Notes on Orthogonal Frequency Division Multiplexing (OFDM)

Notes on Orthogonal Frequency Division Multiplexing (OFDM). Discrete Fourier ransfor As a reinder, the analytic fors of Fourier and inverse Fourier transfors are X f x t t, f dt x t exp j2 ft dt (.) where we have assued that Fourier transfor direction is fro tie axis (doain) to frequency axis (doain). In this sense X f is the frequency signal or rather the frequency spectru of xt is our tie signal, while xt. Finally t, f exp j2 ft is the operator that enables the act of transforation fro tie axis to frequency axis. he inverse Fourier transfor will be given by x t X f f, t df X f exp j2 ft df (.2) (.2) allows the return to the tie axis fro the frequency axis, where the operator is now * f, t t, f exp j2 ft, i.e. the conjugate of the operator in (.). Sybolically (.) and (.2) can be represented as below X f F x t, x t F X f (.3) Soe coents are in order regarding Fourier transfor and its inverse In general Fourier transfor is applicable to transforations between any axes and doains, for instance to go fro spatial axis to spatial frequency axis, again Fourier transfor and its inverse can be used. In these lecture notes, our axes will continue to be tie and frequency as given in (.) and (.2). Fourier transfor can be ultidiensional. It eans that the single integrals in (.) and (.2) can be extended to double or triple or N nuber of integrals. As shown by (.) and (.2), to include all behaviours along one axis, the integral liits range fro to. In general these liits will be finite. f, t t, f. hus no scaling is he transforation operators have the property that applied during transforations. his way, we can expect that the energies or the powers (of the sae signal) in the two doains to be exactly equivalent. We now introduce the discrete Fourier transfor by associating the tie and frequency functions, variables and the operators as follows,,, t f n k f t k n t n f k x t x n X f X k,,,,, (.4) Notes OFDM HE March 23 Sayfa

n and k will be integer variables, liited to the range n N, k N (.5) herefore N ay be regarded to represent the infinity of the discrete world. Bearing in ind that integration of the analytic world will correspond to suation in discrete doain, the discrete Fourier transfor (DF) of x n into X k will be N X k x n exp j2 kn / N (.6) n / N in the arguent of the exponential operator in (.6) acts as noralization factor. he inverse discrete Fourier transfor (IDF) of X k into x n will be hen sybolically, (.6) and (.7) can be cobined as N xn X k exp j2 kn / N (.7) N n DF X k (.8) x n IDF In analytic world, (.) and (.2) work without any probles. When switching to discrete case, care ust be taken, because there are three ain points to observe there a) Infinity of analytic world ust suitably be converted to the appropriate lengths in discrete world. b) Grid spacing of the two doains of the discrete world ust be taken into account. In analytic case this is properly covered since dt and df correspond to the sae infinitesially sall intervals, thus creating no scaling probles. c) he rules of sapling theore should be properly adhered to. his is because n and k are effectively the sapling rates in tie and frequency doains. Hence the sallest tie resolution that can be achieved is 2n and correspondingly the sallest frequency resolution that we can attain is 2k. Exaple. : Let xn be a delta (ipulse) function such that x n n n N (.9) Find X k for this discrete tie delta function. Solution : Graphically, (.9) will be as shown in Fig... Using (.6), we get Notes OFDM HE March 23 Sayfa 2

N exp 2 / X k x n j kn N n exp exp 2 / 2 exp 4 / x n x n j k N x n j k N x n N exp j k (.) So the result is X k which agrees with the analytic result of x ( n ) (.) X f F x t t n n = N - Fig.. he graph of discrete (tie) delta function given by (.9). Fig..2 displays the graph of X k. X ( k ) k k = N - Fig.2 Discrete Fourier transfor of discrete tie delta function of (.9). Exaple.2 : Let xn be tie shifted a delta function such that Notes OFDM HE March 23 Sayfa 3

x n n elsewhere (.2) Find Xk for this discrete tie delta function. Solution.2 : Graphically (.2) will be as shown in Fig..3. Again using the general forulation given in (.6), we get N exp 2 / X k x n j kn N n exp exp 2 / 2 exp 4 / x x j k N x j k N x exp j2 k / N x exp j2 k / N x exp j2 k / N exp exp 2 / x N j k j k N (.3) x ( n ) n n = n = N - Fig..3 he graph of discrete tie shifted delta function given by (.2). Again the finding of (.3) agrees with the analytic result of X f F x t t t exp j2 ft (.4) Since (.3) is coplex, it can be plotted in two parts as split below he plot associated with (.5) is given in Fig..4. k X k X k exp j2 k / N (.5) 2 k / N Notes OFDM HE March 23 Sayfa 4

