Fourier Transform and Convolution

Transcription

1 CHAPTER 3 Fourier Transform and Convolution 3.1 INTRODUCTION In this chapter, both the Fourier transform and Fourier series will be discussed. The properties of the Fourier transform will be presented and the concept of impulse function will be introduced. The definition of convolution and its relation with Fourier transform will be presented. Examples of some commonly used Fourier transforms are given and the results are presented in a table for quick reference. There are many good books on this subject. A few of the books are listed at the end of this chapter. Readers with a background in this area can skip this chapter. However, Examples 3.10 and 3.11 may be of interest, because the former one is related to the radar pulse train and the latter will be used in the Hilbert transform. 3.2 FOURIER TRANSFORM [1 4] Fourier series was introduced in 1807 by a French engineer, Jean Baptiste Joseph de Fourier ( ). He suggested that any arbitrary function defined over a finite interval by any piecewise graph, continuous or discontinuous, could be represented as an infinite sum of continuous functions such as sine and cosine. Although almost all the members of the French Academy questioned its validity, it turned out to be one of the most powerful tools in signal processing. The basic concept of the Fourier transform is that any function in the time domain can be represented by an infinite number of sinusoidal functions. The Fourier transform is defined as follows. A function x(t) in the time domain t has a corresponding function X( f ) in the frequency domain related by F [x(t)] X(f )= x(t) e j2 ft dt (3.1) 39

2 40 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS where j = ( 1) is the imaginary number and f represents frequency. The inverse transform is F 1 [X( f )] x(t) = X( f ) e j2 ft df (3.2) Since an exponential term can be decomposed into sine and cosine functions as (3.1) can be written as X( f )= e j = cos j sin (3.3) x(t) cos(2 ft) dt j x(t) sin(2 ft)dt (3.4) This equation can be used to find the properties of even and odd functions under Fourier transform. A simple example is given below to illustrate the applications of the above equations. Example 3.1 A simple example of Fourier transform is to find the Fourier transform of x(t) =A for T 2 < t < T 2 (3.5) = 0 elsewhere The function x(t) is often referred to as the rectangular function. This function is an even function of t (symmetric with respect to t). However, its product with a sine function is antisymmetric with respect to t, because the sine function is antisymmetric with respect to t. The integral of an antisymmetric function of t from t = to is zero, because the positive and negative parts cancel each other. Thus, the Fourier transform of x(t) is X( f )=A T/2 = e j2 ft dt = A T/2 T/2 T/2 AT sin ( ft) T f cos(2 ft) dt = A sin( ft) f = AT sinc( ft) (3.6) This formula is known as the sinc function, which is defined as

3 Fourier Transform and Convolution 41 Sinc(x) = sin(x) x (3.7) The functions x(t) and X( f ) are shown in Figure 3.1. In this figure A = 2 and T = 1. From (3.6) and (3.7), the first zero of X( f ) occurs at f =1/T, (or at sin( ) = 0). Thus the width of the mainlobe is 2/T. Figure 3.1 A rectangular function and its Fourier transform: (a) time domain, (b) frequency domain.

4 42 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS 3.3 IMPULSE FUNCTION [5 11] An impulse function is represented by the symbol (t). Sometimes, it is also referred to as the Dirac delta function. The output of a system with an impulse function as the input is called the impulse response of the system. The impulse function is only conceptual, (t) cannot be generated in practice. However, it is very useful in digitizing input signals; one can think of sampling as multiplication of the input function by a periodic train of impulses. An impulse function (t) can be considered as a very narrow rectangular function at t = 0, as shown in Figure 3.1(a) with A =1/T and T approaching zero. The area of this rectangular function is 1. The function (t t o ) is the impulse function at t = t o. Integrating the impulse function, the result is b a (t t o) dt = 1 for a t o b = 0 otherwise (3.8) If the integral includes the impulse function, the integral is unity; otherwise, the integral is zero. The impulse function can also be defined more generally by its sampling property as b a x(t) (t t o) dt = x(t o ) a t o b = 0 otherwise (3.9) which represents a specific value of x(t) att = t o. From the definition in (3.9), the Fourier transform of the impulse function can be written as X( f )= (t) e j2 ft dt = e 0 = 1 (3.10) In this derivation, (t) can be considered as (t 0). This result is shown in Figure 3.2. The inverse Fourier transform of the impulse function can be found either from (3.2) or (3.4), and the imaginary part of (3.4) is zero because it is an odd function (or antisymmetry). The result is (t) = e j2 ft df = e j2 ft j2 t = sin(2 ft) 2 t f f = lim f = cos(2 ft) +j sin(2 ft) j2 t 2 sin(2 ft) 2 t (3.11)

5 Fourier Transform and Convolution 43 Figure 3.2 An impulse function and its Fourier transform: (a) time domain, (b) frequency domain. One can consider this as another definition of the impulse function; that is, sin(2 ft) lim = (t) (3.12) f t A similar result can be obtained for a time-shifted impulse function as F [ (t t o )] = (t t o)e j2 ft dt = e j2 fto or (t t o )= e e j2 f(t to) j2 f(t to) df = j2 (t t o ) = sin[2 f(t t o)] f 2 (t t o ) f = lim f sin[2 f(t t o )] (t t o ) (3.13) In this equation, the impulse function (t t 0 ) can be considered as defined by (t t o ) = lim f sin[2 f(t t o )] (t t o ) (3.14) More examples of the impulse function will be illustrated in later sections. 3.4 PROPERTIES OF FOURIER TRANSFORM [5 12] Some of the properties of Fourier transform will be discussed in this section. If the relation is obvious, no proof will be provided. These properties are:

6 44 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS Linearity The Fourier transform is a linear operator, which means F [ x(t) + y(t)] = X( f )+ Y( f ) (3.15) Using the definition of Fourier transform, this property can be proven easily Even and Odd Functions This property can be obtained by observing (3.4). If x(t) is an even function, the second term on the right (the imaginary part) equals zero and the output is real as X( f )= x(t) cos(2 ft) dt (3.16) Since cos( ) = cos( ), X( f ) is symmetrical with respect to f =0. If x(t) is an odd function in (3.4), the first term on the right (the real part) equals zero and the output is imaginary as X( f )= j x(t) sin(2 ft) dt (3.17) Since sin( ) = sin( ), X( f ) is antisymmetric with respect to f = 0. Let us use cos(2 f o t) and sin(2 f o t) to demonstrate these properties. Example 3.2 Find the Fourier transforms of cos(2 f o t) and sin(2 f o t). The Fourier transform of cos(2 f o t) can be written by the definition of (3.1) as X( f )= cos(2 f ot)e j2 ft dt = 1 2 [e j2 ( f fo)t + e j2 ( f+fo)t ]dt (3.18) = 1 2 [ ( f f o)+ ( f + f o )] In obtaining the final result, the relation in (3.14) is used. The Fourier transform of the cosine function is two impulse functions at f = ±f o ; thus, they are real and symmetrical with respect to f = 0. These results are shown in Figure 3.3.

7 Fourier Transform and Convolution 45 Figure 3.3 Fourier transform of cosine wave: (a) time domain, (b) frequency domain. To perform the Fourier transform on the sine function, (3.1) is used and the result is X( f )= sin(2 f ot)e j2 ft dt = 1 2j [e j2 ( f fo)t e j2 ( f+fo)t ]dt (3.19) = j 2 [ ( f f o) ( f + f o )] The outputs are two imaginary impulse functions and they are antisymmetrical with respect to f = 0. These results are shown in Figure 3.4. These results agree with the conclusions drawn from (3.16) and (3.17) Duality Duality means that if the function in time domain is x(t) and the Fourier transform is X( f ), then X(t) has a Fourier transform of x( f ). The proof of this property can be accomplished from the inverse Fourier equation (3.2). The proof is very simple, it only requires three steps and the change of notations of two variables. In (3.2), one can treat both variables t and f as dummy variables. The first step is to replace f by t ; the result is

8 46 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS Figure 3.4 Fourier transform of sine wave: (a) time domain, (b) frequency domain. The second step is to replace t with f; the result is x(t) = X(t ) e j2 t t dt (3.20) x( f )= X(t ) e j2 ft dt (3.21) Since t is a dummy variable it can be written as t; thus, this equation can be written as x( f )= X(t)e j2 ft dt (3.22) which is the desired result. Let us use an example to demonstrate this property. Example 3.3 Find the Fourier transform of x(t) where x(t) =A sin( Ft) t (3.23)

9 Fourier Transform and Convolution 47 This is the result obtained in (3.6) with t replacing f and F replacing T. The direct approach to finding the Fourier transform is quite tedious. This procedure is shown as follows. Since this x(t) is an even function, (3.16) will be used and the imaginary part of X( f ) is equal to zero. The Fourier transform can be calculated as X( f )= A sin( Ft) t = A sin( F +2 f )t 2 t cos(2 ft)dt + sin( F 2 f )t t dt (3.24) This integral can be found in the integral table as [12] sin mx dx = x m > 0 = m < 0 (3.25) Equation (3.24) can be evaluated in three regions in terms of f. If F/2 < f < F/2, then F +2 f and F 2 f are positive. Both integrals result in, thus X( f )=A. Iff < F/2, then F +2 f is negative and F 2 f is positive. Thus, the first integral is and the second integral is, which results in X( f )=0.Iff > F/ 2, then F +2 f > 0, and F 2 f < 0. The first integral is and the second is ; thus, X( f ) = 0. If all the three regions of f are considered, the result is the desired result. X( f )=A for F 2 < f < F 2 (3.26) = 0 elsewhere The same result can be obtained from the duality property of Fourier transform in a much simpler way. Since x(t) matches the output X( f ) of Example 3.1, the result can be directly written as X( f )=A = X(f ) for = 0 elsewhere F 2 < f < F 2 (3.27) In this equation, the result of X( f )=A is not a function of f; thus, the Fourier transform X( f ) has the same form Scaling If in a function x(t), the time t is replaced by t/a, the Fourier transform can be found by replacing t/a with t, and the result is

10 48 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS F a x t = x a t e j2 ft dt = a x(t )e j2 aft dt (3.28) = ax(af ) This expression means that when time is divided by a factor a, the amplitude and the frequency of the Fourier are expanded by the factor a. Similarly, if the frequency is divided by a factor a, the inverse Fourier transform can be written as 1 F a X f = ax(at) (3.29) The proof of this equation is similar to the proof of (3.28) Time Shift When the time of the input signal x(t) is shifted by t 0 as x(t t 0 ), the Fourier transform can be obtained by replacing t t 0 with t. The result is F [x(t t 0 )] = x(t t 0) e j2 ft dt = x(t ) e j2 f(t +t0) dt = e j2 ft0 X( f ) (3.30) This means that a shift in the time domain causes a phase shift in the frequency domain and the amplitude of the output does not change. Example 3.4 Let us find the Fourier transform of x(t) = cos[2 f 0 (t t o )]. From (3.30), the result can be written directly as X( f )= cos[2 f 0(t t 0 )] e j2 ft dt = 1 2 e j2 ft0 [ ( f f 0 )+ ( f + f 0 )] = 1 2 {cos(2 ft 0)[ ( f f 0 )+ ( f + f 0 )] j sin(2 ft 0 )[ ( f f 0 )+ ( f + f 0 )]} (3.31) The output is no longer real, but becomes complex because the function x(t) is not symmetric with respect to t = 0. However, the real part is still symmetric and

11 Fourier Transform and Convolution 49 the imaginary part is antisymmetric. The amplitudes of the impulse functions are still 1/ Frequency Shift When there is a frequency shift f 0 in the Fourier output X( f )asx( f f 0 ), the effect on the input signal can be found by substituting f = f f 0 as F 1 [X( f f 0 )] = X( f f 0) e j2 ft df = X( f ) e j2 ( f +f0)t df = e j2 f0t x(t) (3.32) The corresponding effect is a frequency shift in the time domain, as expected Derivative The derivative of the input x(t) with respect to t can be represented by x (t). Its Fourier output can be found through integration by parts as F [x (t)] = F dx(t) dt = dx(t) e dt j2 ft dt = x(t)e j2 ft ( j2 f ) x(t) e j2 ft dt (3.33) = j2 fx( f ) In the above expression, it is assumed that x(t) is zero at t = ±. This assumption makes the first term equal to zero Integral If the time domain function (t) is (t) = t x( )d (3.34) and ( ) = = 0, the Fourier transform can be found as follows. This equation can be written as d (t) dt = x(t) (3.35)

12 50 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS From the relation in (3.33), the Fourier transform of d (t)/dt is F d (t) dt = j2 f ( f ) (3.36) This equation can be written as ( f )=F [ (t)] = F d (t) dt j2 f (3.37) Substituting the results of (3.34) and (3.35) into this equation, the result is F t x( )d = F [x(t)] j2 f (3.38) This is the desired result. 3.5 FOURIER SERIES [5 11] A periodic function x(t) can be represented by the sum of an infinite number of sinusoidal waves with discrete frequencies. The Fourier series can be written in several forms. One of these is x(t) =A 0 + n=1 A n cos 2 nt T + n=1 B n sin 2 nt T (3.39) where T is the period of x(t). A 0 is the average value of x(t), which can be found as The constants A n and B n can be found as A 0 = 1 x(t)dt (3.40) T A n = 2 T/2 2 nt x(t) cos T T/2 T dt (3.41) B n = 2 T/2 x(t) sin2 nt T T/2 T dt Another form of the Fourier series is combining the sine and cosine terms into exponential terms as

13 Fourier Transform and Convolution 51 x(t) = j2 nt T C n e (3.42) where C n can be found from C n = 1 T/2 j2 nt T x(t)e T dt (3.43) T/2 In the above equation, when T approaches infinity, one can think that f = n/t represents a continuous frequency and the amplitude of the integral is normalized to one. Then, the equation is actually the Fourier transform. In order to have a physical picture, one can imagine that a periodic function in the time domain can be represented by an infinite number of frequencies in the frequency domain and these frequencies are discrete. This is the Fourier series. If the function in the time domain is no longer a periodic function (or, in other words, the period is infinite), it can be represented by an infinite number of frequencies and these frequencies are continuous. 3.6 COMB FUNCTION [5 11] A comb function is defined as the sum of an infinite number of uniformly distributed impulse functions, and they are shaped like a comb. A comb function is shown in Figure 3.5. The function can be defined in either the time or frequency domain. This function can be used to represent digitized data. Mathematically, the comb function in the time domain can be written as Figure 3.5 Fourier transform of a comb function: (a) time domain, (b) frequency domain.

14 52 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS x(t) = comb(t) = (t nt) (3.44) where T is the separation between the delta functions. The Fourier transform of the comb function can be obtained from directly integrating the above equation. The following manipulation demonstrates the approach [5]: n Using the relation X( f )= (3.45) can be written as = n= n= (t nt)e j2 ft dt (t nt)e j2 ft dt = N N z n = z N+1 z N n= N z 1 e j2 fnt n= N X( f )= e j2 f(n+1)t e j2 fnt e j2 ft 1 = e j2 ft/2 [e j2 f(n+1/2)t e j2 f(n+1/2)t ] e j2 ft/2 [e j2 f1/2t e j2 f1/2t ] = sin 2 N ft sin( ft) N (3.45) (3.46) (3.47) In this function, f is the variable. The result of this equation is shown in Figure 3.6. In this figure, N = 7 and f ranges from 3.9 to 3.9. Where f = n/t, whenever n is an integer, there is a peak. This figure has the rough shape of a comb. The last step is to prove that when N, the result of (3.47) approaches a delta function. The result of the above function can be written as sin[2 (N +1)fT] = sin( ft) N sin 2 (N +1) f n T T f n T f n T sin( ft) N (3.48) Using the result of (3.13), one of the definitions of delta function, the first part of the right side of the above equation is

15 Fourier Transform and Convolution 53 Figure 3.6 Plot of sin[2 (N +1)fT]/sin( ft). lim N sin 2 (N +1) f n T T f n T = f n T (3.49) The second part of the right-side equation can be calculated from the L Hospital s rule as f n T = sin( ft) f = n T d(f n/t) df d[sin( ft)] df f= n T (3.50) Thus, any peak can be written as 1 = T cos( ft) f = n T = 1 T

16 54 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS sin[(n + 1)2 ft] sin( ft) = f n T T (3.51) Thus, the Fourier transform of the comb function is another comb function with amplitude of 1/T and separation of 1/T. This result can be mathematically written as X( f )=F n= (t nt) = 1 T f n (3.52) n= T A different approach to obtain the Fourier transform of the comb function is through a Fourier series. The comb function is a period function with a period of T; therefore, it can be represented by a Fourier series as shown in (3.39). The constants can be obtained as A 0 = 1 T/2 T (t) dt = 1 T/2 T A n = 2 T T/2 (t) cos2 nt T/2 T dt = 2 T (3.53) B n = 2 T T/2 T/2 (t) sin2 ntdt =0 T It should be noted that (t) exists only inside the range of one period (T/2 t T/2) and all the other delta functions are outside this period. Substituting these constants into (3.44), the result is x(t) = 1 T + 2 T cos 2 nt n=1 T (3.54) This equation is still in the time domain, but in a different form. Taking the Fourier transform of this function, the first term can be obtained from the symmetry property of the Fourier transform. The result is ( f )/T, which represents a delta function at f = 0. The Fourier transform of the rest of the terms can be obtained by using the result of (3.18) with f o = n/t. Thus, the result is X( f )=F [comb(t)] = 1 T n= f n T (3.55) This is the same result as obtained from (3.52). This result is shown in Figure 3.5. This result is very important for processing radar signals because a radar signal with stable pulse repetition frequency (PRF) can be represented by a comb function.

17 Fourier Transform and Convolution CONVOLUTION [5 11] If an input signal is applied to a linear and time-invariant system, the output can be found through two approaches. The time domain approach is through convolution and the frequency domain approach is through Fourier transform. The results obtained from these two approaches are identical. Let us assume that the input signal to the system is a delta function (t) and the corresponding output is h(t), where h(t) is called the impulse response of the system. Because the system is time-invariant, an input (t ) will produce an output of h(t ). This means that when the input is delayed time, the output is also delayed time, and the system response does not change with time. An input function x(t) can be imagined as consisting of many rectangular functions, each with the same infinitesimal width but different amplitude and delay time. Thus, the input signal x(t) and the output y(t) can be written as x(t) = lim i x(t i ) i 0 i y(t) = lim i x(t i )h( i ) = lim i x( i )h(t i ) (3.56) i 0 i i 0 i = x(t )h( )d = x( )h(t )d x(t) h(t) where i is the width of the rectangular function, t is time, and can be considered as a dummy variable. After the integration, the variable disappears and the output is a function of t as expected. The notation in the equation is the conventional way to represent convolution. The term h(t ) means inverting the h(t) inthe time domain and delay by time. From the above equation, one can see that either the input signal or the impulse response can be inverted in time. The convolution is graphically illustrated in Figure 3.7. Figures 3.7(a) and 3.7(b) show the function of x( ) and h( ), respectively. Figure 3.7(c) shows h( ) reversed in time and represented by h( ). Figure 3.7(d) shows h( ) is shifted by time t and represented by h(t ). Figure 3.7(e) represents the result of x( )h(t ) and Figure 3.7(f) represents the result obtained from the convolution. The point a on the curve (or y(t) =a) of Figure 3.7(f) represents the total area shown in Figure 3.7(e). If x( ) is reversed instead of h( ), the same result will be obtained. Another important example is to find the convolution of an impulse function with function x(t). If system response is an impulse function (i.e., h(t) = (t t o )), the output is y(t) = h( )x(t )d = ( t o)x(t ) d = x(t t o ) (3.57)

