ECE431 Digital Signal Processing

Transcription

1 ECE431 Digital Signal Processing Bruce Francis Course notes, Version 104, September 2009

2 Preface These notes follow the topics in the text Some sections of these notes are complete in the sense of being self-contained Others are not; they are to be supplemented with the lecture notes and the text These incomplete sections are marked frag, meaning fragment Even the complete sections frequently ask you to fill things in, such as to complete examples Usually, we do this fill-in in class The chapter numbering doesn t coincide with that in the text, but the order is the same The main topics in this course are these: the sampling theorem and its applications; processing continuous-time signals using discrete-time components; digital filters and multirate digital filters for block-by-block processing; the discrete Fourier transform and its applications and the FFT; spectral analysis of signals, that is, determining the Fourier transform of a signal The course begins with the basic tools of discrete-time signals and systems, namely, Fourier transforms and z transforms The main chapters in the text are 2 4, 7 10 The treatment in these notes is on the same level as the text neither more nor less difficult But the notes use matrix algebra whereas the text does not DSP systems are entirely linear (except for quantizers), so it s only natural to use linear algebra For example, here s the definition of the discrete Fourier transform of the N samples {x[0],, x[n 1]}: X[k] = n x[n]e j2πkn/n, k = 0, 1,, N 1 (1) If you stack the samples x[0],, x[n 1] into a vector x and the DFT coefficients X[0],, X[N 1] into a vector X, then equation (1) is simply X = F x, (2) where F is an N N matrix, called the Fourier matrix Obviously, (2) is simpler in appearance than (1), but more importantly (2) is ready for computations: X = F x is a MATLAB line command, whereas (1) requires two for loops to implement, one for n and one for k I hope you find the treatment in these notes simple You may have to review matrix algebra don t worry, it s easy to pick up There are several computer applications for solving numerical problems in this course The most widely used is MATLAB, but it s expensive I like Scilab, which is free Others are Mathematica (expensive) and Octave (free) 1

3 Contents Preface 1 The Most Important Things 4 1 Introductory Example: Digital Storage of Music 6 2 Review of Continuous Time 8 21 Elementary Signals 8 22 Transforms 9 23 Laplace vs Fourier LTI Systems 11 3 Discrete-time Signals and Systems Elementary Signals z Transforms Fourier Transforms System Concepts Matrix Representations Difference Equations FIR, IIR, and Block Diagrams 34 4 The Sampling Theorem The Underlying Idea Sampling a Discrete-time Signal Sampling a Continuous-time Signal Ideal, Non-causal Reconstruction Summary of Formulas Causal Reconstruction Discrete-time Processing of Continuous-time Signals Discrete-time Random Signals (frag) A/D and D/A Converters 57 5 Multirate DSP Components of a Multirate System Block Filtering Changing the Sampling Rate 78 2

4 CONTENTS 3 54 Subband Coding MPEG Audio Compression Efficient Implementation 85 6 Filter Design (frag) 88 7 The DFT Definition of DFT Circular Convolution Vectors and Matrices Circular Convolution via the Circulant Matrix Ordinary Convolution via DFT Digital Filter Implementation via DFT Summary The FFT The Algorithm Complexity Spectral Analysis (frag) Using the DFT 109 A Proof of the Sampling Theorem using Fourier Series 111 B The Fourier Chart 113 Review Problems 115

5 The Most Important Things Here s a summary of the most important things we ll cover Read this list several times during the term At the end, you should understand all the points 1 DSP is a signals and systems subject Every component is linear: C/D, D/C, H(z), L, M, DFT The only exception is the quantizer, Q, which is nonlinear and memoryless What makes DSP a richer subject than your previous signals and systems courses are two things: 1) A DSP system usually has both continuous-time and discrete-time signals co-existing; 2) many DSP systems are multirate, with down- and up-samplers; this makes them not time-invariant 2 And because almost everything is linear, linear mathematical tools are central to the subject In particular, we emphasize vectors, matrices, and linear algebra A discrete-time signal can be represented as a vector (of infinite dimension), and as such the set of all signals forms a vector space A discrete-time linear system has a matrix representation Time invariance and causality are then simple properties of the matrix, namely, constant along diagonals and lower triangular, respectively 3 As with all signals and systems subjects, the frequency domain is where we get the most insight Thus, Fourier transforms and z transforms are the most important tools The z transform can be inverted by partial fraction and power series expansion or by the residue theorem The region of convergence (ROC) of a z transform can be a disk, the exterior of a disk, an annulus, the punctured plane, or the whole plane A system is causal iff the ROC of the transfer function is the exterior of a disk and the transfer function is proper (deg num deg den) The system is stable iff the ROC contains the unit circle If x[n] is absolutely summable, the Fourier transform converges absolutely and is continuous If x[n] is only square summable, the Fourier transform is guaranteed to converge only in the mean-square sense If x[n] is a sinusoid, its Fourier transform is an impulse Other periodic signals must be handled carefully too 4 The sampling theorem: A signal bandlimited to f 0 Hz can be reconstructed from its periodically sampled values provided the sampling frequency f s satisfies f s > 2f 0 The reason for the factor of 2 is that the bandwidth of the signal is really ( f 0, f 0 ), which has width 2f 0 You should be able to state the discrete-time sampling theorem in a similar way The sampling theorem provides a perfect reconstruction system: C/D followed by D/C The input to C/D is perfectly reconstructed at the output of D/C The reconstruction system can be derived via impulse-train modulation 4

6 CONTENTS 5 5 Applications of the sampling theorem: A signal sampled at one frequency can be converted entirely in discrete time to the same signal sampled at another frequency provided the two frequencies are rationally related and the signal is bandlimited to less than half both sampling frequencies A continuous-time filter can be implemented by C/D + digital filter + D/C provided the input is bandlimited 6 There are a variety of digital filter design methods A popular one is by windowing: Truncating the ideal impulse response and multiplying by a window function Digital filters are normally implemented in the time domain via convolution If the input is long, it can be broken up into blocks (segments); this then necessitates an overlap-add combiner The block processing involves multiplication by a finite Toeplitz matrix; this can be speeded up via converting to a circulant matrix, which can then be diagonalized via FFT 7 MPEG audio coding is a form of subband coding It is implemented by a multirate filter bank 8 Real A/D converters are based on oversampling, sigma-delta modulation, low-bit quantizers, and noise shaping This is because a high SQNR can be achieved The analysis of these systems is difficult and common practice is to approximate the quantizers by additive white noise 9 The DFT is the simplest of all the Fourier transforms The DFT is an orthogonal transform of a vector of sample values In practice, the DFT is computed via X = F x, where F is the DFT matrix Use of the DFT is widespread because of the fast algorithm of Cooley and Tukey 10 The DFT is also used for spectral analysis of, for example, biological signals, such as an electrocardiogram This involves windowing too

