Frequency Response of a LTI System
Lecture 8
We have seen that LTI system response to x(t)=est is H(s)est. We
represent such input-output pair as:
st
st
e ⇒ H (s)e
Frequency Response
Instead of using a complex frequency, let us set s = jω, this yields:
e jωt ⇒ H ( jω )e jωt
cos ωt = Re(e jωt ) ⇒ Re[ H ( jω )e jωt ]
(Lathi 4.8 – 4.9)
Peter Cheung
Department of Electrical & Electronic Engineering
Imperial College London
It is often better to express H(jω) in polar form:
Therefore
E2.5 Signals & Linear Systems
Amplitude
Response
Lecture 8 Slide 1
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Frequency Response Example (1)
Substitute s=jω
H ( jω ) =
H ( jω ) =
ω 2 + 0.01
2
ω + 25
Phase
Response
L4.8 p423
E2.5 Signals & Linear Systems
Lecture 8 Slide 2
Frequency Response Example (2)
Find the frequency response of a system with transfer function:
s + 0.1
H ( s) =
s+5
Then find the system response y(t) for input x(t)=cos2t and x(t)=cos(10t-50°)
Frequency
Response
cos ωt ⇒ H ( jω) cos[ωt + ∠H ( jω)]
URL: www.ee.imperial.ac.uk/pcheung/teaching/ee2_signals
E-mail: p.cheung@imperial.ac.uk
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H ( jω) = H ( jω) e j∠H ( jω )
H ( jω ) =
ω 2 + 0.01
2
ω + 25
⎛ω ⎞
−1 ⎛ ω ⎞
Φ( jω ) = ∠H ( jω ) = tan −1 ⎜
⎟ − tan ⎜ ⎟
0.1
⎝
⎠
⎝5⎠
jω + 0.1
jω + 5
⎛ω ⎞
−1 ⎛ ω ⎞
and ∠H ( jω ) = Φ( jω ) = tan −1 ⎜
⎟ − tan ⎜ ⎟
⎝ 0.1 ⎠
⎝5⎠
for input x(t)=cos2t and x(t)=cos(10t-50°)
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E2.5 Signals & Linear Systems
Lecture 8 Slide 3
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E2.5 Signals & Linear Systems
Lecture 8 Slide 4
Frequency Response Example (3)
For input x(t)=cos2t, we have:
H ( j 2) =
Frequency Response Example (4)
22 + 0.01
22 + 25
⎛ 2 ⎞
−1 ⎛ 2 ⎞
Φ ( j 2) = tan −1 ⎜
⎟ − tan ⎜ ⎟ = 65.3
⎝ 0.1 ⎠
⎝5⎠
= 0.372
Therefore
For input x(t)=cos(10t-50°), we will use the amplitude and phase response
curves directly:
H ( j10) = 0.894
Φ( j10) = ∠H ( j10) = 26
y(t ) = 0.372cos(2t + 65.3)
Therefore
y(t ) = 0.894cos(10t − 50 + 26) = 0.894cos(10t + 24)
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E2.5 Signals & Linear Systems
Lecture 8 Slide 5
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Frequency Response of delay of T sec
H(s) of an ideal differentiator is:
H (s) = s and H ( jω ) = jω = ωe jπ / 2
(Time-shifting property)
Therefore
Therefore
H ( jω ) = ω
H ( jω) = e− jωT = 1
and
Φ( jω) = −ωT
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:
−
Lecture 8 Slide 6
Frequency Response of an ideal differentiator
H(s) of an ideal T sec delay is:
H (s) = e− sT
E2.5 Signals & Linear Systems
and
∠H ( jω ) =
π
2
This agrees with:
d
(cos ωt ) = −ω sin ωt = ω cos(ωt + π / 2)
dt
d Φ(ω )
= τ g =T
dω
That’s why differentiator is not a nice
component to work with – it amplifies high
frequency component (i.e. noise!).
is known as Group Delay.
