PowerPoint Presentation
Prof. Eliathamby Ambikairajah, School of EE&T Term 3, 2021
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ELEC3104: Mini-Project – Cochlear Signal Processing
Modified SOLO taxonomy framework
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✓ The Structure of Observed Learning Outcomes (SOLO) taxonomy is a framework for analysing students’ depth of
knowledge.
✓ It describes 6 Hierarchical levels (Levels 0 to 5) of increasing complexity in student’s understanding of topics studied.
✓ The taxonomy encourages students to think about which level they are currently with their learning and what they need
to do in order to progress to the next level.
✓ SOLO taxonomy has been adapted for this course to include Pass, Credit, Distinction and High Distinction levels to help
the students to understand the different levels (Levels 0 to 5) on the learning curve and what they need to do to progress.
✓ Students may choose to stay on in a particular level without moving to the next level.
✓ The mini-project is designed to provide an increasing complexity from Pass (Level 2) to High distinction (Level 5) levels
as shown below.
Different Stages of the Modified SOLO Taxonomy
ELEC3104: Project Outline
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✓ This mini project (individual) will focus on understanding and modelling the spectral analyses carried out by
the human cochlea.d modelling the spectral analyses carried out by the human cochlea.
SOLO – Level 1
▪ Introduction to Human Auditory System and MATLAB coding fundamentals
SOLO – Level 2 (Pass Level)
▪ Implementation of a parallel FIR filter bank model of the cochlea for analysis and
synthesis purposes.
SOLO – Level 3 (Credit Level)
▪ Implementation of a parallel IIR filter bank model of the cochlea for spectral analysis.
SOLO – Level 4 (Distinction Level)
▪ Implementation of a parallel IIR filter bank model of the cochlea for pitch detection.
SOLO – Level 5 (High Distinction Level)
▪ Incorporate mechanisms into the parallel IIR filter bank cochlea model that makes
the filterbank adaptive.
Additional Information: In addition to the information provided to you in these slides, you are strongly encouraged
to find and view animations and videos that describe the functioning of the peripheral auditory system and the
cochlea in particular. Visualisation in the form of these animations will be very helpful in understanding cochlear
signal processing.
Eg: Cochlear Animation – https://www.youtube.com/watch?v=dyenMluFaUw
Prof. Eliathamby Ambikairajah, School of EE&T Term 3, 2021
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ELEC3104: Mini-Project – Cochlear Signal Processing
SOLO – Level 1: Introduction to Human Auditory System
and MATLAB coding fundamentals
Introduction to the Human Auditory System
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✓ The human auditory system is responsible for
converting pressure variations caused by the
sound waves that reach the ear into nerve
impulses that are interpreted by the brain.
✓ The Human Auditory System is designed to
assess frequency (pitch) and amplitude
(loudness).
✓ The peripheral auditory system is divided into
the Outer Ear, Middle Ear, and Inner Ear.
✓ The peripheral auditory system and in particular
the cochlea can be viewed as a real-time
spectrum analyser.
✓ The primary role of the cochlea is to transform
the incoming complex sound wave at the ear
drum into electrical signals.
✓ The human ear can respond to minute pressure
variations in the air if they are in the
audible frequency range, roughly 20 Hz – 20 kHz
L
Electrical
signals.
Sounds Level
Faint 20dB (A faint Whisper is 30dB)
Soft (Quiet) 40dB
Moderate 60dB (normal conversation)
Loud 80dB (alarm clocks, vacuum cleaners)
Very Loud 90dB(Blenders);110dB (Concerts, car horns)
Uncomfortable 120dB (jet planes during take off)
Painful and
dangerous
130dB(Jackhammers); 140dB(Gunshots)
*Use hearing protection
✓ Over 85 dB for extended periods can cause permanent hearing loss
✓ Zero decibels (0 dB) represent the absolute threshold of human
hearing, below which we cannot hear a sound.
Outer Ear (Air Vibration): A resonator
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✓ The pinna surround the ear canal and functions
as sound wave reflectors and attenuators .
✓ The sound waves enter a tube-like structure
called ear canal and it serves as a sound
amplifier.
✓ The sound waves travel through the canal and
reach the eardrum and cause it to vibrate
✓ The length (L) of the human ear canal is 2.8 cm
(and 7 mm in diameter)
✓ Speed of sound (c) = 340.3 m/sec ;
✓ The resonant frequency (f) of the canal is =
𝑐
4𝐿
= 3,038Hz.