7 Chapter 1 Introductory Example: Digital Storage of Music An interesting application of DSP is the digital storing of music on, say, a CD player Storage onto a hard-disk or flash memory is similar This chapter, meant to be an introduction to the course, discusses how this is done Notation is introduced that we ll use throughout 1 Imagine sitting at a piano and striking A above middle C What happens? A hammer strikes the string for that note, the string vibrates at 440 Hz, air molecules move, a sound wave travels to your eardrum, your eardrum vibrates, an electrical signal carries that information to your brain, and you hear that note If you strike the A key an octave higher, the same thing happens but at Hz An octave higher, at Hz; an octave higher, at Hz And then the keyboard ends But imagine you could continue to listen to tones of higher and higher frequency Since the eardrum has mass, eventually the amplitude of oscillation of the eardrum would be so small that no signal would be sent to the brain and you wouldn t hear the tone In this sense, the human auditory system is bandlimited, to about 20 khz 1 And it is for this reason that digital storage of music is possible 2 Now we turn to recording a piece of piano music onto a CD We need a microphone and a computer Let s say the piano piece lasts 60 minutes We can represent the music as a continuous-time signal x c (t) With t in seconds, 0 t 60 2 Physically, x c (t) represents the pressure that the sound wave applies to the microphone The output of the microphone is a voltage signal, say w c (t) We want to store w c (t), 0 t 60 2 on the hard-disk of the computer Because the ear is bandlimited to 20 khz, we can lowpass filter w c (t) to this frequency Let s suppose we do this with an analog filter and that the output is v c (t) According to the sampling theorem, that we will study in detail, the signal v c (t) can be sampled at frequency f s > 2 20 khz without any loss of information The CD standard is f s = 441 khz This produces the sample values v[n] = v c (nt ), T = 1/( ), 0 n Notice the notation: square brackets, [n], around the discrete-time variable The signal v[n] is still an analog voltage, so to be stored in a finite capacity memory it has to be quantized, and 1 Note that the system is not exactly a brickwall filter 6

8 CHAPTER 1 INTRODUCTORY EXAMPLE: DIGITAL STORAGE OF MUSIC 7 this is normally done using 16 bits of resolution So the number of bits needed to be stored is = , ie, 25 Gigabits The block diagram representation of the storage process is this: x c (t) w c (t) v c (t) v[n] y[n] C/D Q microphone lowpass filter The block C/D is the ideal continuous-to-discrete transformation, namely, periodic sampling Then Q represents the quantizer We ll study these blocks in detail The output signal y[n] is burned onto the CD Notice the arrow convention: a continuous arrow for a continuous-time signal and a dashed arrow for a discrete-time signal 3 The reverse process is to take y[n] and produce v c (t); neither w c (t) nor x c (t) can be reconstructed The sampling theorem gives a formula for v c (t) in terms of v[n] But this relationship is not causal it requires forward interpolation Since the CD is spinning on playback, it is customary to generate v c (t) causally The standard way is zero-order-hold followed by smoothing ZOH is what it says hold every y[n] for exactly T seconds So the playback block diagram is this: y[n] ZOH ˆv c (t) ˆx c (t) lowpass filter speaker The hat on ˆv c (t) indicates that the signal is some sort of approximation of v c (t); likewise for ˆx c (t) As we will see, in practice the sampling rate is a multiple of 441 khz This is called oversampling

9 Chapter 2 Review of Continuous Time 21 Elementary Signals 1 Let us begin by remembering the most important signal for an electrical engineer, the sinusoid When we write sin(θ), the angle θ is always in radians So in the signal sin(ωt), Ωt has units of radians, and hence the units of Ω are radians/second, assuming t is in seconds We could also write Ω = 2πf, where f is in Hz We usually write sin(ωt) instead of sin(2πft) More useful than sin(ωt) or cos(ωt) is the complex sinusoid, e jωt By Euler s formula, Thus e jωt = cos(ωt) + j sin(ωt) cos(ωt) = 1 2 ( e jωt + e jωt), cos(ωt) = 1 2j ( e jωt e jωt) And why are sinusoids so important? You learned the answer in electric circuits: If you excite an electric circuit by a sinusoid (voltage or current), in steady state all voltages and currents are sinusoids too, of the same frequency We abbreviate this by saying sinusoid in produces sinusoid out 2 The unit step is denoted by u(t) 3 The impulse δ(t) is not a real function in mathematics it s called a distribution The idea behind δ is the sifting formula Let f(t) be a signal that is smooth near t = 0 The sifting formula is f(t)δ(t)dt = f(0) We pretend there is a function δ(t) satisfying this equation for every f(t), and we proceed from there For example, by change of variables f(t)δ(t τ)dt = f(t τ)δ(t)dt = f(τ) 8

10 CHAPTER 2 REVIEW OF CONTINUOUS TIME 9 Also, f(t)δ(at)dt = 1 f(t/a)δ(t)dt = 1 a a f(0), and so δ(at) = (1/ a )δ(t) The product δ(t) 2 is not defined, but u = δ in the sense that, if lim t f(t) = 0, then f(t) u(t)dt = f(0); this follows by integration by parts 22 Transforms 1 Let x(t) be a signal defined either for all t or just for t 0 The one-sided Laplace transform (LT) of x(t) is X(s) = 0 x(t)e st dt The region of convergence (ROC) is a right-half plane, that is, the real part of s has to be large enough Within the ROC X(s) has no poles 2 The Fourier transform (FT) of x(t) is X(jΩ) = The inversion formula is x(t) = 1 2π x(t)e jωt dt X(jΩ)e jωt dω If the absolute value of x(t) is integrable, that is, x(t) dt <, then X(jΩ) is a continuous function of Ω An example is x(t) = e t cos t If the absolute value of x(t) is only square-integrable, that is, x(t) 2 dt <, then X(jΩ) may not be a continuous function of Ω An example is x(t) = sin(t)/t The constant signal 1(t) that equals 1 for all t is neither absolutely-integrable nor squareintegrable Its FT is defined to be 2πδ(Ω) We are convinced of this by the inversion formula, which is an instance of sifting: 1 = 1 2π 2πδ(Ω)e jωt dω

