A1 2011 1 / 1 A1 Time-Frequency Analysis David Murray [email protected] www.robots.ox.ac.uk/∼dwm/Courses/2TF Hilary 2011 A1 2011 2 / 1 Course Content 1 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 A1 2011 3 / 1 Introduction In this lecture we will described the mathematic operation of the convolution of two continuous functions f (t) ∗ g(t) Two functions are blended or folded together. We then discuss the impulse response h(t) of a system, and show how it is related to the transfer function H(s) of the system. First we define a the δ-function or unit impulse. Like the Heaviside step function u(t), it is a generalized function or “distribution”, and is best defined by considering another function in conjunction with it. A1 2011 4 / 1 The δ-function (1) Consider a function g(t) 1/ 1=w 0− < t < w w g(t) = 0 otherwise 0 w + As w becomes very small, w ! 0 , δ (t) the function g(t) turns into a δ-function δ(t) indicated by the arrowed spike. 0 0− is a infinitesimally small amount less than zero, and 0+ is a infinitesimally small amount more than zero. A1 2011 5 / 1 The δ-function (2) One thing of note about g(t) is that g(t) Z w 1/ w g(t)dt = 1: 0− This property is retained as w ! 0+, 0 w so that δ Z 1 Z 0+ (t) δ(t)dt = δ(t)dt = 1 −∞ 0− Akin to the derivative of the Heaviside unit step 0 Z t δ(t)dt = u(t) : −∞ A1 2011 6 / 1 The δ-function (3) More formally the δ−function is δ defined in association with another f(t) (t) arbitrary function f (t), as Z 1 f (t)δ(t)dt = f (0) : 0 −∞ Picking out values of a function is δ (t−τ ) called sifting of f (t) by δ(t). We can also see that f(t) Z 1 f (t)δ(t − τ)dt = f (τ) t −∞ 0 τ How? Subst p = t − τ: then Z 1 f (p + τ)δ(p)dp = f (0 + τ) = f (τ) −∞ A1 2011 7 / 1 The δ-function (4) Although the δ-function is infinitely A δ (t−τ ) high, very often you will see a described as the unit δ-function This is to denote a delta-function t where τ Z 1 0 δ(t)dt = 1 −∞ If you see a δ-function spike with an amplitude A by it this denotes Z 1 Aδ(t)dt = A −∞ A1 2011 8 / 1 Properties of the δ-function #1 Fourier transform of the delta function: FT [δ(t)] = 1 Proof: Use the definition of the δ-function and sift the function f (t) = e−i!t : Z 1 δ(t)e−i!t dt = e−i!0 = 1 : −∞ NB!! Check the minus sign is in the notes ... This make some sense. We know from FS that describing abrupt changes requires many terms in the series — i.e. the high frequencies are important. The δ-function is the mother of all abrupt changes. A1 2011 9 / 1 Properties of the δ-function #2 Symmetry: The δ-function has even symmetry, δ(t) = δ(−t) Proof: Consider sifting an even function E(t) with E(0)=0 and substitute p = (−t) Z 1 E(t)δ(t)dt = E(0) −∞ Z −∞ ) − E(−p)δ(−p)dp = E(0) 1 Z 1 ) E(p)δ(−p)dp = E(0) −∞ Z 1 ) E(t)δ(−t)dt = E(0) −∞ So δ(−t) = δ(t) and δ(t) is even. A1 2011 10 / 1 Properties of the δ-function #3 Parameter Scaling: 1 δ(at) = δ(t) jaj R 1 Proof: Use the fundamental sifting definition −∞ f (t)δ(t)dt = f (0) The jaj again hints dealing with things in two stages ... If a ≥ 0 substitute (at) for t (no change in limits) Z 1 Z 1 f (at)δ(at)d(at) = a f (at)δ(at)dt = f (0) −∞ −∞ Z 1 But f (at)δ(t)dt = f (0) )a δ(at) = δ(t): −∞ A1 2011 11 / 1 Properties of the δ-function #3 /ctd Proof: continued Now if a < 0 substitute (at) for t (but need to change limits as a negative) Z −∞ Z 1 f (at)δ(at)d(at) = −a f (at)δ(at)dt = f (0) 1 −∞ Z 1 But f (at)δ(t)dt = f (0) ) − a δ(at) = δ(t): −∞ So to cover both cases: 1 δ(at) = δ(t) jaj A1 2011 12 / 1 Fourier Transforms that involve the