Math Methods I Lia Vas Fourier Series. Fourier ransform Fourier Series. Recall that a function differentiable any number of times at x = a can be represented as a power series n= a n (x a) n where the coefficients are given by a n = f (n) (a) n! hus, the function can be approximated by a polynomial. Since this formula involves the n-th derivative, the function f should be differentiable n-times at a. So, just functions that are differentiable any number of times have representation as a power series. his condition is pretty restrictive because any discontinuous function is not differentiable. he functions frequently considered in signal processing, electrical circuits and other applications are discontinuous. hus, there is a need for a different kind of series approximation of a given function. he type of series that can represent a a much larger class of functions is called Fourier Series. hese series have the form f(x) = a + (a n cos πnx n= + b n sin πnx ) he coefficients a n and b n are called Fourier coefficients. Note that this series represents a periodic function with period. o represent function f(x) in this way, the function has to be () periodic with just a finite number of maxima and minima within one period and just a finite number of discontinuities, () the integral over one period of f(x) must converge. If these conditions are satisfied and one period of f(x) is given on an interval (x, x + ), the Fourier coefficients a n and b n can be computed using the formulas a n = x + x f(x) cos nπx dx b n = x + x f(x) sin nπx When studying phenomena that are periodic in time, the term π in the above formula is usually replaced by ω and t is used to denote the independent variable. hus, f(t) = a + (a n cos nωt + b n sin nωt) n= dx
If the interval (x, x + ) is of the form ( L, L) (thus = L), then the coefficients a n and b n can be computed as follows. a n = L L L f(x) cos nπx L dx b n = L f(x) sin nπx L L L f(x) = a + (a n cos nπx L n= + b n sin nπx L ) As a special case, if a function is periodic on [ π, π], these formulas become: a n = π π π f(x) cos nx dx b n = π π π f(x) sin nx dx f(x) = a + (a n cos nx + b n sin nx) Obtaining the formulas for coefficients. If f(x) has a Fourier series expansion f(x) = a + (a n cos πnx n= + b n sin πnx ) one can prove the formulas for Fourier series coefficients a n by multiplying this formula by cos πnx and integrating over one period (say that it is (, )). ( / m= / / f(x) cos πnx dx = a m cos πmx cos πnx / dx + / a / / / cos πnx dx + b m sin πmx n= dx ) cos πnx dx All the integrals with m n are zero and the integrals with both sin and cosine functions are zero as well. hus, / / f(x) cos πnx / dx = a n cos πnx / cos πnx dx = a n a n = / / f(x) cos πnx dx
if n >. If n =, we have that a = / f(x)dx and so a / = / he formula for b n is proved similarly, multiplying by sin πnx f(x)dx. / instead of cos πnx. Symmetry considerations. Note that if f(x) is even (that is f( x) = f(x)), then b n =. Since b n = L nπx f(x) sin dx = nπx f(x) sin dx + L nπx f(x) sin dx. Using the substitution L L L L L L L L u = x for the first integral we obtain that it is equal to nπu f( u) sin du. Using that L L L f( x) = f(x), that sin nπu = sin nπu, and that = L, we obtain L nπu f(u)( sin ) du = L L L L L L nπu f(u) sin du. Note that this is exactly the negative of the second integral. hus, the first L L and the second integral cancel and we obtain that b n =. Similarly, a n = L nπx f(x) cos dx = nπx f(x) cos dx + L nπx f(x) cos dx. Using the L L L L L L L L substitution u = x for the first integral we obtain that it is equal to nπu f( u) cos du. Using L L L that f( x) = f(x), that cos nπu = cos nπu, and that = L, we obtain L nπu f(u) cos. Note L L L L L that this is equal to the second integral. hus, the two integrals can be combined and so a n = L L f(x) cos nπx L dx and f(x) = a + If f(x) is odd, using analogous arguments, we obtain that a n = and b n = L L n= n= a n cos nπx L. f(x) sin nπx L dx and f(x) = b n sin nπx L. hese last two power series are called Fourier cosine expansion and Fourier sine expansion respectively. Example. he input to an electrical circuit that switches between a high and a low state with time period π can be represented by the boxcar function. { x < π f(x) = π x < he periodic expansion of this function is called the square wave function. More generally, the input to an electrical circuit that switches from a high to a low state with time period can be represented { by the general square wave function with the following formula x < on the basic period. f(x) = x < 3