11 CHAPTER 2 REVIEW OF CONTINUOUS TIME 10 The forward FT equation is 2πδ(Ω) = e jωt dt and it has no meaning in the sense of ordinary functions Likewise, the sinusoidal signal e jω 0t is neither absolutely-integrable nor square-integrable Its FT is defined to be 2πδ(Ω Ω 0 ) In general, Fourier transforms where either x(t) or X(jΩ) is not a function (ie, has an impulse) must be treated with care to ensure the result is correct 23 Laplace vs Fourier Here s an interesting question: Suppose you have an LTI system with impulse response function h(t) Let s assume h(t) = 0 for t < 0 Let H(s) be the Laplace transform of h(t) and then substitute jω for s in H(s), so you now have H(jΩ) Is H(jΩ) the Fourier transform of h(t)? The answer is, not necessarily To see this, let us temporarily, for this section only, denote the Laplace transform of h(t) by H LT (s) and the Fourier transform by H F T (jω) The Laplace transform H LT (s) exists provided h(t) satisfies some conditions; for example, if it grows without bound, the growth is at most exponential As we said before, the ROC is a right half-plane Let us look at two examples: 1 The unit step function is { 0, t < 0 u(t) = 1, t 0 The Laplace transform and ROC are U LT (s) = 1, ROC : Re s > 0 s Thus, the ROC is the open right half-plane, and U LT (s) has a pole on the boundary of that region, namely at s = 0 2 Consider the causal signal h(t) = e t u(t) Then H LT (s) = 1, ROC : Re s > 1 s + 1 Here the ROC contains the imaginary axis We turn now to the FT: H F T (jω) = 0 h(t)e jωt dt Again, the lower limit could be but it doesn t matter, because h(t) is causal Let us do the two examples from above:

12 CHAPTER 2 REVIEW OF CONTINUOUS TIME 11 1 The Fourier transform of the unit step is U F T (jω) = πδ(ω) + 1 jω Thus the Fourier transform is a distribution, having an impulse transform and the Laplace transform are not equivalent: Notice that the Fourier U F T (jω) U LT (jω) Indeed, setting s = jω in U LT (s), when the imaginary axis is not in the ROC, requires some justification 2 For h(t) = e t u(t) H F T (jω) = 1 jω + 1 In this case the Fourier transform is a function Moreover, the Fourier and Laplace transforms are equivalent: H F T (jω) = H LT (jω) In general, the Fourier and Laplace transforms are equivalent, that is, H F T (jω) = H LT (jω), when the ROC of the Laplace transform includes the imaginary axis 24 LTI Systems 1 Now we turn to the basic notions of a system and its properties A system has an input x(t) and an output y(t) We will assume these are real-valued functions of the time variable t and that t runs over all time, < t < An alternative is 0 t < There may also be initial conditions if the system arises from a differential equation, but we won t review that topic because it s not central in communication systems 2 The system is linear if it has two properties: if x 1 produces y 1 and x 2 produces y 2, then x 1 + x 2 produces y 1 + y 2 ; if x produces y, then cx produces cy for any real number c Which of the following systems are linear? (a) y(t) = tx(t) (b) y(t) = t 2 x(t) (c) y(t) = t 0 x(τ)dτ (d) y(t) = x(t 2) (e) y(t) = e x(t) (f) y(t) = ax(t) + b (g) y(t) = x( t)

13 CHAPTER 2 REVIEW OF CONTINUOUS TIME 12 3 Many (probably most) linear systems have a model like this: y(t) = h(t, τ)x(τ)dτ The function h(t, τ) is the impulse response; h(t, t 0 ) is the output at time t when the input is an impulse at time t 0 The linear system is causal if y(t) depends on x(τ) only for t τ, that is, h(t, τ) = 0 for τ > t The linear system is time-invariant if it has this property: if x(t) produces y(t), then for every T, x(t T ) produces y(t T ) Equivalently, h(t, τ) depends only on t τ Then we write h(t τ) instead of h(t, τ) and we have the convolution equation y(t) = h(t τ)x(τ)dτ which is also written y(t) = h(t) x(t), although more properly y(t) = (h x)(τ) Thus a linear time-invariant (LTI) system is modeled by a convolution equation y(t) = h(t τ)x(τ)dτ If it is causal, then h(t) = 0 for t < 0 and y(t) = t h(t τ)x(τ)dτ 4 The transfer function of an LTI causal system is the Laplace transform H(s) of h(t) The frequency-response function of an LTI system is the Fourier transform H(jΩ) of h(t) 5 A signal x(t) is said to be bounded if (obviously) there is a bound B such that x(t) B for all t Examples: Sinusoids are bounded, but a ramp x(t) = tu(t) is not A linear system with input x(t) and output y(t) is said to be stable if every bounded input x(t) produces a bounded output y(t) Examples: An RLC circuit is stable, but an LC circuit is not It is a theorem that the LTI system y(t) = h(t) x(t) is stable if and only if the impulse response function is absolutely integrable, ie, h(t) dt < This is valid even for non-causal systems Finally, if a causal LTI system has a transfer function H(s), then the system is stable if and only if the ROC of H(s) includes the imaginary axis, that is, all the poles of H(s) have negative real part

14 Chapter 3 Discrete-time Signals and Systems Here we blend the topics of Chapters 2 and 3 in the text 31 Elementary Signals 1 In discrete time, the time variable is denoted n and is an integer; depending on the application, it could range over all positive and negative integers, < n <, or just the non-negative integers, n 0, or just a finite number of integers, 0 n N 1 A discrete-time signal is written x[n], with square brackets The integer n does not stand for real time, but rather for sample number For example, consider the sinusoid x(t) = sin(10t) Suppose starting at t = 0 we sample it every 02 seconds Then we have the samples x[n] = x(02n) = sin(2n), n = 0, 1, 2, You can see that n indicates the sample number, and the samples x[0], x[1], x[2], occur at the real times 0, 02, 04, So sinusoids in discrete time look like x[n] = sin(ωn) or x[n] = e jωn 2 The unit step is u[n] and the unit impulse δ[n] The latter equals 1 when n = 0 and zero otherwise 32 z Transforms 1 The z transform is to discrete time as the Laplace transform is to continuous time The z transform of x[n] is X(z) = n= x[n]z n The region of convergence (ROC) is in general an annulus (ring, doughnut) The function X(z) has no poles inside the ROC 2 Example: Let u[n] denote the unit step in discrete times Thus u[n] = 1 for n 0 and u[n] = 0 for n negative The series U(z) = n= u[n]z n = z + 1 z

