Definitions, Theorems, Examples, Topics for Asmar 2nd edition Chapter 7: 7.2 only 7.2 The Fourier Transform ==== We start with the Fourier integral representation of f(x). THEOREM 1. [Fourier Representation Theorem] Assume f(x) is piecewise smooth on every finite interval and the integral of |f(x)| over (-inf,inf) is finite. Define A(omega) = (1/Pi) integral of f(t)cos(omega t) over (-inf,inf) B(omega) = (1/Pi) integral of f(t)sin(omega t) over (-inf,inf) Then (f(x+)+f(x-))/2 = integral of A(w)cos(wx)+B(w)sin(wx) for w=0 to infinity. ================= The Fourier Transform results by changing from a superposition of sines and cosines to a superposition of complex exponential functions. It is possible because of Euler's formula exp(i u) = cos(u) + i sin(u) and the specific combinations cos(u) = (1/2)( exp(i u) + exp(-i u) ) sin(u) = (1/2i)( exp(i u) - exp(-i u) ) plus the standard trig identity cos(a-b) = cos(a)cos(b) + sin(a)sin(b) FACT 1. f(x) = (1/Pi)(integral of A(w)cos(wx)+B(w)sin(wx), w=0..inf) This follows from the Fourier integral representation FACT 2. Define h[w](x)=convolution of f(t) and cos(wt) =integral of f(t)cos(w(x-t)) on t=-inf..inf Then f(x) = (1/Pi)(integral of h[w](x) on w=0..inf) = a superposition of convolutions FACT 3. Define k[w](x)=convolution of f(t) and exp(iwt) =integral of f(t)exp(iw(x-t)) on t=-inf..inf Then f(x) = (1/2Pi)(integral of k[w](x) on w=-inf..inf) FOURIER TRANSFORM It is defined by the latter equation's right side. We use upper case F for the Fourier transform of lower case f. DEF. F(w)=(1/sqrt(2Pi))(integral of f(x)exp(-iwx) on x=-inf..inf) THEOREM. [Fourier Inverse Transform] f(x) = (1/sqrt(2Pi))(integral of exp(iwx)F(w) on w=-inf..inf) PROOF. This is from FACTS 1,2,3 above. DEF. The symbol FT will stand for the Fourier Transform. This means FT[f](w) = (1/sqrt(2Pi))(integral of f(x)exp(-iwx) on x=-inf..inf) The Fourier inverse transform formula is then f(x) = (1/sqrt(2Pi))(integral of exp(iwx)FT[f](w) on w=-inf..inf) DEF. The MAGNITUDE of the Fourier Transform is |FT[f](w)|. We can write FT[f](w) = R(w) + i C(w) = |FT[f]|epx(i phi) into real and complex parts. Then the MAGNITUDE and PHASE are defined as follows: MAGNITUDE = |FT[f]| = sqrt(R(w)^2 + C(W)^2) PHASE = phi = arctan(C(w)/R(w)) EXAMPLE 1a. Compute FT[f] where f(x) is the unit pulse on |x| < a. ANSWER: FT[f](w) = sqrt(Pi/2) sin(aw)/w for -inf < w < inf EXAMPLE 1b. Express the unit pulse f(x) on |x| < a as an inverse Fourier transform ANSWER: f(x)=(1/Pi)(integral of cos(wx)sin(aw)/w on w=-inf..inf) EXAMPLE 1c. A unit impulse applied to an electric circuit has impulse response f(t) = exp(-t/5)u(t) where u(t) is the unit step, u(t)=1 for t>=0, else u(t)=0. Find FT[f] and plot the signal, its Fourier magnitude and its Fourier phase. MAPLE SOLUTION u:=t->piecewise(t<0,0,1):f:=unapply(exp(-t/5)*u(t),t);cv:=1/sqrt(2*Pi): FT[f]:=inttrans[fourier](cv*f(t),t,w); assume(w,real):R:=unapply(Re(FT[f]),w);C:=unapply(Im(FT[f]),w); Mag:=unapply(simplify(sqrt(R(w)^2+C(w)^2)),w); Phase:=unapply(arctan(C(w)/R(w)),w); plot(f(x),x=-10..10);plot(Phase(w),w=-10..10);plot(Mag(w),w=-10..10); inttrans[invfourier](Mag(w)/cv,w,x); MagInv:=unapply(%,x);plot(MagInv(x),x=-10..10); # Not f(x)! # Answer check. inttrans[invfourier](Mag(w)*exp(I*Phase(w))/cv,w,x); THEOREM. The