Find the Fourier series of the square wave and the general square wave. Solutions. Graph the square wave function and note it is odd. hus, the coefficients of the cosine terms will be zero. Since L = π ( = π), the coefficients of the sine terms can be computed as b n = π f(x) sin nxdx = π sin nxdx = cos π π π nπ nx π = nπ (( )n ). Note that ( ) n = = if n is even (say n = k) and ( ) n = = if n is odd (say n = k + ). hus, b k = and b k+ = ( ) = 4. Hence, nπ (k+)π f(x) = 4 π n=odd sin nx = 4 sin(k+)x n π k= = 4 sin 3x sin 5x (sin x + + +...). k+ π 3 5 For the general square wave, analogously to this previous consideration you obtain that a n =, b k = and b k+ = 4 (k+)πx sin. hus, f(x) = 4 (k+)πx (k+)π π k= sin. k+ Even and odd extensions. If an arbitrary function f(x), not necessarily even or odd, is defined on the interval (, L), we can extend it to an even function { f(x) < x < L g(x) = f( x) L < x < and consider its Fourier cosine expansion: a n = L L f(x) cos nπx L dx and f(x) = a + Similarly, we can extend f(x) to an odd function { f(x) < x < L h(x) = f( x) L < x < and consider its Fourier sine expansion: b n = L L n= n= a n cos nπx L. f(x) sin nπx L dx and f(x) = b n sin nπx L. Useful formulas. o simplify the answers, sometimes the following identities may be useful sin nπ = cos nπ = ( ) n In particular, cos nπ = and cos(n + )π =. If n = k + is an odd number, sin nπ = sin (k+)π = ( ) k cos nπ = cos (k+)π = Example. Find the Fourier cosine expansion of f(x) = { x < x x < x < 4
Solutions. First extend the function symmetrically with respect to y-axis so that it is defined on basic period [-,] and that it is even. hus = 4 and L =. he coefficients b n are zero in this case and the coefficients a n can be computed as follows. a n = nπx f(x) cos dx = nπx x cos dx + ( x) cos nπx nπx dx. Using integration by parts with u = x, v = sin for the first and u = x and same nπ v for the second, you obtain a n = ( ) ( ) x nπx sin + 4 cos nπx nπ n π + ( x) sin nπx 4 cos nπx nπ n π = nπ sin nπ + 4 cos nπ n }{{ 4 4 cos nπ } π n π n π nπ sin nπ + 4 cos nπ n }{{ = 8 cos nπ 4 4 cos nπ = } π n π n π n π 4 n π ( cos nπ cos nπ). 4 If n = k + is odd, a n = ( + ) =. If n = k is even, a (k+) π n = 4 (( ) k ). (k) π Because of the part with ( ) k, we can distinguish two more cases depending on whether k is even or odd. hus, if k = l is even, a n = 4 ( ) =. If k = l + is odd, a (4l) π n = 4 (( ) ) = ((l+)) π 6 = 4. (4l+) π (l+) π If n =, a = f(x)dx = xdx + So, f(x) = 4 π l= ( x)dx = + =. cos (4l+)πx = 4 (l+) π l= Example 3. Consider the function f(x) = x for x. cos(l + )πx. (l+) (a) Sketch the graphs of the following () the periodic extension of f(x), () the even periodic extension of f(x), (3) the odd periodic extension of f(x) and write the integrals computing the coefficients of the corresponding Fourier series in all three cases. (b) Find the Fourier cosine expansion for f(x). Solutions. (a) he periodic extension of f(x) is neither even nor odd. You can obtain the graph of it by replicating f(x) on intervals... [ 4, ], [, ], [, ], [, 4], [4, 6]... of length =. he coefficients of the corresponding Fourier series can be calculated by a n = b n = x cos nπx dx x sin nπx dx. he even extension of f(x) is obtained by extending f(x) from [, ] to [, ] by defining f(x) = ( x) = x on [, ]. hus, the result is the function x defined on interval [, ]. hen, replicate this function on intervals... [ 6, ], [, ], [, 6], [6, ],... of length = 4. hus, = 4 and L =. he coefficients of the corresponding cosine Fourier series can be calculated by a n = x cos nπx dx b n =. 5
