Lecture 8 - Frequency Response

Lecture 8 Slide 1 PYKC 8-Feb-11 E2.5 Signals & Linear Systems

Lecture 8

Frequency Response (Lathi 4.8 – 4.9)

Peter Cheung Department of Electrical & Electronic Engineering

Imperial College London

URL: www.ee.imperial.ac.uk/pcheung/teaching/ee2_signals E-mail: [email protected]


Frequency Response of a LTI System

  We have seen that LTI system response to x(t)=est is H(s)est. We represent such input-output pair as:

  Instead of using a complex frequency, let us set s = jω, this yields:

  It is often better to express H(jω) in polar form:

  Therefore

L4.8 p423

( )st ste H s e⇒

( )j t j te H j eω ωω⇒

cos Re( ) Re[ ( ) ]j t j tt e H j eω ωω ω= ⇒

( )( ) ( ) j H jH j H j e ωω ω ∠=

cos ( ) cos[ ( )]t H j t H jω ω ω ω⇒ +∠

Frequency Response

Amplitude Response

Phase Response


Frequency Response Example (1)

  Find the frequency response of a system with transfer function:

  Then find the system response y(t) for input x(t)=cos2t and x(t)=cos(10t-50°)

  Substitute s=jω

0.1( )5

sH ss+

=+

0.1( )5

jH jjω

ωω+

=+

21 1

2

0.01( ) and ( ) ( ) tan tan0.1 525

H j H j jω ω ωω ω ω

ω− −+ ⎛ ⎞ ⎛ ⎞= ∠ =Φ = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+



2

2

0.01( )25

H j ωω

ω

+=

+

1 1( ) ( ) tan tan0.1 5

j H j ω ωω ω − −⎛ ⎞ ⎛ ⎞Φ =∠ = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

for input x(t)=cos2t and x(t)=cos(10t-50°)



  For input x(t)=cos2t, we have:

  Therefore

2

2

2 0.01( 2) 0.3722 25

H j += =

+

1 12 2( 2) tan tan 65.30.1 5

j − −⎛ ⎞ ⎛ ⎞Φ = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) 0.372cos(2 65.3 )y t t= +



  For input x(t)=cos(10t-50°), we will use the amplitude and phase response curves directly:

  Therefore

( 10) 0.894H j =

( 10) ( 10) 26j H jΦ =∠ =

( ) 0.894cos(10 50 26 ) 0.894cos(10 24 )y t t t= − + = +


Frequency Response of delay of T sec

  H(s) of an ideal T sec delay is:

  Therefore

  That is, delaying a signal by T has no effect on its amplitude.

  It results in a linear phase shift (with frequency), and a gradient of –T.

  The quantity:

is known as Group Delay.

( ) (Time-shifting property)sTH s e−=

( ) 1 and ( )j TH j e j Tωω ω ω−= = Φ = −

( ) =Tgd

dω

τω

Φ− =


Frequency Response of an ideal differentiator

  H(s) of an ideal differentiator is:

  Therefore

  This agrees with:

  That’s why differentiator is not a nice component to work with – it amplifies high frequency component (i.e. noise!).

/ 2( ) and ( ) jH s s H j j e πω ω ω= = =

( ) and ( )2

H j H j πω ω ω= ∠ =

(cos ) sin cos( / 2)d t t tdt

ω ω ω ω ω π= − = +


Frequency Response of an ideal integrator

  H(s) of an ideal integrator is:

  Therefore

  This agrees with:

  That’s why integrator is a nice component to work with – it suppresses high frequency component (i.e. noise!).

/ 21 1 1( ) and ( ) jjH s H j es j

πωω ω ω

−−= = = =

1( ) and ( )2

H j H j πω ω

ω= ∠ = −

1 1cos sin cos( / 2)t dt t tω ω ω πω ω

= = −∫


Bode Plot – Sketching frequency response .. without (much) calculation (1)

  Consider a system transfer function:

  The POLES are roots of the denominator polynomial. In this case, the poles of the system are: s=0, s=-b1 and the solutions of the quadratic

which we assume to be complex conjugates values.

  The ZEROS are roots of the numerator polynomial. In this case, the zeros of the system are: s =-a1, s=-a2.

1 21 2 1 22 2

1 2 3 1 3 2

1 2 3

1 1( )( )( )

( )( )1 1

s sa aK s a s a Ka aH s

s s b s b s b b b bs ss sb b b

⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟

+ + ⎝ ⎠⎝ ⎠= =+ + + ⎛ ⎞⎛ ⎞

+ + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

(s 2 +b2s +b3) = 0

L4.9 p430


Bode Plot – Sketching frequency response .. without (much) calculation (2)

  Now let s=jω,

  Express this as decibel (i.e. 20 log(…)):

  Now amplitude response (in dB) is broken into building block components that are added together.

1 21 22

1 3 2

1 3 2

1 1( )

( )1 1

j ja aKa aH j

b b bj jj jb b b

ω ω

ωωω ω

ω

+ +

=

+ + +

1 2zeros at and a a− −

poles at 0

1poles at b−conjugate poles

constant term


Building blocks for Bode Plots – amplitude (1)

  Pole term

  Zero term

  Pole term

20log 1 jaω

− +

20log jω−

20log jω+

at ,aω =

20log 1 20log( 2) 3j dB− + = − ≈ − 1 decade

20dB

For s+a

2aω =/ 2aω =for ω a ,

for ω a ,

1For s+a



  Now consider the quadratic poles:   Better to express this as:

  The log magnitude response is:

22 3s b s b+ +

2 22 n ns j sςω ω+ +

damping factor

natural frequency



  Elsewhere, the exact log amplitude is:

1 decade

40dB

weak dampling ς

strong dampling ς


Bode Plots Example – amplitude (1)

  Consider this transfer function:

  We re-write this as

  Step 1: Establish where x-axis crosses the y-axis •  Since the constant term is 100 = 40dB, x-axis cut the vertical axis at 40.

  Step 2: For each pole and zero term, draw an asymptotic plot. •  We need to draw straight lines for zeros at origin and ω=100. •  We need to draw straight line for poles at ω=2 and ω=10.

  Step 3: Add all the asymptotes.   Step 4: Apply corrections if necessary.


Bode Plots Example – amplitude (2)


  Now consider phase response for the earlier transfer function:

  Therefore:

  Again, we have three type of terms.

1 21 22

1 3 2

1 3 2

1 1( )

( )1 1

j ja aKa aH j

b b bj jj jb b b

ω ω

ωωω ω

ω

+ +

=

+ + +

Building blocks for Bode Plots – Phase (1)



  Pole term

  Pole term

for ω a ,

for ω a ,

For s+a

10aω =/10aω =

1For s+a



  2nd order poles:   Phase response is:

2 22 n ns j sςω ω+ +


Bode Plots Example – phase (1)

  Consider this again:


  You will be applying frequency response in various areas such as filters and 2nd year control. You have also used frequency response in the 2nd year analogue electronics course. Here we explore this as a special case of the general concept of complex frequency, where the real part is zero.

  You have come across Bode plots from 2nd year analogue electronics course. Here we go deeper into where all these rules come from.

  We will apply much of what we done so far in the frequency domain to analyse and design some filters in the next lecture.

Relating this lecture to other courses

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Lecture 8 - Frequency Response