Sinusoidal signal. Arbitrary signal. Periodic rectangular pulse. Sampling function. Sampled sinusoidal signal. Sampled arbitrary signal

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1 Techniques o Physics Worksheet 4 Digital Signal Processing 1 Introduction to Digital Signal Processing The ield o digital signal processing (DSP) is concerned with the processing o signals that have been converted into digital data. DSP techniques have ound application in many areas including image processing, instrumentation and control, data compression, audio, telecommunications and biomedicine to name but a ew. Although primarily a subject area o electronic engineering, an understanding o DSP can be extremely valuable to anyone involved in science and technology, including physicists. Modern physics experiments invariably use a data acquisition (DAQ) system to collect and record data in digital orm. The methods o DSP are then used to extract the required inormation rom the data. This might be done in real-time as the data is acquired or `o-line' as the irst step in the analysis o the data. 1.1 Discrete-time Systems A signal processor can be pictured as a `black box' with an input that receives a signal and an output that transmits a version o that signal ater it has been transormed in some way. A digital signal processor receives and processes discrete-time signals i.e. signals that have been sampled at regular time intervals 1. A sampled signal is basically a representation o a continuous signal d(t) by its values at discrete instants o time, x n = d(nt ), where T is the interval between regular samples, and n =; 1; 2;:::. Sampled signals are usually digitised, i.e. each sample is converted to a numerical value representing the magnitude o the signal at the instant o the sample, which enables them to be processed digitally. With both sampling and digitisation there is a loss o inormation and there are resulting side eects. An understanding o the side eects is essential when considering DSP systems. A system that process discrete-time signals is said to be linear i it obeys the principle o superposition, i.e. the response o the system to two or more inputs is equal to the sum o the separate responses to each input in the absence o other inputs. For example, i a discrete-time input signal a n gives rise to the output signal c n and b n gives rise to d n then the input signal a n + b n produces c n + d n. A discrete-time system is said to be 1 Signals need not be time-based but or the purposes o this document we are going to always talk in terms o time-based signals 1

2 } Figure 1: The convolution o a sampled signal with a real-time system. time invariant i its output does not depend on the time that the input is applied, i.e. the input signal x n gives output y n then the time-shited input signal x n+i will give the output y n+i. When a discrete-time system is both linear and time invariant its output y n or an input signal x n is given by the convolution sum y n = 1X k= h k x n k (1) where h k is the impulse response o the system. The convolution operation is shown graphically in igure 1; each output is generated by a sum o products between one input x n k and one coeicient rom the impulse response o the discrete-time system, h k. As a side remark, one might expect the convolution to be symmetrically centred around a given x n value, using input data rom both x n+i and x n i. However, in a real-time data processing system, where one wants to generate an output with minimum delay ater acquiring each new input sample, the x n+i data are not available at the time o the acquisition o x n, and so the convolution sum must work backwards rom the most recently acquired data. I the input signal is in the orm o an impulse i.e. x n = :::;; ; 1; ; ;:::the output o the system is h k (although shited in time.) Any input signal can be considered as a sequence o impulses with dierent amplitudes so, by the principle o superposition, the output is a sum o the responses to the individual input impulses. Hence, the output is the convolution o the input signal with impulse response described by equation 1. It is important to appreciate that the values o impulse response completely deine the system. Once you know the impulse response you know everything about the system and can calculate the output or any input signal. An alternative and complementary way o looking at signals and the response o systems is in the requency domain. Signals can be described by the amplitude and phase o their requency components and systems can be described by their requency response. 2