18 56 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS Figure 3.7 A graphic display of convolution of x(t) and h( ): (a) x( ), (b) h( ), (c) h( ), (d) h(t ), (e) x( )h(t ), (f) y(t). The operation shifts the function x(t) tox(t t o ). Figure 3.8 shows the result graphically. Now we try to solve the output y(t) from the frequency domain approach. The input and impulse response of the system are still x(t) and h(t), respectively. This operation can be found from taking the Fourier transform of y(t). The result is Y( f )= x( )h(t ) d e j2 ft dt (3.58)

19 Fourier Transform and Convolution 57 Figure 3.8 Convolution of (t t o ) with x(t): (a) (t t o ), (b) x(t), (c) y(t). Changing the order of integration, this equation can be written as Y(f )= x( ) h(t )e j2 ft dt d (3.59) Changing the variable by letting t = u, then Y( f) = x( ) h(u)e j2 fu du e j2 f d = x( )H(f )e j2 f d (3.60) = H(f ) x( )e j2 f d = H(f )X( f ) In order to find y(t), an inverse Fourier transform on Y( f ) is required. This relation can also be written as

20 58 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS x(t) h(t) =F 1 [X( f )H( f )] (3.61) From the above discussion, the frequency domain solution can be accomplished as follows. Find the Fourier transforms of both x(t) and h(t) asx(f ) and H( f ). Multiply X( f ) and H( f ) to obtain Y( f ) and find the inverse Fourier transform of Y( f ) to obtain y(t). The above derived relation also implies that the convolution in time domain is equivalent to multiplication in the frequency domain. This relation is often expressed as x(t) h(t) X( f )H( f ) (3.62) and is referred to as the convolution theorem. One can take the convolution in the frequency domain in the same manner as in the time domain. This is shown as X( f ) H( f )= X( )H( f )d (3.63) where is a dummy variable. The inverse Fourier transform of this result is F 1 [X( f ) H( f )] = X( )H( f )d e j2 ft df = X( ) H(u)e j2 ut du e j2 t d = h(t)x(t) (3.64) The proof is identical to (3.60). The above relation can be expressed as X( f ) H( f ) x(t)h(t) (3.65) Equations (3.62) and (3.65) are referred to as the duality of convolution and Fourier transform. 3.8 PARSEVAL S THEOREM [5 8] The Parseval s theorem states that for a given signal, the total energy in the time domain equals the total energy in the frequency domain. This relation can be written as x(t) 2 dt = X( f ) 2 df (3.66)

21 Fourier Transform and Convolution 59 This relation can be proven as follows. From (3.64) or (3.65), the result can be written as F [h(t)x(t)] = X( f ) H( f ) h(t)x(t)e j2 t dt = H( f )X( f )df (3.67) In this equation both sides are functions of, a dummy variable. If = 0, this equation becomes h(t)x(t)dt = H( f )X(f )df (3.68) Let us assume that h(t) = x*(t) where * represents a complex conjugate; then H( f )= h(t)e j2 ft dt = x*(t)e j2 ft dt = x(t)e j2 ft dt * = X*( f ) (3.69) Substituting this relation into (3.68), the result of (3.66) can be obtained. This completes the proof of the Parseval s identity. 3.9 EXAMPLES [3, 5, 8, 12, 13] This section will present some examples that are often encountered in signal processing and wideband receivers. Most of the problems are not solved directly, but through the duality of convolution and Fourier transform. Example 3.5 Find the Fourier of e j2 fot. This result can be obtained directly from the Fourier transform or from the following relation since the Fourier transform of sine and cosine are already known. e j2 fot = cos(2 f o t)+j sin(2 f o t) (3.70) Using the linearity property of (3.15) and combining with the results of (3.18) and (3.19), the result is F (e j2 fot )= 1 2 [ ( f f o)+ ( f + f o )+ ( f f o ) ( f + f o )] = ( f f o ) (3.71)

22 60 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS which is a single-sided response. This result is important in discussing I and Q channels of frequency downconverters. This result is shown in Figure 3.9. It should be noted that Figure 3.9(a) is not in the time domain but is in the complex plane, and the time is represented by the rotating vector. Example 3.6 Find the Fourier transform of a rectangular windowed cosine wave. This waveform is often used to represent the output of pulsed radar. If the window is from T/2 to T/2 with amplitude A, the windowed cosine wave can be generated by multiplying the window function and a cosine wave as x 1 (t) = cos(2 f o t) w(t) =A for T 2 < t < T 2 (3.72) = 0 elsewhere x(t) =w(t)x 1 (t) where f 0 is the frequency of the input pulse. Since multiplication in time domain is equivalent to convolution in the frequency domain, the Fourier transforms of the rectangular window and cosine wave are needed. The Fourier transform of the window, from Example 3.1, is W( f )=A sin( ft)/ f. The Fourier transform of a cosine wave, from Example 3.2, is X 1 ( f )=[ ( f f o )+ ( f f o )]/2. The Fourier transform of x(t) in the above equation is Figure 3.9 Fourier transform of e j2 f0t : (a) complex plane, (b) frequency domain.

23 Fourier Transform and Convolution 61 X( f )=W( f ) X 1 ( f )= A sin( T) = A 2 sin[2 ( f f o)t] ( f f o ) + sin[2 ( f + f o)t] ( f + f o ) [ ( f f o ) + ( f + f o )] d 2 (3.73) The results are shown in Figure In this figure, A =1,T =1,andf 0 = 10. Since the Fourier transform of a cosine function has two delta functions, the convolution with them will generate two outputs. One shifts by f o, the other one shifts by f o. Example 3.7 Find the Fourier transform of an isosceles triangle. Instead of finding the transform directly, the following approach is used. An isosceles triangle with height (AT) 2 and width 2T (from T to T) can be generated by convolving two identical rectangular functions of height A and width T (from T/2 to T/2) as shown in Figure The Fourier transform of such a rectangular function of unit height is given by (3.6) as X 1 ( f )=A sin( ft)/ f. Since convolving in the time domain is equivalent Figure 3.10 Fourier transform of a rectangular windowed cosine wave.

24 62 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS Figure 3.11 Convolving two rectangular functions to generate an isosceles triangle and its Fourier output: (a) two rectangular functions, (b) isosceles triangle, (c) Fourier transform output. to multiplication in the frequency domain, the Fourier output of the isosceles triangle is the square of X 1 ( f ). Mathematically, the equations can be written as x(t) =A 2 T t 2 1 for t <T T The Fourier output is shown in Figure 3.11(c). = 0 elsewhere (3.74) X( f )= A2 sin 2 ( ft) ( f ) 2 Example 3.8 Find the Fourier transform of x(t) =a +(1 a) cos 2 t T for T 2 < t < T 2 = 0 elsewhere (3.75)

25 Fourier Transform and Convolution 63 where a < 1. The x(t) is often referred to as a generalized cosine window function. Different window functions are often used in signal processing. When a = 0.54, x(t) is called the Hamming window. The limited response in time domain ( T/2 < t < T/2) can be achieved by multiplying a +(1 a)cos(2 t/t) by a rectangular function. The duality of the convolution theorem can be used to obtain the desired transform. Let x 1 (t) =a +(1 a)cos(2 t/t) and x 2 (t) represent the rectangular function; their Fourier transforms are X 1 ( f )=a (t) + 1 a 2 [ ( f f o)+ ( f + f o )] where f o = 1 T (3.76) X 2 (f )= sin( ft) f The Fourier transform of x(t) equals the convolution of X 1 ( f ) and X 2 ( f )as X( f )=X 2 ( f ) X 1 ( f )= sin( ft) f = a sin( ft) f + 1 a 2 sin[ ( f f o)t] ( f f o ) {a ( f )+ 1 a 2 [ ( f f o)+ ( f + f o )]} + sin[ ( f + f o)t ] ( f + f o ) (3.77) The time and frequency domains are shown in Figure In this figure, a = 0.5 and T = 5; thus, 2.5 < t < 2.5 and f is from 10/T < f < 10T. Example 3.9 Find the Fourier transform of the following function: x(t) = 1 2 e t (3.78) This is a Gaussian (or normal) function of t. The solution of this equation can be obtained from direct integration. The result is X( f )= = 1 2 e e t e j2 ft dt t cos(2 ft)dt = e f 2 (3.79)

26 64 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS Figure 3.12 Fourier of a general cosine window with a = 0.5: (a) time domain, (b) frequency domain. This integral can be found from [3] (861.20). This result shows that the Fourier transform of a Gaussian function in time domain is a Gaussian function in the frequency domain also. These results are shown in Figure In this figure =.8. Example 3.10 Find the Fourier transform of a radio frequency (RF) pulse train. This problem can be solved directly, as shown in [13]. The approach provided here is a step-bystep combination of desired signals to provide the pulse train. At each step the duality theorem of convolution is used. There are four steps to generate the pulse train in the time domain. In each step the time domain operations and the corresponding frequency domain operations will be represented. The last step produces the desired spectrum distribution. To keep this discussion simple, let us assume the pulse train has unit amplitude. In order to keep this approach clear, the

27 Fourier Transform and Convolution 65 Figure 3.12 (continued). selected functions are represented by c(t), s i (t), and r(t); the resulting functions are represented by x i (t) where i is an integer. 1. Choose a comb function c(t) with period T as c(t) = n= C(f )= 1 T n= (t nt ) f n T (3.80) where T represents the pulse repetition interval (PRI) of the pulse train and n is an integer. 2. Choose a rectangular function s 1 (t) with width and unit amplitude to represent an individual pulse. The time and frequency domains are

28 66 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS Figure 3.13 A Gaussian function and its Fourier transform: (a) time domain, (b) frequency domain. s 1 (t) =1 for t < 2 = 0 elsewhere (3.81) S 1 ( f )= sin( f ) f In order to create a pulse train, x 1 (t), s 1 (t) can be used to convolve with the comb function c(t); thus x 1 (t) =c(t) s 1 (t) X 1 ( f )=C(f ) S 1 (f )= sin( f ) f f n n= T (3.82) In the above operation, f = n/t is substituted into S 1 (f ) because the delta functions in C( f ) only exist at these values. 3. Choose a second rectangular function s 2 (t) with width NT and unit amplitude to represent the total length of the pulse train; thus

29 Fourier Transform and Convolution 67 Figure 3.13 (continued). s 2 (t) =1 for t < NT 2 = 0 elsewhere (3.83) S 2 (f )= sin( fnt) f The function s 2 (t) is multiplied by x 1 (t) to create the pulse train, which is represented by x 2 (t). The frequency domain result is obtained by convolving X 1 ( f ) with S 2 ( f ) as x 2 (t) =x 1 (t)s 2 (t) x 2 ( f )=X 1 ( f ) S 2 ( f ) (3.84) = sin( f ) 2 f n= sin f n T NT f n T

30 68 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS It should be noted that convolving with a delta function shifts the function by n/t. 4. The last step is to add the RF in the pulse train. Let us choose r(t) to represent a cosine wave; then r(t) = cos(2 f o t) R( f )= 1 2 [ (f f o)+ ( f + f o )] (3.85) where f 0 is the RF of the pulse. The function x 2 (t) is multiplied by r(t) to produce the RF pulse train. The final results are x(t) =x 2 (t)r(t) X( f )=X 2 ( f ) R( f ) (3.86) = sin( f ) 2 f n= sin f n fo T sin NT f + f o T n NT f f o n + f + f T o n T The time domain signal generated through the four steps of operations is shown in Figure 3.14(a). It is a series of RF pulses. This signal in radar is called coherent because there is a certain phase relation between the individual pulses. In other words, a continuous sinusoidal wave that is gated by a pulse train generates a coherent pulse train because every pulsed RF signal is part of the same sinusoidal wave. Now let us discuss the result of (3.86). The first term represents the envelope of the spectrum. The first zero of the envelope can be found from f = 1 (3.87) The width of the mainlobe from null to null is 2/, and this result is shown in Figure 3.14(b). The second and third terms represent upper and lower band spectra. They are located at ±f o and are symmetrical with respect to zero frequency. The upper group of spectrum lines occur only at f = f o + n T (3.88) The spectrum lines are separated by 1/T. The fine structure of the individual line can be found from the second term in (3.86) as

31 Fourier Transform and Convolution 69 Figure 3.14 Fourier transform of an RF pulse train: (a) RF pulse train, (b) spectrum lines, (c) details of individual lines.

32 70 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS X( f )=K sin f f o n T NT f f 0 n T NT (3.89) where K is a constant. The first zero occurs at NT = 1; thus, the width of the individual spectrum line is 2/NT. This result is shown in Figure 3.14(c). Some radars measure the velocity of a target through the Doppler frequency shift f d. In a radar system, this quantity can be written as [13] f d = 2f 0V C (3.90) where f 0 is the operating frequency, V is the velocity of the target, and C is the speed of light. The factor 2 in this equation is caused by the round trip of the radar signal. This type of radar is referred to as a pulse Doppler radar. In order to reduce the ambiguity in frequency measurement, the separation between frequency lines must be wide. This requires high PRF or low pulse repetition interval (PRI), which means short T. In order to have fine spectrum line width, NT must be large or N must be large. In other words, a large number of pulses must be integrated. Therefore, pulse Doppler radar can generate very high pulse density. Example 3.11 Find the inverse Fourier transform for a quadrature phase shift. Before starting this problem, let us define a sign function as sgn( f )= 1 f > 0 1 f < 0 (3.91) When f > 0, the function is a positive one and when f < 0, the function is a negative one. The quadrature phase shift is defined in terms of the sign function as X( f )= j sgn( f ) (3.92) The phase relation is shown in Figure 3.15(b), which is referred to as a step function. The approach to find the inverse transform of this function is to use the derivative of X( f ). The derivative can be obtained through a similar way as shown in (3.33), and the result is dx( f ) F 1 df = j2 tx(t) (3.93)

33 Fourier Transform and Convolution 71 The next step is to obtain the derivative of X( f ). The derivative of a step function is a delta function at the transition position multiplied by a constant. The constant depends on the amplitude change of the step function. In this special case, the amplitude change is 2j (from +j to j). Thus, the derivative is j2 ( f ). Substituting this relation in the above equation, the result is x(t) = 1 dx( f ) F j2 t 1 df = j 2 t [ j2 ( f )]e j2 ft df = j 2 t ( j2) = 1 t (3.94) This equation has a singularity at t = 0. This result will be used in the Hilbert transform in Chapter 8. The time and frequency domain plots are shown in Figure SUMMARY This section will list some of the results obtained previously as a quick reference. The following are some of the properties of the Fourier transform. 1. Linearity 2. For even functions x(t) 3. For odd function x(t) 4. Duality F [ x(t) + y(t)] = X( f )+ Y( f ) (3.15) X( f )= x(t) cos(2 ft)dt (3.16) X( f )= j x(t) sin(2 ft)dt (3.17) If X( f )=F [x(t)] then x( f )=F [X(t)] (3.22) 5. Scaling F x t a = ax(af ) (3.28)

34 72 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS Figure 3.15 A quadrature phase shift: (a) time domain, (b) frequency domain.

35 F 1 X f a Fourier Transform and Convolution 73 = ax(at) (3.29) 6. Time shifting F [x(t t o )] = e j2 fto X( f ) (3.30) 7. Frequency shifting F 1 [X( f f o )] = e j2 fot x(t) (3.32) 8. Derivative F [x (t)] = j2 fx( f ) x(t) t=± = 0 (3.33) 9. Integral F t x( )d = F [x(t)] j2 f t x( )d t= = 0 (3.38) 10. Convolution The properties of impulse function are x(t) y(t) X( f ) Y( f ) (3.62) x(t)y(t) X( f ) Y( f ) (3.65) (t) dt = 1 (3.8) x(t) (t t o)dt = x(t o ) (3.9) (t) = lim f (t t o ) = lim f sin(2 ft) t sin[2 f(t t o )] (t t o ) (3.11) (3.13) Next, some of the commonly used Fourier transforms are listed below.

36 74 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS 1. Rectangular function X(t) =A for T 2 < t < T X(f )=A sin ( ft) (3.5) 2 f = 0 elsewhere = AT sinc( ft) (3.6) (See Figure 3.1.) 2. Impulse function x(t) = (t) X( f ) = 1 (3.10) (See Figure 3.2.) 3. Cosine function x(t) = cos(2 f o t) X( f )= 1 2 [ ( f f o)+ ( f + f o )] (3.18) (See Figure 3.3.) 4. Sine function x(t) = sin(2 f o t) X( f )= j 2 [ ( f f o) ( f + f o )] (3.19) (See Figure 3.4.) 5. Comb function x(t) = (t nt) (3.44) X( f )= 1 T f n T (3.55) (See Figure 3.5.)

37 Fourier Transform and Convolution x(t) =e j2 fot (See Figure 3.9.) 7. Windowed cosine function x(t) =e j2 fot (3.70) x( f )= ( f f o ) (3.71) x(t) =A cos(2 f o t) X( f )= A 2 sin[2 ( f f o)t ] ( f f o ) for T 2 < t < T 2 = 0 elsewhere + sin[2 (f + f o)t] ( f + f o ) (3.73) (See Figure 3.10.) 8. For an isosceles triangle t x(t) =A 2 T 2 1 for t <T T X( f sin 2 ( ft) )=A2 ( f) 2 = 0 elsewhere = [AT sinc( ft)] 2 (3.74) (See Figure 3.11(b,c).) 9. Generalized cosine window function x(t) =a +(1 a) cos 2 t T X( f )= a sin( ft) f for T 2 < t < T + 1 a 2 2 sin[ ( f f o)t] ( f f o ) = 0 elsewhere + sin[ (f + f o)t] (f + f o ) (3.75) (3.77) (See Figure 3.12.) 10. Gaussian function (See Figure 3.13.) t 1 x(t) = 2 e 2 2 (3.78) X( f )=e 2 2 2f 2 (3.79)

38 76 DIGITAL TECHNIQUES FOR WIDEBAND RECEIVERS 11. Coherent RF pulse train X( f )= sin( f ) 2 f x(t) = n= (t nt) s 1(t) s 2(t) cos(2 f o t) where s 1 (t) =1 for t < 2 s 2 (t) =1 for t < NT 2 n= sin f n f0 T NT f f 0 n + T sin f + f o n T NT f + f 0 n T (3.86) (See Figure 3.14(a,b).) 12. Quadrature phase shift x(t) = 1 t x(f )= jsgn( f ) (3.92) (See Figure 3.15.) REFERENCES [1] Campbell, G. A., and Foster, R. M. Fourier Integrals for Practical Applications, New York, NY: Van Nostrand Reinhold, [2] Erdilyi, A. Table of Integral Transforms, Vol. 1, New York, NY: McGraw Hill Book Co., [3] Dwight H. B. Tables of Integrals and Other Mathematical Data, 4th Edition, New York, NY: MacMillan Co., [4] Robinson, E. A. A Historical Perspective of Spectrum Estimation, IEEE Proc., Vol. 70, Sept. 1982, pp [5] Papoulis, A. The Fourier Integral and its Applications, New York, NY: McGraw Hill Book Co., [6] Papoulis, A. Probability, Random Variables, and Stochastic Processes, New York, NY: McGraw Hill Book Co., [7] Bracewell, R. The Fourier Transform and its Applications, New York, NY: McGraw Hill Book Co., [8] Brigham, E. O. The Fast Fourier Transform, Englewood Cliffs, NJ: Prentice-Hall, [9] Carlson, A. B. Communication Systems, New York, NY: McGraw Hill Book Co., [10] Stremler, F. G. Introduction to Communication Systems, 3rd Edition, Reading, MA: Addison-Wesley Publishing Co [11] Ziemer, R. E., and Tranter, W. H. Principles of Communications: System, Modulation, and Noise, 3rd Edition, Boston, MA: Houghton Mifflin Publishing Co., [12] Stimson, G. W. Introduction to Airborne Radar, El Segundo, CA: Hughes Aircraft Co., [13] Skolnik, M. I. Introduction to Radar Systems, New York, NY: McGraw Hill Book Co., 1962.