A1 Time-Frequency Analysis

Transcription

1 A 20 / A Time-Frequency Analysis David Murray david.murray@eng.ox.ac.uk dwm/courses/2tf Hilary 20

2 A 20 2 / Content 8 Lectures: 6 Topics... From Signals to Complex Fourier Series 2 From Complex Fourier Series to the Fourier Transform 3 Convolution. The impulse response and transfer functions 4 Sampling, Aliasing 5 Power & Energy Spectra, Autocorrelation, and Spectral Densities 6 Random Processes and Signals

3 A 20 3 / Topic : From Signals to Complex Fourier Series. The gap in your knowledge what you know, and what you don t.2 Signal definitions.3 Fourier Series (largely revision) Orthogonal Basis Functions Dirichlet conditions Discontinuities Symmetry Completion Gibbs phenomenon Parseval s Theorem.4 Complex Fourier Series

4 A 20 4 / You know about linear systems... In a linear system one described by a linear differential equation Output y(t) occurs at same frequency as the input x(t) If x(t) s amplitude changes by a factor y(t) s amplitude will change by the same factor Outputs combine linearly... Input Output x (t) y (t) x 2 (t) y 2 (t) α x (t) + α 2 x 2 (t) α y (t) + α 2 y 2 (t) Faced with non-linear systems, you often linearize the system by considering incremental inputs and outputs that occur around a fixed operating point... another story

5 A 20 5 / You know about frequency response You also have a firm understanding of the importance of a system s frequency response described by the transfer function H(ω) H(ω) relates input and output in the frequency domain Y (ω) = H(ω)X(ω) What are X(ω) and Y (ω)? Up to now you have considered single frequency harmonic signals x(t) and y(t), where X and Y are phasors: x(t) = Re ( X(ω)e iωt) y(t) = Re ( Y (ω)e iωt)

6 A 20 6 / For example... Input voltage 0V, but leading the reference phase by 45 X(ω) = 0e jπ/4 = 0 2 ( + j) The transfer function is H(ω) = R (R + jωl) X (ω) L R Y (ω) The output phasor is Y (ω) = 0 2 ( + j)r (R + jωl) ( 0 ( + j)re jωt ) The actual output voltage is y(t) = Re 2 (R + jωl)

7 A 20 7 / You know about linear superposition in frequency space Suppose the input is a sum at different frequencies X(ω) = α X(ω ) + α 2 X(ω 2 ) + Y (ω) = α Y (ω ) + α 2 Y (ω 2 ) + = α H(ω )X(ω ) + α 2 H(ω 2 )X(ω 2 ) +. But one would guess that inputs to systems are rarely nice sums of pure harmonic signals... Wrong! Last year you learned that a periodic signal of frequency ω can be represented as the sum of a set of harmonic signals at frequencies ω, 2ω, 3ω, and so on. Provided the Dirichlet conditions are satisfied. See later.

8 A 20 8 / You know about Fourier Series Eg, the Fourier series of a unit square wave with a zero at t = 0 and period T = 2π/ω is f (t) = 4 [sin ωt + 3 π sin 3ωt + 5 ] sin 5ωt + Armed with (i) a system s transfer function H(ω), (ii) the principle of linear superposition, and (iii) a Fourier Series... you can work out the output of the system corresponding to a periodic input. Fourier Series X ( ω) X ( 2 ω) X (3 ω) H( ω) H ( 2 ω) ( H 3 ω) Y Y ( ω) ( 2 ω) Y (3 ω) Σ

9 A 20 9 / What you don t know :-( You can make a signal of finite duration periodic by pretending it repeats but this is useless for infinite duration signals. The periodic functions that you can handle give rise only to discrete frequency spectra. But surely there must be signals that have a continuous frequency spectrum. f(t) F( ω) + +/3 +/5 T= 2 π/ω 0 ω 0 3ω 0 5ω 0 ω F( ω) ω This course fills the gap with FOURIER TRANSFORMS

10 A 20 0 / Fourier Transforms? Fourier transforms provide a method of transforming infinite duration signals both non-periodic and periodic from the time domain into the continuous frequency domain. They provide an entire language with which to work and think in terms of frequency. Involves new vocabulary and new mathematical techniques, like convolution correlation modulation sampling spectral density aliasing Most of these involve rather daunting-looking integrals. You must look beyond the grunt practice the mathematics, but also practice fixing the concepts in your head by using simple physical examples.