3 Converting between time domain and requency domain descriptions involves the Fourier transorm, which allows a periodic signal d(t) with period T to be decomposed as a set o components: d(t) = C n = 1 T 1X n= 1 Z T=2 C n e j2ßnt=t = T=2 1X n= 1 C n (cos 2ßnt=T + j sin 2ßnt=T ) d(t) e j2ßnt=t dt (2) A time domain signal x(t) and its requency domain equivalent X(!) orm a Fourier transorm pair and are related by: X(!) = F [x(t)] x(t) = F 1 [X(!)] (3) where F represents the Fourier transorm and F 1 represents the inverse Fourier transorm. In general, X(!) is complex and can be written in the amplitude-phase orm jx(!)j exp(i X (!)) where X (!) is the requency dependent phase angle. Similarly, a system's impulse response h(t) orms a Fourier transorm pair with the system's requency response or transer unction, H(!): H(!) = F(h(t)) h(t) = F 1 (H(!)): (4) Since H(!) is a complex unction it can modiy both the amplitude spectrum and the phase spectrum o the input signal. In the requency domain, the output o a system, Y (!), is calculated by multiplying the input, X(!), by the transer unction H(!) i.e. Y (!) =H(!)X(!): (5) Equation 5 is the requency domain equivalent o convoluting the input signal in the time domain with the impulse response, as can be seen by taking the Fourier transorm o y(t) = h(t) Λ x(t) where the Λ operator represents convolution: F [y(t)] = F [h(t) Λ x(t)] F [y(t)] = F [h(t)] F [x(t)] Y (!) = H(!)X(!): (6) In this working we have made use o the convolution theorem which states that convolution in the time domain is equivalent to multiplication in the requency domain.e. F [x 1 (t) Λ x 2 (t)] = X 1 (!)X 2 (!): (7) 3

4 Similarly, multiplication in the time domain is equivalent to convolution in the requency domain i.e. F [x 1 (t)x 2 (t)] = X 1 (!) Λ X 2 (!): (8) When thinking about signal processing problems it can be very helpul to keep these two relationships in mind. Figures 2a) d) show a number o kinds o signals in the time domain, and their representation in the requency domain, which can be obtained by the application o the Fourier Transorm. The requency domain diagrams show only the magnitude o the requency components, without any phase inormation; in general, the C n coeicient o any component is complex. A pure sine wave has only one requency component, so appears in a) as a delta unction at the requency o the signal. Note that a corresponding requency component exists with negative requency, but the coeicient is o opposite sign. The more general signal shown in b) is made up o a range o requency components, and in the arbitrary (non-periodic) signal case these components will orm a continuous distribution. For a periodic signal, such as the rectangular pulse shown in c), then only discrete requency components are required. The components are spaced by 1=T, where T is the period o the signal, and or the rectangular wave the coeicients o the requency components are given by the sinc() unction, with nodes spaced by 1=i, where tau is the width o each pulse. A special case o the periodic rectangular pulse is the sampling unction shown in d), where the width o the pulse has been reduced to zero. The components are still spaced by 1=T, where T is now the sampling period, and hence s =1=T is the sampling requency. The components are all o uniorm height since the nodes o the envelope are spaced at ±n1 realistic sampling involves non-zero sampling times, and so the distribution o components is not completely uniorm. 1.2 Sampling A sampled signal can be considered as being non-zero only at the regular sampling instants and zero at other times. In the time domain, this is equivalent to multiplying the continuous analogue signal by the ininite sequence o Dirac delta unctions (spaced by the sampling period T ) which make up the sampling unction. As has already been stated in Section 1.1, multiplying in the time domain is equivalent to convolution in the requency domain so we need to take the Fourier transorms o the input signal and the sampling unction and convolute them. First, consider again the pure sine wave; when multiplied in the time domain with the sampling unction, one obtains the sampled signal shown in Figure 2.e). The convolution o the delta unction o the signal with the sampling unction results in the signal appearing with an oset o d on either side o each integer multiple o the sampling requency, s. This pattern o signal appearing in upper and lower sidebands around repeated harmonics o the sampling requency, s, can be understood by expanding the 4

5 a) Time Domain Frequency Domain Fourier Transorm Sinusoidal signal t b) Arbitrary signal d t c) T Periodic rectangular pulse 1/T d) T t -3/ -2/ -1/ Sampling unction 1/ 2/ 3/ t -s s=1/t e) Multiplication Convolution Sampled sinusoidal signal ) t -s-d -s+d -d d Sampled arbitrary signal s-d s+d t Figure 2: Time-domain Frequency-domain equivalence and the eect o sampling a signal. 5 -s s