15 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 14 converges for 1/z < 1, ie, z > 1 Thus the ROC is the exterior of the unit disk For z in this region, the series converges to 1 1 (1/z) = z z 1 Thus there s a pole at z = 1, on the boundary of the ROC 3 Example: Let x[n] = u[ n] Then X(z) = 1 + z + z 2 + The ROC is the disk z < 1, inside which X(z) = 1 1 z Thus there s a pole at z = 1, on the boundary of the ROC 4 For the next example, it s convenient to have a formula for the finite series 1+a+a 2 + +a N 1 Since so (1 + a + a a N 1 )(1 a) = 1 a N, 1 + a + a a N 1 = 1 an 1 a Notice on the right-hand side the numerator and denominator are both zero at a = 1 Thus the roots of the polynomial 1+a+a 2 + +a N 1 are the points a such that a N = 1 excluding the point a = 1 That is, the points e j2πk/n, k = 1,, N 1 5 Example: Let x[n] = u[n] u[n 10] Thus X(z) = z z 9 = 1 (1/z10 ) 1 (1/z) = z10 1 z 9 (z 1) The ROC is z 0, the extreme form of an annulus: 0 < z < There are 9 poles at z = 0 and 9 zeros on the unit circle, at z = e j2πk/10, k = 1,, 9 6 Example: Take the continuous-time sinusoid x(t) = cos(t)u(t) and sample it at 1 sample per second: x[n] = cos(n)u[n] (u is the unit step) Find X(z) and the ROC Locate the poles and zeros 7 The signal x[n] = cos(n) does not have a z transform The reason is that z has to be large enough, namely z > 1, for n=0 x[n]z 1 to converge, but also small enough, namely z < 1, for 1 n= x[n]z 1 to converge The regions z > 1 and z < 1 have empty intersection

16 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 15 8 Consider the following picture: x(t) Im s t Re s The graph on the left is of a decaying exponential signal in continuous time; x(t) = 0 for t < 0 The figure on the right is the s-plane Shown is the ROC as a shaded right half-plane that includes the imaginary axis Extend to all other cases For signals that are nonzero in negative time, you ll have to use the two-sided Laplace transform: X(s) = x(t)e st dt For example, here s another pair: x(t) Im s t Re s The signal is blowing up in positive time but decaying in negative time; the ROC is a vertical strip in the right half-plane 9 Do all the analogous pictures for discrete time The ROC may be a disk, an annulus, etc 10 What about inversion? How do we get x[n] if we have X(z)? One way is to expand X(z) as a series and then read off x[n] from the coefficients That is, if we can expand X(z) = + a 2 z a 1 z + a 0 + a 1 z + a 1 2 z 2 +, then x[n] = a n Such an expansion is called a Laurent series Example: X(z) = 1/(z 1) If the ROC is z < 1, then the series expansion must be X(z) = 1 1 z = (1 + z + z2 + )

17 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 16 Read off x[n] But if ROC is z > 1, then X(z) = 1 z Read off x[n] Another example: X(z) = 1 1 z 1 = 1 z (1 + z 1 + z 2 + ) z 1 (z 2)(z 3) There are three possible ROCs You have to know which one Then you do a partial fraction expansion, like this: X(z) = 1 z z 3 Then do the appropriate Laurent series for both terms Then read off the coefficients You finish 11 There s another approach to inversion, based on complex function theory, in particular, Cauchy s theorem and its corollary, the residue theorem This is an advanced topic but it s mentioned in case you have taken a course on complex variables Lemma 1 Let C be a circle centred at the origin and contained within the ROC of X(z) Then x[n] equals the sum of the residues of X(z)z n 1 at all poles encircled by C Example X(z) = 2z + 1 z(z 2) This has two poles, at z = 0 and 2 By partial fraction expansion we can write X(z) = c 1 z + c 2 z 2 for some constants c 1 and c 2 The constant c 1 is called the residue of X(z) at the pole z = 0; likewise c 2 is the residue of X(z) at the pole z = 2 For this example c 1 = 1/2 and c 2 = 5/2 The ROC could be either z > 2 or 0 < z < 2 Imagine a circle C in the complex plane, centred at the origin, and contained within the ROC So for the former ROC, the radius of the circle is greater than 2; for the latter ROC, less than 2 First We have X(z) = 2z + 1, ROC : z > 2 z(z 2) X(z)z n 1 = zn 2 (2z + 1) z 2

18 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 17 The radius of C is greater than 2 Inside C, X(z)z n 1 has a pole at z = 2 and also a pole at z = 0 if n < 2 For n = 0 we have x[0] = sum of the residues of 2z + 1 z 2 (z 2) Now 2z + 1 z 2 (z 2) = c 1 z 2 + c 2 z + c 3 z 2 and the sum of the residues is c 2 + c 3, which equals 0 (The residue at the pole z = 0 is c 2, not c 1 ) For n = 1 x[1] = sum of the residues of Then for n 2, there s only a pole at z = 2: 2z + 1 z(z 2) = 2 x[n] = the residue of zn 2 (2z + 1) z 2 = 5 2 n 2 = 5 4 2n Now let s turn to the other ROC: X(z) = 2z + 1, ROC : 0 < z < 2 z(z 2) Then inside C there s at most a pole at z = 0, so x[n] = the residue of zn 2 (2z + 1) z 2 For n 2 there is no pole and so x[n] = 0 For n = 1 x[1] = the residue of z 1 (2z + 1) z 2 = 1 2 And so on Finally, here s a formula for residues: If X(z) has a simple pole at z = p, then its residue is lim(z p)x(z) z p If the pole has multiplicity m, then the residue is lim z p 1 (m 1)! dz m 1 (z p)m X(z) d m 1 12 The convolution theorem says that convolution in the time domain corresponds to multiplication in the transform domain: If y[n] = h[n] x[n], then Y (z) = H(z)X(z) The proof is quite direct and is similar to the analogous result in continuous time

