SIGNALS AND SYSTEMS LABORATORY 10: Sampling, Reconstruction, and Rate Conversion

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1 SIGNALS AND SYSTEMS LABORATORY : Sampling, Reconsrucion, and Rae Conversion INTRODUCTION Digial signal processing is preerred over analog signal processing when i is easible. Is advanages are ha he qualiy can be precisely conrolled (via wordlengh and sampling rae), and ha changes in he processing algorihm are made in soware. Is disadvanages are ha i may be more expensive and ha speed or hroughpu is limied. Beween samples, several muliplicaions, addiions, and oher numerical operaions need o be perormed. This limis sampling rae, which in urn limis he bandwidh o signals ha can be processed. For example, he sandard or high ideliy audio is, samples per second. This limis bandwidh o below he hal-sampling requency o abou / 2 = 22 khz, and gives he processor only 22.7 microseconds beween samples or is compuaions. I should be noed here ha he hal-sampling requency / 2 may also be reerred o as he Nyquis requency and he sampling rae may be reerred o as eiher he Nyquis sampling rae, Shannon sampling rae, or criical sampling rae. In order o do digial signal processing o analog signals, one mus irs sample. Aer he processing is compleed, a coninuous ime signal mus be consruced. The pariculars o how his is done are addressed in his lab. We will examine i. aliasing o dieren requencies separaed by ineger muliples o he sampling requency, ii. iii. echniques o digial inerpolaion o increase he sampling rae (upsampling) and conver o coninuous ime, sample rae conversion using upsampling ollowed by downsampling. The highes usable requency in digial signal processing is one hal he sampling requency. In pracice, however, undesirable eecs creep in a requencies below he hal sampling requency. (This ypically happens near 9% o / 2. This is why he sandard or audio is khz insead o khz.) We shall observe his phenomenon and explain i. SAMPLING IN TIME, ALIASING IN FREQUENCY x() = samples/second x[k] Figure One Sampling Figure One is a represenaion or he sampling operaion. The inpu signal is analog and coninuous ime. The oupu signal is ypically digial, i.e. quanized, bu we will ignore he ac ha he oupu samples are inie words o digial daa and no real numbers. The par ha ineress us is ha he signal x[k] is discree ime. The inpu signal x() is coninuous ime and has a Fourier ransorm. Fourier () x( ) ( ) The discree ime oupu signal x[k] has a DTFT. The ime and requency represenaions or his signal are DTFT 2π x[ k] = x( k ) ( e ) = j( ω nω ), ω = = 2π. n= (2) ( ) Page o 9

2 The ime domain relaion on he le-hand side o equaion (2) is very simple, bu he requency domain relaion on he righ hand side is no. I is ar more ineresing. All he specral componens o () a requencies ha dier by an ineger muliple o he sampling requency are added ogeher. Thus one canno disinguish hese specral componens rom he samples x[k]. I he sampling requency is, hen or every sinusoid o requency here is anoher sinusoid o requency +, which has he same samples. The same is rue or + 2, + 3, and so on. All o hese sinusoids go by he same alias, namely he sinusoid o requency. We will observe his phenomenon o aliasing experimenally. This is demonsraed in Figure Two below. x().5 () [ k] = x( k ), =.5 sec. -2π 2π ω [rad/sec] ( e ) = j( ω nω ) n= 5 x ( ) π -π -2π 2π π 6π ω [rad/sec] Figure Two Sampling In Time, Aliasing In Frequency A ew commens abou Figure Two. The coninuous-ime signal x() is a dampened sinusoid deined by (3) x( ) = e sin() u( ), were u() is he uni sep uncion. This signal could be an example o a naural response o an underdamped parallel RLC circui. The specrum o x() is a commonly known Laplace ransorm given by () ( ) = ( s) s= = s 2 2s + s= which is shown in he upper righ panel o Figure Two. From his graph, i can be seen ha () is essenially bandlimied o he requency band 2π ω 2π. This is o say ha he requency componens ω > 2π conain much less energy han he requency componens o ω 2π. In pracice, he signal x() will be guaraneed o be Page 2 o 9

