0:00:13you
0:00:14a fine
0:00:15we work
0:00:17uh in which he that issue
0:00:19oh well will be C i E
0:00:22oh a function of
0:00:23yeah
0:00:24mean
0:00:26oh what
0:00:27we action
0:00:28a a yeah for had a
0:00:31a function
0:00:32oh would you that of a function i three
0:00:36oh the future it actual P in a total of not
0:00:41it's a a a a single the issue and uh it's all those a you know because
0:00:46uh one thing to have a different difference
0:00:49oh it's was five so you know have and total
0:00:52um
0:00:54it's not yeah
0:00:55um can be a change just you three five regiment
0:00:58oh
0:00:59of of of a should a shot yeah
0:01:02i i but is only
0:01:04um
0:01:05a response and
0:01:07uh sum
0:01:08some
0:01:09they uh some spatial position and
0:01:12the whole procedure is
0:01:14oh we see time as you me
0:01:17so you want to have a but and uh
0:01:20can you model to recreate then
0:01:22in we a charge yeah
0:01:24oh but a higher order to C
0:01:27yeah
0:01:28i'm going to uh uh and for the use the three D five you
0:01:32a a continuous hrtf model
0:01:35uh
0:01:36here's the
0:01:37a a a a reference
0:01:38with with the house
0:01:40uh that's we can see that
0:01:42um in the model
0:01:44model of is the frequency components to from
0:01:48spatial component
0:01:49and because call it so that the you know
0:01:52oh function
0:01:53and it
0:01:54a a very easy to reconstruct a shot yeah
0:01:57at a at a a a three that they shouldn't and in the and
0:02:01a and G O frequency point
0:02:05a
0:02:05we have
0:02:06to question is for this model
0:02:08oh well as to how should
0:02:10it this model
0:02:12and the seconds
0:02:13i also and that's if you are are also if representation is that we
0:02:19to here
0:02:21oh oh one of this question
0:02:23oh first the we you can have a you know that he man to match to they should be and
0:02:29then oh we use the true
0:02:31a he met to evaluate the function able
0:02:36there are very
0:02:37oh we have you
0:02:39a random or it can be expressed as a linear combination
0:02:43oh on it and then no determined that
0:02:47deterministic all normal functions is weighted a random variables
0:02:51oh the um that C functions are orthonormal with
0:02:56read that to and in a product
0:02:59in fact is
0:03:00we use a rate be free small approximate this is
0:03:04using a an it's number of of the old normal function
0:03:08then and O
0:03:10the expected value of the squared norm of error of the phoenix representation
0:03:16can be given by these uh by prominent a two
0:03:20where a is the random process
0:03:22and has a a it
0:03:25representation of the random four
0:03:28uh the core work if they don't be
0:03:31also also no expansion the or
0:03:35so that he man
0:03:37a now we focus on a team for at least for illustration
0:03:42but we have a a random you find on them
0:03:46and that interval with zero mean and and in in the G
0:03:50you can handle
0:03:52um all normal compression be in a four inches rule that
0:03:59um
0:04:00the random draw that can be right yeah
0:04:03um
0:04:04and the
0:04:06you combination or you know
0:04:09uh uh another
0:04:12uh
0:04:13or in finance
0:04:14uh
0:04:15the complete
0:04:17a basis
0:04:18and way by a very well is a random variable is given by the end of all that so at
0:04:24the end
0:04:25the complete basis
0:04:28and the truncation
0:04:30um
0:04:31for for five
0:04:32or shows the arrow for uh
0:04:36oh of in a representation
0:04:39and the fact that the right along
0:04:43oh oh error
0:04:45a we a good a part of um that thing and so we can see that
0:04:50that was that so will change
0:04:52and and it you can do
0:04:55oh in that that is all of the
0:04:57a a representation
0:04:59so to the meeting my the that have that
0:05:02oh oh we need to make my
0:05:04uh the second time
0:05:07now
0:05:07oh you got a a a a a second term you take
0:05:12that second so
0:05:14that second so um
0:05:15a a you know by
0:05:17a by a far
0:05:19as as useful we a random variable is
0:05:22uh a given by a
0:05:24you know but that so at the end i T so from the calculation
0:05:29uh
0:05:31they
0:05:34they fire
0:05:35a can be right yeah
0:05:38a a problem that
0:05:40uh where yeah
0:05:41see the autocorrelation function of the random process
0:05:46and
0:05:47the uh a a a at the one change
0:05:50so
0:05:51we can say that
0:05:52a a a a a a a mean it's on the choice of the complete a base
0:05:59okay
0:06:00but to
0:06:01uh uh any a complete or normal because can be used
0:06:05so we have a question which is a that
0:06:08now a a a a a just five C
0:06:11uh uh i i think of of the for
0:06:14in will in creation with for no
0:06:17the also correlation function of the random drug that's
0:06:20and then we use each
0:06:23um i think function
0:06:24uh to we spend the random for that
0:06:28at the formal or the and and the
0:06:30oh random for a random variable
0:06:33a a is given by you inner product of that
0:06:36and the and the these
0:06:38oh i can function
0:06:41now
0:06:42oh
0:06:43compute
0:06:44a a screen play five
0:06:47from the calculation we can see that
0:06:50oh okay "'cause" i was to
0:06:53uh that some model
0:06:54a a a a and i i is the eigenvalue
0:06:58and we know that
0:07:00a a a that was to i can function account
0:07:03i well as i my of error a uh variability of the random process at was able
0:07:10so we
0:07:11we say that
0:07:12we a a "'cause" i have three
0:07:15you or greater than that you buy out the
0:07:18a a complete basis
0:07:22and
0:07:23then all with a a a a that can be a a a a similar if then a long uh
0:07:29i can function
0:07:31but
0:07:32quite well and the distribution of property in the random i thought
0:07:37i a i don't know all too complicated to be described as think they
0:07:42all in close to form
0:07:44so at to read about it
0:07:46extremely you can go to find a
0:07:48oh okay L expression for the random problem
0:07:53i
0:07:54uh
0:07:54given a a a a a set of real i they shouldn't of of the random process
0:07:59oh
0:08:00and it's and i three are all normal be we can calculate
0:08:05uh the random error
0:08:06oh variable able for each realisation
0:08:09and then we apply that there
0:08:11random variable and use
0:08:13a a of various to the them that so
0:08:17the that the energy of the random or
0:08:20and then the ah
0:08:22a a a a a a hell of a a a is a random variable so are part
0:08:27uh
0:08:27screen a five in
0:08:29we know that it's of ones that are then
0:08:32a a sign which is given by a
0:08:34oh of to more expansion
0:08:38so we
0:08:39for a T max
0:08:40oh which is that um well
0:08:43and i think the sample variance
0:08:45are are among random variable
0:08:47oh which is
0:08:48involved in the model
0:08:52now still we use the
0:08:53i um i don't five three made to evaluate
0:08:57the function model
0:09:00i
0:09:00oh the kl expansion is unavailable
0:09:04oh
0:09:05there are two seconds to is are and a consideration
0:09:09oh the
0:09:10uh the
0:09:11representation is not offer more
0:09:14redundant you must at this to the in the mall then
0:09:18how do we measure the redundancy or equivalence is the H E
0:09:24i we know that is small so in the
0:09:27oh model uh imply that error of the representation is a there is a trade-off efficient the end and ri
0:09:36uh
0:09:38you can uh uh uh and six fashion
0:09:40oh we can calculate a very is for and coefficient
0:09:45then
0:09:46we find that
0:09:47only only
0:09:48uh
0:09:51or i coefficients are a very small which used
0:09:56the corresponding sounds to be removed them from the model
0:10:00that's in a large than the error
0:10:02and i
0:10:03so only uh and i a uh so that the model
0:10:08and uh
0:10:11a so uh we can you to me
0:10:14the efficiency of the model in terms of that so
0:10:18which are given by a and the it
0:10:23then uh
0:10:24that i is one is
0:10:26oh there are many a point is all orthonormal basis
0:10:31so will which one is a bass
0:10:34oh
0:10:36and uh
0:10:37oh we know that a human
0:10:39uh well i think that all of normal basis we can calculate a
0:10:43oh but and all coefficients
0:10:46and
0:10:47a i
0:10:48screen a a
0:10:49i friends
0:10:51so we compare them at all three play
0:10:55then the lack of that bad has so um
0:10:58the well known be up by thing a like to create a a a a is the best choice
0:11:07okay
0:11:08oh
0:11:09we use the uh
0:11:11proposed a map to be evaluate to each you yeah no
0:11:15for the efficiency in
0:11:18a a spatial component is there's
0:11:22in
0:11:23a a a there is an um
0:11:25each each they show the spatial components it
0:11:28a a a a a a a very low the harmonic
0:11:32and uh a a i i the
0:11:34coefficients still we hold
0:11:37spherical harmonic
0:11:38uh coefficients
0:11:39which is a given by a
0:11:41of the you know what are so it not yeah a and a very money
0:11:46and uh
0:11:48we might not be way and have to a a and is given by a
0:11:53uh
0:11:55uh right
0:11:56um
0:11:57and
0:11:58it is a need to me by a a a a wave number of K
0:12:04okay and not a
0:12:06oh
0:12:08oh of for that model we have
0:12:10oh have they have a
0:12:12capital and coefficient
0:12:15then oh we can do
0:12:17the are there the whole
0:12:19all these there will have money
0:12:21oh oh coefficient
0:12:23oh we find that
0:12:25and right oh okay efficient
0:12:28a a a a a nine one nine then
0:12:31of the total energy
0:12:33so uh for me looking and
0:12:36uh we define the H I T N as the initial efficient the a a and the whole key
0:12:44oh i shows
0:12:46um um
0:12:47the uh very is all all their a harmonic
0:12:51coefficients for the frequency and
0:12:53for a you know that
0:12:54and with C
0:12:56that
0:12:57the was the C coefficients of hand
0:13:00oh most of the energy
0:13:03and then uh
0:13:04think of be shows that the speaking a at
0:13:09okay so and and the capital and wrong
0:13:12calculate and in these oh to try to go and O
0:13:17a a and one the red
0:13:19a down
0:13:21and a problem if we should both he we can
0:13:25uh as you can be a the of they show of component expansion is a wrong
0:13:31seventeen then
0:13:33and then
0:13:34a a a a think good shows a migration
0:13:37a model the man
0:13:38a H I yeah the blue one
0:13:41and then
0:13:42a a a a a a a one in a a a a we constructed
0:13:46each each yeah using
0:13:48and and for a capital and to themselves
0:13:50spherical a a harmonic
0:13:53and that really one still
0:13:55uh
0:13:56the reconstructed hrtf
0:13:59oh using and trying in terms of spherical
0:14:02harmonic
0:14:04that we want to different this speaking writing one and right one is
0:14:08oh or wrong to one percent
0:14:12i
0:14:13oh let's say which are normal with select uh a base it is that a frequency components there is
0:14:22a a a a a a in it is a three D uh hrtf model
0:14:28we use
0:14:28uh
0:14:29spherical be L to expend the of frequency component
0:14:34and uh
0:14:36i
0:14:37i is that there will be a more efficient of which is given by be formula
0:14:44i still we have a other classes
0:14:47that's a
0:14:47a complex exponentials
0:14:50oh and agenda probably and T V share on a because
0:14:54uh are they are all you can have but it's in expanding of frequency components use or on the in
0:15:01term
0:15:03and uh
0:15:04collocation can be given by
0:15:06oh these three
0:15:08you cliche
0:15:09respectively
0:15:10then
0:15:11a here it would treat the and the
0:15:14a property of H T of components
0:15:17oh models using
0:15:18these four sets of orthonormal basis
0:15:21and
0:15:24but that and so on i shows the screen play
0:15:29and we can see that
0:15:31a a a a new value
0:15:32for for a a a a a uh
0:15:34according to compress C spun that actually is quite close
0:15:38to that there will be a function
0:15:40but still we model the
0:15:42note that
0:15:44uh i think he said all
0:15:46a a will be out in a sample
0:15:49and i with and still of the total energy of we're
0:15:53oh while uh i think in a comp simple and now shows only a have for three presents no the
0:16:01oh
0:16:02and G
0:16:03so
0:16:03with i
0:16:04here are
0:16:06spread will be found at most and is the passage voice C
0:16:10it's yeah frequency expansion
0:16:14okay
0:16:15uh motion
0:16:16uh and i right know we a a a a set the question
0:16:19how to we if they really the efficiency of a and
0:16:23oh
0:16:24a function of hrtf model
0:16:26and oh we use a
0:16:29we use the some of the
0:16:31where of random variable in in the model to uh evaluate the efficiency and the read that still
0:16:39efficiency of the model
0:16:41oh a spatial component expansion is around seventy percent and the
0:16:45but what is still
0:16:47uh a frequency component
0:16:49pension is the spherical bessel don't function
0:16:53thanks for attention
0:16:55you
0:17:01i
0:17:02i
0:17:17i
0:17:19okay
0:17:23i
0:17:24and
0:17:30all
0:17:32i
0:17:34i
0:17:34the that
0:17:35do that but um
0:17:37i think it it can be
0:17:39uh
0:17:40for
0:17:41the D C U and was to for all the prosody
0:17:45oh
0:17:46a a representation
0:17:49mm
0:17:50so i think it's can
0:17:52four
0:17:54hmmm
0:17:55i
0:17:56yeah
0:17:57hmmm hmmm
0:18:00hmmm
0:18:03yeah
0:18:06i
0:18:07i