19 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS Fourier Transforms 1 The Fourier transform (FT) of x[n] is X(e jω ) = n= x[n]e jωn This is the analysis equation getting the sinusoidal components of the signal The inversion formula (IFT) is x[n] = 1 2π π π X(e jω )e jωn dω This is the synthesis equation combining the sinusoidal components into the signal 2 Given a signal x[n], let s say we have its z transform X(z) and ROC, and we have its Fourier transform X(e jω ) So it seems the FT is just the z transform with z replaced by e jω, a point on the unit circle Is this correct? Yes, if the ROC of the z transform includes the unit circle For example, it s not true for the unit step, u[n] Even worse, the sinusoid cos n has a FT but not a z transform 3 The meaning of the FT formulas is different depending on the properties of the signal this is important There are three different cases, as we now discuss 4 Case 1: If the absolute value of x[n] is summable, that is, x[n] dt <, then X(e jω ) is a continuous function of ω Example: x[n] = δ[n] + δ[n 1] + δ[n 2] This signal has a z transform and the ROC contains the unit circle Therefore all we need to do is find the z transform of x[n] and then set z = e jω to get the FT 5 Case 2: If x[n] is only square-summable, that is, x(t) 2 dt <, then X(e jω ) may not be a continuous function of ω Example: Let s start with the FT: { X(e jω 1, ω π/2 ) = 0, π/2 < ω < π Then by the inversion formula x[n] = 1 π X(e jω )e jωn dω = 1 π/2 e jωn dω 2π π 2π π/2

20 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 19 The indefinite integral equals 1 2πjn ejωn Thus x[n] = 1 ( e jπn/2 e jπn/2) = 1 2πjn πn sin(πn/2) This is not absolutely summable, because 1/n does not converge to zero fast enough as n goes to π So in this example, X(e jω ) and x[n] are both well defined The anomaly concerns the analysis equation X(e jω ) = n= x[n]e jωn In what sense does the right-hand side converge to the discontinuous function on the left? It s like this Fix N and define X N (e jω ) = N n= N x[n]e jωn Then the question we just asked is, in what sense does X N (e jω ) converge to X(e jω )? In the integral-squared sense: lim N π π XN (e jω ) X(e jω ) 2 dω = 0 6 There s a beautiful theory for the functions in Case 2 It goes like this Let x[n] and y[n] be square-summable and let X(e jω ) and Y (e jω ) be their FTs Define x, y = n x[n]y[n] This is called an inner product It s exactly like the dot product of two vectors Also, define the inner product of the two FTs: X, Y = 1 2π π π X(e jω )Y (e jω )dω The overbar denotes complex conjugate So we have an inner product in the time domain and an inner product in the frequency domain In fact, they re equal for every x and y This is called Parseval s Theorem In the special case where x[n] = y[n], we have n x[n] 2 = 1 2π π π X(e jω ) 2 dω

21 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 20 7 Case 3: The sinusoidal signal e jω 0n is neither absolutely-summable nor square-summable Its FT is defined to be 2πδ(ω ω 0 ) The FT equation 2πδ(ω ω 0 ) = e jω0n e jωn has no meaning in the sense of ordinary functions However, the inversion formula e jω 0n = 1 2π π π 2πδ(ω ω 0 )e jωn dω is consistent with the sifting property of the impulse Another example: x[n] = k δ[n 3k], which is an upsampled impulse, the output of 3 for the input δ[n]] 8 You can prove these properties of the DTFT: (a) X(e jω ) is periodic, of period 2π We restrict the angle ω to [ π, π) (b) If x[n] is real, then X(e jω ) has even magnitude and odd phase 9 The convolution theorem holds for Fourier transforms too Do this example: x[n] = k δ[n 3k], h[n] = (δ[n] + δ[n 1])/2 Find y[n] = h[n] x[n] 10 This is an aside: Consider a causal signal x[n] that has a FT Causal means x[n] = 0 for n < 0 It turns out the real part of X(e jω ) is uniquely determined by the imaginary part, and vice versa This isn t important in our course, but it is in electromagnetics, in the form of the Kramers-Kronig relations 34 System Concepts 1 A system has an input x[n] and an output y[n] There may also be initial conditions if the system arises from a difference equation The system is linear if it has two properties: if x 1 produces y 1 and x 2 produces y 2, then x 1 + x 2 produces y 1 + y 2 ; if x produces y, then cx produces cy for any real number c Which of the following systems are linear? (a) y[n] = nx[n] (b) y[n] = n 2 x[n] (c) y[n] = n k=0 x[k] (d) y[n] = n+2 k=n 2 x[k] (e) y[n] = x[n 2]

22 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 21 (f) y[n] = e x[n 2] (g) y[n] = ax[n] + b (h) y[n] = x[ n] (i) y[n] = x[n] + 3u[n + 1] 2 Most linear systems have a model like this: y[n] = m= h[n, m]x[m] The function h[n, m] is the impulse response; h[n, m] is the output at time n when the input is a pulse at time m, ie, δ[n m] The linear system is causal if y[n] depends on x[m] only for n m, that is, h[n, m] = 0 for m > n 3 The linear system is time-invariant if it has this property: if x[n] produces y[n], then for every k, x[n k] produces y[n k] Equivalently, h[n, m] depends only on n m Then we write h[n m] instead of h[n, m] and we have the convolution equation y[n] = m= h[n m]x[m] which is also written y[n] = h[n] x[n] Thus a linear time-invariant (LTI) system is modeled by a convolution equation y[n] = m= h[n m]x[m] This is frequently abbreviated y[n] = h[n] x[n], although it s not quite correct because it suggests is an operator that takes the two inputs h[n], x[n] If we were fussy we d write y[n] = (h x)[n] If the system is causal, then h[n] = 0 for n < 0 and y[n] = n m= h[n m]x[m] 4 The transfer function of an LTI system is the z transform H(z) of h[n] Convolution in the time domain is equivalent to multiplication in the frequency domain: y[n] = h[n] x[n] is equivalent to Y (z) = H(z)X(z) The frequency-response function of an LTI system is the Fourier transform H ( e jω) of h[n] 5 For an LTI system with transfer function H(z), when is the system causal? It s easy to answer when H(z) is rational, the ratio of two polynomials, H(z) = N(z)/D(z) Lemma 2 The LTI system with transfer function H(z) = N(z)/D(z) is causal iff the ROC is the exterior of a disk and degree D(z) degree N(z)

