3.2 Complex Sinusoids and Frequency Response of LTI Systems

Transcription

1 3. Introduction. A signal can be represented as a weighted superposition of complex sinusoids. x(t) or x[n]. LTI system: LTI System Output = A weighted superposition of the system response to each complex sinusoid. y(t) or y[n] 3. Four distinct Fourier representations 3. Complex Sinusoids and Frequency Response of LTI Systems Frequency response The response of an LTI system to a sinusoidal input. Discrete-time LTI system. Impulse response of discrete-time LTI system = h[n], input = x[n] = e j n. Output: y j n k n hk xn k hk e k k

2 Figure 3. (p. 96) The output of a complex sinusoidal input to an LTI system is a complex sinusoid of the same frequency as the input, multiplied by the frequency response of the system. j n j k j j n y n e h k e H e e k 3. Frequency response: Complex scaling factor j j k H ( e ) h[ k ] e (3.) k A function of frequency Continuous-time LTI system. Impulse response of continuous-time LTI system = h(t), input = x(t) = e j t. Output: j ( t ) j t j y ( t ) h( ) e d e h( ) e d (3.) H ( j ) e j t

3 3. Frequency response: j H ( j ) h( ) e d (3.3) Polar form complex number c = a + jb: c j arg{ c} b c e where c a b and arg{ c} tan a Polar form for H (j): H ( j) H ( j) e j arg{ H ( j )} where H j H j ( ) M agnitude response and arg ( ) Phase respnse j t argh j y t H j e Example 3. RC Circuit: Frequency response The impulse response of the system relating to the input voltage to the voltage across the capacitor in Fig. 3. is derived in Example. as t / RC h( t ) e u( t ) RC Find an expression for the frequency response, and plot the magnitude and phase response. 3

4 <Sol.> Frequency response: H RC j j e u e d j RC e d RC RC RC RC j j RC RC e j RC Magnitude response: H j Figure 3. (p. 97) RC circuit for Example 3.. RC RC RC j RC Phase response: arg H j arctan RC Fig. 3.6 (a) and (b) 4

5 H arg H j j 4 4 Figure 3.3 (p. 98) Frequency response of the RC circuit in Fig. 3.. (a) Magnitude response. (b) Phase response. Eigenvalue and eigenfunction of LTI system [A] Continuous-time case:. Eigenrepresentation: Fig. 3.4 (a) H t t 5

6 e j jn j t j t ( t ) ( t ) e H ( j ) e j n H e e H H H [ n ] [ n ] Figure 3.4 (p. 98) Illustration of the eigenfunction property of linear systems. The action of the system on an eigenfunction input is multiplication by the corresponding eigenvalue. (a) General eigenfunction (t) or [n] and eigenvalue. (b) Complex sinusoidal eigenfunction e jt and eigenvalue H(j). (c) Complex sinusoidal eigenfunction e jn and eigenvalue H(e j ). Eigenfunction: () t j t e Eigenvalue: H ( j ). If e k is an eigenvector of a matrix A with eigenvalue k, then Ae e k k k Arbitrary input = weighted superpositions of eigenfunctions Convolution operation Multiplication Ex. Input: M Output: M j k t j k t xt a e y t a H k j k e k k k 6

7 [B] Discrete-time case:. Eigenrepresentation: Fig. 3.4 (c) H ( [ n]) [ n] Eigenfunction: [ n] j Eigenvalue: H ( e ). Input: M j k x[ n] a e k k n e j n Output: M k k y[ n] a H e e k k j j n 3.3 Fourier Representations for Four classes of Signals Four distinct Fourier representations: Table 3.. 7

8 8

9 3.3. Periodic Signals: Fourier Series Representations. x[n] = discrete-time signal with fundamental period N. DTFS of x[n] is jk n x[ n] A[ k ] e (3.4) o = /N Fundamental frequency of x[n] k. x(t) = continuous-time signal with fundamental period T. FS of x(t) is jk x( t ) A[ k ] e (3.5) k t jk t j t e is the k th harm onic of e. o = /T Fundamental frequency of x(t) ^ denotes approximate value. A[k] = the weight applied to the k th harmonic. 3. The complex sinusoids exp(jk o n) are N-periodics in the frequency index k. <pf.> j N k n jn n jk n j n jn n e e e e e jk n e There are only N distinct complex sinusoids of the form exp(jk n) should be used in Eq. (3.4). Eq. (3.4) N x[ n] [ ] A k e (3.6) k jk n 9

10 4. Continuous-time complex sinusoid exp(jk t) with distinct frequencies k are always distinct. FS of continuous-time periodic signal x(t) becomes jk x( t ) A[ k ] e (3.7) k t Mean-square error (MSE) between the signal and its series representation: Discrete-time case: N M SE x[ n] x[ n] dt (3.8) N n Continuous-time case: T M SE x( t) x( t) dt (3.9) T 3.3. Nonperiodic Signals: Fourier-Transform Representations. FT of continuous-time signal: j t x t X je d X(j)/() = the weight applied to a sinusoid of frequency in the FT representation.

11 . DTFT of discrete-time signal: j jn x n X e e d X(e j )/() = the weight applied to the sinusoid e j n in the DTFT representation. 3.4 Discrete-Time Periodic Signals: The Discrete-Time Fourier Series. DTFS pair of periodic signal x[n]: N x[ n] [ ] X k e (3.) k N X [ k ] [ ] N jk n x n e (3.) n jk n Notation: DTFS ; x n X k Fundamental period = N; Fundamental frequency = o = /N Fourier coefficients; Frequency domain representation Example 3. Determining DTFS Coefficients Find the frequency domain representation of the signal depicted in Fig <Sol.>. Period: N = 5 o = /5

12 . Odd symmetry 3. Fourier coefficient: X k x n e 5 n Figure 3.5 (p. 3) Time-domain signal for Example 3.. jk n / 5 n = to n = jk 4 / 5 5 x jk 4 / 5 jk / 5 j jk e x e x e x e / 5 x jk / 5 jk / 5 X [ k ] { e e } 5 { j sin( k / 5)} 5 (3.) e

13 One period of the DTFS coefficients X[k], k = to k = : X X X X 5 5 j j sin sin j.e 5 5 j sin 4 / 5 j e / 5 j e / 5 j e 4 / 5 j. 53 sin X j.3 e Calculate X[k] using n = to n = 4: X Eq. (3.) jk 8 / 5 j jk / 5 jk 4 / 5 jk 3 6 k x e x e x e x e / 5 x 4 5 jk / 5 jk 8 / 5 5 e e Fig jk 8 / 5 jk jk / 5 jk / 5 e e e e e 3

14 X [ k] X [ k ] M agnitude spectrum of x[ n] Even function arg X [ k] Odd function arg X [ k ] P hase spectrum of x[ n] Figure 3.6 (p. 4) Magnitude and phase of the DTFS coefficients for the signal in Fig

15 Example 3.3 Computation of DTFS by Inspection Determine the DTFS coefficients of x[n] = cos (n/3 + ), using the method of inspection. <Sol.>. Period: N = 6 o = /6 = /3. Using Euler s formula, x[n] can be expressed as j ( n ) j ( n ) 3 3 j n j n 3 j 3 e e j x[ n ] e e e e (3.3) 3. Compare Eq. (3.3) with the DTFS of Eq. (3.) with o = /3, written by summing from k = to k = 3: x n 3 k X [ k ] e jk n /3 j n / 3 j n / 3 j n / 3 j n / 3 j n X [ ] e X [ ] e X [] X [] e X [] e X [3] e j e /, k D TFS ; 3 j x n X k e /, k, otherw ise on k 3 (3.4) 5

16 Example 3.4 DTFS Representation of An Impulse Train Find the DTFS coefficients of the N-periodic impulse train x n n ln l as shown in Fig <Sol.> It is convenient to evaluate Eq.(3.) over the interval n = to n = N to obtain N jkn / N X k n e N n N Figure 3.9 (p. 7) A discrete-time impulse train with period N. 6

17 The Inverse DTFS The similarity between Eqs. (3.) and (3.) indicates that the same mathematical methods can be used to find the time-domain signal corresponding to a set of DTFS coefficients. Example 3.5 The Inverse DTFS Use Eq. (3.) to determine the time-domain signal x[n] from the DTFS coefficients depicted in Fig. 3.. <Sol.>. Period of DTFS coefficients = 9 o = /9. It is convenient to evaluate Eq.(3.) over the interval k = 4 to k = 4 to obtain 4 x n X k e k 4 jk n / 9 j / 3 j 6 n / 9 j / 3 j 4 n / 9 j / 3 j 4 n / 9 j / 3 j 6 n / 9 e e e e e e e e n n cos 6 / 9 / 3 4 cos 4 / 9 / 3 Example 3.6 DTFS Representation of A Square Wave Find the DTFS coefficients for the N-periodic square wave given by 7

18 /3 Figure 3. (p. 8) Magnitude and phase of DTFS coefficients for Example 3.5. /3 8

19 , M n M xn, M n N M That is, each period contains M + consecutive ones and the remaining N (M +) values are zero, as depicted in Fig. 3.. Note that this definition requires that N > M +. <Sol.> Figure 3. (p. 9) Square wave for Example Period of DTFS coefficients = N o = /N. It is convenient to evaluate Eq.(3.) over the interval n = M to n = N M. N M X [ k ] x[ n ] e N N nm M nm e jk n jk n 9

20 X [ k ] N e N M m e jk ( m M ) M jk M jk m m 3. For k =, N, N,, we have e (3.5) Change of variable on the index of summation: m = n + M e jk jk o e o (3.5) M M X k, k, N, N, N N m 3. For k, N, N,, we may sum the geometric series in Eq. (3.5) to obtain jk M jk M ( M ) e e X [ k ], k, N, N,... jk N e X k (3.6) jk M / jk M jk M / jk M / e e e e jk / jk jk / jk / N e e N e e k, N, N,...

21 k sin k M / N, k, N, N, X k N sin k / N M / N, k, N, N, sin k M / X k, k, N, N, N sin / 4. Substituting o = /N, yields lim k, N, N, sin k M / N M N sin k / N N The numerator and denominator of above Eq. are divided by j. For this reason, the expression for X[k] is commonly written as X k N sin k M sin Fig. 3.. k / N / N L Hôpital s Rule The value of X[k] for k =, N, N,, is obtained from the limit as k.

22 Figure 3. (p. ) The DTFS coefficients for the square wave shown in Fig. 3., assuming a period N = 5: (a) M = 4. (b) M =.

23 Symmetry property of DTFS coefficient: X[k] = X[ k].. DTFS of Eq. (3.) can be written as a series involving harmonically related cosines.. Assume that N is even and let k range from ( N/ ) + to N/. Eq. (3.) becomes x n N / X k N / x k e jk n / jk n e jk n jk n n X X N / e X m e X m N / m 3. Use X[m] = X[ m] and the identity N o = to obtain N / jm n jm n j n e e x n X X N / e X m m N / / cos cos X X N n X m m n m e jn = cos(n) since sin(n) = for integer n. B k X k, k, N / X k, k,,, N / N / x[ n] B[ k ] cos( k n) k 3

24 3.5 Continuous-Time Periodic Signals: The Fourier Series Fourier series pair Exponential FS. Continuous-time periodic signals are represented by the Fourier series (FS).. A signal with fundamental period T and fundamental frequency o = /T, the FS pair of x(t) is jk x( t ) X [ k ] e (3.9) k T jk t X [ k ] x( t) e dt (3.) T 3. Notation: FS; x t X k t Frequency domain representation of x(t) The variable k determines the frequency of the complex sinusoid associated with X[k] in Eq. (3.9). Mean-square error (MSE). Suppose we define jk o xˆ t X k e k t and choose the coefficients X[k] according to Eq.(3..) 4

25 . If x(t) is square integrable, T T x t dt then the MSE between x(t) and ˆx t is zero, or, mathematically, T T x t x x( t ) t dt xˆ t at all values of t; it simply implies that there is zero power in their difference. 3. Pointwise convergence of x( t ) xˆ t is guaranteed at all values of t except those corresponding to discontinuities if the Dirichlet conditions are satisfied: x(t) is bounded. x(t) has a finite number of maximum and minima in one period. x(t) has a finite number of discontinuities in one period. If a signal x(t) satisfies the Dirichlet conditions and is not continuous, then ˆx t converges to the midpoint of the left and right limits of x(t) at each discontinuity. 5

26 Example 3.9 Direct Calculation of FS Coefficients Determine the FS coefficients for the signal x(t) depicted in Fig <Sol.>. The period of x(t) is T =, so o = / =.. One period of x(t): x(t) = e t, t. 3. FS of x(t): t jk t jk t X k e e dt e dt Figure 3.6 (p. 6) Time-domain signal for Example jk t 4 jk e X k e e e jk 4 jk 4 jk Fig The Magnitude of X[k] the magnitude spectrum of x(t); the phase of X[k] the phase spectrum of x(t). e jk = 6

27 Figure 3.7 (p. 7) Magnitude and phase spectra for Example

28 Example 3. FS Coefficients for An Impulse Train Determine the FS coefficients for the signal x(t) defined by 4 x t t l l <Sol.>. Fundamental period of x(t) is T = 4, each period contains an impulse.. By integrating over a period that is symmetric about the origin < t, to obtain X[k]: jk / t X k t e dt The magnitude spectrum is constant and the phase spectrum is zero. Example 3. Calculation of FS Coefficients By Inspection Determine the FS representation of the signal x t 3 cos t / / 4 <Sol.> Fundamental period of x(t) is T = 4, and o = /4 = /. Eq. (3.9) is written as j t / / 4 j t / / 4 e e 3 j / 4 j t / 3 j / 4 j t / x t 3 e e e e 8

29 Time-domain representation obtained from FS coefficients Example 3. Inverse FS Find the time-domain signal x(t) corresponding to the FS coefficients k / k jk / X e Assume that the fundamental period is T =. <Sol.>. Fundamental frequency: o = /T =. From Eq. (3.9), we obtain jk / jk t k jk / jk t k / / x t e e e e k k k k jk / jk t / e e / l l e jl / e jl t x t / e j t / / e j t /. Putting the fraction over a common denominator results in x t 5 4 cos 3 t / 9

30 Example 3.3 FS for A Square Wave Determine the FS representation of the square wave depicted in Fig. 3.. Figure 3. (p. ) Square wave for Example 3.3. <Sol.>. Fundamental frequency: o = /T.. X[k]: by integrating over T/ t T/ T / T jk t jk t T X k x t e dt e dt T T / T T jk T jk T sin k T Tjk Tk j Tk jk t e e e k T, For k =, we have X dt T T T T T 3

31 3. By means of L Hôpital s rule, we have lim k sin k T Tk X k sin T T k T Tk 4. X[k] is real valued. Using o = /T gives X[k] as a function of the ratio T o /T: X[ k] sin( k T / T ) (3.3) k 5. Fig. 3. (a)-(c) depict X[k], 5 k 5, for T o /T =/4, T o /T =/6, and T o /T = /64, respectively. Definition of Sinc function: sin( u ) sinc( u ) (3.4) Fig u ) Maximum of sinc function is unity at u =, the zero crossing occur at integer values of u, and the amplitude dies off as /u. ) The portion of this function between the zero crossings at u = is manilobe of the sinc function. 3

32 Figure 3.a&b (p. ) The FS coefficients, X[k], 5 k 5, for three square waves. (see Fig. 3..) (a) T o /T = /4. (b) T o /T = /6. (c) T o /T = /64. 3

33 Figure 3.c (p. ) Figure 3.3 (p. 3) Sinc function sinc(u) = sin(u)/(u) 33

34 3) The smaller ripples outside the mainlobe are termed sidelobe. 4) The FS coefficients in Eq. (3.3) are expressed as T T X k k T T sinc Example 3.5 RC Circuit: Calculating The Output By Means of FS Let us find the FS representation for the output y(t) of the RC circuit depicted in Fig. 3. in response to the square-wave input depicted in Fig. 3., assuming that T o /T = ¼, T = s, and RC =. s. Figure 3. (p. ) Square wave for Example 3.3. Figure 3. (p. 97) RC circuit for Example

35 <Sol.>. If the input to an LTI system is expressed as a weighted sum of sinusoid, then the output is also a weighted sum of sinusoids.. Input: 4. Frequency response of the system H(j): t k k jk t x X e 3. Output: t H jk X k k jk t y e FS; y t Y k H jk X k 5. Frequency response of the RC circuit: H j / RC j / 6. Substituting for H(jk o ) with RC =. s and o =, and T o /T = ¼, gives Y k j k sin k / k 7. We determine y(t) using the approximation jk y ( t ) Y [ k ] e (3.3) k t RC The magnitude and phase spectra for the range 5 k 5: Fig. 3.6 (a) and (b). Waveform of y(t): Fig. 3.6(c). 35

36 Figure 3.6 (p. 8) The FS coefficients Y[k], 5 k 5, for the RC circuit output in response to a square-wave input. (a) Magnitude spectrum. (b) Phase spectrum. c) One period of the input signal x(t) dashed line) and output signal y(t) (solid line). The output signal y(t) is computed from the partial-sum approximation given in Eq. 3.3). 36

37 Figure 3.6 (p. 8) The FS coefficients Y[k], 5 k 5, for the RC circuit output in response to a square-wave input. (a) Magnitude spectrum. (b) Phase spectrum. c) One period of the input signal x(t) dashed line) and output signal y(t) (solid line). The output signal y(t) is computed from the partial-sum approximation given in Eq. (3.3). 37

38 3.6 Discrete-Time Nonperiodic Signals: The Discrete-Time Fourier Transform (DTFT). The DTFT is used to represent a discrete-time nonperiodic signal as a superposition of complex sinusoids.. DTFT representation of time-domain signal x[n]: j jn x[ n] X ( e ) e d j j n X ( e ) x[ n] e (3.3) 3. Notation: n (3.3) DTFT j x n X e DTFT pair Frequency-domain representation x[n] 4. Condition for convergence of DTFT: If x[n] is of infinite duration, then the sum converges only for certain classes of signals. If x[n] is absolutely summable, i.e., n x n The sum in Eq. (3.3) converges uniformly to a continuous function of. 38

39 If x[n] is not absolutely summable, but does satisfy (i.e., if x[n] has finite energy), It can be shown that the sum in Eq. (3.3) xn converges in a mean-square error sense, n but does not converge pointwise. Example 3.7 DTFT of An Exponential Sequence Find the DTFT of the sequence x[n] = n u[n]. <Sol.>. DTFT of x[n]: X j n j n n j n e une e n. For <, we have n n n j j X ( e ) e, (3.33) j e 3. If is real valued, Eq. (3.33) becomes X e j cos j sin This sum diverges for Euler s Formula 39

40 4. Magnitude and phase spectra: X arg e j / cos sin cos sin j X e arctan cos Fig. 3.9 for =.5 and =.9. Example 3.8 DTFT of A Rectangular Pulse Let, n M xn, n M as depicted in Fig. 3.3 (a). Find the DTFT of x[n]. <Sol.>. DTFT of x[n]: M j j n. X e e nm Odd function / Even function 4

41 Figure 3.9 (p.3) The DTFT of an exponential signal x[n] = () n u[n]. (a) Magnitude spectrum for =.5. (b) Phase spectrum for =.5. (c) Magnitude spectrum for =.9. (d) Phase spectrum for =.9. 4

42 Figure 3.3 (p. 33) Example 3.8. (a) Rectangular pulse in the time domain. (b) DTFT in the frequency domain. p 4

43 M M j j ( m M ) j M j m X e e e e m m Change of variable m = n + M j ( M ) j M e e,,, 4, j e M, =,, 4,. The expression for X(e j ) when,, 4, j jm X ( e ) e e j M / j M / j M / e e e j / j / j / e e e j M / j M / e e j / j / e sin M / L Hôptital s Rule lim M ;,, 4,, sin With understanding that X(e sin M / j j ) for X ( e ),, 4, is obtained as sin limit. 43

44 3. Graph of X(e j ): Fig. 3.3(b). Example 3.9 Inverse DTFT of A Rectangular Pulse Find the inverse DTFT of X e j,, W, W n which is depicted in Fig. 3.3 (a). <Sol.> Figure 3.3 (p. 34) Example 3.9. (a) Rectangular pulse in the frequency domain. (b) Inverse DTFT in the time domain.. Note that X(e j ) is specified only for <.. Inverse DTFT x[n]: W j n x n e d W p 44

45 nj e j n W W, n sin W n, n. n 3. For n =, the integrand is unity and we have x[] = W/. Using L Hôpital s rule, we easily show that W lim sin Wn, n n and thus we usually write sin n W n x n as the inverse DTFT of X(e j ), with the understanding that the value at n = is obtained as limit. 4. We may also write W sin /, c W n x n Fig. 3.3 (b). 45

46 Example 3. DTFT of The Unit Impulse Find the DTFT of x[n] = [n]. <Sol.>. DTFT of x[n]: j j n X e n e n DTFT n.. This DTFT pair is depicted in Fig Figure 3.3 (p. 35) Example 3.. (a) Unit impulse in the time domain. (b) DTFT of unit impulse in the frequency domain. Example 3. Inverse DTFT of A Unit Impulse Spectrum Find the inverse DTFT of X(e j ) = (), <. <Sol.>. Inverse DTFT of X(e j ): p 46

47 jn x n e d. Sifting property of impulse function DTFT,. We can define X(e j ) over all by writing it as an infinite sum of delta functions shifted by integer multiples of. p. This DTFT pair is depicted in Fig Figure 3.33 (p. 36) Example 3.. (a) Unit impulse in the j X e k. frequency domain. (b) Inverse DTFT in k the time domain. Dilemma: The DTFT of x[n] = /() does not converge, since it is not a square summable signal, yet x[n] is a valid inverse DTFT! 47

48 Example 3. Moving-Average Systems: Frequency Response Consider two different moving-average systems described by the input-output equations y n x n x n and y n x n x n The first system averages successive inputs, while the second forms the difference. The impulse responses are h n n n and h n n n Find the frequency response of each system and plot the magnitude responses. <Sol.>. The frequency response is the DTFT of the impulse response.. For system # h [n] : Frequency response: j H ( e ) e j j j j ( j ) e e H e e j e cos. 48

49 Magnitude response: j H e cos, 3. For system # h [n] : Frequency response: Phase response: j arg H ( e ). j j j j j j e e H ( e ) e je je sin. j Magnitude response: j H e sin, Phase response: j arg H ( e ) 4. Fig (a) and (b) depict the magnitude responses of the two systems on the interval <.,, 49

50 p Figure 3.36 (p. 39) The magnitude responses of two simple discrete-time systems. (a) A system that averages successive inputs tends to attenuate high frequencies. (b) A system that forms the difference of successive inputs tends to attenuate low frequencies. Example 3.3 Multipath Communication Channel: Frequency Response The input-output equation introduced in Section. describing a discrete-time model of a two-path propagation channel is y n x n ax n. In Example.3, we identified the impulse response of this system as h[n] = [n] + a[n ] and determined that the impulse response of the inverse system was h inv [n] = ( a) n u[n]. The inverse system is stable, provided that a<. 5

51 Compare the magnitude responses of both systems for a =.5 e j/3 and a =.9 e j/3. <Sol.>. The frequency response is the DTFT of the impulse response.. Frequency response of the system modeling two-path propagation: j j. H e ae 3. Magnitude response: a j j arg H e a e. Using a a e j H e a cos arg a j a sin arg a. jarg a Apply Euler s formula j cos arg sin arg H e a a a a a a cos arg a, cos sin 4. The frequency response of the inverse system may be obtained by replacing α with a in Eq. (3.33). (DTFT of h inv [n] = ( a) n u[n]) 5

52 inv j H e, a j ae inv j cos arg sin arg H e a a a a a a cos arg a, Expressing a in polar form The frequency response of the inverse system is the inverse of the frequency response of the original system. 5. Magnitude response of the inverse system: H inv e j a a cos arg a 6. Fig depicts the magnitude response of H(e j ) for both a =.5 e j/3 and a =.9 e j/3 on the interval <. 7. The magnitude response approaches a maximum of + a when = arg{a} and a minimum of a when = arg{a}. 8. The magnitude response of the corresponding inverse system: Fig

53 Figure 3.37 (p. 4) Magnitude response of the system in Example 3.3 describing multipath propagation. (a) Echo coefficient a =.5e j/3. (b) Echo coefficient a =.9e j/3. 53

54 Figure 3.38 (p. 4) Magnitude response of the inverse system for multipath propagation in Example 3.3. (a) Echo coefficient a =.5e j/3. (b) Echo coefficient a =.9e j/3 If a =, then the multipath model applies zero gain to any sinusoid with frequency = arg{a}. The multipath system cannot be inverted when a=. 54

55 3.7 Continuous-Time Nonperiodic Signals: The Fourier Transform (FT). The Fourier transform (FT) is used to represent a continuous-time nonperiodic signal as a superposition of complex sinusoids.. FT representation of time-domain signal x(t): j t x( t) X ( j) e d (3.35) j t X ( j ) x( t ) e dt (3.36) 3. Notation for FT pair: FT. x t X j Example 3.4 FT of A Real Decaying Exponential Find the FT of x(t) = e a t u(t), shown in Fig. 3.39(a). <Sol.>. For a, since x(t) is not absolutely integrable, i.e., Frequency-domain representation of the signal x(t) at e dt, a The FT of x(t) does not converge for a. 55

56 . For a >, the FT of x(t) is at j t a j t X j e u t e dt e dt a j t e a j a j 3. Magnitude and phase of X(j): X j a X j a arg arctan, Fig (b) and (c). 56

57 Figure 3.39 (p. 43) Example 3.4. (a) Real time-domain exponential signal. (b) Magnitude spectrum. (c) Phase spectrum. X(j) arg{x(j)} /4 /4 /4 / 57

58 Example 3.5 FT of A Rectangular Pulse Consider the rectangular pulse depicted in Fig. 3.4 (a) and defined as, T t T x t, t T Find the FT of x(t). <Sol.>. The rectangular pulse x(t) is absolutely integrable, provided that T o <.. FT of x(t): T jt jt X j x t e dt e dt T j t T e sin T, T j 3. For =, the integral simplifies to T o. lim sin T T. 4. Magnitude spectrum: sin, X j T With understanding that the value at = is obtained by evaluating a limit. 58

59 X j sin T, 5. Phase spectrum: arg X j,, sin T sin T Fig (b). 6. X(j) in terms of sinc function: X j T sinc T. As T o increases, the nonzero time extent of x(t) increases, while X(j) becomes more concentrated about the frequency origin. v ; p Figure 3.4 (p. 44) Example 3.5. (a) Rectangular pulse in the time domain. (b) FT in the frequency domain. 59

60 Example 3.6 Inverse FT of A Rectangular Pulse Find the inverse FT of the rectangular spectrum depicted in Fig (a) and given by, W W Figure 3.4 (p. 46) X j Example 3.6. (a) Rectangular spectrum in, W the frequency domain. <Sol.> (b) Inverse FT in the time domain. v ; p. Inverse FT of x(t): W jt jt W x t e d e W sin( W t ), t W j t t. When t =, the integral simplifies to W/, i.e., 6

61 lim sin W t W, t t 3. Inverse FT is usually written as sin, W t or x t sinc, x t t W W t With understanding that the value at t = is obtained as a limit. Fig (b). As W increases, the frequency-domain representation becomes less concentrated about =, while the time-domain representation X(j) becomes more concentrated about t =. Example 3.7 FT of The Unit Impulse Find the FT of x(t) = (t). <Sol.>. x(t) does not satisfy the Dirichlet conditions, since the discontinuity at the origin is infinite.. FT of x(t): j t X j t e dt 6

62 Notation for FT pair: t, Example 3.8 Inverse FT of An Impulse Spectrum Find the inverse FT of X(j) = (). <Sol.>. Inverse FT of X(j): j t x t e d. Notation of FT pair: FT FT Duality between Example 3.7 and 3.8 Example 3.9 Characteristic of Digital Communication Signals In a simple digital communication system, one signal or symbol is transmitted for each in the binary representation of the message, while a different signal or symbol is transmitted for each. One common scheme, binary phase-shift keying (BPSK), assumes that the signal representing is the negative of the signal and the signal representing is the positive of the signal. Fig depicts two candidate signals for this approach: a rectangular pulse defined as 6

63 x r t A, t T /, t T / and a raised-cosine pulse defined as x c t A / cos t / T, t T / c, t T / Figure 3.44 (p. 49) Pulse shapes used in BPSK communications. (a) Rectangular pulse. (b) Raised cosine pulse. The transmitted BPSK signals for communicating a sequence of bits using each pulse shape are illustrated in Fig Note that each pulse is T o seconds long, 63

64 so this scheme has a transmission rate of /T o bits per second. Each user s signal is transmitted within an assigned frequency band, as depicted in Chapter 5. In order to prevent interference with users of other frequency bands, governmental agencies place limits on the energy of a signal that any user transmits into adjacent frequency bands. Suppose the frequency band assigned to each user is khz wide. Then, to prevent interference with adjacent channels, we assume that the peak value of the magnitude spectrum of the transmitted signal outside the -khz band is required to be 3 db below the peak in-band magnitude spectrum. Use the FT to determine the maximum number of bits per second that can be transmitted when the rectangular and raised-cosine pulse shapes are utilized. <Sol.>. FT of the rectangular pulse x r (t): sin T X j r In terms of Hz rather than rad/sec ( = f): X ( j ) X ( jf ) X r jf sin ft f r r 64

65 Figure 3.46 (p. 5) Spectrum of rectangular pulse in db, normalized by T. 65

66 . From Fig. 3.46, we must choose To so that the th zero crossing is at khz in order to satisfy the constraint that the peak value of the magnitude spectrum of the transmitted signal outside the -khz band allotted to this user be less than 3 db. f = k/t o = /T o T o = 3 Data transmission rate = bits/sec. 3. FT of raised-cosine pulse x c (t): 8 T j t X j cos c t T e dt 3 T Euler s formula T T T j t j T t j T t X j c e dt e dt e dt 3 T 3 T 3 T Each of three integrals is of the form T j t sin T Substituting the appropriate e dt value of for each integral T X c j T sin sin T T sin T T 3 3 T 3 T 66

67 In terms of f (Hz), the above equation becomes X ft sin sin f T T sin f T T jf.5.5 c 3 f 3 f T 3 f T 4. In this case, the peak value of the first sidelobe is below 3 db, so we may satisfy the adjacent channel interference specifications by choosing the mainlobe to be khz wide: f = k/t o, = /T o T o = 4 seconds. Data transmission rate = 5 bits/sec. Figure 3.48 (p. 5) Spectrum of the raised-cosine pulse in db, normalized by T. 67