11 A 20 / Yo Fourier If at any point you become really cross just remember that Fourier seems a really jolly sort of chap He had an interesting life not only in mathematics but also in politics in France during Napoleon s time. Joseph Fourier Read his biography by following the links from www-history.mcs.st-and.ac.uk/ history/

12 A 20 2 / Signal definitions The emphasis of our introductory discussion has drifted from systems, and towards signals this course could be titled An introduction to analogue signal processing. A signal might be a function of one or of several variables. Space and time are very common variables, but here we will tend to stick to one variable, and choose time t more often than not. Again more often than not, we will think about electrical signals but do remember that the variation in temperature during the day is just as much a signal as the voltage output of a thermocouple sensing it. Let s define some signal types...

13 A 20 3 / Signal definitions () An Analogue signal is one whose amplitude covers a continuous range. It may be bounded (e.g. 0 5 V) but it can just as easily take the value V as V and anything in between. A continuous-time analogue signal is one that has a value for a continuous sweep of values of its parameter, t. Notice the value is allowed to be zero. f(t) f(t) f(t) Examples of continuous-time analogue signals Note that continuous-time does not mean the function is continuous.

14 A 20 4 / Signal definitions (2) A discrete-time analogue signal is one that has an analogue value at certain times only. Typically these will be at regular intervals, arising from regular sampling. The signal is not defined between times. NB this type of signal is labelled f (nt ), where n is an integer, and T is the sampling interval. f(t) Sampling Period T f(nt) Sampling Period T

15 A 20 5 / Signal definitions (3) A digital signal is one whose amplitude can only take one of a discrete set of values represented by a binary coding scheme. Suppose we used 4 bits {0000, 000, 000,..., }. Could represent {0,,..., 5} V, or { 0.2, 0., 0.0,...,.3} V, or whatever. But we cannot properly represent any value not in the discrete set of values. A digital signal is always sampled. In this course we are not concerned with digital signals, and consider only analogue continuous-time, and analogue discrete-time.

16 A 20 6 / Signal definitions (4) A causal signal is one that is finite only for t 0 ie f (t) = 0 for all t < 0 A deterministic signal is one that can be described by a function, mapping, or some other recipe or algorithm. If you know t, you can work out f (t). A random signal is determined by some underyling random process. Although its statistical properties might be known (e.g. you might know its mean and variance) you cannot evaluate its value at time t. You might by able to say something about the likelihood of its taking some value at time t, but not more.

17 A 20 7 / Orthogonal basis functions Consider vectors v, v 2, v 3, which are of different magnitudes but lie at right angles to each other. They form a set of orthogonal basis vectors in 3D. Any 3D vector f can be expressed uniquely f = A v + A 2 v 2 + A 3 v 3. How to find these unique coefficients for a particular f? To find A, take the scalar or inner product with v f v = A v v + A 2 v 2 v + A 3 v 3 v, Then exploit orthogonality... v 2 v = 0 and v 3 v = 0... A = f v v v.

18 A 20 8 / Orthogonal basis functions (2) Possible to treat functions f in the same way using a set of orthogonal basis functions. Pretend there is a set of orthogonal v-functions, so that f (t) = A v (t) + A 2 v 2 (t) + A 3 v 3 (t) +. How do we find A and so on? Need to define the scalar or inner product between functions but for now just assume we can, and denote it f, f. Then, by analogy f (t), v (t) = A v (t), v (t) + A 2 v 2 (t), v (t) + But v 2 (t), v (t) = 0 and so on so that A = f (t), v (t) v (t), v (t) and A n = f (t), v n(t) v n (t), v n (t).