23 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 22 Proof Necessity: Assume causality Then h[n] = 0 for n < 0 and so from the definition of H(z), H(z) = h[0] + h[1] 1 z + h[2] 1 z 2 + Thus the ROC has the form z > r Also, N(z) D(z) = h[0] + h[1]1 z + h[2] 1 z 2 + Letting z, we get the degree inequality Sufficiency: Since the ROC is the exterior of a disk, H(z) must have the form H(z) = n=n 0 h[n]z n for some finite n 0, that is, h[n] can be nonzero for only finitely many negative values of n From the degree inequality, n 0 must be 0 Thus the system is causal Examples: Which of these could be causal: 1 z, z, z 2 z 1, z z A signal x[n] is said to be bounded if there is a bound B such that x[n] B for all n Examples: Sinusoids are bounded, but a ramp x[n] = nu[n] is not A linear system with input x[n] and output y[n] is said to be stable if every bounded input x[n] produces a bounded output y[n] (text, same section) Here s the time-domain test for stability of an LTI system: Lemma 3 The LTI system y[n] = h[n] x[n] is stable iff the impulse response function is absolutely summable, ie, n= h[n] < Here s the frequency-domain test for stability of an LTI system: Lemma 4 The LTI system with transfer function H(z) is stable iff unit circle ROC Here s the frequency-domain test for stability of a causal LTI system: Lemma 5 The causal LTI system with transfer function H(z) is stable iff all poles are in z < 1

24 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 23 Examples: Which of these could be stable: 1 z, z, 1 (2z 1)(z 2) Why must H(z) = 1/(z 1) be unstable? 7 LTI and FIR = stable 8 Consider an LTI system with impulse response h[n] Let s discuss sinusoid in implies sinusoid out Suppose the input is the sinusoid x[n] = e jωn Is there an output? We have to be careful, because the expression x[n] = e jωn suggests x[n] was applied starting at n =, that is, it suggests the system is in steady state Will there be a steady state if the input is a sinusoid? Yes, if the system is stable So let s start again: Consider an LTI, stable system with impulse response h[n] and suppose the input is the sinusoid x[n] = e jωn Then the output is as follows: y[n] = h[n] e jωn = m = m h[n m]e jωm h[m]e jω(n m) = e jωn m h[m]e jωm = e jωn H(e jω ) So we see that the steady-state output is a sinusoid too, of the same frequency By the way, the equation h[n] e jωn = H(e jω )e jωn has the same form as the equation Ax = λx where A is a square matrix, x is a nonzero vector, and λ is a complex number We say λ is an eigenvalue of A and x is an eigenvector Likewise, in the preceding equation, we say that sinusoids are eigenfunctions of LTI systems and Fourier transforms H(e jω ) are eigenvalues 9 As examples, find H(e jω ) in the following cases: (a) y[n] = x[n 1] (b) y[n] = x[ n] (c) y[n] = x[2n] (d) y[n] = x[n + 1] + x[n] + x[n 1] 35 Matrix Representations Now we introduce linear algebra into the picture We ll see how a discrete-time linear system can be represented by a matrix

25 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 24 Vectors and Matrices For those students who haven t used matrices since first-year linear algebra, let us review the elements 1 A vector is an ordered list of numbers, real or complex We usually write the list as an array with one column Example: 1 x = 2 3 But we may occasionally write it as an n-tuple: x = (1, 2, 3) 2 An array of the form A = is a 4 3 matrix It has 4 rows and three columns 3 If x is a column vector, the multiplication Ax is defined if x has dimension 3 (it has 3 rows) Let y = Ax Then y has dimension 4 Its first component is the dot product of the first row of A with x: y[0] = [ ] x[0] x[1] x[2] = x[0] + 2x[1] Its second component is the dot product of the second row of A with x: y[0] = [ ] x[0] x[1] = x[0] + 8x[2] x[2] Thus the equation y = Ax, for this A, is a concise way of writing the 4 equations y[0] = x[0] + 2x[1] y[1] = x[0] + 8x[2] y[2] = 3x[1] + 4x[2] y[3] = 6x[1] 5x[2] Each component of y is a linear combination of the components of x

26 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 25 4 To multiply two matrices AB, the number of columns of A must be equal to the number of rows of B When this is true, the first column of C = AB equals, by definition, Ab 1, where b 1 is the first column of B Likewise for the other columns of C In general, matrix multiplication is not commutative, that is, AB BA in general 5 Matrices represent linear transformations Example Consider the linear transformation that takes a vector in the plane and rotates it clockwise by π/2, There is a matrix A satisfying y = Ax, where x is the given vector in the plane and y is the rotated vector To find A, first let x be the unit vector on the horizontal axis, ie, [ ] 1 x = 0 Rotating by π/2 gives the vector [ ] 0 y = 1 Thus [ 0 1 ] [ 1 = A 0 ] Thus the first column of A equals [ ] 0 1 Likewise, the second column is the second basis vector [ ] 0 1 rotated by π/2, namely, [ ] 1 0 Thus [ A = ] A Signal is a Vector 1 Consider a discrete-time signal x[n] and let us suppose the time variable n runs over the range 0, 1, 2,, that is, the non-negative integers We could write the signal as the text does in (21), x = {x[n]},

27 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 26 or we could write the components out as an ordered list, x = (x[0], x[1], x[2], ) Another way to write the signal is to stack the components up to form a column vector: x[0] x[1] x = x[2] Of course, this vector has an infinite number of components, that is, it s an infinite dimensional vector For example, here are the unit impulse and unit step: δ = 0, u = Signals form a vector space, that is, you can add two of them and you can multiply a signal by a scalar real number Of course, to add them you add their components x[0] y[0] x[0] + y[0] x[1] x = x[2], y = y[1] y[2] = x + y = x[1] + y[1] x[2] + y[2] Likewise with scalar multiplication: x[0] cx[0] x[1] cx[1] x = x[2] = cx = cx[2] 3 If the time variable n ranges over all integers, both negative and positive, ie, < n <, then the components of a signal x are, x[ 2], x[ 1], x[0], x[1], x[2], x[3], In writing such a signal as a column vector, we need a marker that divides time into n < 0 and n 0 We use a horizontal line: x[ 2] x[ 1] x = x[0] x[1] x[2]