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E2.5 Signals & Linear Systems
Lecture 8 Slide 7
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E2.5 Signals & Linear Systems
Lecture 8 Slide 8
Bode Plot – Sketching frequency response..
without (much) calculation (1)
Frequency Response of an ideal integrator
H(s) of an ideal integrator is:
H ( s) =
⎛ s
⎞⎛ s
⎞
⎜ + 1⎟ ⎜ + 1⎟
a
a
K ( s + a1 )( s + a2 )
Ka a
⎝ 1 ⎠⎝ 2 ⎠
= 1 2
H (s) =
s ( s + b1 )( s 2 + b2 s + b3 )
b1b3 ⎛ s
⎞
⎞ ⎛ s2 b
s ⎜ + 1⎟ ⎜ + 2 s + 1⎟
b
b
b
⎝ 1 ⎠⎝ 2
3
⎠
Therefore
1
ω
and
∠H ( jω ) = −
π
2
This agrees with:
1
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
1
( s 2 + b2 s + b3 ) = 0
∫ cos ωt dt = ω sin ωt = ω cos(ωt − π / 2)
Consider a system transfer function:
1
1 − j 1 − jπ / 2
and H ( jω ) =
=
= e
s
jω ω ω
H ( jω ) =
That’s why integrator is a nice component to
work with – it suppresses high frequency
component (i.e. noise!).
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.
L4.9 p430
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Lecture 8 Slide 9
E2.5 Signals & Linear Systems
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Bode Plot – Sketching frequency response..
without (much) calculation (2)
Now let s=jω,
Ka a
H ( jω ) = 1 2
b1b3
1+
jω 1 +
Building blocks for Bode Plots – amplitude (1)
jω
jω
1+
a1
a2
b ω ( jω ) 2
jω
1+ j 2 +
b1
b3
b2
Express this as decibel (i.e. 20 log(…)):
Lecture 8 Slide 10
E2.5 Signals & Linear Systems
zeros at − a1 and − a2
Pole term
Zero term +20log jω
Pole term −20 log 1 +
for ω a ,
−20log jω
jω
a
ω = a/2
poles at 0
ω = 2a
For s+a
for ω a ,
constant term
conjugate poles
poles at − b1
Now amplitude response (in dB) is broken into building block components
that are added together.
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E2.5 Signals & Linear Systems
Lecture 8 Slide 11
For
at ω = a,
−20log 1 + j = −20log( 2) ≈ −3dB
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20dB
1
s+a
1 decade
E2.5 Signals & Linear Systems
Lecture 8 Slide 12
Building blocks for Bode Plots – amplitude (2)
Now consider the quadratic poles: s 2 + b2 s + b3
Better to express this as:
s 2 + 2 jςωn s + ωn2
damping
factor
Building blocks for Bode Plots – amplitude (3)
Elsewhere, the exact log amplitude is:
natural
frequency
weak dampling ς
The log magnitude response is:
strong dampling ς
40dB
1 decade
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E2.5 Signals & Linear Systems
Lecture 8 Slide 13
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Bode Plots Example – amplitude (1)
Consider this transfer function:
We re-write this as
Step 1: Establish where x-axis crosses the y-axis
Step 2: For each pole and zero term, draw an asymptotic plot.
E2.5 Signals & Linear Systems
Lecture 8 Slide 14
Bode Plots Example – amplitude (2)
•  Since the constant term is 100 = 40dB, x-axis cut the vertical axis at 40.
•  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.
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E2.5 Signals & Linear Systems
Lecture 8 Slide 15
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E2.5 Signals & Linear Systems
Lecture 8 Slide 16
Building blocks for Bode Plots – Phase (1)
Now consider phase response for the earlier transfer function:
Ka a
H ( jω ) = 1 2
b1b3
1+
jω 1 +
Building blocks for Bode Plots – Phase (2)
Pole term
Pole term
jω
jω
1+
a1
a2
b ω ( jω ) 2
jω
1+ j 2 +
b1
b3
b2
ω = a /10
ω = 10a
For s+a
Therefore:
for ω a ,
for ω a ,
For
Again, we have three type of terms.
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E2.5 Signals & Linear Systems
Lecture 8 Slide 17
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Building blocks for Bode Plots – Phase (3)
2nd order poles: s 2
Phase response is:
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+ 2 jςωn s + ωn2
E2.5 Signals & Linear Systems
E2.5 Signals & Linear Systems
Lecture 8 Slide 18
Bode Plots Example – phase (1)
Lecture 8 Slide 19
1
s+a
Consider this again:
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E2.5 Signals & Linear Systems
Lecture 8 Slide 20
Relating this lecture to other courses
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.
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E2.5 Signals & Linear Systems
Lecture 8 Slide 21