δ-function Fourier Transform of ei!0t Z 1 FT ei!0t = ei(!0−!)t dt −∞ = 2πδ(! − !0) : How could you prove this? Fourier Transform of ein!0t Z 1 FT ein!0t = ei(n!0−!)t dt −∞ = 2πδ(! − n!0) : A1 2011 13 / 1 Fourier Transforms that involve the δ-function Fourier Transform of 1 FT [1] = 2πδ(!) : Proof Either i! t put !0 = 0 into FT e 0 = 2πδ(! − !0) or use the dual property FT [δ(t)] = 1 )FT [1] = 2πδ(−!) )FT [1] = 2πδ(!) as δ is an even function. A1 2011 14 / 1 Fourier Transforms that involve the δ-function Fourier Transform of cos !0t 1 FT [cos ! t] = FT ei!0t + e−i!0t 0 2 = π (δ(! − !0) + δ(! + !0)) : Fourier Transform of sin !0t 1 FT [sin ! t] = FT ei!0t − e−i!0t 0 2i = −iπ (δ(! − !0) − δ(! + !0)) : A1 2011 15 / 1 Fourier Transforms that involve the δ-function Fourier Transform of Complex Fourier Series " n=1 # n=1 X in!0t X FT Cne = 2π Cnδ(! − n!0) : n=−∞ n=−∞ If you are wondering ... ... yes, this will prove very useful! A1 2011 16 / 1 Convolution (1) This is a key technique in signal processing The convolution of continuous functions f (t) and g(t) is defined by an integral The convolution integral. The value of f ∗ g at t is Z 1 f ∗ g (t) = f (τ)g(t − τ)dτ −∞ where τ is a dummy variable. Note that one function is “reversed” ... But the process is commutative, so it doesn’t matter which: Z 1 Z 1 (f ∗g)(t) ≡ (g ∗f )(t) ) f (τ)g(t −τ)dτ ≡ f (t −τ)g(τ)dτ −∞ −∞ A1 2011 17 / 1 | Convolution Example #1 [Q] Find and sketch the convolution of f (t) = u(t)e−at with g(t) = u(t)e−bt , where both a and b are positive. [A] The “unhelpful” answer must be R 1 −aτ −b(t−τ) f ∗ g = −∞ u(τ)e u(t − τ)e dτ : f(τ ) g(τ ) To make sense of this... Make sketches of the functions f (τ) and g(t − τ) as τ τ τ varies. 0 0 Function f (τ) looks just like g(−τ ) g(t−τ ) f (t) of course. But g(t − τ) is a reflected (“time reversed”) and shifted version of g(t). τ τ 0 0 t A1 2011 18 / 1 | Convolution Example #1 Now multiply the two τ ) τ ) functions g(t−τ ) f( g(t−τ ) f( BUT we must worry τ τ about the fact that t is t 0 0 t a variable. f(τ )g(t−τ ) f(τ )g(t−τ ) t <0 t >0 In this case there are two different regimes, τ τ one when t < 0 and 0 t 0 t the other when t ≥ 0. A1 2011 19 / 1 | Convolution Example #1 For t < 0, the function to be integrated is everywhere zero, and the result is zero. g(t−τ ) f(τ ) For t ≥ 0 1 Z τ u(τ)e−aτ u(t − τ)e−b(t−τ)dτ 0 −∞ t t Z τ ) = e−bt e(b−a)τ dτ f(τ )g(t− 0 t >0 e−bt = e(b−a)t − 1 b − a τ 0 t So e−at − e−bt =(b − a) for t ≥ 0 f ∗g(t) = 0 for t < 0 A1 2011 20 / 1 | Convolution Example #1 Important to realize that the convolution function is the integral evaluated for the complete range of t values. e−at − e−bt =(b − a) for t ≥ 0 f ∗ g(t) = 0 for t < 0 This shows the t > 0 part of the convolution for b = 2 and a = 1. 0.25 (exp(-x)-exp(-2*x)) 0.2 0.15 0.1 0.05 0 0 1 2 3 4 5 6 A1 2011 21 / 1 | Convolution Example #2 [Q] Derive an expression for the convolution of an arbitrary signal f with the function g shown in the figure. Determine the convolution when f (t) = A, a constant, and when f (t) = A + (B − A) u(t). g(t) f(t) g(t− τ ) f(t) 1 a t 1 −a t t−a t τ −1 t t+a −1 [A] Function f (τ) looks exactly like f (t), but g(t − τ) is reflected and shifted. Multiply and integrate over τ from −∞ to 1. Because g only has finite range, we can pinch in the limits of integration, and the convolution becomes Z t Z t+a f ∗ g(t) = − f (τ)dτ + f (τ)dτ t−a t A1 2011 22 / 1 | Convolution Example #2 ctd Z t Z t+a f ∗ g(t) = − f (τ)dτ + f (τ)dτ t−a t When f (t) = A, f ∗ g(t) = 0 for all t.