X ( k ) k = N - k k = N - k - 2 ( N - ) / N ( k ) Fig..4 Discrete Fourier transfor of tie shifted delta function (fro (.5)). Exaple.3 : Now take a single exponential at a specific frequency k, thus we ay write So we try to find the discrete Fourier transfor of Solution.3 : Again by using (.6), we get xn as given in (.6). N N exp 2 / exp 2 / exp 2 / X k x n j kn N j n N j kn N n n N n By using the identity N n exp j 2 n k / N (.7) N if k pn, p integer exp j 2 n k / N (.8) if k, integer, k integer Xk given by (.7) becoes x n exp j2 n / N (.6) X k N if k otherwise (.9) (.9) eans a delta function at k. he corresponding relation of the analytic world is Fig..5 displays the graph of (.9). X f F x t exp j2 f t f f (.2) Notes OFDM HE March 23 Sayfa 5

X ( k ) N k k = k = N - Fig..5 Discrete Fourier transfor of a single exponential as given by (.6). Exercise. : Prove the identity in (.8). Note that if and k are liited to a single N range, so that N and k N, then (.8) can be converted into N n N if k, N exp j 2 n k / N (.2) if k, k N Exercise.2 : Find the discrete Fourier ransfor of the su of two exponentials as given below. x n Aexp j2 n / N A exp j2 n / N (.22) 2 2 Check that the discrete Fourier ransfor that you have found for (.22) resebles the corresponding analytic expression. 2. Basis for Orthogonal Frequency Division Multiplexing (OFDM) Conventionally we place a essage signal on a single carrier. For instance, to have double side band AM type of odulation, if stis the essage signal, we siply ultiply by a sinusoidal at the frequency of f to obtain the odulated signal as c y t s t exp j2 f t or y t s t cos 2 f t or y t s t sin 2 f t (2.) Due to the frequency spectru of st, i.e., S c c c f occupying a finite bandwidth, we have to take into account, the frequency response of the channel for the transitted signal to arrive at the receiver without distortions. Suppose that S f has a bandwidth of 2W, then Y f F yt 2W each centered around has the two sided spectru of f f as shown in Fig. 2.. Assuing that the counication channel c that our odulated signal is going to pass through has the following frequency response Notes OFDM HE March 23 Sayfa 6

C f or any constant if f f W c arbitrary elsewhere (2.) 2 W C ( f ) 2 W f f = - f c f = f c Y ( f ) Fig. 2. A counication channel that introduces no distortion. then confined to the frequency range of f f W at the output of the channel, we will have c r t F C f Y f F Y f y t (2.2) he iplication of (2.2) is that we have been able to receive the exact copy of the transitted signal and no distortions due to channel ipairents have been experienced. Note (2.2) is as a result of the channel response being frequency independent within the band of our transitted signal. Next, let s assue that our transitted signal reains the sae, but the channel response turns into the one shown in Fig. 2.2. > 2 W C ( f ) > 2 W f f = - f c f = f c Flattened bandwidth slices Y ( f ) Flattened bandwidth slices Fig. 2.2 A counication channel that will introduce distortion. Notes OFDM HE March 23 Sayfa 7

Note that in the case of Fig. 2.2, the spectru of the transitted signal is still accoodated within C f no longer satisfies (2.), the transitted signal will the channel response, but since definitely be received definitely with soe distortion. But if we iagine that we slice the channel into finer frequency intervals, then it is possible to obtain nearly flat channel (frequency independent) responses like the one given in (2.). For each slice, we can assign a different carrier rather than using a one single carrier centered at f. Slicing into the source signal band into c narrower frequency slots eans that we can use extended sybol durations. 3. Construction of Orthogonal Frequency Division Multiplexing (OFDM) OFDM is historically based on frequency shift keying (FSK). In OFDM, the nuber of carriers called subcarriers, are too any and they are arranged to be orthogonal over one sybol interval. For the derivation of the basic relation for orthogonality, we first consider the following integral covering two sinusoidal signals p p I cos 2 f t cos 2 f t dt (3.) It is easy by hand derivation that the integral in (3.) will evaluate to zero under the following conditions (3.2) can also be interpreted as p f, f p, integers p, arbitrary (3.2) p p f f p,, (3.3) p p f f p p he proof of (3.) becoing zero with the setting of (3.2) is given below A B A B A B using cos cos.5 cos cos I.5cos 2 f f t dt.5 cos 2 f f t dt p p p p sin 2 f f t sin 2 f f t p p p p.5.5 2 f f 2 f f p p sin 2 p sin p p.5 sin 2 p sin p p.5 2 f f 2 f f p p (3.4) Notes OFDM HE March 23 Sayfa 8

Note that the validity of (3.4) can also be shown in Matlab as follows sys Fip Fi t =.;p = 5; = 2;fp = p/;f = /; I = int(cos(2*pi*fp*t + Fip)*cos(2*pi*f*t + Fi),t,,) After running this piece of Matlab code, we get the result of I =. (3.4) can also be established using exponentials in a shorter way. p p I exp j2 f t exp j exp j2 f t exp j dt p p exp j exp j2 f f t dt exp j if f f or p p p exp j2 f f t exp j2 f f p p exp j exp j p p j2 f f j2 f f p for pinteger exp j2 p exp j if p (3.5) p j2 f f p Note that for the coplex case, orthogonality integral is written by taking the coplex conjugate of the second function. he iplications of (3.2) to (3.5) are that if cos 2 ft p and cos 2 ft p are to be selected as OFDM subcarriers, then the orthogonality condition stated in (3.2) or (3.3) ust hold with the relative phase difference between the subcarriers being uniportant. Exaple 3. : In Fig. 3., we illustrate the siple case of four OFDM subcarriers 4 3 =.8 sec Saple OFDM subcarriers 2 Relative aplitude - c = A cos ( 2 f t + ) c 4 = A cos ( 2 f 4 t + 4 ) -2 c 2 = A cos ( 2 f 2 t + 2 ) -3 c 3 = A cos ( 2 f 3 t + 3 ) -4..2.3.4.5.6.7.8.9. t - ie (seconds) Fig. 3. Four saple OFDM subcarriers. Notes OFDM HE March 23 Sayfa 9

We can copile the following nueric data fro Fig. 3..8 sec 8 sec First subcarrier : c t Acos 2 f t, f 2.5 Hz /, / 2 Second subcarrier : c t Acos 2 f t, f 25 Hz 2 /, 2 2 2 2 2 hird subcarrier : c t Acos 2 f t, f 3 3 3 37.5 Hz 3/, / 2 3 3 Fourth subcarrier : c t Acos 2 f t, f 5 Hz 4 /, (3.6) 4 4 2 4 4 As seen fro (3.6) and Fig. 3., the OFDM subcarriers despite their differing (initial) phases, c t, c t, c t, c t have coplete (integer) nuber of cycles within one whole sybol 2 3 4 duration,. Siilarly the differences between the frequencies of the subcarriers are integer ultiples of the inverse of the sybol duration,. his way, orthogonality conditions of (3.2) c t, c t, c t, c t. and (3.3) are perfectly satisfied by 2 3 4 Exercise 3. : Using the Matlab file orthogonality_sys. or otherwise, confir that the c t, c t, c t, c t do indeed satisfy the orthogonality, when any of the two subcarriers 2 3 4 subcarriers are integrated over the interval, that is c t c t dt, p 4, 4, p (3.7) p At the transitter side OFDM signal will be obtained by odulating the OFDM subcarriers by a PSK or QAM signals. he siplified block of an OFDM transitter is given in Fig. 3.2. Fig. 3.2 Siplified block diagra of an OFDM transitter. As seen in Fig. 3.2 for an M ary PSK or QAM, K nuber of subcarriers is reserved. It could be that K M, which will be the setting in these notes. o exeplify the operations carried out in Fig. 3.2, we take the constellation diagra of 4 PSK and a typical flow of the first four PSK sybols along tie axis as illustrated in Fig. 3.3. Notes OFDM HE March 23 Sayfa

4 PSK Constellation 2 s ie flow of 4 PSK wavefors s 2 s 2 s 3 s 2 s 2 j s 3 - s 4 - j s t - ie s 2 s 2 s 3 s s 2 s 3 s 4 s Fig. 3.3 4 PSK constellation and related first four sybols. Note that on the constellation diagra of Fig. 3.3, we also show the Matlab notation of nueric values for the 4 PSK signals, s s. hese are 4 s, s j, s, s j (3.8) 2 3 4 As indicated in Fig. 3.2, odulation of subcarriers will be established by the ultiplication of the PSK sybols of (3.8) by the subcarriers. his way, k th odulated subcarrier will be obtained fro s s y t c t cos 2 f t, k K (3.9) k k k k k where K is the total nuber of subcarriers. We show in Fig. 3.4 how the four PSK signals of Fig. c t c t y t y t with 3.3 are assigned to the four OFDM subcarriers of, to constitute 4 4 the initial phases of the subcarriers given in (3.6) all being zero. As deonstrated by Fig. 3.4, the assignent operation also allows the extension of sybol duration fro to 4 s s. his is equivalent to dividing the signal bandwidth into finer slices so that the response of the counication channel will appear nearly flat, i.e. the act illustrated in Fig. 2.. Notes OFDM HE March 23 Sayfa

s s 2 s 2 s 3 s s s 2 2 s 3 s y = s 2 c = s 2 cos ( 2 f t ) y 2 = s 2 c 2 = s 2 cos ( 2 f 2 t ) t t t y 3 = s 3 c 3 = s 3 cos ( 2 f 3 t ) t y 4 = s c 4 = s cos ( 2 f 4 t ) t s 2 s 3 s 4 s = Fig. 3.4 he assignent of the four PSK sybols to the four OFDM subcarriers. Bearing in ind that the OFDM subcarriers odulated by the PSK sybols are going to be added to each other within one OFDM sybol duration, i.e., within M as shown in Fig. 3.2, then we will have the following atheatical expression for the OFDM signal, K K K k s k k s k k k k k s yt y t y t c t cos 2 f t, M, t (3.) where k s is one of the th sybol of PSK, i.e., s, placed onto the k subcarrier. Considering the nueric values of s in (3.8), we see that as yt will be coplex. hen yt can be represented real part of, iaginary part of r i r i 2 2.5 y t i r i y y t y t y t y t y t y t y t y t y t y t y t t hus, yt t, y tan (3.) r and their cobination can be plotted as shown in Fig. 3.5, where soe nueric values of the agnitude and the phase are given as well. hus we can visualize whose length (agnitude) and phase are changing with the progress of tie. yt as a vector Notes OFDM HE March 23 Sayfa 2

y 2 2.3 t y / 2 / 3 t y i 2 / 2 ) y r y r y r 2 / 3.3 ) t 4 s = Fig. 3.5 he agnitude and phase plots of OFDM signal. By converting the OFDM subcarriers into exponentials, (3.) will becoe K y t s exp j2 f t, M, t (3.2) k k k Written in discrete Fourier transfor notation, (3.2) will be K yn s exp j2 f n / N (3.3) k k N k where we have let t n / N. Coparing (3.3) to (.7), we conclude that (3.2) is in the for of inverse discrete Fourier transfor (IDF). his eans that the transitter of Fig. 3.2 can ipleent the OFDM odulation via IDF in a fast anner even if the nuber of subcarriers reaches nubers like 52. We know that the job of the receiver will be to deodulate and recover the sybols k s placed on the subcarriers. Fro the relations given in (.6) and (.7), we expect that the deodulation of the essage sybols can be perfored by the discrete Fourier transfor (DF) operation, that is N d y n exp j2 f n / N s (3.4) k k k n We shall prove later that the operation in (3.4) will indeed deliver the k s sybols. It is instructive to exaine the frequency spectrus of OFDM signal. For this, we select the siple case of 4 PSK and four OFDM subcarriers given in (3.6). As apparent fro Fig. 3.4 and (3.8), the sinusoidal OFDM subcarriers are going to be odulated by rectangular shaped tie liited (PSK) sybols. herefore it is natural to anticipate that, the frequency spectrus of thus odulated OFDM subcarriers will be in the for of sinc functions as shown in Fig. 3.6. Here the relative phases introduced by PSK sybols are ignored. As seen fro Fig. 3.6, the orthogonality condition ebedded into these subcarriers ensures that at the spectral peak of one odulated subcarrier, the Notes OFDM HE March 23 Sayfa 3

spectrus of all other subcarriers pass through zero. his point is iportant fro the point of view of successful deodulation and detection of the transitted PSK sybols. Y ( f ) Y 2 ( f ) Y 3 ( f ) Y 4 ( f ) Peaks of individual spectrus Zero crossings of three other spectrus f f = f f = f 2 f = f 3 f = f 4 Fig. 3.6 Spectrus of odulated OFDM subcarriers. We now ove onto OFDM receiver. Fig. 3.7 gives the siplified block diagra of such a receiver. Fig. 3.7 Siplified block diagra of an OFDM receiver. As understood fro Fig. 3.7, the counication channel that exists between the transitter of Fig. 3.2 and receiver of Fig. 3.7 is assued to be band unliited. he way, at the input of the, receiver, there is an exact copy of the transitted OFDM signal. On each ar of the receiver of Fig. 3.7, ultiplication of the received signal by one of the subcarriers corresponds to the act of correlation. hus the output of k th ar of the correlator is d y t cos 2 f t dt (3.5) k k Notes OFDM HE March 23 Sayfa 4

It is easy to prove that the atheatical operation of (3.5) will deliver the individual PSK sybols placed on the subcarriers sequentially at the transitter. Below we do this for the saple case of 4 PSK of Fig. 3.4. Initially we write the coplete expression of the transitted signal with the PSK sybols assigned to the subcarriers as shown in Fig. 3.4. 4 4 4 cos 2 k s k k s k k y t y t c t f t k k n f t f t f t f t s cos 2 s cos 2 s cos 2 s cos 2 2 2 2 3 3 4 j cos 2 f t j cos 2 f t cos 2 f t cos 2 f t (3.6) 2 3 4 Now based on (3.5), we evaluate the correlator output for the first ar of the receiver in Fig. 3.7. cos2 d y t f t dt 2 2 j cos 2 f t cos 2 f t dt j cos 2 f t cos 2 f t dt I I 3 4 cos 2 f t 3 cos 2 f t dt cos 2 f t 4 cos 2 f t dt (3.7) Due to the selection of frequencies f, f, f, f, the orthogonality of (3.7) is satisfied, thus 2 3 4 p p cos2 f tcos 2 f tdt, if p and f f p p cos 2 f t cos 2 f t dt.5, if p (3.8) then the individual integrals in (3.6) will be I I, due to f / sin 4 ft cos2 cos2.5.5 8 f I j f t f t dt jt j I j cos 2 f t cos 2 f t dt, I cos 2 f t cos 2 f t dt 2 2 3 3 I cos 2 f t cos 2 f t dt (3.9) 4 4 So, eventually the output of the first ar of the correlator will becoe d.5 j.5s which 2 eans that we have been able to deodulate successfully the PSK sybol (the scaled version of) s placed on the subcarrier c t. 2 Notes OFDM HE March 23 Sayfa 5

