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Signals and Systems
Lecture 10 Final
DR TANIA STATHAKI READER (ASSOCIATE PROFFESOR) IN SIGNAL PROCESSING IMPERIAL COLLEGE LONDON
β’ We have seen in a previous lecture that a LTIβs system response to an
everlasting exponential π₯ π‘ = ππ 0π‘ is π»(π 0)ππ 0π‘.
[Proof: If β(π‘) is the unit impulse response of an LTI system then:
π¦ π‘ = β π‘ β ππ 0π‘ = β(πβ
ββ)ππ 0 π‘βπ ππ = ππ 0π‘ β(π
β
ββ)πβπ 0πππ = ππ π‘π»(π 0)]
β’ We represent such input-output pair as:
ππ 0π‘ β π»(π 0)ππ 0π‘
β’ We set π 0 = ππ0. This yields the so called frequency response evaluated
at π = π0. πππ0π‘ β π»(ππ0)π
ππ0π‘
cos π0π‘ = Re{πππ0π‘} β Re{π»(ππ0)π
ππ0π‘} β’ It is often better to express π»(ππ) in polar form as:
π» ππ = π» ππ ππβ π» ππ
β’ Therefore, π» ππ ππππ‘ = π» ππ ππβ π» ππ ππππ‘ = π» ππ ππ(ππ‘+β π» ππ )
cos π0π‘ β π» ππ0 cos[π0π‘ + β π» ππ0 ] π» ππ : Frequency response
π» ππ : Amplitude response
β π» ππ : Phase response
Frequency response of a LTI system to
an everlasting exponential or a cosine
β’ We can also show that a LTIβs system response to an everlasting
exponential π₯ π‘ = πππ0π‘+π0 is πππ0π‘+π0π»(ππ).
Proof:
If β(π‘) is the unit impulse response of a LTI system then:
π¦ π‘ = β π‘ β πππ0π‘+π0 = β(πβ
ββ
)πππ0 π‘βπ +π0ππ
= πππ0π‘+π0 β(πβ
ββ
)πβππ0πππ = πππ0π‘+π0π»(ππ0)
β’ Therefore,
cos π0π‘ + π0 β π» ππ0 cos[ π0π‘ + π0 + β π» ππ0 ]
π» ππ : Frequency response
π» ππ : Amplitude response
β π» ππ : Phase response
Frequency response of a LTI system to
an everlasting exponential or a cosine cont.
β’ Find the frequency response (amplitude and phase response) of a
system with transfer function:
π» π =π + 0.1
π + 5
Then find the system response π¦(π‘) for inputs π₯ π‘ = cos2π‘ and
π₯ π‘ = cos(10π‘ β 50o).
β’ We substitute π = ππ. Then, we obtain π» ππ =ππ+0.1
ππ+5.
Amplitude response: π» ππ =π2+0.01
π2+25 .
Phase response: β π» ππ = Ξ¦ π = tanβ1π
0.1β tanβ1
π
5.
Example
β’ Amplitude response: π» ππ =π2+0.01
π2+25 .
β’ Phase response: β π» ππ = Ξ¦ π = tanβ1π
0.1β tanβ1
π
5.
Example cont.
β’ Consider the input π₯ π‘ = cos2π‘. We have:
cos ππ‘ + π β π» ππ cos[ππ‘ + π + β π» ππ ] In that case we haveπ0 = 2 and π0 = 0.
Amplitude response: π» π2 =22+0.01
22+25= 0.372.
Phase response: β π» π2 = Ξ¦ π2 = tanβ12
0.1β tanβ1
2
5= 65.3o.
β’ Therefore,
π¦ π‘ = 0.372cos(2π‘ + 65.3o) π₯ π‘ = cos2π‘
Example cont.
β’ Consider the input π₯ π‘ = cos(10π‘ β 50o). We have:
cos ππ‘ + π β π» ππ cos[ππ‘ + π + β π» ππ ] In that case we have π0 = 10 and π0 = β50.
Amplitude response:
π» π10 = 0.894.
Phase response:
β π» π10 = Ξ¦ π10 = 26o. β’ Therefore,
π¦ π‘ = 0.894cos(10π‘ β 50o + 26o)
Example cont.
β’ The transfer function of an ideal delay is π» π = πβπ π.
β’ Therefore,
Amplitude response: π»(ππ) = πβπππ = 1.
Phase response: β π» ππ = Ξ¦ ππ = βππ.
β’ Therefore:
Delaying a signal by π has no effect
on its amplitude.
It results in a linear phase shift
(with frequency) with a gradient of βπ.