✓ The human outer ear is most sensitive at about
3kHz and provides about 20dB (decibels) of
gain to the eardrum at around 3000Hz.
Outer ear is a low-Q bandpass filter
(Representative figure only)
Middle Ear: An Impedance Matcher & an Amplifier
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✓ Middle ear transforms the vibrating motion of
the eardrum into motion of the stapes via the
two tiny bones, the malleus and incus .
✓ The pressure of the sound waves on the oval
window is around 25 times higher than on the
eardrum.
✓ Since the sound Intensity (𝐼) is proportional (∝)
to the square pressure (𝑃2) , the sound intensity
increases 625 times (or 28dB)
✓ Middle ear converts acoustic energy to
mechanical energy and mechanical energy to
hydraulic energy
Outer Ear Middle Ear
To
cochlea
Sound
input
dB
f
dB
f
The combined frequency response of the outer
and middle ear is a band-pass response, with
its peak dominated near 3 kHz
Middle Ear Gain function
d
B
Inner Ear
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✓ The inner ear consists of the cochlea responsible for converting the
vibrations of sound waves into electrochemical impulses which are
passed on to the brain via the auditory nerve.
✓ The cochlea is a spiral shaped structure which is about 3.5 cm in length
if uncoiled.
✓ The cochlea is divided along its length by the basilar membrane (BM)
which partitions the cochlear into two fluid canals (scala vestibuli and
scala tympani).
✓ The BM terminates just reaching the helicotrema, so there is a passage
way between the scala vistibuli and the scala tymapni equalising the
difference in pressure at the ends of the two scalas.
A longitudinal
section of an
uncoiled cochlea
Basilar Membrane (Hydro Dynamical process)
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✓ The Basilar Membrane varies in width and stiffness along
its length.
✓ At basal end it is narrow and stiff where as towards the
apex it is wider and more flexible.
✓ Each point along the basilar membrane has a
characteristic frequency, 𝑓𝑝(𝑥), to which it is most
responsive.
✓ The maximum membrane displacement occurring at the
basal end for high frequencies (20 kHz) and at the apical
end for low frequencies (70Hz) .
✓ When the vibrations of the eardrum are transmitted by
the middle ear into movement of the stapes, the
resulting pressure differences between the cochlear fluid
chambers, generate a travelling wave that propagates
down the cochlea and reach maximum amplitude of
displacement on the basilar membrane at a particular
point before slowing down and decaying rapidly
✓ The location of the maximum amplitude of this travelling
wave varies with the frequency of the eardrum
vibrations
3.5 cm
Base
Apex
= 3.5 cm
Basilar membrane
0.05155 cm
If 𝑥 is the distance of a point on the basilar membrane from
the stapes, then the frequency, 𝑓𝑝(𝑥), that produces a peak
at this point is given by:
𝑓𝑝 𝑥 = 20000.0 10
−0.667 𝑥 𝐻𝑧 0 ≤ 𝑥 ≤ 3.5 𝑐𝑚
• It is evident that a 20 kHz tone at the stapes will cause the
BM to vibrate at a point 𝑥 = 0.
• A 70 Hz tone will excite the BM at a point x = 3.5 cm (i.e. at
the apex)
The basilar membrane is a resonant structure that vibrates, vertically in
sympathy with pressure variations in the cochlear fluid.
Basilar Membrane
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✓ Different frequencies stimulate different areas of the
basilar membrane
✓ When a tone (single sinusoid) is applied, the
cochlear fluid oscillates in phase with the stimulating
frequency causing a travelling wave pattern of the
vibration on the basilar membrane
✓ There will be one place where the resonant
frequency of the membrane matches the stimulus
frequency and this place will show the maximum
amount of vibration
✓ By measuring vibration at particular points on the
membrane for a range of stimulus frequencies we
can plot the frequency response of each place on
the membrane
✓ The essential function of the basilar membrane is to
act as a frequency analyser (a set of band-pass filters
each responding to a different frequency region)
resolving an input sound at the eardrum into its
constituent frequencies
Organ of Corti
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✓ Attached to the basilar membrane and running its entire
length is the organ of corti containing some 30,000 sensory
hair cells.
✓ The hairs (cilia) of these cells stick up from the organ of corti
and are in contact with overlying Tectorial Membrane
✓ There are two types of sensory hair cells:
▪ One row of inner hair cells, whose cilia float freely in the
fluid-filled region called subtectorial space
▪ Three rows of outer hair cells whose cilia are attached to
the tectorial membrane
✓ Most of the afferent fibres (neurons which carry signals to
the brain) come from inner hair cells,
✓ The efferent fibres (which receive signals from the brain) go
mainly to outer hair cells.