19 A 20 9 / Orthogonal basis functions (3) Numerous famous sets of orthogonal basis functions many derived by French mathematicians... But it was Fourier who noticed that the cosines c and sines s made up such a set where the scalar or inner product is defined as c m, c n = T s m, s n = T c m, s n = T T /2 T /2 T /2 T /2 T /2 T /2 cos mωt cos nωtdt = sin mωt sin nωtdt = cos mωt sin nωtdt = 0 0 m=n /2 m = n > 0 m = n = 0 0 m=n /2 m = n > 0 0 m = n = 0

20 A / Fourier Series A periodic function f (t), with period T and ω = 2π/T is Fourier Series f (t) = 2 A 0 + A n cos(nωt) + B n sin(nωt). n= n= A n and B n are found from the orthogonality conditions as Fourier Coefficients For n = 0,,... A n = 2 T + T 2 T 2 For n =, 2,... B n = 2 T + T 2 T 2 f (t) cos(nωt)dt A 0 = 2 T f (t) sin(nωt)dt + T 2 T 2 f (t)dt

21 A 20 2 / Verifying those coefficient expressions To verify the stated expressions for the coefficients, we use the inner products as for the v-functions. The only complication is that we have both c n and s n in the basis set, so we need two coefficients with the subscript n. Calling these A n and B n we find A n = and B n = f (t), cn(t) c n(t), c n(t) = f (t), sn(t) s n(t), s n(t) = T T T /2 T T /2 f (t) cos nωtdt T /2 cos nωt cos nωtdt = 2 T T /2 T /2 T T /2 f (t) sin nωtdt T /2 sin nωt sin nωtdt = 2 T T /2 T /2 T /2 T /2 T /2 f (t) cos nωtdt f (t) sin nωtdt

22 A / A more explicit approach If you felt there was sleight of hand in the above, you may prefer to write down the series, and then, to find the B m for example, multiply it by sin(mωt) and average over a period T /2 f (t) sin(mωt)dt = [ T /2 T T /2 T T /2 2 A 0 + The three terms on the right are T /2 2 A 0 sin(mωt)dt = 0 T T /2 T /2 A n cos(nωt) sin(mωt)dt T T /2 n= = T /2 B n sin(nωt) sin(mωt)dt T T /2 n= = Hence, in agreement with the earlier statement, T /2 T T /2 ] A n cos(nωt) + B n sin(nωt) sin(mωt)dt n= n= n= n= T /2 A n cos(nωt) sin(mωt)dt = 0 T T /2 T /2 B n sin(nωt) sin(mωt)dt = B m T T /2 2 f (t) sin(mωt)dt = B m 2

23 A / Eg Square Wave Always SKETCH the function. Then FIND ω: Period is T, ω = 2π/T. A m = 2 +T /2 f (t) cos(mωt)dt = 2 [ 0 ] T /2 cos(mωt)dt + cos(mωt)dt = 0 T T /2 T T /2 0 B m = = = 2 +T /2 f (t) sin(mωt)dt = 2 [ 0 ] T /2 sin(mωt)dt + sin(mωt)dt T T /2 T T /2 0 4 T /2 sin(mωt)dt = 4 /2 [ cos( mωt) T 0 T 0 T mω 2 2 [ ] [ cos( mπ) + ] = ( ) m+ + mπ mπ This gives the series stated earlier: [sin(ωt) + 3 sin(3ωt) + 5 sin(5ωt) +... ] f(t) f (t) = 4 π T/2 +T/2 t Period T

24 A / Square wave /ctd. The FS of a square wave built up over,3,5 terms; then and 0 terms; then 00 terms..5 f(t) 0.5 y(t) 0 T/2 +T/2 t 0.5 Period T t y(t) 0 y(t) t t Overshoot at discontinuity is the Gibbs phenomenon.