The quantity βπΞ¦ π
ππ= ππ = π is
known as Group Delay.
Frequency response of delay of π sec
β’ The transfer function of an ideal differentiator is π» π = π .
β’ For π = ππ we have π» ππ = ππ = ππππ
2.
β’ Therefore, for π > 0 we obtain:
Amplitude response: π»(ππ) = π.
Phase response: β π» ππ =π
2.
Example
β’ Consider the input of an ideal differentiator
to be π₯ π‘ = cosππ‘. β’ The output is:
π
ππ‘cosππ‘ = βπsinππ‘ = πcos(ππ‘ +
π
2)
β’ Thatβs why differentiator is not
a nice component to work with;
it amplifies high frequency
components (i.e., noise).
Frequency response of an ideal differentiator
β’ The transfer function of an ideal integrator
is π» π =1
π .
β’ For π = ππ, π» ππ = 1
ππ= βπ
1
π=1
ππβπ
π
2.
β’ Therefore, for π > 0 we obtain:
Amplitude response: π»(ππ) =1
π.
Phase response: β π» ππ = βπ
2.
β’ Consider the input of an ideal integrator
to be π₯ π‘ = cosππ‘. β’ The output is:
πππ ππ‘ ππ‘ =1
ππ ππππ‘ =
1
πcos(ππ‘ β
π
2)
β’ Thatβs why an integrator is a nice
component to work with; it supresses
high frequency components (i.e., noise).
Frequency response of an ideal integrator
β’ Consider a system with transfer function:
π» π =πΎ(π +π1)(π +π2)
π (π +π1)(π 2+π2π +π3)
= πΎπ1π2
π1π3
(π
π1+1)(
π
π2+1)
π (π
π1+1)(
π 2
π3+π2π3π +1)
β’ The poles are the roots of the denominator polynomial. In this case, the poles of the system are π = 0, π = βπ1 and the solutions of the quadratic
π 2 + π2π + π3 = 0
Which we assume to form a complex conjugate pair.
β’ The zeros are the roots of the numerator polynomial. In this case, the zeros of the system are: π = βπ1, π = βπ2.
Bode Plots: Asymptotic behaviour of amplitude and phase response
β’ Now let π = ππ.The amplitude response π»(ππ) can be rearranged as:
π»(ππ) = πΎπ1π2
π1π3
1+ππ
π11+ππ
π2
ππ 1+ππ
π11+π
π2π
π3+(ππ)2
π3
β’ We express the above in decibel (i.e., 20log(β)):
20log π»(ππ) = 20log πΎπ1π2π1π3
+20log 1 +ππ
π1+ 20log 1 +
ππ
π2
β20 log ππ β 20log 1 +ππ
π1β 20 log 1 + π
π2π
π3+(ππ)2
π3
β’ By imposing a log operation the amplitude response (in dB) is broken
into building block components that are added together.
β’ We have three types of building block terms: A term ππ, a first order
term 1 +ππ
π and a second order term with complex conjugate roots.
Bode Plots: Asymptotic behaviour of amplitude and phase response
β’ They are desirable in several applications, where the variables considered
have a very large range of values.
β’ The above is particularly true in frequency response amplitude plots since
we require to plot frequency response from 10β6 to 106 or higher.
β’ A plot of such a large range on a linear scale will bury much of the useful
information at lower frequencies.
β’ In humans the relationship between stimulus and perception is logarithmic.
This means that if a stimulus varies as a geometric progression (i.e.,
multiplied by a fixed factor), the corresponding perception is altered in
an arithmetic progression (i.e., in additive constant amounts). For
example, if a stimulus is tripled in strength (i.e., 3 x 1), the
corresponding perception may be two times as strong as its original
value (i.e., 1 + 1).
There is behind the above observations developed by Weber and
Frechner.
Advantages of logarithmic units
β’ A pole at the origin 1/π contributes to the frequency response with the term
β 20 log ππ = β20 logπ.
We can effect further simplification by using the logarithmic function for the
variable π itself. Therefore, we define π’ = logπ.
β’ Therefore, β20 logπ = β20π’. The above is a straight line with a slope of β20.
A ratio of 10 in π is called a decade. If π2 = 10π1 then
π’2 = logπ2 = log10π1 = log10 + logπ1 = 1 + logπ1 = 1 + π’1 β20logπ2 = β20log10π1 = β20log10 β 20logπ1 = β20 β 20logπ1
A ratio of 2 in π is called an octave. If π2 = 2π1 then
π’2 = logπ2 = log2π1 = log2 + logπ1 = 0.301 + logπ1 = 0.301 + π’1 β20logπ2 = β20log2π1 = β20log2 β 20logπ1 = β6.02 β 20logπ1
β’ Based on the above, equal increments in π are equivalent to equal
ratios in π.