28 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 27 So for example 0 δ = 1 0, u = A Linear System is a Matrix 1 Consider the system whose output is the sum of the current and past input: y[n] = x[n] + x[n 1] (31) Let us suppose the time set is n 0 To determine y[0] from (31), we have to assign a value to x[ 1]; let us take 0 Then as n ranges over 0, 1, 2,, (31) generates the equations y[0] = x[0] y[1] = x[1] + x[0] y[2] = x[2] + x[1] etc These equations can be assembled into one vector-matrix equation: y[0] x[0] y[1] x[1] y[2] = x[2] y[3] x[3] Let us denote the matrix by H, so that the preceding equation is y = Hx Thus the notation is: boldface capital letter for matrix representation of a system To be consistent with the time set, we choose to number the rows as 0,1,2 etc Likewise the columns Think of the equation y = Hx like this: To get the output at time, say, n = 2, take the dot product of row number 2 of the matrix H with the input vector x For the above example, y[2] = [ ] x[0] x[1] x[2] x[3] = x[2] + x[1] 2 Going beyond the example, let us consider a general case y = Hx, still with the time set n 0 Then the matrix H has components that it is convenient to write as h[m, n] the component in row m, column n Then the equation y = Hx in component form is y[m] = h[m, n]x[n], m 0 n=0

29 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 28 ie, the dot product of row m of H with x 3 Some other examples are as follows A memoryless pure gain: y[0] a x[0] y[1] 0 a 0 0 x[1] y[n] = ax[n], y[2] = 0 0 a 0 x[2] y[3] a x[3] The unit time delay: y[n] = x[n 1], y[0] y[1] y[2] y[3] = x[0] x[1] x[2] x[3] The unit time advance: y[n] = x[n + 1], y[0] y[1] y[2] y[3] = x[0] x[1] x[2] x[3] The downsampler, also called the compressor (Example 29): y[0] x[0] y[1] x[1] y[n] = x[2n], y[2] = x[2] y[3] x[3] 4 When the time set is < n <, we have to place markers in the matrix that divide time into n < 0 and n 0 We do it like this, shown for the unit time delay: y[ 2] x[ 2] y[ 1] x[ 1] y[n] = x[n 1], y[0] y[1] = x[0] x[1] y[2] x[2] y[3] x[3] We continue to write this equation as y = Hx We number the rows with respect to the horizontal line: The rows are numbered 0,1,2 etc going down from the horizontal line, and

30 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 29 1, 2 etc going up from the horizontal line Likewise, the columns are numbered 0,1,2 etc going right from the vertical line and 1, 2 etc going left Then the matrix H has components that it is convenient to write as h[m, n] the component in row m, column n Again, we have y[m] = n= h[m, n]x[n] (32) Notice that if x[n] = δ[n], then y[m] = h[m, 0], so h[m, 0] is the output at time m when an impulse is applied at time 0 More generally h[m, n] is the output at time m when an impulse is applied at time n The preceding matrix is important and we give it the special symbol U, for unit delay: U = To recap: A linear discrete-time system can be represented by the equation y = Hx 6 Now we look at two properties the linear system may have, causality and time invariance, and we see how those properties are reflected in the matrix H We do this for the time set < n < First, causality Causality means that y[n] depends only on x[n], x[n 1], This is true iff the equation y[m] = takes the form y[m] = n= m n= and this is equivalent to h[m, n]x[n] h[m, n]x[n], h[m, n] = 0, n > m Equivalently, H is lower triangular For example, the unit time advance is not causal Time invariance is this property: If the input is delayed by 1 time unit, so is the output Lemma 6 The following conditions are equivalent: (a) System (32) is LTI

31 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 30 (b) For every n, m, h[m 1, n 1] = h[m, n] (c) H is constant along diagonals (d) H and U commute For example, if the impulse δ[n] at time 0 is applied, the output is y[m] = h[m, 0]; if the impulse δ[n 1] at time 1 is applied, the output is y[m] = h[m, 1] For the system to be LTI, the second output must be the shifted first output: h[m, 1] = h[m 1, 0] 7 Let s run through the examples: memoryless pure gain: time invariant, causal unit time delay: time invariant, causal unit time advance: time invariant, not causal downsampler: not time invariant, not causal 8 Consider an LTI system The input-output equation in component form is (32) Since h[m k, n k] = h[m, n], by setting k = n we see that h[m, n] depends onlyl on the difference m n, and not m and n individually We may therefore redefine h[m, n] to be h[m n], the output at time m when an impulse is applied at time n So finally, the function h[n] is the impulse-response function, the output at time n when an impulse is applied at time 0 To recap, an LTI system can be modeled by the equation y[n] = m= h[n m]x[m], < n < This is the familiar convolution equation, usually written y[n] = h[n] x[n] The matrix form is y = Hx, or in more detail y[ 2] h[0] h[ 1] h[ 2] h[ 3] h[ 4] h[ 5] y[ 1] h[1] h[0] h[ 1] h[ 2] h[ 3] h[ 4] y[0] y[1] = h[2] h[1] h[0] h[ 1] h[ 2] h[ 3] h[3] h[2] h[1] h[0] h[ 1] h[ 2] y[2] h[4] h[3] h[2] h[1] h[0] h[ 1] y[3] h[5] h[4] h[3] h[2] h[1] h[0] x[ 2] x[ 1] x[0] x[1] x[2] x[3] Notice that H is constant along its diagonals h[0] on the main diagonal, etc Such a matrix is called a Toeplitz matrix 9 Among the following systems, for every linear system find H

32 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 31 (a) y[n] = nx[n] (b) y[n] = n 2 x[n] (c) y[n] = n k=0 x[k] (d) y[n] = n+2 k=n 2 x[k] (e) y[n] = x[n 2] (f) y[n] = e x[n 2] (g) y[n] = ax[n] + b (h) y[n] = x[ n] (i) y[n] = x[n] + 3u[n + 1] 10 Here s a method to find H if you know the system is linear Apply the input δ[n] and figure out what y[n] is: That s column 0 of H Apply the input δ[n 1] and figure out what y[n] is: That s column 1 of H And so on 11 Finally, interconnections of systems If two systems H 1 and H 2 are connected in series, the resulting matrix is the product of the two matrices Of course, there s usually a difference between this way H 1 H 2 and this way H 2 H 1 You get H 1 H 2 if you connect the output of H 2 to the input of H 1 If the two systems are LTI, then the order doesn t matter If two systems H 1 and H 2 are connected in parallel, that is, a common input is applied and the outputs added, then the resulting matrix is the sum of the two matrices 36 Difference Equations The purpose of this section is to review and unify the different ways we have of describing an LTI system matrix representation, convolution equation, transfer function and to introduce a fourth way, the difference equation It is the difference equation that is actually implementable in hardware or software 1 Let s start with the simple time delay, y[n] = x[n 1] This is in fact the difference equation representation of the system There is no ambiguity in the equation, that is, there s just one way to compute the output from the input, and the system is causal The matrix representation is y = Hx, H = ,