25 A / Example: Triangular waveform () SKETCH (2) Period is 2π ω =. +π A m = 2 f (t) cos(mωt)dt 2π π = [ 0 π ] t cos(mt)dt + t cos(mt)dt π π 0 Use integration by parts with u = t, dv/dt = cos mt: A m = B m = 2 [ 0 2π f (t) = π π 2 [ [ cos mπ] = πm2 + ( ) m ] = πm 2 t sin(mt)dt + π π 0 π f(t) π Period 2π π for m = 0 0 for m EVEN 4 πm 2 ] t sin(mt)dt = grind = 0. [ cos(t) + 9 cos(3t) + 25 cos(5t) +... ] π t for m ODD

26 A / Triangular waveform /ctd f(t) π "tri.5" "tri.".5 π π t 0.5 Period 2π Triangle wave form Using then 5 then terms Looks better with few terms than square wave. Why?

27 A / The Dirichlet Conditions The Dirichlet conditions determine whether or not a function can be written as a Fourier Series. A function must be periodic, or be of finite extent so that it can be made periodic by extension; 2 have only a finite number of discontinuities within a period; 3 the discontinuities must be of finite size; and 4 have a finite number of maxima and minima. + f (t) = e t t < is OK + f (t) = /t t < is BAD.

28 A / Fourier Series at Discontinuities Provided the Dirichlet conditions are satisfied, at a point of discontinuity in the orginal function, a Fourier series converges to FS(t) 2 [f (t ) + f (t + )] = average of value below and above.5 f(t) 0.5 y(t) 0 T/2 +T/2 t 0.5 Period T t Note that a large number of terms is required to reproduce the function at a discontinuity.

29 A / Symmetry properties () The task of deriving series coefficients is made a little easier by exploiting symmetries in the signal f (t) and the basis functions The sine basis functions all have have odd /2-wave symmetry If the signal f (t) is even, then 0 T /2 f (t) sin(nωt)dt = sin(nωt) = sin( nωt) T /2 So all the B n coefficients are zero. 0 f (t) sin(nωt)dt T /2 T /2 Therefore an even signal contains only cosine terms Similarly an odd signal contains only sine terms. f (t) sin(nωt)dt = 0

30 A / Symmetry properties (2) One notes that for odd signals f (t) T /2 f (t) sin(nωt)dt = 2 T /2 0 T /2 f (t) sin(nωt)dt sin ωt cosωt /2 wave reflection /2 wave reflection and similarly for an even signals f (t) T /2 f (t) cos(nωt)dt = 2 T /2 0 T /2 f (t) cos(nωt)dt

31 A 20 3 / Symmetry properties (3) Further use can be made of symmetries about the /4-wave points. Any sin(nωt) with n-even has odd symmetry about these points. Thus if a signal f (t) has even symmetry about the /4-wave points, any n-even sine terms will vanish. sin ωt sin 2 ωt /4 wave /2 wave /4 wave /4 wave /2 wave /4 wave reflection reflection reflection reflection reflection reflection

32 A / Symmetry properties (3) However if the signal s symmetry is odd about the /4-wave points, the n-odd sine terms vanish. sin ωt sin 2 ωt /4 wave /2 wave /4 wave /4 wave /2 wave /4 wave reflection reflection reflection reflection reflection reflection One could consider higher symmetries, but they get increasingly difficult to recognize.

33 A / Examples: Symmetry properties f(t) T/2 +T/2 t Period T [ 4 π sin(ωt) + 3 sin(3ωt) + 5 sin(5ωt) + ] f(t) π π π t Period 2π π 2 + [ 4 π cos(t) + 9 cos(3t) + 25 cos(5t) +...]