β’ The amplitude response plot has a slope of β20dB/decade or β20 0.301 =β 6.02dB/octave.
β’ The amplitude plot crosses the π axis at π = 1, since π’ = logπ = 0 for π = 1.
Building blocks for Bode plots
a pole at the origin: amplitude response
β’ A zero at the origin π contributes to the frequency response with the term
20 log ππ = 20 logπ = 20π’.
β’ The amplitude plot has a slope of 20πB/ decade or 20 0.301 =6.02πB/octave.
β’ The amplitude plot for a zero at the origin is a mirror image about the π
axis of the plot for a pole at the origin.
β’ As you can see the horizontal axis depicts π’ = logπ and not π.
Amplitude responses of a pole (solid line) and a zero (dotted line) at the origin
Bode plots β a zero at the origin: amplitude
β’ The log amplitude of a first order pole at βπ is β20log 1 +ππ
π.
π βͺ π β β20log 1 +ππ
πβ β20log1 = 0
π β« π β β20log 1 +ππ
πβ β20log(
π
π) = β20logπ + 20logπ
This represents a straight line (when plotted as a function of π’, the log
of π) with a slope of β20ππ΅/decade or β20 0.301 = β6.02ππ΅/octave.
When π = π the log amplitude is zero. Hence, this line crosses the π
axis at π = π. Note that the asymptotes meet at π = π.
β’ The exact log amplitude for this pole is:
β20log 1 +ππ
π= β20log 1 +
π2
π2
12
= β10log 1 +π2
π2
β’ The maximum error between the actual and asymptotic plots occurs at
π = π called corner frequency or break frequency. This error is:
err π =exact_amplitude π βasymptotic_amplitude π
= βπππ₯π¨π π +ππ
ππβ π = βπππ₯π¨π π = βππ π.
Bode plots β first order pole: amplitude response
β’ A first order zero at βπ gives rise to the term 20log 1 +ππ
π.
π βͺ π β 20log 1 +ππ
πβ 20log1 = 0.
π β« π β 20log 1 +ππ
πβ 20log(
π
π) = 20logπ β 20logπ.
This represents a straight line with a slope of 20πB/decade.
β’ The amplitude plot for a zero at βπ is a mirror image about the π axis of
the plot for a pole at βπ.
Bode plots β first order zero: amplitude
β’ Pole term: β20 log ππ
= β20 logπ
β’ Zero term: 20 log ππ
= 20 logπ
β’ Pole term: β20log 1 +ππ
π
π βͺ π β β20log 1 +ππ
π
β β20log1=0
π β« π β β20log 1 +ππ
π
β β20log(π
π)
= β20logπ + 20logπ
π = π
β20log 1 + π = β20log 2 β β3ππ
Summary of first order building blocks for Bode plots: amplitude
β’ The error of the approximation as a function of π is shown in the figure
below.
β’ The actual plot can be obtained if we add the error to the asymptotic plot.
β’ Problem: Find the error when the frequency is equal to the corner
frequency and 2, 5 and 10 times larger or smaller.
(Answers: β3ππ΅,β1ππ΅,β0.17ππ΅, negligible. See subsequent slides.)
Error in the asymptotic approximation of amplitude
due to a first order pole
β’ The error of the approximation of the true response with an asymptotic line
at a frequency π for the case of a first order pole is:
err π = true π β asymptotic π
= β10log 1 +π2
π2β β20logπ + 20logπ
= β10log 1 +π2
π2+ 20logπ β 20logπ
β’ For π = π (also shown at bottom of Slide 16)
err π = β10log 1 +π2
π2+ 20logπ β 20logπ = β10log 2 = β10 β 0.3
= β3dB
β’ For π = 2π,
err π = β10log 1 +4π2
π2+ 20log2π β 20logπ =
= β10log 5 + 20log 2 = β10 β 0.7 + 20 β 0.3 = β1dB
Error in the asymptotic approximation of amplitude
due to a first order pole
err π = β10log 1 +π2
π2+ 20logπ β 20logπ
β’ For π = 5π,
err π = β10log 1 +25π2
π2+ 20log5π β 20logπ = β10log 26 + 20log 5
= β10 β 1.41497334 + 20 β 0.69897 = β0.17dB
For π = 10π,
err π = β10log 1 +100π2
π2+ 20log10π β 20logπ
= β10log 101 + 20log 10 = β20.0432 + 20 β 1 β 0
Bode plots β error of the approximation cont.