Fourier transform of a piecewise smooth f(x) is always continuous in w. EXAMPLE 2. The Fourier Transform can be complex even if f(x) is real. Compute FT[f] for f(x) = exp(-x)u(x), where u(x) is the unit step, defined by u(x)=1 for x>=0, u(x)=0 for x<0. ANSWER. FT[f](w) = (1/sqrt(2Pi))(1-iw)/(1+w^2) DEF. The CAUCHY PRINCIPAL VALUE is the symmetric limit at a=infinity for an integral over [-a,a]. This limit is by default used for integrals involving Fourier Transforms FT[f], because they may not be integrable over the line. THEOREM 1. [Linearity] FT[f+g] = FT[f] + FT[g] FT[cf] = c FT[f] THEOREM 2. [Fourier Transform of x-Derivatives] (i) FT[f'] = iw FT[f] (ii) FT[f''] = (iw)^2 FT[f] (iii) FT[(d/dx)^n f] = (iw)^n FT[f] THEOREM 3. [w-Derivative of a Fourier Transform] (i) FT[xf(x)](w) = i (d/dw)FT[f](w) (ii) FT[x^nf(x)](w) = i^n (d/dw)^n FT[f](w) CONVOLUTION DEF. f*g (x) = (1/sqrt(2Pi))(integral of f(t)g(x-t) on t=-inf..inf) THEOREM. f*g=g*f EXAMPLE 3. Convolution with the cosine Compute f*g when f is integrable and even (f(-x)=f(x)) and g(x)=cos(ax). ANSWER. f*g(x)=cos(ax)FT[f](a) EXAMPLE 4a. Convolution is an average. Let g_n(x) = n sqrt(Pi/2) times the unit pulse on |x|<1/n. Show that f*g_n is the average value of f(x) over the interval [x-1/n,x+1/n]. PROOF. f*g_n(x)=(n/2)(integral of f(t) over |t-x|<1/n)=average value. EXAMPLE 4b. Show that limit f*g_n = f as n approaches infinity. PROOF. Because the interval [x-1/n,x+1/n] has length convergent to zero, then the average value converges to f(x). DEF. The function g_n is an approximation to an IMPULSE FUNCTION. This is because the integral of g_n equals sqrt(2Pi). The integral is the force involved in the impulse. View the function as being a realization of c*Dirac(x) [Dirac delta] where the force is c=sqrt(2Pi). COMPUTING CONVOLUTIONS Because a Fourier Transform is a superposition of convolutions, there is a demand for being able to easily compute a convolution integral. THEOREM 4. A convolution f*g can be computed as the inverse Fourier transform of the product FT[f]FT[g]. Succinctly, FT[f*g] = FT[f]FT[g] EXAMPLE 5. Fourier transform of a convolution Let f(x) be the unit pulse on |x|<1. Find f*f. ANSWER. Compute FT[f]FT[f]=(2/Pi)(sin(w)/w)^2. The f*f is the inverse Fourier transform of this answer. We can look it up in a table or use a computer algebra system to look up the answer. # Warning: Maple uses a different definition of Fourier Transform cv:=1/sqrt(2*Pi);f:=x->piecewise(abs(x)<1,1,0); G:=inttrans[fourier](cv*f(x),x,w); inttrans[invfourier](G^2/cv,w,x); expand(%);convert(%,piecewise); # ANS: f*f(x)=sqrt(2/Pi)*(1-|x|/2)*pulse(x,-2,2) DEF. GAUSSIAN function is f(x)=exp(-x^2). LEMMA. The integral of the Gaussian is sqrt(Pi). THEOREM 5. Fourier Transform of the Gaussian Let a>0. Then FT[exp(-a x^2/2)] = (1/sqrt(a)) exp(-w^2/(2a)) PROOF. Check that f'+axf=0. FT[f'] + a FT[xf(x)] = FT[0] Take transform of the DE. w FT[f](w) + a (d/dw) FT[f](w) = 0 Apply theorems. FT[f](w) = A exp(-w^2/(2a)) Solve wy + ay' = 0 for y The ans is y=constant/integ. fac. A=1/sqrt(a) Set w=0, change vars in integral. THEOREM 5a. FT[exp(-a x^2)] = (1/sqrt(2a)) exp(-w^2/(4a)) FT[exp(-x^2/2)] = exp(-w^2/2) EXAMPLE 6. Clever identity using THEOREM 5. f(x) = exp(-x^2)cos(x) integral of f(x) equals sqrt(Pi) exp(-1/4)