33 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 32 the convolution equation is y[n] = h[n] x[n], h[n] = δ[n 1], and the transfer function is H(z) = 1, ROC : z 0 z Notice that for the function 1/z there can be only one ROC 2 The next example is y[n] = x[n + 1] + x[n] + x[n 1] Again, the output is uniquely defined in terms of the input by this equation The system is not causal The matrix representation is y = Hx, H = the convolution equation is , y[n] = h[n] x[n], h[n] = δ[n + 1] + δ[n] + δ[n 1], and the transfer function is H(z) = z , ROC : z 0 z Again, for the function z z there can be only one ROC 3 The next example is y[n] = y[n + 1] + x[n] How are we to view this? To compute the output y[n] given the input it seems we have to have stored the future output y[n + 1] But suppose we re-arrange y[n + 1] = y[n] x[n] and then replace n by n 1: y[n] = y[n 1] x[n 1]

34 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 33 Now it seems we don t need the future output in the computation of y[n], we need to have stored the past value y[n 1] We conclude that the difference equation could represent either of two systems, one causal and one not The situation is clarified by taking z transforms of the original equation: Y (z) = zy (z) + X(z) = H(z) = 1 1 z So for this H(z) there are two possible ROCs: z < 1 and z > 1 The system is not causal for the former, and is causal for the latter 4 Here s a more elaborate example Consider the causal linear system modeled by We have y[n] = n k= y[n] = x[n] + x[k] n 1 k= x[k] and thus the input and output satisfy the difference equation Now consider y[n] = x[n] + y[n 1] y[n] = k=n+1 x[k] This system is anticausal in the sense that the output at any time depends entirely on future inputs We have y[n] = x[n + 1] k=n+2 x[k] and thus the input and output satisfy the difference equation y[n] = x[n + 1] + y[n + 1] Since n is just a dummy time variable, we can replace n by n 1 in the latter equation to get the difference equation or equivalently y[n 1] = x[n] + y[n], y[n] = x[n] + y[n 1] So the two systems have identical difference equations You can work out the transfer functions in the two cases

35 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 34 5 Consider an LTI system with transfer function H(z) and assume H(z) is rational the ratio of polynomials: H(z) = B(z) A(z) Suppose the numerator and denominator are completely general: B(z) = b M z M + + b 0, A(z) = a N z N + + a 0 Then the input and output satisfy the equation A(z)Y (z) = B(z)X(z), which in the time domain is a N y[n + N] + + a 0 y[n] = b M x[n + M] + + b 0 x[n] This is the difference equation model of the system How to implement it, that is, how to compute the output y[n] for each n, depends on the ROC of H(z) For example, suppose the system is causal Then the ROC is the exterior of a disk, N M, and we can write the difference equation (assuming a N 0) as y[n] = 1 a N ( a N 1 y[n 1] a 0 y[n N] + b M x[n (N M)] + + b 0 x[n N]) Thus y[n] can be computed from stored past values of y and x, and also the current input, x[n], if N = M Implementation in the general case, where H(z) is not causal, is more complicated We may come back to that later Or, maybe you can figure it out 6 In conclusion: (a) An LTI system is completely specified by the matrix H (b) It is completely specified by the convolution equation, equivalently, the impulse response h[n] (c) The transfer function H(z) needs the ROC to completely specify the system (d) The difference equation needs details about what is stored in memory to produce an update rule and therefore to completely specify the system 37 FIR, IIR, and Block Diagrams The purpose of this section is to introduce a simple way to derive a block diagram implementation of an LTI system The material is essentially the direct form realization of Chapter 6, but it s simplified by the use of matrices

36 CHAPTER 3 DISCRETE-TIME SIGNALS AND SYSTEMS 35 1 Before we come to block diagrams, a few words about how systems are implemented How to implement a continuous-time system G(s): It can be done by an analog integrated circuit From G(s) draw the block diagram in terms of (1/s); then implement each integrator via an op amp The disadvantage of an analog circuit is its sensitivity to resistor values These may vary by 10% or more, resulting in large, undesirable bandwidth variations In addition, resistors are hard to build on integrated circuits (they take up a lot of room) Recently many active filters with resistors and capacitors have been replaced with a special kind of filter called a switched capacitor filter The switched capacitor filter allows for very sophisticated, accurate, and tuneable analog circuits to be manufactured without using resistors Instead, equivalent resistances can be made to depend on ratios of capacitor values (which can be set accurately), and not absolute values (which vary between manufacturing runs) Switched capacitor filters operate in discrete time The inputs are sampled and held This is therefore an approximation to the original analog system As for a discrete-time system with a TF H(z), there s no op amp for (1/z) Instead, it could be implemented again by a switched capacitor filter But more relevantly, it could be implemented in software in a DSP Block diagrams are then useful for implementing an H(z) Typically one starts with the transfer function, derives the block diagram, and then programs the DSP unit 2 FIR Systems FIR stands for finite impulse response The more proper term is finiteduration impulse response The definition is this: An LTI system with impulse response function h[n] is FIR if h[n] is nonzero for only finitely many values of n Equivalently, H(z) has the form Examples are H(z) = (polynomial in z) + (polynomial in z 1 ) (a) H(z) = 1 + z 1 + z 2, ROC : z 0 (b) H(z) = z 2 + z + 1, ROC : all z (c) H(z) = z 2 + z z 1 + z 2, ROC : z 0 Notice that for an FIR transfer function, we don t have to be told the ROC: It can only be one thing since the only poles can be at z = 0; so the ROC is either all z, or all nonzero z It follows that every FIR system is stable Proof 1: h[n] is absolutely summable since there are only finitely many nonzero ones Proof 2: The unit circle is in the ROC FIR filters are widely used in practice for that reason: They are automatically stable There are practical filters in current use that have up to order 4000 (the degree of the polynomial H(z)) A special case is a causal FIR system Then H(z) is a polynomial in z 1 there are no terms of the form z, z 2, Of the three examples above, only the first is causal 3 Now we turn to the block diagram realization of a causal FIR system Consider this figure: