Response and Bode Plots (For use by ME2143 only)

Thus, the time response, from equation (24), is

c(t) = 0.3846e−t + 0.392 sin(5t − 1.373) . (29)

Figure 3 shows the input and the response.

0 1 2 3 4 5 6-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Time (sec)

Amplitude

Input Output

Figure 3: Sinusoidal input and response of a first-order system.

From Figure 3, we see that the response reaches the steady state after about

two periods (at around 2.3 second). The steady-state response (i.e., the frequency

response), from equation (26), is

css = 0.392 sin(5t − 1.373) . (30)

✷

From Example 3.1, we see that the frequency response of a first-order system

differs from the input signal only in amplitude and phase angle. Thus, we can

represent the frequency response of a system in terms of these differences.

The difference in amplitude is defined as the ratio between the amplitude of

the input and the amplitude of the steady-state response. This ratio is called the

gain of the system. The difference in phase angle is defined as the phase angle of

the response relative to the phase angle of the input. This difference is called the

phase shift. For instance, with reference to Example 3.1, the gain (denoted by the

symbol M(ω)) of the first-order system shown in Figure 2 is

M(ω) = amplitude of steady-state output

amplitude of input= A

A

= AA

A= A = 1√

1+ ω2(from equation (19)), (31)

Peter Chen 7

and the phase shift (denoted by the symbol φ(ω)) is

φ(ω) = phase angle of steady-state output− phase angle of input

= φ1 − 0 = φ1 = − tan−1(ω) (from equation (20)). (32)

We claim that the gain M(ω) and the phase shift φ(ω) as determined in equa-

tions (31) and (32) are in fact equal to the magnitude and the angle (respectively)

of the system transfer function G(s) = 1s+1 if we just let s = jω. We verify this

claim as follows. The transfer function G(s) as shown in Figure 2 can be written,

with s = jω, as

G( jω) = 1

jω + 1. (33)

The magnitude of G( jω) is

|G( jω)| = |1|| jω + 1| =

1√1+ ω2

. (34)

The angle of G( jω) is

� G( jω) = ��

1

jω + 1

�= � (1+ j0) − � (1+ jω) = − tan−1(ω) . (35)

Comparing equation (31) with equation (34), and equation (32) with equation

(35), we see that

M(ω) = |G( jω)| , (36)

and

φ(ω) = � G( jω) . (37)

The relationships as shown in equations (36) and (37) are true for linear time-

invariant systems. These are fundamental results in the theory of frequency re-

sponse analysis. We will not present the proof for these results here. We will

simply use these results as facts in our subsequent analysis.

We summarize the above discussion as follows. The frequency response of a

system is defined as the steady-state response of the system to a sinusoidal input.

Frequency response is characterized by two entities, namely, the gain M(ω) and

the phase shift φ(ω). The gain and the phase shift are equal to the magnitude and

the angle of the system transfer function G(s) (with s = jω) respectively. In other

words, G( jω) represents the frequency response of G(s).

We can determine the frequency response of a system analytically or graphi-

cally. We have seen, from Example 3.1, that the analytical approach could be quite

tedious. The graphical approach (discussed below), on the other hand, appears to

be simpler and more intuitive.

The frequency response of a system can be shown graphically in three types

of plots. The first is called the Bode diagram (or Bode plots), the second is called

the log-magnitude-phase diagram, and the third is called the polar plot or Nyquist

plot. In these notes, we only discuss Bode plots.

8 Frequency Response and Bode Plots (For use by ME2143 only)

4 Bode plots

A Bode diagram is a graphical representation of the frequency response of a trans-

fer function G(s). It consists of the following two plots.

1. Magnitude plot: A plot of the magnitude of G( jω) v.s. the frequency ω. The

vertical axis represents the quantity 20 log10(M(ω)). The horizontal axis

represents the quantity log10 ω. For brevity, we will simply use log to mean

log10 in the sequel.

2. Phase-angle plot: A plot of the phase angle of G( jω) v.s. the frequency ω.

The vertical axis represents the phase angle φ(ω) in a linear scale and nor-

mally in the unit of degrees. The horizontal axis represents the frequency ω

in an logarithmic scale.

In Bode plots, the ratio between two frequencies is expressed in terms of oc-

taves or decades. A ratio of 2 is called an octave. A ratio of 10 is called a

decade. For example, the ratio between ω2 = 50 rad/sec and ω1 = 25 rad/sec

is ω2

ω1= 50

25= 2 ≡ 1 octave, and the ratio between ω4 = 100 rad/sec and ω3 = 10

rad/sec is ω4ω3

= 10010

= 10 ≡ 1 decade. We also say that the frequency band

between ω1 and ω2 is 1 octave, and the frequency band between ω3 and ω4 is 1

decade.

Consider the first-order system G(s) = 1s+1 discussed earlier in Example 3.1.

The magnitude of G( jω) is M(ω) = |G( jω)| = 1√1+ω2

. To generate the mag-

nitude plot for a certain frequency range, say, between 0.01 and 102 rad/sec, we

calculate the quantity 20 log(M(ω)) = 20 log

�1√1+ω2

�for a set of values for

ω in that range. We then plot 20 log

�1√1+ω2

�against log(ω). The unit of the

quantity 20 log(M(ω)) is decibel (denoted by dB). Thus one decibel corresponds

to a value of 10120 for M(ω), since

20 log�10

120

�= 1 dB . (38)

Table 1 lists some of the data points for constructing the magnitude plot of G(s) =1

s+1 . The resulting magnitude plot is shown in the top graph in Figure 4. Note that

in the graph, the horizontal axis is in log scale and labeled with the value of ω.

To generate the phase-angle plot of G(s) = 1s+1 , we simply plot − tan−1(ω)

against ω for a set of ω in the chosen frequency range. Table 1 lists some data

points for constructing the phase-angle plot. The resulting phase-angle plot is

shown in the bottom graph in Figure 4.

Peter Chen 9

ω M(ω) = 1√1+ω2

20 logM(ω) φ(ω) = − tan−1(ω)

0.01 1.0 0.0 −0.6◦0.02 1.0 0.0 −1.1◦0.05 1.0 0.0 −2.9◦0.1 1.0 0.0 −5.7◦0.2 0.98 −0.18 −11.3◦0.5 0.89 −1.0 −26.6◦1 0.71 −3.0 −45.0◦2 0.45 −6.9 −63.4◦5 0.2 −14.0 −78.7◦10 0.1 −20 −84.3◦20 0.05 −26.0 −87.1◦50 0.02 −34.0 −88.9◦100 0.01 −40 −89.4◦

Table 1: Magnitude and phase angle data for Bode plots.

Frequency (rad/sec)

Phase (deg) Magnitude (dB)

-40

-35

-30

-25

-20

-15

-10

-5

0

10-2

10-1

100

101

102

-100

-80

-60

-40

-20

0

Figure 4: Bode diagram of 1s+1 .