3 bandlimied by an analog lowpass iler called an ani-aliasing iler. The Shannon sampling heorem hen says ha we need o sample a a rae wice he maximum requency componen o our bandlimied signal. In our case, his ω 2 means ha our sampling requency will be ω = π. (Also, 2, = 2π = Hz and = π = =.5sec.) In ω accordance wih he Shannon sampling heorem, x() was sampled a rae ω = π rad/sec. The sample daa signal is shown in he lower le panel o Figure Two and he accompanying discree-ime requency response is shown in he lower righ panel. Noice he marks on he requency axis o he lower righ panel. These marks show where he aliases o he original signal are cenered. As he sampling rae increases, he cener requencies o he () aliases will move arher away rom he original signal s specrum. In Figure Two, he sampling requency ω prevens a noiceable overlap o aliases, leaving ( e ) looking essenially like (). Thus aliasing eecs are minimized. Aliasing eecs due o undersampling amoun o a se o samples ha appear o be he samples o a signal wih a lower requency, hence he erm alias. This is o be avoided since hese eecs canno be removed in digial processing. To preven his, one should sample a he Nyquis sampling rae or higher. THE DISCRETE-TIME TO CONTINOUS TIME RECONSTRUCTION PROCESS Le us sae an unwelcome ac. Given he inpu x(), one can easily ge he oupu samples x[k], bu he reverse is no rue. Given he oupu samples, we canno reconsruc he inpu wihou knowing exra inormaion, because we have hrown ou everyhing beween he samples. Obviously here are ininiely many ways o ill in he gaps beween samples. The sampling operaion is many o one, and is no generally inverible. However, i he inpu is bandlimied o he hal sampling requency, hen here will be exacly one inpu signal x() or he given se o samples. This is he Shannon sampling heorem, and he bandlimied reconsrucion is (5) x ) = x[ k] h( k ) k= (, where ( ) = sinc ( / ) h π. Equaion (5) represens he ideal. I is he goal o a good analog reconsrucion design o approximae his relaion. Thereore, make noe o wo hings: irs, equaion (5) is a pulse ampliude modulaion (PAM) equaion, and second, he PAM pulseshape is ha o he sinc uncion, or Shannon wavele. The sinc uncion in quesion is he impulse response o an ideal lowpass iler wih gain equal o he sampling period, and bandwidh one hal he sampling requency, hence he erm bandlimied reconsrucion. One deparure rom he ideal is allowed in audio sysems. Time delay is o no consequence, and hereore one can use a delayed sinc pulse. In digial eedback conrol sysems however, ime delay can presen sabiliy problems and is o be avoided. The ime domain process o reconsrucing x() rom is samples x[k] is demonsraed in Figure Three below. Page 3 o 9

4 x[k] h() = sin(π/ ) / (π/ ) x() = Σ x[k] h( - k ) Figure Three: A PAM implemenaion o Bandlimied Reconsrucion In Figure Three above, a random discree ime sequence x[k] was creaed ha is seven samples long. This inie sequence was hen reconsruced using Shannon reconsrucion. This migh be called ideal consrucion. Noice in he upper righ panel ha = sec, since he zero crossings o h() occur a second inervals. From his, i can be inerred ha a coninuous ime signal x() was sampled a a rae o one sample per second ( = Hz and ω = 2π rad/sec) o generae x[k]. PRACTICAL RECONSTRUCTION METHODS: THE DIGITAL PIECE The basic elecronic elemen used in convering a discree ime digial signal o a coninuous ime analog volage signal is called a digial o analog converer or DAC. I he conversion is very as, hen he oupu o he DAC will be essenially consan beween sample imes, and look like a sequence o up/down sair seps. We shall see ha his signal has undesirable high requency ariacs. For a sample rae o khz, or example, he DAC oupu will sound ariicial wih high requency inny sounds ha aren in he righ harmonic relaion o he original signal. These high requency componens are a he aliases o he baseband signal. Thereore hey move up in requency wih he sample rae. The process o qualiy reconsrucion o signals involves a sample rae increase beore using a DAC. I he samples, which are insered beween he known samples, x[k] are smoohly inerpolaed, hen he original high requency ariacs can be compleely suppressed. I is imporan o undersand how his whole process works. An incoming coninuous signal, x(), is bandlimied by an ani-aliasing iler and hen sampled (analog o digial conversion) o orm x[n]. I he samples x[n] were sampled Page o 9