he odd extension of f(x) is obtained by extending f(x) from [, ] to [, ] by defining f(x) = (x) = x on [, ]. hus, the result is the function { x x x x < defined on interval [, ]. Replicate this function on intervals... [ 6, ], [, ], [, 6], [6, ],... of length = 4. hus, = 4 and L =. he coefficients of the corresponding since Fourier series can be calculated by b n = x sin nπx dx a n =. (b) By part (a), the coefficients a n can be calculated by a n = x cos nπx dx. Using integration by parts with u = x and v = cos nπxdx = nπx sin, we obtain that a nπ n = nπ x sin nπx 4 nπx x sin dx. he first term is zero and the second term requires another integration by parts, nπ this time with u = x and v = sin nπx nπx dx = cos. hus a nπ n = 8 x cos nπx n π 6 sin nπx n 3 π 3. he first term is zero in the lower bound and the second term is zero at both bounds. hus a n = 6 cos nπ = 6( )n. Note that this formula works just for n > so a n π n π has to be computed separately by a = x dx = x3 3 = 8. 3 hus, the Fourier series is x = 4 3 + 6 π n= ( ) n n cos nπx. Complex Fourier Series. he complex form of Fourier series is the following: f(x) = n= c n e nπix = n= ( c n cos nπx + i sin nπx ) where c n = x + x f(x)e nπix If f(x) is a real function, the coefficients c n satisfy the relations c n = (a n ib n ) and c n = (a n + ib n ) for n >. hus, c n = c n for all n >. In addition, a n = c n + c n and b n = i(c n c n ) for n > and a = c. hese coefficients c n are further associated to f(x) by Parseval s heorem. his theorem is related to conservation law and states that x + x f(x) dx = n= c n = a 4 + (a n + b n). Note that the integral on the left side computes the average value of the moduli squared of f(x) over one period and the right side is the sum of the moduli squared of the complex coefficients. he proof of this theorem can be your project topic. Symmetry Considerations. If f(x) is either even or odd function defined on interval ( L, L), the value of f(x) on ( L, ) is the same as the value of f(x) on (, L). hus, n= dx. 6
L L L f(x) dx = L L f(x) dx. In this case, Parseval s heorem has the following form. L L f(x) dx = n= c n = a 4 + (a n + b n). Example 4. Using the Fourier series for f(x) = x for < x < from Example 3 and Parseval s heorem, find the sum of the series n=. n 4 Solutions. Recall that = 4, L = and that the function is even. In Example 3 we have find that a n = 6( )n for n >, a n π = 8. hus, Parseval s heorem applied to this function results in the 3 following x 4 dx = a 4 + Note that integral on the left side is Practice Problems. n= 5 = 9 + 8 π 4 n= (a n + b n) = 6 9 + 6 π 4 n= ( ) n +. 4 x4 dx = 6. Dividing the equation above by 6 produces 5 n= n 4 n= n 4 = π4 9.. Use the Fourier series you obtain in Example to find the sum of series n= ( ) n n +.. Note that the boxcar function from Example represents the odd extension of the function f(x) = for x < π. Consider the even extension of this function and find its Fourier cosine expansion. 3. Find the Fourier sine expansion of f(x) = series n= { x < x x < x < (n + ). Use it to find the sum of 4. he voltage in an electronic oscillator is represented as a sawtooth function f(t) = t for t that keeps repeating with the period of. Sketch this function and represent it using a complex Fourier series. hen use this Fourier series and Parseval s heorem to find the sum of the series n. 5. Find the Fourier series of f(x) = Solutions. n= { x x < + x < x < 7
. In Example, we obtain that f(x) = 4 sin(n+)x π n= Choosing x = π, we have that = 4 π n= where f(x) is the square wave function. n+ ( ) n ( ) n n+ n= = π. n+ 4 sin(n+) π n+ = 4 π n=. he even extension of f(x) is f(x) = for π < x < π. he periodic extension of this function is constant function equal to for every x value. Note that this is already in the form of a Fourier series with b n = for every n, a n = for n > and a =. Computing the Fourier coefficients would give you the same answer: a n = π cos nx dx = if n > and π a = π dx =. hus, f(x) = + π n= =. his answer should not be surprising since this function is already in the form of a Fourier series = + n= ( cos nx + sin nx). 3. Extend the function symmetrically about the origin so that it is odd. hus = 4 and L =. Since this extension is odd, a n =. Compute b n as b n = nπx f(x) sin dx = nπx x sin dx + nπx ( x) sin dx = ( ) ( ) x nπx cos + 4 sin nπx nπ n π + ( x) cos nπx 4 sin nπx nπ n π = cos nπ + 4 sin nπ + cos nπ + 4 sin nπ = 8 sin nπ. his is if n is even. If n = k +, nπ n π nπ n π n π 8( ) this is k. So, f(x) = 8 ( ) k (k+) π π k= sin (k+)πx. (k+) ( ) n n+ o find the sum of series n= and its Fourier sine expansion is equal to 8 8 π n= (n+) so 8 π n= (n+), note that when x = the function f() is equal to ( ) n π n= sin (n+)π = 8 ( ) n (n+) π n= ( ) n = (n+) = n= 4. =, f(t) = n= c ne nπit and c n = te nπit = t nπi e nπit + e nπit 4n π = nπi e nπi + e nπi. 4n π 4n π (n+) = π 8. Note that e nπi = cos( nπ) + i sin( nπ) =. hus c n = + = nπi c n = i = c nπ n. c = t dt =. his gives us f(t) = + n=,n a = c =, a n = for n > and b n = Parseval s heorem gives us that x dx = + 4 n= n= = π. n 6 nπ. i. Note that nπ i nπ enπit. Note also that = + n π 3 4 π n= 5. Graph the function and note it is even. hus, b n =. Since = and L = a n = ( x) cos(nπx) dx. Using the integration by parts with u = x and v = cos(nπx)dx = sin(nπx), obtain that a nπ n = ( x) nπ sin(nπx) + sin(nπx) dx = cos(nπx) nπ n π = (( ) n ). his last expression is if n is even and equal to = if n = k+ n π n π (k+) π is odd. Note that the formula (( ) n ) does not compute a n π because of the n in the denominator so calculate a from the formula a = x ( x) dx = x =. his gives you the Fourier series expansion f(x) = + 4 4 k= cos(k + )πx. (k+) π n 8
Fourier ransformation he Fourier transform is an integral operator meaning that it is defined via an integral and that it maps one function to the other. If you took a differential equations course, you may recall that the Laplace transform is another integral operator you may have encountered. he Fourier transform represents a generalization of the Fourier series. Recall that the Fourier series is f(t) = n= c ne nπit. he sequence c n can be regarded as a function of n and is called Fourier spectrum of f(t). We can think of c(n) being another representation of f(t), meaning that f(t) and c(n) are different representations of the same object. Indeed: given f(t) the coefficients c(n) can be computed and, conversely, given c(n), the Fourier series with coefficients c(n) defines a function f(t). We can plot c(n) as a function of n (and get a set of infinitely many equally spaced points). In this case we think of c as a function of n, the wave number. We can also think of c as a function of ω = nπ, the frequency. Note that if is large, then ω is small and the function c n becomes a continuous function of ω for. Note also that if we let, the requirement that f(t) is periodic can be waved since the period becomes (, ). he Fourier ransform F (ω) of f(t) is the limit of the continuous function π c n when. Since we have that π c n = / nπit π f(t)e dt = / / π / f(t)e iωt dt, c n F (ω) = π π f(t)e iωt dt Let us consider now the Fourier series expansion of f(t) with the substitution ω = nπ. wo consecutive n values are length apart so dn =. hus, dω = π π dω dn dω = and =. Hence π f(t) = n= c n e nπit = ω/(π)= ωit dω c n e π = π ω/(π)= c n e ωit dω π π F (ω)e ωit dω. hus, for a given function F (ω) of ω, the above expression computes a function f(t) of t. his formula computes the Inverse Fourier ransform f(t) for a given function F (ω). We can still think of f(t) and F (ω) being the same representations of the same object: F (ω) = π f(t)e iωt dt F (ω) is the Fourier transform of f(t) F (ω) the coefficients of the Fourier series for f(t) f(t) = π F (ω)eiωt dω f(t) is the inverse Fourier transform of F (ω) f(t) the expansion having coefficients F (ω) he significance of the constant π in the two formulas above is just in the fact that the substitution constant π in the numerator of the formula ω = nπ is evenly distributed in both formulas since 9
π = π π making the formulas for Fourier and Inverse Fourier transforms symmetric and certain calculations easier. o understand the significance of Fourier transform, it is helpful to think of f(t) as of a signal which is measured in time and is represented as a function of time t but which needs to be represented as a function of frequency, not time. In this case, Fourier transform produces representation of f(t) as a function F (ω) of frequency ω. he Fourier transform is not limited to functions of time and temporal frequencies. It can be used to analyze spatial frequencies. If the independent variable in f(t) stands for space instead of time, x is usually used instead of t. In this case, the independent variable of the inverse transform is denoted by k. t x and ω k Besides the fact that it generalizes Fourier series, the mathematical importance of the Fourier transform lies in the following:. If the initial function f(t) has the properties that are not desirable in a particular application (e.g. discontinuous, non smooth), we can consider the function F (ω) instead which is possible better behaved.. Fourier transform represents a function that is not necessarily periodic and that is defined on infinite interval. he only requirement for the Fourier transform to exist is that the integral f(t) dt is convergent. One of the most important applications of Fourier transform is in signal processing. We have pointed out that the Fourier transform presents the signal f(t), measured and expressed as functions of time, as a function F (ω) of frequency. he transform F (ω) is also known as the frequency spectrum of the signal. Moreover, Fourier transform provides information on the amplitude and phase of a source signal at various frequencies. he transform F (ω) of a signal f(t) can be written in polar coordinates as F = F e θi. he modulus F represents the amplitude of the signal at respective frequency ω, while θ (given by arctan( Im F/ Re F ) computes the phase shift at frequency ω. his gives rise to various applications in cryptography, acoustics, optics and other areas. In addition, Fourier transform can be used similarly to Laplace transform: in converts a differential equation into an ordinary (algebraic) equation that is easier to solve. After solving it, the solution of the original differential equation can be obtained by using the inverse Fourier transform. Example. Find the Fourier transform F (ω) of the boxcar function. Express your answer as real function. { < t < f (t) = otherwise Generalize your calculations to find the Fourier transform{ Fa A (ω) for the general boxcar function A a < t < a fa A (t) =. otherwise he boxcar function is said to be normalized if A = so that the total area under the function a is. We shall denote the normalized boxcar function that is non-zero on interval [ a, a] by f a.
Sometimes the values of f a at x = a and x = a are defined to be. For example, if 4a a =, the following graph represents f a. Consider how changes in value of a impact the shape and values of F a (ω). Solution. F (ω) = π f(t)e iωt dt. Since f(t) = for t < and t >, and f(t) = for t, we have that F (ω) = π e iωt dt = πiω e iωt = F (ω) = even and sin( ω) = sin ω since sine is odd function, the cosine terms cancel and we obtain that F (ω) = πiω ( i sin ω) = πω sin ω. πiω (e iω e iω ). Use the Euler s formula to obtain πiω (cos( ω) + i sin( ω) cos ω i sin ω). Using that cos( ω) = cos ω since cosine is Similarly, if a boxcar function of height A is nonzero on interval [ a, a], the Fourier transform is Fa A (ω) = a π a Ae iωt dt = A πiω (e iaω e iaω ) = πω sin aω. A e iaω e iaω πω = A i Function sin x x is known as sinc function. sinc(x) = sin x x Sometimes the normalized sinc function sin πx is used. Note that lim πx x sinc(x) = and lim x sinc(x) =. Using the sinc notation, we can represent the Fourier transform of the boxcar function f A a (t) as the sinc function F A a (ω) = aa π sinc(aω).
If the box function is normalized, A = so F a a(ω) = π sinc(aω). Since sinc()=, the peak of this function has value π. Let us compare the graphs of f a and F a for several different values of a. You can notice that larger values of a correspond to f a having smaller hight and being more spread out. In this case, F a has a narrow, sharp peak at ω = and converges to faster. If a is small, f a has large height and very narrow base. In this case, F a has very spread out peak around ω = and the convergence to is much slower. In the limiting cases a represents a constant function and a represents the Dirac delta function δ(t). his function, known also as the impulse function is used to represent phenomena of an impulsive nature. For example, voltages that act over a very short period of time. It is defined by δ(t) = lim a f a (t). Since the area under f a (t) is, the area under δ(t) is as well. hus, δ(t) is characterized by the following properties: () δ(t) = for all values of t () δ(t)dt =. Since no ordinary function satisfies both of these properties, δ is not a function in the usual sense of the word. It is an example of a generalized function or a distribution. Alternate definition of Dirac delta function can be found on wikipedia. We have seen that F a for a small a is very spread out and almost completely flat. So, in the limiting case when a =, it becomes a constant. hus, the Fourier transform of δ(t) can be obtained as limit of F a (ω) when a. We have seen that this is a constant function passing π. hus, the Fourier transform of the delta function δ(t) is the constant function F (ω) = π. Conversely, when a f a π. he Fourier transform of that is limit of F a for a. We have seen that this limit function is zero at all nonzero values of ω. So, we can relate this limit to δ(ω). Requiring the area under F a to be can explain why normalized sinc function is usually considered instead of regular sinc. Example. Find the inverse Fourier transforms of boxcar and Dirac delta functions. Solutions. he inverse transform of the general boxcar function can be computed by f(t) = π a a Aeiωt dω = A πit (e iat e iat ) o express this answer as a real function, use the Euler s formula and symmetries of sine and cosine functions similarly as in Example to obtain the following.
f(t) = A πit (cos at + i sin at cos( at) i sin( at)) = = aa π sin at A πit (i sin at) = A πt sin at = aa sin at πt a at boxcar function is the sinc function. his also illustrates that A πit (cos at + i sin at cos at + i sin at) = = aa π sinc(at). hus, the inverse transform of the the Fourier transform of the sinc function is the boxcar function. aking the limit when a in the normalized case, we obtain that the inverse Fourier of the delta function is the limit of π sinc(at) when a which is the constant function π. hus, the Fourier transform of the constant function f(t) = π is the delta function δ(ω). Example 3. Consider the Gaussian probability function (a.k.a. the bell curve ) f(t) = Ne at where N and a are constants. N determines the height of the peak and a determines how fast it decreases after reaching the peak. he Fourier transform of f(t) can be found to be F (ω) = N a e ω /4a. his is another Gaussian probability function. Examine how changes in a impact the graph of the transform. Solutions. If a is small, f is flattened. In this case the presence of a in the denominator of the exponent of F (ω) will cause the F to be sharply peaked and the presence of a in the denominator of N a, will cause F to have high peak value. If a is large, f is sharply peaked and F is flattened and with small peak value. All the previous examples are related to Heisenberg uncertainty principle and have applications in quantum mechanics. he following table summarize our current conclusions. Function in time domain Fourier ransform in frequency domain boxcar function sinc function delta function constant function Gaussian function sinc function boxcar function constant function delta function Gaussian function 3
Symmetry considerations. he table on the right displays the formulas of Fourier and inverse Fourier transform for even and odd functions. Fourier sine and cosine functions. Suppose that f(t) is defined just for t >. We can extend f(t) so that it is even. hen we get the formula for F (ω) by using the formulas for even function above. F (ω) is then called Fourier cosine transformation. f(t) even F (ω) = F (ω) even f(t) = f(t) odd F (ω) = F (ω) odd f(t) = π π π π f(t) cos ωt dt F (ω) cos ωt dω f(t) sin ωt dt F (ω) sin ωt dω Similarly, if we extend f(t) so that it is odd, we get the formula for F (ω) same as for an odd function above. F (ω) is then called Fourier sine transformation. Relation of Fourier transform and Fourier Series. We illustrate this relation in the following example. Consider the Fourier series of a boxcar function f a (t). Let s n denote the Fourier coefficient in the complex Fourier series. he value of a determines sampling period and spacing ω in frequency-domain. w = π and ω = nπ ω = nω. In the following figures, we consider the boxcar function with a = (so = and ω = π). he first graph on each figure displays the Fourier transform, sinc function, in the frequency-domain. he highlighted frequency on the first graph determines the value of n. he second graph displays the harmonic function corresponding to n-th term of the Fourier series of the function f(t) in the time-domain. he third graph is the sum of the first n terms of the Fourier series of f(t). Note that as n the sum of Fourier series terms converge to the boxcar function. he following figures are from the presentation on Fourier transform in Magnetic Resonance Imaging he Fourier ransform and its Applications by Branimir Vasilić. he author is grateful to Branimir for making the slides available. 4
5
Discrete versus Continuous Functions. In the previous examples, the function in frequency-domain was a continuous function. In applications, it is impossible to collect infinitely many infinitely dense samples. As a consequence, certain error may come from sampling just finite number of points taken over a finite interval. hus, it is relevant to keep in mind what effects in time (resp. frequency) domain may have finite and discrete samples of frequencies (resp. time). frequency-domain function in time-domain effects in time-domain infinite continuous set of frequencies non-periodic function none finite continuous set of frequencies non-periodic function Gibbs ringing, blurring infinite number of discrete frequencies periodic function aliasing finite number of discrete frequencies periodic function aliasing, Gibbs ringing, blurring 6
In the following consideration, assume that the sample consists of frequencies and that all the values lie on the graph of a sinc function. he goal is to reconstruct the boxcar function in the time-domain. he following four figures illustrate each of the four scenarios from the table. Application in Magnetic Resonance Imaging. he Fourier transform is prominently used in Magnetic Resonance Imaging (MRI). he fist figure represents a typical MRI scanner. he scanner samples spacial frequencies and creates the Fourier transform of the image we would like to obtain. he inverse Fourier transform converts the measured signal (input) into the reconstructed image (output). In the following two figures, the image on the right represents the amplitude of the scanned input signal. he image on the left represents the output - the image of a human wrist, more precisely, the density of protons in a human wrist. he whiter area on the image correspond to the fat rich regions. he darker area on the image correspond to the water rich regions. 7
Each point on the input images corresponds to certain frequency. wo smaller figures on the right side represent the components of the inverse transforms at two highlighted frequencies. he output image is created by combining many such images - one for each sampled frequency in fact. he last figure represent the output with high (first pair of images) and low frequencies (second pair of images) removed. If high frequencies are removed (low-pass filter), the image becomes blurred and only shows the rough shape of the object. If low frequencies are removed (high-pass filter), the image is sharp but intensity variations are lost. Practice Problems.. Find the Fourier transform of f (t) = e t, t >, f (t) = otherwise.. Find Fourier cosine transformation of the function from the previous problem. 3. Find cosine Fourier transform of f (t) = t 3 for < t < 3/, f (x) = otherwise. 8
4. If f a (t) = a a + t the graph of f has a peak at. Since f() =, the height is conversely proportional to a. he a Fourier transform of f can shown to be F a (ω) = π/e a ω. Graph F a (ω) for several values of a and make conclusion how values of a impact the graph of F a. 5. Solve the equation { ω ω f(t) cos tω dt = ω > Solutions.. F (ω) = π e t e iωt dt = π(+iω) e (+iω)t = π(+iω).. F (ω) = π F (ω) = ( π π ω ω 3. F (ω) = e t cos ωt dt. Using two integration by parts with u = e t, one obtains that e t sin ωt dt) = ( e t cos ωt π ω e t cos ωt dt) = ω π ( ω e t sin ωt + ω F (ω). Solving for F (ω) gives your F (ω)( + ) = ω get F (ω)(ω + ) = F (ω) =. π π ω + 3/ ( t 3 π π ω π ( + ω cos ωt 3/ ) = (t 3) cos ωt dt = π π sin ωt 3/ 3/ sin ωt dt) = ω (cos 3ω ) = ω ω (cos 3ω ). π. Multiply by ω to ω 4. If a is small, then f a has a larger peak. In this case F a is more spread out and flattened. If a is large, f a is spread out and the height of the peak is not large. In this case, F a has a sharp peak and converges to zero fast. 5. Use inverse cosine Fourier transform. First multiply with π to match the definition of the transform. hen the left side is exactly the Fourier cosine transform of f(t). So we can get f(t). π as the inverse transform of the left side of the equation multiplied by f(t) = ( ω) cos ωt dω = ( ω sin ωt π π π t + sin ωt dt) = ( cos ωt t π t ) = ( cos t + ) = ( cos t). πt πt 9