It is easy to deonstrate that, the output of the reaining ars, i.e., d, d, d, in the 4 PSK 2 3 4 c t, c t, c t. equivalent of Fig. 3.7 will deliver the other sybols placed onto subcarriers 2 3 4 Exercise 3.2 : Bearing in ind the assignent of 4 PSK sybols in Fig. 3.4, prove that the correlation operation of Fig. 3.7 and (3.5) will correctly deodulate the sybols on the c t, c t, c t. subcarriers of 2 3 4 4. Orthogonal Frequency Division Multiplexing (OFDM) Analysis in Discrete Doain Now we describe odulation and deodulation processes of OFDM in ters of discrete Fourier and inverse Fourier transfors. For this, we again take the siplified case of 4 PSK and write for c t, c t, c t, c t the discrete equivalent of subcarriers 2 3 4 exp 2 / 4, exp 2 / 4 c n j f n c n j f n 2 2 c n exp j2 f n / 4, c n exp j2 f n / 4 (4.) 3 3 4 4 In (4.) we have intentionally changed the tie range into N 4, since the subcarriers are liited to that tie range, which can also be denoted by 4. Keep in ind that in discrete applications, the frequencies can only take integer values, hence the appropriate choice is f, f 2, f 3, f 4 (4.2) 2 3 4 As seen fro (4.), the subcarriers are coplex, hence, we revise the orthogonality test of (3.7) as follows * c t c t dt, p 4, 4, p (4.3) p where * takes into account the possibility of subcarriers being coplex. he discrete equivalence of (4.3) is N N *, p, orthogonal c n c n exp 2 / 4 exp 2 / 4 p j f n j f n p n n 4, p, otherwise p 4, 4, n N 3 (4.4) For instance after setting, p, 2, (4.4) evaluates to j f n j f n 2 j n j n N 3 exp 2 / 4 exp 2 / 4 exp 2 / 4 exp 4 / 4 n n 3 n n n n2 n3 j n j j j exp 2 / 4 exp exp / 2 exp exp 3 / 2 j j (4.5) Notes OFDM HE March 23 Sayfa 6

Hence we conclude that with the choice of nueric settings of (4.) and (4.2), are indeed orthogonal. c n and c2 n Exercise 4. : Prove that the validity of the first line of (4.4) by inserting for the subcarriers fro (4.) and (4.2) for the cases other than the one evaluated in (4.5). Now we consider the operations on the side of the receiver. We recollect fro (3.3) and (3.4), in OFDM odulation corresponds to inverse discrete Fourier transfor operation, while OFDM deodulation corresponds to discrete Fourier transfor operation. hen after odulation, for the siple 4 PSK case, we will have K 4 y n s exp j2 f n / N exp 2 / k k s j f n N k k N k N k exp j2 f n / N exp 2 j2 f n / N 2 exp 3 j2 f n / N 3 exp 4 j2 f n / N 4 N s s s s exp 2 j2 f n / N exp 2 j2 f n / N 2 exp 3 j2 f n / N 3 exp j2 f n / N 4 N s s s s j exp j2 f n / N j exp j2 f n / N 2 exp j2 f n / N 3 exp j2 f n / N 4 (4.6) N On the receiver side, perforing the correlation on the first ar of Fig. 3.7, we will get N exp 2 / d y n j f n N n N s exp 2 / exp 2 / exp 2 / exp 2 / 2 2 N n j f n N j f n N s j f n N j f n N j f n N j f n N s j f n N j f n N s exp 2 / exp 2 / exp 2 / exp 2 / 3 3 4 4 s s j (4.7) 2 As seen fro the result of (4.7), deodulation is successful, hence the PSK the sybol placed on the first subcarrier c n at the transitter, is correctly recovered as s or j at the receiver. 2 Exercise 4.2 : Prove that the sybols placed onto other subcarriers, i.e.,, also deodulated successfully, as it is done for the first sybol in (4.7). c n c n c n are 2 3 4 Exercise 4.3 : Available on the course webpage, the Matlab file OFDM_exp_siple. ipleents the odulation and deodulation for an OFDM syste whose sybols are obtained fro an 8 rectangular QAM constellation. o increase the nuber of subcarriers to 8, four higher frequencies are selected in addition to those listed in (3.6). Run and observe the outputs of OFDM_exp_siple.. Find the additional subcarriers, observe the orthogonality of all subcarriers both fro the tie wavefors and the Ctat atrix written onto the workspace. Coent on the spectrus of odulated subcarriers. Make several runs to see if deodulation is successful. Repeat this by changing the targeted sybol of deodulation on line 44 of the code. By odifying the sybol duration on line 4 of the code, observe that the orthogonality is destroyed and the transitted sybol is not deodulated correctly. Notes OFDM HE March 23 Sayfa 7

References. John G. Proakis, Masoud Salehi, Counication Systes Engineering 2 nd Ed., Prentice Hall 22, ISBN : -3-6793-8, Chapter 9. 2. S. K. Mitra, Digital Signal Processing, 3 rd Ed. McGraw Hill 26, ISBN : -3-84788-7, pp 233-244. 3. Matlab help - DF 4. My own lecture notes. Notes OFDM HE March 23 Sayfa 8