34 A / Train of tophats The period is T, so ω = 2π/T. The function is even, so only A n coefficients exist. A 0 = 2 T A n = 4 T T /2 T /2 a/2 0 f (t) dt = 2 T = 2 nπ sin(nωa/2) cos nωtdt = 4 T f (t) = a T + n= a/2 a/2 dt = 2a T sin nωt a/2 0 nω f(t) a/2 2 sin(nωa/2) cos nωt nπ a/2 Period T t

35 A / Train of tophats /ctd Of particular interest later are the values of the A n. Suppose we set the (on/off) ratio to be α = a/t, A n = 2 nπ sin(nωa 2 ) = 2 nπ sin(n2πa 2T ) = 2αsin(nπα) (nπα) If the on/off ratio α = /π then the A n are taken from the sin(x)/(x) curve as shown on the left. If we reduce α, here by /2, then the A n values are sampled from the curve more closely, as on the right..0 sin(x)/(x).0 sin(x)/(x) x x

36 A / Completing Functions Suppose you need derive the Fourier series of a nontime-continuous, non-period function. For example, f (t) = t for 0 t. First, you MUST make the function periodic but exactly how is a matter of choice. Defined Completed as sine as cosine as mixed sin, cos Double period sine completion Which is best? Because the cosine series has no discontinuities it will require fewer terms to make an overall decent approximation. However, the kink at t = 0 will not be accurate. If a good approximation with few terms is required close to t = 0, it is probably best to use the sine series completion in this case.

37 A / The Gibbs phenomenon At a discontinuity, residual oscillations lead to an overshoot, whose size is a characterstic of the underlying function. (It can be large! For a square wave of amplitude A, the overshoot is around δ = 0.8A.).5 "gibbs.00" "gibbs.000" y(t) t As more terms are added to the series, the oscillations get squeezed into a shorter and shorter region around the discontinuity, but their characteristic amplitude remains constant!

38 A / Mean square value of an FS: Parseval s Theorem () The instantaneous power in a signal is proportional to its modulus squared. For a periodic signal, derive the average power by integrating over a period, and dividing by the period... +T /2 Average signal power = f (t) 2 dt T T /2 +T /2 2 = T T /2 2 A 0 + A n cos(nωt) + B n sin(nωt) dt n= n= Nightmarish? Squaring introduces an infinite number of cross terms... +T /2 T T /2 { ( 2 A 0 ) 2 + A 2 cos2 ωt + A 2 2 cos2 2ωt B 2 sin2 ωt + B 2 2 sin2 2ωt A 0 A cos ωt +... A 0 B sin ωt A B cos ωt sin ωt + 2A B 2 cos ωt sin 2ωt +...} dt

39 A / Mean square value of an FS: Parseval s Theorem (2) To repeat that expression { +T /2 ( ) 2 T T /2 2 A 0 + A 2 cos2 ωt + A 2 2 cos2 2ωt B 2 sin2 ωt + B 2 2 sin2 2ωt A 0 A cos ωt A 0 B sin ωt A B cos ωt sin ωt + 2A B 2 cos ωt sin 2ωt +...} dt Good news! Average values of cos nωt and sin nωt are zero, orthogonality kills off all the cross terms, and the average values of cos 2 nωt and sin 2 nωt are /2. We are left with the clean result that the mean square is T +T /2 T /2 f (t) 2 dt = ( ) 2 2 A (A n ) 2 + (B n ) 2 2 n= n=

40 A / Mean square value of an FS: Parseval s Theorem (3) To summarize... T +T /2 T /2 f (t) 2 dt = ( ) 2 2 A (A n ) 2 + (B n ) 2 2 n= n= Remember this as (true for d.c. & single frequency a.c too) Mean Square = (d.c. amplitude) 2 + (a.c. amplitude) 2 2 The RMS value of a periodic signal is therefore ) 2 Root Mean Square = ( 2 A 0 + (A n ) 2 + (B n ) n= n=