5 a or above he Nyquis sampling rae, hey will conain enough inormaion o reconsruc he analog signal x ˆ( ), which is an esimae o he original x(). These samples, x[n], may hen be sored on a CD (compac disc) or ransmied wirelessly rom one cellular phone user o anoher or example. Once his digial signal x[n] needs o be recovered back ino an analog signal, i is processed digially and hen convered back o an analog signal. This is shown below in Figure Four. Since ideal reconsrucion is no pracical, digial signal processing is required beore he analog reconsrucion process can ake place. Le us discuss he elemens o Figure Four. x() Ani- Aliasing Filer = x[n] M w[k] PAM iler P(z) y[k] DAC v() Ani- Imaging Filer xˆ ( ) Analog o Digial Conversion Digial Signal Processing Digial o Analog Conversion Figure Four Digial Signal Processing block diagram Upsampling or Zero-Fill The irs block in he Digial Signal Processing secion o Figure Four above is a rae conversion o he digial signal x[k]. The sampling rae has been ariicially increased by he acor M, by placing M - zeros beween every wo samples x[k]. This is known as upsampling (or zero-ill). The oupu o he upsampler is DTFT / M j( Mn) ω / M (5) w[ k] = x[ n] δ [ k Mn] W ( e ) = x[ n] e = ( e ) n= The ineger M is he upsampling raio and M / is he new sampling rae aer he inserion o M - zeros. We say he sampling period or w[k] is / M, and he sampling requency is M. Wha eec does his have in he requency domain? Since he non-zero samples w[mn] = x[n] occur only a muliples o M, he erms in he DTFT o w[k] are Mπ Mπ exacly he same as he erms in he DTFT o x[n]. Thus he Nyquis band o W is < ω, and on his band / M (6) W ( e ) = ( e ), n= The graph o his as a uncion o ω is unchanged excep ha i exends M imes as ar o ge o he new halsampling requency. This is illusraed in Figure Five below. There is anoher way o look a his. By illing he gaps wih zeros, we have inroduced no inormaion ha wasn already here. This demonsraed in Figure Five below. Page 5 o 9

6 (e ). The Hal Sampling Frequency ω / 2 is Marked. -π -2π -π -8π -6π -π -2π 2π π 6π 8π π 2π π ω [rad/sec] Nyquis Band W(e / M ). M =. The Hal Sampling Frequency Mω / 2 is Marked. -π -2π -π -8π -6π -π -2π 2π π 6π 8π π 2π π ω [rad/sec] Nyquis Band Figure Five Zero-Fill Upsampling Inerpreed in he Frequency Domain Noice ha he original signal (e ) has a Nyquis band o 2π (-π o π) and W(e ) has a Nyquis band o *2π = 8π (-π o π), ye he requency conen is he same in boh graphs. Only he Nyquis band has changed as a resul o w[k] having a higher sample rae han x[n]. Upsampling wih a discree-ime PAM Filer The second block in he Digial Signal Processing secion o Figure Four above is a digial iler, which is used or digial inerpolaion. This iler is also called a discree-ime PAM iler. The oupu o his iler is he convoluion o he zero-illed sequence w[k] wih he pulse response o a PAM (or smoohing) iler p[k]: = n= n= DTFT / M / M / M / M (8) y[ k] w[ n] p[ k n] = x[ n] p[ k Mn] Y ( e ) = W ( e ) P( e ) = ( e ) P( e ) where P(z) is he ranser uncion o he PAM iler: (9) P ( z) = p[ k] z k k Noe here ha he discree-ime PAM iler, p[k], runs a a rae M imes aser han x[n], meaning i oupus values y[k] a rae M in response o nonzero inpus a rae. In order or he convoluion in equaion (8) o be Page 6 o 9

7 convenional, he rae o x[k] had o be increased o he rae o p[k], hence w[k] was creaed o mach he rae o p[k]. From equaion (8) we can see ha he ollowing noaions are equivalen: x[n] w[k] y[k] x[n] y[k] M P(z) M P(z) y [k] = w [ n] p[ k n] n y [k] = x [ n] p[ k Mn] Figure Six Equivalen Implemenaions o he Same Digial Inerpolaion Scheme Figure Six illusraes wo equivalen digial inerpolaion schemes ha ake a low rae inpu signal, x[n], and oupu a higher rae signal y[k]. On he le side o Figure Six, a more radiional and mahemaically convenional implemenaion is given. In his case, he signal rae o x[n] is increase o he rae o he inerpolaion iler by he use o a zero-ill upsampler or rae converer. The oupu o his upsampler, w[k], is hen passed hrough he inerpolaion iler p[k] where a sraigh convoluion is used o generae he oupu y[k]. In his case, p[k] is running a he same signal rae as w[k]. This is he way ha many exbooks explain he digial inerpolaion process. The righ side o Figure Six uses a dieren represenaion o digial inerpolaion. A low rae signal, x[n], is running hrough a high rae iler. This iler is oupuing samples y[k] a a rae M imes aser han he inpus x[n] are coming in. Given ha boh o he sysems are linear, an inpu o zero mus oupu a zero. For he case on he le, w[k] and p[k] are running a a rae ha is M imes aser han he inpu x[n], bu only he low rae inpus rom x[n] produce non-zero oupus rom he high rae iler p[k]. From his, i should be clear ha boh implemenaions produce he same exac oupu, y[k]. Figure Seven depics wha happens in his upsampling process, or a paricularly uninspired kind o inerpolaion. Suppose ha we increase he sampling rae by a acor o our by simply repeaing each sample our imes (known as zero-order hold upsampling by a acor o our). Clearly his won conribue anyhing o he DAC sair sep oupu problem, since i will lead o he same evenual analog signal as we would have had wihou he sample rae increase, bu we wan o analyze he general case, and his is one example. All ive o he signals in Figure Seven have he same ime base. The irs signal is he original coninuous ime signal x(). We will use a simple pulse shape. (This pulse was consruced in MATLAB using he saemen x=pulse(2,9,3). The uncion pulse.m can be ound on he webpage.) The sequence o samples is shown in he second panel o Figure Seven. The upsampled sequence w[k], which is our imes as as as x[n], is shown in he hird panel. The impulse response o p[k] is shown he ourh panel and he inal oupu y[k] is shown in he las panel. n Page 7 o 9

8 x() x[n], 25 samples/sec w[k], samples/sec Oupu o a Discree-ime Zero-Order Hold p[k], samples/sec y[k], samples/sec Figure Seven Simple repeiion o known samples Even hough Figure Seven does no help wih DAC problem, i gives invaluable insigh o he PAM idea ha was illusraed in Figure Six. Assume ha w[k] was never creaed and ha p[k] is running a a rae imes aser han he x[n]. As each sample o x[n] eners he iler, he ull impulse response o p[k] is oupued beore he nex impulse o x[n] excies he iler. This is a special case where he iler is only oupuing a response based on a single inpu sample. In oher words, he iler does no have o perorm any non-zero summing operaions beore i oupus is curren sample. Even hough his is a rivial case, i helps o demonsrae he idea o how a high rae discree-ime PAM iler responds o a low rae inpu signal, wihou he signal rae conversion. See he m-ile demo_pam_iler.m locaed on he webpage under Demos or Lab. Down load his ile and run i a ew imes o ge he eel o how discree-ime PAM ilers work. For your own amusemen, re-wrie demo_pam_iler.m o model he le hal o Figure Six. Noe, only he signal rae o he inpu will change. The oupu should look exacly he same. The requency responses o each o he ive panels in Figure Seven above are given in Figure Eigh below. Anoher common way o graph a complex requency response is o use he ideniy ω = 2π. Now () = (2π ), and he requency axis is now in Hz insead o radians/second. This is he echnique used in he ollowing complex requency response graphs. The same ive signals are shown wih a common requency axis. The hal sampling Page 8 o 9

9 requency or he our discree-ime signals are marked on he horizonal axis. Wihou his mark, he second and / M hird panels, which show ( e ) and W ( e ) would look exacly he same, because o equaion (7). Indeed he only dierence in he requency domain is he sampling rae. The signals in he hree lowes panels are / M relaed by equaion (8), i.e. Y ( e ) is he produc o he aliased specrum ( e ) and he inerpolaion / M / M iler requency response P( e ). In Figure Eigh P( e ) is he DTFT o a square pulse and is a sinc / M pulse. Thereore he high requency aliases in ( e ) appear in he inal signal Y ( e ). () ( e ). The Hal Sampling Rae / 2 is Marked. W ( e 2 / M ) = 2 2. Hal Sampling Rae = * / 2. P( e / M ). Hal Sampling Rae = * / 2. M = 2 2 Y ( e j 2π / M ). Hal Sampling Rae = * / 2. 2 Figure Eigh The Frequency domain equivalen o Figure Seven. 2 / M You should be able o compue he complex requency response Y ( e ) looks righ. o veriy ha panel o Figure Six Page 9 o 9

10 A more inspiring ype o digial inerpolaion can be seen when p[k] is a sinc uncion. The advanage o using his is ha a sinc uncion acs as a lowpass iler in he requency domain, which will do a beer job o suppressing he / M higher requency ariacs o Y ( e ) ha can be seen in he ih panel o Figure Eigh. This is demonsraed in Figures Nine and Ten below. Figure Nine conains he ime domain graphs and Figure Ten he requency domain graphs. The same pulse x() was used in Figures Seven hrough Ten. x() x[k], 25 samples/sec w[k], samples/sec Linear Phase FIR Shannon Reconsrucion p[k], samples/sec y[k], samples/sec Figure Nine Smoohing wih a digial lowpass iler Noice ha y[k] now looks more like he original analog signal x(), exceped ha i is delayed in ime. This is he price ha one mus pay or beer digial inerpolaion. Page o 9

11 () ( e ). The Hal Sampling Rae / 2 is Marked. W ( e 2 j 2π / M ) = 2 2. Hal Sampling Rae = * / 2. / M Linear Phase FIR Shannon Reconsrucion P( e ).Hal Sampling Rae = * / 2. M = Y ( e / M ). Hal Sampling Rae = * / 2. 2 Figure Ten The Frequency domain equivalen o Figure Nine. In Figure Ten, he high requency aliases are suppressed. I he iler P(z) were an ideal lowpass iler wih sharp cuo, hen he whole baseband < / 2 would be usable. Bu his is no pracical, and hereore a compromise mus be made. In high ideliy digial audio, he analog signals are bandlimied o 2 khz in he recording sudio wih a high qualiy lowpass iler. Then he sampling is done a. khz, which is o course percen above he Nyquis rae o wice 2 khz. The exra band above 2 khz and below 22 khz is used or a ransiion region in he design o he digial PAM inerpolaion iler. Such a ransiion can be seen in he ourh panel o Figure Ten. In he aliasing measuremens ha will be made, we will see a degradaion in he reconsrucion o sinusoids abou percen below he hal sampling requency. Page o 9

12 THE FINAL STAGE OF CONVERSION: THE ANALOG PIECE The upsampled and inerpolaed sequence y[k] is now passed hrough a DAC. The high requency ariacs due o his operaion will be muliples o M or he upsampling raio imes he original sampling requency. Here hey can easily be removed wih a cheap analog lowpass iler. The process is shown in Figure Eleven. Noe ha his is he las secion o Figure Four. y[k] Common DAC Q() v() Analog LPF F() x ˆ( ) The inal coninuous ime oupu signal is Figure Eleven Analog oupu sage () ˆ ( ) ( ) [ ] ( DTFT ) ˆ / ( ) ( M x τ y k q τ k dτ = Y e ) Q( ) F( j ω ) M =. k Here, Q() represens he DAC. In ime he DAC iler will have a uni sep response, which is a square pulse o widh / M. The Fourier ransorm will have magniude () ( ) ( / M ) sinc( πω Mω ) Q =. Le he oupu o he DAC be / (2) v ( ) = y[ k] q( k ), where wih he ollowing specrum k M, q( ) =, M, oherwise DTFT j ωτ d (3) v( ) V ( ) = y[ k] q( τ k ) e τ = k k = Y ( e y[ k] e M k M M ) Q( ) Q( ) Noice ha he analog reconsrucion in equaion (2) is a PAM equaion jus as i was in equaion (5). The dierence is ha equaion (2) uses a causal square pulse, and equaion (5) uses a non-causal sinc uncion. Le us examine he requency responses o each o he boxes in Figure Eleven. Page 2 o 9

13 Complex Frequency Response o a Common DAC Q( ). -M / 2 - / 2 / 2 M / 2 Complex Frequency Response o an Analog Low Pass Filer F( ). -M / 2 - / 2 / 2 M / 2 Complex Frequency Response Q( )F( ). -M / 2 - / 2 / 2 M / 2 Figure Twelve Complex Frequency Response o Figure Eleven In Figure Twelve, we can see ha he DAC iler Q( )is a very poor lowpass iler whose irs zero crossing is a M / 2. Noice also ha Q( ) does no ully suppress high requency componens o y[k], which are images or suppressed aliases o he original signal ( ). To compensae or his, he signal v() in Figure Eleven is passed hrough an analog lowpass iler F( ). This iler could be considered eiher a smoohing iler (or he eecs ha are seen in he ime domain) or an ani-imaging iler (or he eecs seen in he requency domain). Eiher way, he produc Q( )F( ) is hen a beer lowpass iler which is essenially consan on he requency band rom - / 2 o / 2, bu eecively aenuaes all requency componens above M / 2. This leaves a long gap or a ransiion band rom / 2 o M / 2. This is crucial because our inormaion lies in he requency band - / 2 o / 2, so his is he only par o he specrum ha we wan. Since he alised specrum o Y(e j2 π /M ) rom / 2 o M / 2 was aken ou wih digial signal processing, he inal sage o analog signal reconsrucion is no criical. Any reasonable iler will work because he diicul par has been done by he digial iler P(z). We will no simulae he inal oupu sage in Figure Nine, because (i M is reasonably large, say or more) hen he analog oupu will be indisinguishable rom a graph o he discree ime signal y[k] wih he poins conneced by sraigh line segmens. This is wha he MATLAB plo uncion does. We will however demonsrae his las sep in he requency domain. Figure Thireen below shows he reconsrucion o Y(e j2 π /M ) rom Figure Ten. Page 3 o 9

14 Y(e /M ). Hal Sampling Rae = * / 2. -M / 2 - / 2 / 2 M / 2 Complex Frequency Response o a Common DAC Q( ). -M / 2 - / 2 / 2 M / 2 Complex Frequency Response o he Analog Signal V( ). -M / 2 - / 2 / 2 M / 2 Complex Frequency Response o he Analog Low Pass Filer F( ). -M / 2 - / 2 / 2 M / 2 Complex Frequency Response o he oupu ˆ ( ). -M / 2 - / 2 / 2 M / 2 Figure Thireen Analog Reconsrucion o Figure Ten Figure Ten showed only he Nyquis band o Y(e j2 π /M ), bu he Nyquis band aliases every M. This is shown in he irs panel o Figure Thireen. The shorcoming o jus using a common DAC is ha he higher requency aliases o Y(e j2 π /M ) will no be suppressed. The combinaion o a DAC and a reasonable analog LPF will hen be enough o give reasonably accurae analog reconsrucion o a digial signal. You may noice some requency componens in ˆ ( ) ha were no in he original signal ( ). This is due o he ac ha a mid-grade digial iler was used in Figure Ten. In pracice, digial ilers are used ha have a beer complex requency response, bu i helps o show ha he qualiy o he inal analog signal depends almos enirely on he qualiy o he digial signal processing perormed and no he qualiy o he analog componens used. Page o 9

15 MATLAB TOOLS FOR RATE CONVERSION You will ind wo MATLAB uncions or up and down sampling, and wo demonsraion m-iles, which simulae experimens, involving rae conversion. The PAM iler upsampling process described in he ex above can be done wih he uncion d_up.m ound on he web page under Funcions For Lab. Type»help d_up y=d_up(x,raio,ype); upsample x, a an ineger (raio)=_ou/_in using PAM. Time delay is reurned in second oupu. pulse ypes: zero ill preserved square or zero-order hold (ordinary DAC) 2 riangular (linear inerpolaion) 3 FIR linear phase hamming o order 8(raio)+ There are our PAM ilers available, as one can see rom he above help lising. These are zero-ill upsampling wihou any ilering a all, he square pulse used in Figures Seven and Eigh, a linear inerpolaion scheme which uses a riangular pulse response, and he Shannon approximaion used in Figures Nine and Ten. Any ineger upsampling raio, rom o 6 can be used. The original samples will be preserved by any o hese upsampling schemes. However he Shannon approximaion will have a laency (delay) o d. To ge he delay as well as he oupu signal, use he orm [y,d]=d_up(x,raio,ype). The down sampling process (subsampling) uses he ool d_down.m, which is also on he web page under Funcions For Lab. Type»help d_down x=d_down(u,decimae); downsample x, a an ineger (decimae)=_in/_ou The demonsraion m-ile demoa.m produces hree panels o oupu like hose shown in Figure Foureen. In he irs panel, he original signal x() and he (ime delay adjused) inal coninuous ime oupu xˆ ( ) are overlayed. Also he samples o x() which were used o produce x ˆ( ) are shown wih small circles. The specra o hese wo signals are shown in he second and hird panel. This simulaion necessarily uses shor signals wih a small number o samples. One can choose hree kinds o signals, shown in he noe below Figure Foureen. There is a requency parameer alpha ha is normalized. The sampling requency is alpha = 2. The hal sampling requency is alpha =. The PAM iler ype and upsampling raio are arbirary. Since his is an m-ile, one mus speciy all parameers in command mode. The demonsraion m-ile demob.m produces wo panels o oupu like hose shown in Figure Fieen. These are simply he ime domain and requency domain represenaions o he inal coninuous ime oupu x ˆ( ). This simulaion involves longer signals wih many samples. I sound is available, he signal will be played hrough he audio oupu. This is especially revealing o he high requency ariacs o he reconsrucions. In boh hese simulaions, he sampling requency is khz. The oal ime duraion is abou.28 seconds. The ype o reconsrucion iler will be divulged in he iles o he ime domain plos. The upsampling raio can be deeced by dividing he maximum requency o he reconsrucion specrum by 2, which is he original hal sampling requency. For sinusoidal signals, he specra rom demob.m will be much sharper han hose rom demoa.m because he daa lenghs are greaer. Page 5 o 9

16 blue is single sinusoid. black is linear inerpolaion x -3 single sinusoid x -3 linear inerpolaion requency in Herz x requency in Herz Figure Foureen Oupu o demoa.m. This example uses ype=2, or linear inerpolaion, and s= s or a sinusoidal signal. The upsampling raio is 8. The requency o he sinusoid is deermined by he parameer alpha. In his case, alpha =.75. Observe he errors in he ime domain, and he high requency ariacs in he requency domain. To recreae his in MATLAB, ype»s= s ; ype=2;raio=8; alpha=.75;demoa»help demoa demonsraion o aliasing undeined vbls: s,ype,raio,alpha (mus deined in workspace) s= s sinusoid wih normalized requency alpha s= c wo sinusoids wih normalized req. alpha, alpha/3 s= b narrowband process wih normalized cener requency alpha (or ype and raio see d_up.m) Page 6 o 9

17 oupu o common DAC or samples o narrowband noise ime in seconds specrum requency in Herz Figure Fieen Oupu o demob.m. The signal ype is s= b, or a narrowband signal. The cener requency is deermined by he parameer alpha. In his case, alpha =.5. The upsampling raio is 8, and he PAM reconsrucion ype is, or a square pulse response. Observe he seps in he ime domain plo, and he high requency ariacs in he requency domain. To recreae his in MATLAB, ype»s= b ; ype=;raio=8; alpha=.5;demob»help demob demonsrae inerpolaion/reconsrucion o analog signals rom samples undeined vbls: raio, ype, s (sring), alpha (mus deined in workspace) signals: s= s single one a requency alpha*s/2 s= c chord s= u uni pulse s= n noise s= b band noise Page 7 o 9

18 THE DIGITAL CRACKS HYPOTHESIS Several years ago he ollowing Exchange was made wihin a cerain users group on he ne. I involves a heory abou he qualiy o digial audio ha says ha cerain requencies are avored, because aer all, he medium is digial. I have removed all inormaion ha would ideniy he auhors, and I hope ha I am no breaking any laws by reproducing his maerial. (Fan number one.) Now, I know ha CDs have some major advanages over vinyl in erms o wear -n- ear, and background noise due o dus, ec. My riend and I came up wih a heory why digial music will never reproduce sound as well as analog devices. Since music is made up o sound waves which range over a coninuous specrum o requencies, i seems ha digial music will suer in any requencies in beween he ones recognized by he digial equipmen. (Fan number wo.) This is quie rue. The analog-o-digial converers are calibraed o pick up he requencies o _correcly uned_ guiar srings, piano keys, ec. These are reproduced unerringly. However, i is seldom he case ha insrumens are in perec une. I, or example, an A is Hz raher han he sandard, a digial recorder will no reproduce i a ull volume, and i may be slighly muled. This is he key o he o-heard complain ha grungy music (e.g., Sonic Youh) sounds oo hin and no very powerul on CD relaive o LP. In he case o SY, hey oen use non-sandard guiar unings, which, since hey do no go o he expense o using specially-calibraed A/D converers, are poorly convered o digial waveorms. A group like Floyd would cerainly be in perec _relaive_ une in he sudio, bu possibly no in perec _absolue_ une. The eec would be he same -- dull sound rom he alling beween he binary cracks eec. I, during he re-mixing rom wo-rack analog masers o 2- rack digial ape, he masering engineer adjuss he numeric represenaion used in he digial domain, a airly accurae conversion can be accomplished. In pracice, a sraigh ranser is done wih lile care, and he resul sounds hin, as essenial inormaion was los. This is probably he eec ha you have noiced. I may seem odd ha Mobile Fideliy Labs didn ge his righ wih Dark Side o he Moon, bu i is rumored ha his was due o heir use o ixed-poin arihmeic in he D/A converers due o an engineering oversigh a he ime (DSOTM was one o MFSL s earlier eors, alas). The Seely Dan album I don know abou. Analogous o he problem o represening cerain numbers in binary, such as repeaing decimals. The new Sony MD recordable discs avoid his problem enirely by using a coninued-racion represenaion o he signal. This is possible because he MD orma is no purely opical (as wih CDs), bu raher magneo-opical, hereby incorporaing more o he coninuousdomain inormaion rom he analog signal, much as an analog magneic ape gives a perecly coninuous playback o an analog signal. Thus, we can expec he new orma o have much less o a binary problem han curren CDs. I hope his cleared hings up. Assignmen: Beore doing he pars below, amiliarize yoursel wih he demonsraion m-iles demoa.m and demob.m. Exercise hem wih all inpu choices, or ypes and 3 o PAM ilers. You do no need o make any prins.. PAM iler ypes Use demob.m wih a uni pulse inpu o observe he our PAM ilers in he ime and requency domains. The ollowing command line can be used.»s= u ;ype=;raio=8;demob Repea wih ype =, 2, and 3. Make plos, and wrie commens on each. Mark he posiion o he hal sampling requency (2 Hz). Ideniy high requency ariacs in he specrum. Now repea hese excep using a sinusoidal inpu signal. Use he ollowing command.»s= s ;alpha=.7928;ype=;raio=8;demob Page 8 o 9

19 Assignmen: 2. Aliasing Use demob.m, wih s= s, ype=3, and raio=8. Vary he parameer alpha beween and 2 o change he requency, and observe he oupu in boh ime and requency. I is no necessary o make any prins o plos. We wan only o characerize he oupu. I will be eiher a single sinusoid (alpha=.3 or insance), or a mixure o wo (alpha=.97). Ignore he lile ransien a he beginning. The parameer alpha is a racion o s 2 = 2 Khz. Over he range α 2 deermine he oupu requencies (eiher or 2) and skech he specra in he range 2. 5 khz. Use closely spaced alpha near he value alpha=. For he requencies alpha=.85,.973,.27,.9, run demoa.m and observe aliasing. Do you deec any experimenal evidence or he digial cracks hypohesis? I so, ind a heoreical explanaion. I no, ind a heoreical jusiicaion. 3. Wrie a MATLAB m-ile o duplicae Figure Nine. Use he ool d_up.m o do he upsampling. Make sure ha he same ime axis is used or all our plos. In order o do his, i is helpul o compose a ime base or each signal, like w=sw*[:lengh(w)].. Wrie a uncion (m-ile) o do sample rae conversion by a raional number M / N. This uncion should call boh d_up.m and d_down.m o do he rae conversion. Creae an ineresing signal, upsample by M and downsample by N. (Use M=5 and N=7.) Plo he samples, using sem on wo separae panels, agains he same ime axis or comparison. Then repea he process excep downsample beore upsampling. Which is preerable? Page 9 o 9