β’ Now consider the quadratic term: π 2 + π2π + π3. β’ It is quite common to express the above term as:
π 2 + 2ππππ + ππ2 = ππ
2 π 2
ππ2 + 2π
π
ππ+ 1 = ππ
2[1 + 2π(π
ππ) + (
π
ππ)2]
The scalar π is called damping factor.
The scalar ππ is called natural frequency.
β’ The log amplitude response is obtained by setting π = ππ and taking the
magnitude.
β’ For the time being we are not interested in the gain ππ2.
β’ log amplitude = β20log 1 + 2ππ(π
ππ) + (
ππ
ππ)2
π βͺ ππ, log amplitude β β20log1 = 0
π β« ππ, log amplitude β β20log β(π
ππ)2 = β40log (
π
ππ)
= β40logπ + 40logππ = β40π’ + 40logππ
The exact log amplitude is β20log [1 β (π
ππ)2]2+4π2(
π
ππ)2
1/2
Bode plots β second order pole : amplitude
β’ The log amplitude involves a parameter π called damping factor,
resulting in a different plot for each value of π.
β’ It can be proven that for complex-conjugate poles π < 1.
β’ For π β₯ 1, the two poles in the second order factor are not longer
complex but real, and each of these two real poles can be dealt with as
a separate first order factor.
β’ The amplitude plot for a pair of complex conjugate zeros is a mirror
image about the π axis of the plot for a pair of complex conjugate poles.
Bode plots β second order pole : amplitude
β’ The exact log amplitude is = β20log [1 β (π
ππ)2]2+4π»π(
π
ππ)2
1/2
Bode plots β second order pole : amplitude
β’ The error of the approximation as a function of π is shown in the figure
below for various values of πs.
β’ The actual plot can be obtained if we add the error to the asymptotic
plot.
Error in the asymptotic approximation of amplitude
due to a pair of complex conjugate poles
Bode plots example: amplitude
β’ Consider a system with transfer function:
π» π =20π (π + 100)
(π + 2)(π + 10)
π» π =20Γ100
2Γ10 π (1+
π
100)
(1+π
2)(1+
π
10)= 100
π (1+π
100)
(1+π
2)(1+
π
10)
β’ Step 1: Since the constant term is 100 or 20log100 = 40dB, we relabel
the horizontal axis as the 40dB line.
β’ Step 2: For each pole and zero term draw an asymptotic plot.
β’ Step 3: Add all the asymptotes. β’ Step 4: Apply corrections if possible. For corrections we use the
frequencies 1, 2, 10, 100. They are mostly corner frequencies.
Bode plots example: amplitude. Corrections.
β’ Correction at π = π
Due to corner frequency at π = 2 is β1πB.
Due to corner frequency at π = 10 is negligible.
Due to corner frequency at π = 100 is negligible.
Total correction at π = 1 is β1πB.
β’ Correction at π = π
Due to corner frequency at π = 2 is β3πB.
Due to corner frequency at π = 10 is β0.17πB.
Due to corner frequency at π = 100 is negligible.
Total correction at π = 2 is β3.17πB.
β’ Correction at π = ππ
Due to corner frequency at π = 10 is β3πB.
Due to corner frequency at π = 2 is β0.17πB.
Due to corner frequency at π = 100 is negligible.
Total correction at π = 10 is β3.17πB.
Bode plots example: amplitude. Corrections cont.
β’ Correction at π = πππ
Due to corner frequency at π = 100 is 3πB.
Due to corner frequency at π = 2 is negligible.
Due to corner frequency at π = 10 is negligible.
Total correction at π = 100 is 3πB.
β’ Correction at intermediate points other than corner frequencies may be
considered for more accurate plots.
Bode plots example: total amplitude
β’ Observe now the final plot for the previous system with transfer function:
π» π = 100 π (1+
π
100)
(1+π
2)(1+
π
10)
β’ Now consider the phase response for the earlier transfer function:
π»(ππ) = πΎπ1π2
π1π3
(1+ππ
π1)(1+
ππ
π2)
ππ(1+ππ
π1)(1+π
π2π
π3+(ππ)2
π3)
β’ The phase response is:
β π» ππ = β 1 +ππ
π1+ β 1 +
ππ
π2β β ππ
ββ (1 +ππ
π1) β β (1 + π
π2π
π3+ππ 2
π3)
β’ Again, we have three types of terms.
Bode plots: phase
β’ A pole at the origin gives rise to the term βππ.
π» ππ = βππ = ππβππ
2 β β π» ππ = ββ ππ = β90o.
The phase is constant for all values of π.
β’ A zero at the origin gives rise to the term ππ.
β π» ππ = β ππ = 90o. The phase plot for a zero at the origin is a
mirror image about the π axis of the phase plot for a pole at the origin.
Bode plots β a pole or zero at the origin: phase
β’ A pole at βπ gives rise to the term 1 +ππ
π.
β π» ππ = β β 1 +ππ
π= βtanβ1
π
π.
π βͺ π βπ
πβͺ 1 β βtanβ1
π
πβ 0
π β« π βπ
πβ« 1 β βtanβ1
π
πβ β90o
β’ The phase plot for a zero at βπ is a mirror image about the π axis of the
phase plot for a pole at the origin.
Bode plots β a first order pole or zero: phase
π
β π» ππ
β’ We use a three-line segment asymptotic plot for greater accuracy. The
asymptotes are:
π β€ π/10 β 0o
π β₯ 10π β β90o
A straight line with slope β45o /decade connects the above two
asymptotes (from π = π/10 to π = 10π) crossing the π axis at π = π/10.
Bode plots β a first order pole or zero: phase
β’ The maximum error occurs at π = 0.1π and π = 10π and is 5.7o.
β’ The actual phase can be obtained if we add the error to the asymptotic
plot.
err πππ =exact_phase πππ βasymptotic_phase πππ
= βπππ§βππππ
πβ βπππ = βπππ§βπ
πππ
π+πππ= βπππ§βπ ππ +πππ=
β ππ. ππππππππ+πππ= π. ππ¨. err π. ππ =exact_phase π. ππ βasymptotic_phase π. ππ
= βπππ§βππ.ππ
πβ π = βπππ§βπ π. π = βπ. ππ¨.
Bode plots β a first order pole or zero: phase error
β’ Now consider the term:
1 + 2πππ
ππ+
ππ
ππ
2= 1 β
π
ππ
2+ π 2π(
π
ππ)
β π» ππ = βtanβ12π(
πππ)
1 β (πππ)2
π βͺ ππ βπ
ππβ 0,
β π» ππ = βtanβ10
1 β 0= βtanβ1(0) = 0
π β« ππ βπ
ππβ β,
β π» ππ = βtanβ12πβ
1ββ2= βtanβ1 0 = β180o
β’ The phase involves a parameter π, resulting in a different plot for each
value of π.
Bode plots β second order complex conjugate poles : phase
β’ An error plot is shown in the figure below for various values of π.
β’ The actual phase can be obtained if we add the error to the asymptotic
plot.
Bode plots β second order complex conjugate poles : phase error
Bode plots example: phase
β’ Consider the previous system with transfer function:
π» π =20π (π +100)
(π +2)(π +10)= 100
π (1+π
100)
(1+π
2)(1+
π
10)
β’ For the pole at π = β2 (π = β2) the phase plot is:
π β€2
10= 0.2 β 0o
π β₯ 10 β 2 = 20 β β90o
A straight line with slope β45o /decade connects the above two
asymptotes (from π = 0.2 to π = 20) crossing the π axis at π = 0.2.
β’ For the pole at π = β10 (π = β10) the phase plot is:
π β€10
10= 1 β 0o
π β₯ 10 β 10 = 100 β β90o
A straight line with slope β45o /decade connects the above two
asymptotes (from π = 1 to π = 100) crossing the π axis at π = 1.
Bode plots example: phase cont.
β’ Consider the previous system with transfer function:
π» π =20π (π +100)
(π +2)(π +10)= 100
π (1+π
100)
(1+π
2)(1+
π
10)
β’ The zero at the origin causes a 90ophase shift.
β’ For the zero at π = β100 (π = β100) the phase plot is:
π β€100
10= 10 β 0o
π β₯ 10 β 100 = 1000 β 90o
A straight line with slope 45o /decade connects the above two
asymptotes (from π = 10 to π = 1000) crossing the π axis at π =10.
Bode plots example: total phase cont.
β’ Consider the previous system with transfer function:
π» π = 100 π (1+
π
100)
(1+π
2)(1+
π
10)
Relating this lecture to other courses
β’ 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.
β’ You have come across Bode plots from 2nd year analogue electronics
course. Here we go deeper into where all these rules come from.
β’ We will apply much of what we have done so far in the frequency
domain to analyse and design some filters in the next lecture.