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1

Chapter 5

Sinusoidal Input

2

Chapter 5

Examples:

1. 24 hour variations in cooling water temperature.
2. 60-Hz electrical noise (in USA!)

Processes are also subject to periodic, or cyclic, disturbances. They can be approximated by a sinusoidal disturbance:

( ) ( )

sin

0 for 0

sin for 0

t

U t

A ω t t

where: A = amplitude, ω = angular frequency

sin ( )^2

A

U s

s

ω ω

4

( )

(^2 2 2 )

2 2

2 2

( ) sin( ) (^1 )

sin( ) sin( ) 1 where

amplitude ratio 1 arctan( ) phase angle

y t AK^ e t AK t

y AK t B t

B K A

ωτ τ ω φ ω τ ω τ

ω φ ω φ ω τ

ω τ φ ωτ

= − + +

• (^) +

∞ + = +

= =

= − =



Note that the amplitude ratio and phase angle is not a function of t but of τ and ω. For large t, y(t) is also sinusoidal, output sine is attenuated by…

Inverting, (^) this term dies out for large t

1

1 ω^2 τ^2 +

Chapter 5

Figure 13.1 Attenuation and time shift between input and output sine waves ( K = 1). The phase angle of the output signal is given by , where is the (period) shift and P is the period of oscillation.

φ

φ = − ∆ ( t / P )× 360 ^ ∆ t

which can, in turn, be divided by the process gain to yield the normalized amplitude ratio (AR (^) N ) (or magnitude ratio):

N (^ ) 2 2

AR

ω τ 1

Dividing both sides by the input signal amplitude A yields the amplitude ratio (AR)

2 2

AR

ω τ 1

A K

A

1. The output has a phase shift, φ, relative to the input. The amount of phase shift depends on frequency, i.e.,

( ) 2 tan^1

t P



8

Basic Theorem of Frequency

Response

( ) ( )

The frequency response of a system can be found

by substituting for in the system transfer function,

i.e.

Laplace domain ferquency domain

s j

j s

G s G j

ω

ω

ω

Shortcut Method for Finding

the Frequency Response

The shortcut method consists of the following steps:

Step 1. Set s=j ω in G ( s ) to obtain. Step 2. Rationalize G ( j ω); We want to express it in the form. G ( j ω)= R + jI where R and I are functions of ω. Simplify G ( j ω) by multiplying the numerator and denominator by the complex conjugate of the denominator. Step 3. The amplitude ratio and phase angle of G(s) are given by:

G ( j ω)

2 2 1

AR

tan ( / )

R I

ϕ − R I

Memorize ⇒

Example 13.

Find the frequency response of a first-order system, with

τ 1

G s s

Solution

First, substitute s = j ω in the transfer function

ω (13-17) τ ω 1 ωτ 1

G j j j

Then multiply both numerator and denominator by the complex conjugate of the denominator, that is, −^ j ωτ^ +^1

2 2

2 2 2 2

ωτ 1 ωτ 1 ω ωτ 1 ωτ (^1) ω τ 1 1 ωτ (13-18) ω τ 1 ω τ 1

j j G j j j

j R jI

• • Docsity.com

Consider a complex transfer G ( s ),

1 (^ )^2 (^ )^3 (^ )

ω ω ω ω (13-23) ω ω ω

G j G^ a^ j^ Gb^ j^ Gc^ j G j G j G j

From complex variable theory, we can express the magnitude and

angle of G ( j ω) as follows:

1 (^ )^2 (^ )^3 (^ )

ω ω ω ω (13-24a) ω ω ω

G j G^ a^ j^ Gb^ j^ Gc^ j G j G j G j

1 (^ )^2 (^ )^3 (^ )

ω ω ω ω [ω ω ω ] (13-24b)

G j G a j Gb j Gc j G j G j G j

Complex Transfer Functions

1 ( )^2 ( )^3 ( )

G s G^ a^ s Gb^ s Gc^ s G s G s G s Substitute s=j ω,

14

Transfer Functions in Series

( ) ( ) ( )^ ( )

( ) (^ )^ ( ) (^ )^ ( )

( )

1

1

1

1 1

1 1

1

1

ln ln

ln ln

n i (^) i i

n i i

n i i G j n^ G j n^ G^ j i i^ i i n n i i^ i i

n i i n i i

Y s (^) G s G s X s s j G j G j

G j e G j e G j e

G j j G j G 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

=

=

= ∠ ∠ ∠ = =

= =

=

=

= =

=

  ∑ = =    

• ∠ = ^ + ∠   ⇒ =

∠ = ∠

∏

∏

∏ ∏

∏ ∑

∑

∑

16

Bode Diagrams

• A special graph, called the Bode diagram or

Bode plot , provides a convenient display of the

frequency response characteristics of a transfer

function model. It consists of plots of AR and

phase angle as a function of frequency.

• Ordinarily, frequency is expressed in units of

radians/time.

17

Bode Plot of A First-order System

( )

N 2

N

N

Recall:

1 AR φ tan ωτand ω τ 1

ω 0 and ω 1) :

AR 1 and

ω 0 and ω 1) :

ωτ and AR 1/

= = −^ −

• → τ

= ϕ = 0

• → τ

= ϕ = −

At low frequencies (

At high frequencies ( 





• Note that the asymptotes intersect at , known as the break frequency or corner frequency. Here the value of AR (^) N from (13-21) is:

ω = ω (^) b =1/ τ

N^ (^ )

ARω ω 0.707 (13-30) b 1 1

• Some books and software defined AR differently, in terms of decibels. The amplitude ratio in decibels AR (^) d is defined as AR (^) d =20 log AR (13-33)

20

G ( j ω )= K ( ωτ j +1)

2 2

1

ARω ω τ 1

φ ω tan − ωτ

G j K

G j

Negative Zeros

Substituting s=j ω gives

Consider a process zero term,

G s ( ) = K s (τ +1)

Thus:

Note : In general, a multiplicative constant (e.g., K ) changes the AR by a factor of K without affecting φ.