41 A 20 4 / Fourier and Parseval in an electrical circuit [Q] Suppose the full wave rectified voltage is applied to the circuit 2. What is the average power dissipated? 2V t (ms) v(t) 5kΩ µ F [A] From the diagram T = 0 2 s and ω 0 = 2π/T = 200π. By grinding, or from HLT we find the signal v(t), with t in seconds, is v(t) = 24 { + 2 π 3 cos ω 0t 2 } 5 cos 2ω 0t +... We know that power is dissipated in the resistor alone, so P ave = ( ) { ( 2 (2 R v 2 24 = ) 2 ( ) )} π This is a silly circuit by the way...

42 A / Fourier and Parseval in an electrical circuit [Q] What is current drawn from the source? [A] Try this after lecture... The current drawn from the source at a single frequency ω is I(ω) = Y (ω)v (ω), where Y is the admittance. But we have voltage components at ω = 0, ω = ω 0, ω = 2ω 0 and so on. We must work out the admittances at all these frequencies. Y (ω) = R + jωc = (2 0 4 ) + jω( 0 6 ) ( 0 = 0 4 ω ) 4 (2 + j 0), dc j = 4 (2 + j 2π), ω = ω 0 = 200π (2 + j 4π), ω = 2ω 0 = 400π 0 4 (2 + j 2nπ), ω = nω 0

43 A / Fourier and Parseval in an electrical circuit To avoid mixing functions of time with phasors, it makes sense to rewrite v(t) as the real part of ( { 24 v(t) = Re + 2 π 3 ejω 0t 2 }) 5 ej2ω 0t +... Hence π ( 24 0 i(t) = Re 4 π { 2 + (2 + j2π) 2 3 ej200πt (2 + j4π) 2 5 ej400πt +... }) = { + ( + π 2 ) 2 3 cos(200πt + φ ) ( + 4π 2 ) 2 5 cos(400πt + φ 2) +... with φ = tan π and φ 2 = tan 2π. }

44 A / The Complex Fourier Series You will have noticed while working out i(t) that there was a rather unsatisfactory moment when we had to rewrite the Fourier series using an exponential representation. It raises the following question... Would it be possible use an exponential, or complex, form of the Fourier Series from the outset?

45 A / The Complex Fourier Series It turns out perhaps not surprisingly that the set e inωt is a set of orthogonal basis functions. The inner product is defined as integration over a period with the complex conjugate, divided by the period. The orthogonality conditions are simpler than earlier... T /2 e inωt e imωt dt = T T /2 { 0 m=n m = n

46 A / The Complex Fourier Series A periodic function with period T = 2π/ω that satisfies the Dirichlet conditions can be written as The complex Fourier series f (t) = n= C n e inωt NOTE!! The range of n is from to +. The coefficients are determined from f (t), e imωt C m = e imωt, e imωt = T /2 T T T /2 f (t)e imωt dt T /2 T /2 eimωt e imωt dt = T /2 f (t)e imωt dt T T /2

47 A / To summarize: The complex Fourier series f (t) = C m e imωt m= C m = T T /2 T /2 f (t)e imωt dt Note that C m = C m, where denotes complex conjugate. Another way to derive the series is to use the link e imωt = cos mωt i sin mωt. The relationship between the C coefficients and those for the non-complex Fourier series is (A m ib m )/2 for m > 0 C m = A 0 /2 for m = 0 (A m + ib m )/2 for m < 0

48 A / Summary of Lecture We have defined certain terms used to describe signals. 2 We have reviewed how periodic signals can be represented as Fourier Series linear sums of pure harmonic signals and revised their properties. 3 We have introduced the complex Fourier Series, which is often a more convenient representation to use when having to deal with phase shifts. 4 However, there was nothing especially new in the complex Fourier Series. Recall that the real gap in our knowledge is how to cope with non-periodic signals of infinite duration; that is, those which have continuous spectra in the frequency domain. This is where we move next.