6 sampling unction, consisting o delta unctions at times t = ;T;2T :::, as a Fourier series in terms o its cosine components as z(t) =a + a 1 cos! s t + a 2 cos 2! s t + (9) where! s =2ß=T. I d(t) is a sinusoidal signal, sin! d t, then the sampled waveorm x(t) is given by multiplying the two time-domain unctions x(t) = sin! d t (a + a 1 cos! s t + a 2 cos 2! s t + ) = a sin! d t + a 1 2 sin(! s! d )t + a 1 2 sin(! s +! d )t + a 2 2 sin(2! s! d )t + a 2 2 sin(2! s +! d )t + (1) where the n! s ±! d terms ollow rom sin A cos B = [sin(a + B) + sin(a B)]=2. In general the input signal will consist o a range o requencies with an upper limit, which is perhaps determined by the limited requency response o a transducer (see Figure 2.b). When this signal is multiplied in the time domain by the sampling unction the result is the sampled signal shown in Figure 2.). The convolution o the requency domain representations o the input signal and the sampling unction is achieved by the same expansion around harmonics o s applied to every requency component o d(t). This results in the requency domain picture shown, with symmetric images o the input signal around each integer multiple o the sampling requency (only a ew are shown on the igure). Problems occur i the input signal contains requencies greater than s=2, the so-called Nyquist requency. In this case, the repeated spectra start to overlap as illustrated in Figure 3 and it becomes impossible to distinguish in this overlap between input requencies greater than s=2 and those less than s=2. This eect is called aliasing 2. The conclusion o this analysis is that requencies greater than s=2 cannot be recovered once the signal has been sampled at s. Not only is inormation rom requencies greater than s=2 lost, but these requencies appear as requencies below s=2, distorting the description o the requencies which are truly below s=2. To avoid the conusion o lower and higher requencies, it is good practice to remove the higher requencies by applying a low-pass ilter at the input o sampling circuit to remove all requency components above s=2. I the signal contains useul inormation at requencies > s=2 then it is necessary to increase the sampling requency. 1.3 The Fast Fourier Transorm The ast Fourier transorm (FFT) is a very useul tool or estimating the requency content o signals. The algorithm is highly eicient and using modern microprocessors 2 The amiliar eect in ilms o spoked wheels appearing to rotate more slowly than we know tobe the case, or even in reverse, is closely related to aliasing; the rotation o the wheel corresponds to a periodic waveorm and the sampling is provided by the camera shutter. 6

7 -s s Figure 3: Aliasing in the requency domain. it is possible to do real-time requency analysis o signals in the audio requency range. The FFT is a special orm o the more general discrete Fourier transorm (DFT). In turn, the DFT is a discrete-time orm o the Fourier transorm. That is, the input time domain data is in the orm o a sequence o discrete values and the output is a set o discrete requency amplitude-phase values. The DFT and Fourier transorm are thereore related but are not exactly equivalent. The DFT o a sequence o N discrete-time values x n, where n = ; 1;:::N 1, is given by: X k = p 1 N 1 X x n e i2ßkn=n (11) N n= where k =; 1;:::N 1. Hence, the DFT returns N complex values X k which represent the amplitude and phase or the harmonic requencies = ks=n where s is the sampling requency. The inverse DFT is given by: X x n = p 1 N 1 X k e i2ßkn=n : (12) N k= The FFT is mathematically identical to the DFT but the number o data points is restricted to 2 M where M is an integer. FFT's are much aster because the algorithm takes advantage o computational redundancies in the DFT. The discrete nature o the DFT results in side eects which need to be appreciated when using the DFT (or FFT) to analyse signals. The irst problem is aliasing has already been discussed. This problem can be solved by increasing the sampling requency until the requencies o interest are below the Nyquist requency. The second problem occurs because a signal component with requency not exactly equal to one o the harmonic requencies = ks=n cannot be properly represented. The result is that its amplitude is shared between nearby harmonics. This eect can be reduced by increasing the number o data points either by analysing more points or, i that is not possible, by adding zero values to the end o the data. This improves the 7

8 No window - sine sine Rectangular window - sine sine Figure 4: The eect o a rectangular window on a sine wave in the requency domain. spectral resolution o the DFT by reducing the spacing o harmonic requencies,, since or an N-point DFT = s=n. The third problem is spectral leakage and is the result o analysing only the inite time interval N=s. In order to resolve a signal into a inite set o discrete requency components, the DFT assumes that the signal is periodic (non-periodic signals require a continuous spectrum o ininitessimally-spaced components). However, in obtaining a inite number (2 M ) o input samples a signal o ininite duration has eectively been multiplied by a rectangular window unction to give a signal o inite duration. In the requency domain this is equivalent toconvoluting the requency spectrum o the signal with the Fourier transorm o the window unction. The simple rectangular window unction in the time domain transorms to a sinc() unction in the requency domain, with the width and spacing o the sinc unction's lobes being inversely related to the width o the rectangular window. Convolution with the input signal, as shown in Figure 4, with a wide central lobe and the long tails o the sinc unction results in the smearing o each requency component in the signal, so that it leaks" across several requency bands. For requencies or which the sampling duration (NT) is an integer multiple o the true period o the signal, the assumption by the DFT o a periodic signal is valid and the DFT works as i a inite window unction had never been applied, hence no spectral leakage. The spectral leakage problem problem can be improved irstly by increasing the width o the time 8

9 window, and secondly by using window unctions other than a rectangle which shape the input waveorm to look more like a periodic waveorm. Such window unctions have shorter tails in the requency domain, and so introduce less spreading over neighbouring requencies when convoluted with the signal. 1.4 Digital Filters Digital ilters are discrete-time systems that modiy the amplitude and/or phase o signals in a requency-dependent way. Filters are usually used to extract only the requencies o interest rom a signal. Very oten ilters are used to remove noise which contaminates a signal. The great advantage o digital iltering over using analogue ilters is that virtually any kind o digital ilter can be realized and implemented in a lexible and convenient way. The properties o a ilter are determined completely by its impulse response and digital ilters can be designed to have any arbitrary impulse response. I the impulse response is o inite length the ilter is a Finite Impulse Response (FIR) ilter and the output can be calculated directly rom the convolution sum presented earlier. In the context o implementing a digital ilter, the values o the impulse response become the ilter coeicients. Designing FIR ilters is relatively straightorward. The impulse response required to implement the ilter is obtained as a Fourier Transorm o the desired requency response. A simple and useul example is the ideal low-pass ilter response H D (!) shown in Figure 5. A ilter with this ideal response removes all requencies above the cut-o angular requency! c (the requency range has been normalised so that the sampling angular requency is 2ß). Notice that the requency response repeats because o the discrete-time nature o the signals. The impulse response can be calculated by taking the inverse Fourier transorm o H D (!), resulting in h n =! c ß =! c ß sin(n! c ) n! c ; n 6= n = (13) where n is an integer and 1 <n<+1. Immediately we see that we have a problem because an ininite number o values are required. To produce a practical ilter it is necessary to use only a `window' o values o h n around n =. This is equivalent to multiplying the impulse response by a rectangular window and the result is that the requency response deviates rom the ideal low-pass response (in a manner analogous to the spectral leakage caused by multiplying a time-domain signal by a rectangular window). Rather than multiply the requency response by a rectangular window, other window unctions can be used to reduce some o these problems in the same way that spectra rom the FFT can be improved by using a suitable time-domain window unction. 9

10 H D (ω) -2π -ω c ω c 2π ω (normalised) Figure 5: Ideal low-pass ilter requency response. It is highly recommended that you do some supplementary reading on digital signal processing beore attempting this worksheet. Search or keywords Digital Signal Processing" or Digital Filters". Some example titles are listed at the end o the worksheet (any one o these should contain useul analyses o sampling and iltering and there are many similar texts). 2 Exercises Week 5, Session 1/2 2.1 Sampling Use Mathcad to simulate sampling o a sine wave o requency at a sampling requency s = 1 Hz. Use the orm sin(2ßnt) where initially =1Hz and T =1=s and plot the sample values on a graph 3. Note that sampled input data is best stored in an array o samples, rather than as a unction o time or sample number; this approach will be needed or the later exercises (it is also advisable to treat time in the same way, as t n = n t, and not to use integer values o time t = n). Describe the appearance o the sampled signal as you increase in several steps rom a ew Hz to the Nyquist requency s=2 and then above the Nyquist requency (make sure you include requencies just a ew Hz either side o the Nyquist requency). At exactly the Nyquist requency, it is necessary 3 This may sound diicult, but all it means is that you plot the values o a sine wave at discrete intervals o T = :1 seconds. The sampling is done or you by using nt as the time variable in plotting the sine wave, rather than any other arbitrary time-base. Note that whenever you plot a continuous unction with MathCad you are actually plotting its value at a ew discrete points the curve" you see is due to the connection o discrete points by the deault option or displaying a trace. Presentation o discrete or sampled data is made clearer by using other Trace option to explicitly show that the data is discrete, such as bar, point, symbol or stem. 1

11 to introduce a phase shit into the signal in order to see the sampled waveorm. Compare graphs or requencies separated by multiples o the sampling requency i.e. ( + Ns) where N is an integer. Also compare negative and positive requencies in the ormula. At each stage try to justiy what you observe by thinking in the requency domain. 2.2 Discrete-time Systems We are going to investigate the properties o a discrete-time system which has the ollowing inite impulse response: h =:81 h 1 =:247 h 2 =:344 h 3 =:247 h 4 =:81 Using equation 1, plot the output o the system or a unit impulse input i.e. x n = or all values o <n<n (N 1) apart rom one value x m =1,where m =4(or 5 or 6 etc). Veriy that the output o the convolution o x with h is in act the impulse response h given above, but shited in time (remember that Mathcad's arrays run rom zero by deault.) Note that in constructing the convolution ollowing equation 1 an error will be produced i x n k is addressing a non-existent element ox, i.e. n k<. There are a numberoways o tackling this, o which simply starting the convolution at x 4 is the crudest in a real application valuable transient data might be contained in the irst samples o x and by not processing them some inormation may be lost. A little bit o thought and an i(n k ;:::) construction should do the trick (non-existent data rom beore x can be presumed to be zero). Investigate the response o the system to a sampled input sine wave (such as those produced in the irst exercise), changing the requency o the input between zero and the Nyquist requency. Comment on the amplitude and phase delay responses you observe, plotting the amplitude response as a unction o requency. What kind o requency response is this? What unction is h perorming? The extraction o amplitude response can be automated" by using 2D arrays with dimensions o sample number and requency index or both input and output, as or studying orced oscillations and resonance in Worksheet 2. Generate the input data with requencies taken rom an array i with values covering the range 1 Hz. The maximum point on the output waveorm or input requency i can be determined using the max() unction acting on the i th column o output values, which may be extracted using the A <i> operator (see p 156). To avoid transient eects, copy the latter part in time (covering a complete cycle or all requencies) o the output array to a new array beore apply the max() unction. Automation o the phase-delay response is too complex 11

12 to do here, but the approximate behaviour can be determined by visually comparing input and output waveorms or several requencies. Finally, try sampled square waves o dierent requencies, starting with» 5Hz and going up to the Nyquist requency. Comment on what you see, particularly how the shape o the output is dierent to the input (think in terms o the requency components o the square wave input and what eect the transer unction o h will have on each component.) A square wave with period i can be generated using a construction such as: sq n =i(mod(nt; i) <i=2; 1; 1) where the square wave will only be regular i i=2isaninteger multiple o the sampling interval T, so stick to these requencies. Week 6, Session Using the Fast Fourier Transorm Mathcad includes the built-in unctions t() and it() or perorming the Fast Fourier Transorm and Inverse Fast Fourier Transorm respectively (see p 18). Note that t() and it() only work i supplied with a vector o N =2 M samples, and that they then return requency components n = :::N=2. Use the t() unction to analyse a sampled sine wave generated using: x1 n = sin(2ßnt) where n is an integer such that» n» N 1 (N =2 M is the number o samples), is the requency o the input sine wave, and T =1=s (s is the sampling requency and should initially take the same value as N, so that the sampled input covers 1 second o data). Start by using N = 64 and = 6: and comment on what you see. Then try scanning rom 6. to 7. Hz in steps o.25 Hz and comment on what you see. Since the output t() is complex, you should look at the modulus o the individual components. Try changing the number o data points, N, you use without changing the sample requency, s. First ix the input requency to = 6:25 Hz and set the number o samples to 128 and then 256, then ix the requency to =7: Hz and hal the number o samples. Compare how each requency appears to when N = 64 and try to explain any changes you observed. Return to N = 64 and = 7: Hz and try doubling the sample requency, s. Again, comment on what has changed relative to s =64and why. Note that with each new requency it is necessary to avoid the possibility o re-using data in old versions o x rom previous waveorms; this can be done either by renaming the input array or each requency or by explicitly setting its contents to zero beore generating each new sampled sine wave. 12

13 It is particularly important to use a new vector when reducing the number o samples, since once declared, a Mathcad vector can only have its number o elements increased. Think about how the Fourier Transorm is attempting to represent the input in terms o a certain set o requency components. Can you explain why certain values o are ree o spectral leakage? What determines the requency components o the Fast Fourier Transorm? The Fast Fourier Transorm has been presented with samples within a inite time window, but will attempt to analyse the input as i it were an ininte periodic unction which repeats outside the sampling window. Why are some requencies better suited to this treatment than others? I the input waveorm is being multiplied by a rectangular window in the time domain, what is the equivalent picture in the requency domain? It should be clear by now that there are eects which limit the spectral resolution that can be achieved using a FFT. The inite resolution limits our ability to resolve individual spectral components when spectral leakage is present. Look at the FFT spectrum o the sum o two sine waves, given by: z n = sin(2ß 1 nt )+:1 sin(2ß 2 nt ) where 1 = 4:1 and 2 = 7 using N = 64 data points, s = N and T = 1=s. It is very hard to resolve the smaller sine wave because o spectral leakage rom the larger sine wave. So ar, we have eectively been using a rectangular window o width N samples i.e. the sampled sine wave has been multiplied by a unction z n = z n w n, where w n =1 or < n < N 1 and w n = otherwise. Now see what happens when you multiply the data by the Hanning window unction given by: w n =:5+:5 cos( 2ß(n N=2) ): (14) N Plot the windowed version o z n and comment on the dierence between this and the original z n plot (clearest by overlaying them). Then perorm the t() and comment on the changes you see in the requency response and try to explain them, considering the requency domain equivalent o multiplying z n by the windowing unction, w n. You will ind it helpul to compare the requency domain representations o the rectangular and Hanning windows. Week 6, Session Designing a Digital Filter Use equation 13 to calculate the impulse response or a low-pass inite impulse response (FIR) ilter with a normalised cut-o requency w c =2ß=1. Use a rectangular window 13

14 o width 128 to start with. Arrange the centre o the impulse response h n so that it is a maximum at n = 64, and be sure to repair the glitch" in the centre o the sinc unction. Look at the amplitude response o the ilter by using t() to calculate the transer unction corresponding to this impulse response (which has been realised with a inite number o coeicients). Compare this amplitude response with the ideal rom which the impulse response was derived, thinking about the important characteristics o an amplitude response in both the pass band (low requencies), the stop band (high requencies) and the transition between the two. The logarithmic plotting option will allow you to look at the requency response above the cut-o requency. Try changing the width o the rectangular window containing the impulse response (always using 2 M samples), re-centring the impulse response in the new window. Comment on what you see, considering how changes to the rectangular window aect the requency domain. Now return to 128 coeicients and multiply the impulse response by the Hanning window unction Compare the requency response to that o the original 128 coeicient ilter and comment on the changes you see, trying to explain them by thinking in both time and requency domains. 2.5 Using a Digital Filter Finally, add some noise random positive and negative luctuations generated using the rnd() unction to a sine wave. Use your FIR ilter rom the previous exercise to ilter the noisy signal to reduce the noise (in the same way as you iltered various sine waves in exercise 2.2), while preserving the input sine wave (its requency should clearly be below the ilter cut-o requency). Plot the noisy input signal and your iltered output on top o each other and comment on the results. Use the FFT to analyse the requency content o the noisy signal beore and ater it is iltered by the FIR ilter. See i there is any improvement when using the Hanning window unction applied to the impulse response. Suggested reading: It is highly recommended that you do some supplementary reading on digital signal processing beore attempting this worksheet. Search or keywords Digital Signal Processing" or Digital Filters". Some example titles are listed below (any one o these should contain useul analyses o sampling and iltering and there are many similar texts). A. Bateman and W. Yates, Digital Signal Processing Design, Pitman, M. Bellanger, Digital Processing o Signals, 2nd Ed., John Wiley and Sons, J. Dunlop and D.G. Smith, Telecommunications Engineering, 3rd Ed., Chapman and Hall,

15 R.W. Hamming, Digital Filters, 2nd Ed, Prentice-Hall, L.B. Jackson, Digital Filters and Signal Processing, Kluwer, R. Kuc, Introduction to Digital Signal Processing, McGraw-Hill, P.A.Lynn, Introductory Digital Signal Processing with Computer Applications, Wiley, A. Peled, Digital Signal Processing: Theory, Design and Implementation, Wiley,