✓ When the basilar membrane deflects, due to pressure wave
in the cochlear fluid, the tectorial membrane move and
shear which causes the hairs of the outer hair cells to bend
and also cause the fluid flow in the subtectorial space.
✓ This in turn triggers the inner hair cells to transmit nerve
impulses along the afferent fibres and eventually to brain.
✓ The motion of each part of the basilar membrane as
detected by the inner hair cells is transmitted as neural
description to the brain. A simplified Diagram of a Human Auditory System
OUTER &
MIDDLE
EARs
SOUND BASILAR
MEMBRANE
INNER HAIR
CELLS
OUTER
HAIR CELLS
AFFERENT
FIBRES
EFFERENT
FIBRES
AUDITORY
PROCESSING
(Brain)
Inner Ear (Cochlea)
Nerve Fibres
Organ of Corti
Mechanical to Neural Transduction (Electro Chemical)
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✓ The mechanical displacement to electrical energy
transduction process takes place in the inner hair
cells
✓ Bending of the inner hair cell cilia due to basilar
membrane displacement produces a change in the
overall resistance (reduces it) of the inner hair cell,
thus modulating current flow through the hair cell.
✓ The modulation being directly proportional to the
degree of bending of the cilia and the bending of
the cilia is one direction only; in effect a half wave
rectification of the basilar membrane displacement
takes place.
✓ Bending of the cilia releases neurotransmitter
which passes into synapses of one or more nerve
cells which fire to indicate vibration
✓ The amount of firing is thus related to the amount
of vibration
✓ Since the neurotransmitter is only released when
the cilia are bent in one direction, firing tends to be
in phase with basilar membrane movement
Here bending the inner hair cell cilia is simulated by
charging of the capacitor and returning to the initial
position of the cilia is equivalent to discharging the
capacitor.
Inner Hair Cell model
Cochlear Modelling: Cascade and Parallel Models
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✓ The basic model of the cochlea is a transmission line model
(cascade model) in which the basilar membrane is modelled as a
cascade of 128 low pass filters, notch filters and resonators as
shown above.
✓ Each digital filter section in the model above represents a section
of the basilar membrane (tuned to a specific frequency) with 128
sections representing the entire basilar membrane
Magnitude response
of the cascaded filter
bank model
✓ The peripheral auditory system is often
modelled as a bank of 128 bandpass filters
(auditory filters) with overlapping passbands.
✓ Typically modelled using a finite number of
bandpass filters, equally spaced along the
Basilar Membrane.
Magnitude response of the parallel filter bank model
Learning Activity 1: Modelling the Outer Ear and the Middle Ear
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✓ The middle ear may be modelled as a cascade of two complex pairs of zeros (to remove very high and very low
frequencies) and one complex pair of poles (to provide low-Q gain at the middle frequencies). The approximate frequency
response of the middle ear can be seen in the figure below.
Assuming a sampling frequency of 16kHz:
(a) Obtain the transfer function of the middle ear filter, by suitably placing poles and zeros on the z-plane. Verify your
results in MATLAB.
(b) Using placement of poles and zeros, estimate a model for the outer ear and cascade it with your previous model of
the middle ear and show using MATLAB that the overall response matches the one shown in this figure.
The combined frequency response of the outer
and middle ear is a band-pass response (sum –
see the adjacent magnitude response diagram),
with its peak dominated near 3 kHz
Filter Design: Pole zero placement
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✓ Calculate the digital filter coefficients of the resonant pole and resonant zeros using pole zero
placement (e.g. : see diagram below)
✓ Resonant pole frequency = 𝜃𝑝; radius = 𝑟𝑝; 𝜃𝑝 =
2𝜋𝑓𝑝
𝑓𝑠
; 𝑓𝑠 = 16𝑘𝐻𝑧 (or higher)
✓ Resonant zero frequency = 𝜃𝑧; radius = 𝑟𝑧 (𝑟𝑧 > 𝑟𝑝 and closer to unit circle); 𝜃𝑧 =
2𝜋𝑓𝑧
𝑓𝑠
✓ 𝐻𝑝 𝑧 =
𝑧2
𝑧−𝑟𝑝𝑒
𝑗𝜃𝑝 𝑧−𝑟𝑝𝑒
−𝑗𝜃𝑝
=
𝑧2
𝑧2−𝑟𝑝 𝑒
𝑗𝜃𝑝+𝑒
−𝑗𝜃𝑝 𝑧+𝑟𝑝
2
=
𝑧2
𝑧2−𝑟𝑝 2 cos 𝜃𝑝 𝑧+𝑟𝑝
2 =
1
1−2𝑟𝑝 cos 𝜃𝑝𝑧
−1+𝑟𝑝
2𝑧−2
✓ 𝐻𝑝 𝑧 =
1
1−𝑏1𝑧
−1+𝑏2𝑧
−2
(from one section of the digital filter) – Equating to above, we obtain
𝑏1 = 2𝑟𝑝 cos 𝜃𝑝 and 𝑏2 = 𝑟𝑝
2.
✓ Similarly, 𝑎1 = 2𝑟𝑧 cos 𝜃𝑧 and 𝑎2 = 𝑟𝑧
2 for 𝐻𝑧 𝑧 = 1 − 𝑎1𝑧
−1 + 𝑎2𝑧
−2
✓ Both transfer functions can be normalised such that DC gain = 1 as follows:
𝐻𝑝 𝑧 =
1−𝑏1+𝑏2
1−𝑏1𝑧
−1+𝑏2𝑧
−2
and 𝐻𝑧 𝑧 =
1−𝑎1𝑧
−1+𝑎2𝑧
−2
1−𝑎1+𝑎2
✓ 𝑟𝑝 and 𝑟𝑧 can be calculated approximately as follows:
𝑟𝑝 ≈ 1 −
𝐵𝑊𝑝
𝑓𝑠
𝜋 ; 𝑟𝑧 ≈ 1 −
𝐵𝑊𝑧
𝑓𝑠
𝜋
𝑄𝑝 =
𝑓𝑝
𝐵𝑊𝑝
;Q-factors: 𝑄𝑧 =
𝑓𝑧
𝐵𝑊𝑧
Pole – zero plots and magnitude responses of the outer and middle ears
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Outer Ear Filter Middle Ear Filter
The combined frequency response of the outer and middle ears
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Learning Activity 2
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✓The impulse response of an auditory filter can be modelled by:
𝒈 𝒏 = 𝒂 𝒏𝑻 𝑵−𝟏𝒆−𝟐𝝅𝒃 𝟐𝟒.𝟕+𝟎.𝟏𝟎𝟖𝒇𝒑 𝒏𝑻𝒄𝒐𝒔 𝟐𝝅𝒇𝒑𝒏𝑻
where, 𝑓𝑝 is the centre frequency, 𝑇 is the sampling period (𝑓𝑠= Τ
1
𝑇 ) , 𝑛 is the discrete time sample index, 𝑁 is the order of the
filter (𝑁 = 4) and 𝑎 is a constant chosen such that the filter gain at the centre frequency is 0dB ; 𝑏 = 1.14; 𝑓𝑠= 16,000 Hz.;
[Initially, you may choose a=1 and then change the value such that the gain of the filter is normalised to 0dB at the centre
frequency, 𝑓𝑝 .
✓You are required to calculate the impulse response, 𝑔(𝑛), for four auditory filters of your choice from the low, mid and high
frequency regions of the basilar membrane using the equation {𝑓𝑝(x)} given below in MATLAB.
𝒇𝒑 𝒙 = 𝟖𝟎𝟎𝟎. 𝟎 𝟏𝟎
−𝟎.𝟔𝟔𝟕 𝒙 𝑯𝒛 𝟎 ≤ 𝒙 ≤ 𝟑. 𝟓 𝒄𝒎
✓You will notice that the impulse responses have infinite duration, and thus each impulse response will need to be truncated to,
say, 150 to 200 coefficients (i.e., 0 ≤ 𝑛 < 200). Plot the impulse responses of all four filters in the same screen.
(a) What major differences do you see between the impulse responses you have plotted and why?
(b) Using these impulses responses, find the magnitude responses of all four filters and plot them (frequency vs magnitude in dB)
on the same figure so you can compare them.
(c) The gains at the centre frequencies of the filters may not be equal, choose the scaling factor ′𝒂′ for each filter (see equation of
g[n]) such that the gain of each filter is normalised to 0dB at the centre frequency.
(d) Approximately estimate the 3dB bandwidths of all four filters from your plots. Do they vary with the centre frequency? If so,
how do you think they are related?
(e) Explain your understanding of constant-Q filters and constant-Bandwidth filters.
Reflections
Learning Activity 2: Impulse and Magnitude Responses
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Impulse responses of Auditory Filters Magnitude responses of the Auditory Filters
Prof. Eliathamby Ambikairajah, School of EE&T Term 3, 2021
20
ELEC3104: Mini-Project – Cochlear Signal Processing
SOLO – Level 2
SOLO – Level 2 (Pass Level): Implementation of a parallel FIR filter
bank model of the cochlea for analysis and synthesis purposes.
ELEC3104: Mini project (SOLO – Level 2)
21
✓ The Cochlea can be modelled as a parallel filter bank
consisting of overlapping bandpass filters covering the
frequency range from 50Hz to 8kHz (Sampling frequency =
16 kHz).
✓ 128 bandpass filters (Resonant filters) with varying centre
frequencies (𝑓𝑝) and quality factors (𝑄𝑝) are used to model
the cochlea. The bandwidth ( ൗ
𝑓𝑝
𝑄𝑝
) spanned by each of
these filters increases with frequency.
✓ The centre frequency of the bandpass filters can be
calculated using the following equation:
𝑓𝑝 𝑘 = 8000 10
−0.667 kΔ𝑥 Hz 0.0869 ≤ 𝑥 ≤ 2.9985 𝑐𝑚
∆𝑥 =
2.9985−0.0869
𝑁−1
≈ 0.02293 cm; N = 128; k = 1, 2,…., N;
✓ From experiments, it is known that the Quality factors 𝑄𝑝
of the bandpass filters vary linearly from 5 (Low frequency
filter, n=1) to 10 (High frequency filter, n=128)
✓ The bandwidth of the nth bandpass filter is given by:
𝐵𝑊𝑝 𝑛 =
𝑓𝑝 𝑛
𝑄𝑝 𝑛
.
Filter
No. (𝒏)
Resonant Frequency
𝒇𝒑 𝒏 in Hz
𝑸𝒑 𝒏 𝑩𝑾𝒑 𝒏
in Hz
1 𝑓𝑝 1 = 88 5 18
2 𝑓𝑝 2 =91 5.039 18
3 𝑓𝑝 3 =95 5.079 19
4 𝑓𝑝 4 =98 5.118 19
. . . .
128 𝑓𝑝 128 =7723 10 772
Table 1
Note: when n=1, k= 128; n =2, k =127 ….., n =128, k=1.
Step 1:
Bandpass Filter Design: Pole-zero placement method
22
✓ Complete the table 1 using the equations given in the previous slide.
✓ Calculate the digital filter coefficients of the poles (𝑏1 & 𝑏2) and zeros
using pole-zero placement (e.g: see diagram):
▪ Pole frequency = 𝜃𝑝; radius = 𝑟𝑝; 𝜃𝑝 =
2𝜋𝑓𝑝
𝑓𝑠
; 𝑓𝑠 = 16 𝑘𝐻𝑧
▪ 𝐻𝑝 𝑧 =
𝑧2−1
𝑧−𝑟𝑝𝑒
𝑗𝜃𝑝 𝑧−𝑟𝑝𝑒
−𝑗𝜃𝑝
=
1−𝑧−2
1−2𝑟𝑝 cos 𝜃𝑝𝑧
−1+𝑟𝑝
2𝑧−2
----- (1)
▪ One section of the digital filter: 𝐻𝑝 𝑧 =
1
1−𝑏1𝑧
−1+𝑏2𝑧
−2
----- (2)
▪ Equating the denominator of (1) and (2), we obtain
▪ 𝑏1 = 2𝑟𝑝 cos 𝜃𝑝 and 𝑏2 = 𝑟𝑝
2. ----- (3)
▪ 𝑟𝑝 and 𝐵𝑊𝑝 (bandwidth of 𝐻𝑝) can be calculated as follows:
𝑟𝑝 ≈ 1 −
𝐵𝑊𝑝
𝑓𝑠
𝜋 ;
✓ Calculate the coefficients 𝑏1 & 𝑏2 using Equation 3 above.
✓ Repeat the above calculations to obtain all 128 IIR filters {𝐻𝑝 𝑧 }.
𝐵𝑊𝑝 =
𝑓𝑝
𝑄𝑝
; 𝑄𝑝 - Q factor
Complex Pole pair
and zeros placement
Magnitude response of a
bandpass filter with
centre frequency 3171 Hz
Step 1 (Continued):
Analysis/Synthesis Filter Bank Implementation
23
✓ In a filter bank analysis, an input signal is fed into a series of analysis
filters (in this project we use 𝑁 = 128 filters), decomposing the signal
into subbands.
✓ This means that the output signal from each analysis filter represents
different frequency components of the input signal.
✓ The analysis filters 𝐻𝑚(𝑧) (see Figure on the right) comprise a set of
overlapping N bandpass filters used to model the human auditory
filters as explained in the previous slides.
✓ The synthesis filter 𝐺𝑚(𝑧), in each channel employs the time reversed
impulse response of the analysis filter, 𝐻𝑚(𝑧).
✓ The analysis filter 𝐻𝑚(𝑧) (impulse response h[n]) can be implemented
in practice as an FIR filter.
✓ By time reversing the FIR analysis filters’ impulse responses, we can
form the synthesis filters (Impulse response ℎ[−𝑛]} and an overall
zero phase characteristic is obtained.
✓ Perfect reconstruction of the input signal 𝑥[𝑛] is achieved if and only if
the output signal ො𝑥[n] is identical to 𝑥[𝑛] or a delayed version of x[n].
✓ Show that perfect reconstruction is possible if σ𝑚=1
𝑁 𝐺𝑚(𝑧) 𝐻𝑚(𝑧) = K
where K is appositive constant.
Step 2:
Block Diagram of Analysis/Synthesis Filter Bank
Low Frequency
filter (LF)
High Frequency
filter (HF)
Analysis/Synthesis Filter Bank Implementation
24
✓ Once you have calculated the filter coefficients of the 128 IIR filters, use
MATLAB to calculate the impulse responses of your IIR filters .
✓ You will notice that the impulse responses have infinite duration (because they
are all IIR filters), and thus each filter impulse response should be truncated to,
say, 160 coefficients, in order to obtain the FIR filter coefficients of your
analysis filters {𝐻1(𝑧) to 𝐻𝑁(𝑧)} {see an example shown on the right)
✓ Using these coefficients, the filters {𝐻1(𝑧) to 𝐻𝑁(𝑧)} can now be implemented
as a bank of FIR filters.
✓ The analysis filters can be implemented either as IIR or FIR filters. For LEVEL 2
mini-project we will implement analysis/synthesis filters as FIR filters for
perfect reconstruction purposes.
Synthesis filters
✓ If perfect reconstruction property is required, the analysis/Synthesis filter pair
for each filter is by no means independent. In fact, by specifying either the
analysis or the synthesis filter response, the perfect reconstruction
requirement implicitly determines the other.
✓ In this project, we choose the synthesis filters {G1(z) to GN(z)} to be the time-
reversed version of the corresponding analysis filters i.e.
𝑔𝑚[𝑛] = ℎ𝑚[−𝑛] for m = 1, 2….N
✓ To make the synthesis filters causal, a delay is introduced . i.e.
𝑔𝑚[𝑛] = ℎ𝑚[𝐿 − 𝑛] for m = 1, 2….N and L is the delay
Step 3:
Finite Impulse Response
160 coefficients
Truncated IIR
impulse response
Analysis/Synthesis Filter Bank Implementation
25
✓ Show analytically, that the reconstructed signal is a scaled and/or delayed version of the
input signal (i.e. ො𝑥[n] = 𝑥[𝑛 − 𝐿] )
✓ For the filter with centre frequency of 1 kHz, calculate the IIR filter impulse response h[n],
truncate it to M coefficients and plot the magnitude response of the filter using these
coefficients.
✓ On the same figure, plot the magnitude response of the corresponding 1kHz IIR filter. You
may notice that the magnitude responses for the filters are not exactly same. Discuss the
reasons for this. The plot should be given with centre frequency of filter normalised to 0dB.
✓ Using MATLAB, plot the magnitude responses of analysis filters (you may select 25 filters).
The plot should be given with centre frequency of each filter normalised to 0dB.
✓ You may repeat the same for the synthesis filters. Do you notice any differences in the
magnitude responses between the analysis filters and the corresponding synthesis filters. If
so or if not Why? Explain briefly.
✓ Using MATLAB, filter the speech signal (saved in speech.wav from ‘Wav files for DSP Labs’
on Moodle) sampled at 16 kHz using the analysis filter bank. Each analysis filter bank output
should be fed to the input of the corresponding synthesis filter. The outputs of these
synthesis filters can then be added to produce the synthesised speech.
✓ Using the soundsc command in MATLAB, compare the quality of the reconstructed speech
with that of the original. Discuss any noticeable differences.
Step 4:
Note: This mini project involves
a substantial time commitment
to successfully complete all
parts. It is suggested that you
commence work on this project
straight away.