MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems-Fall 2003 PROblem set 3 SOLUTION Home Study Exercise (E1)O&W3.46(a)and(c) (t) and y(t) are continuous-time periodic signals with a period= To and Fourier series representations given b r(t)= akejkw'ot y(t)=>bk k=-0 (a) Show that the Fourier series coefficients of the signal x(t)=r(t)y(t)=ko_o Ckejkwot are given by the discrete convolution Ck=>to_o anbk-n(multiplication property) z()=x(t)y(t)= ∑ bm e a. bmenwoLe3muo n=一 Let k=n+ ∑a Interchange the summations order=>i(t)= ∑ anbk-njejkwot k=-0(n=-∞ 2()=x((t)=∑ k=-∞0 (c) Suppose that y(t)=x*(t). Express bk in terms of ak, and use the result of part(a) to prove Parseval's relation for periodic signals
+� +� +� +� +� +� +� +� +� +� +� +� +� +� MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 3 Solution Home Study Exercise (E1) O&W 3.46 (a) and (c) x(t) and y(t) are continuous-time periodic signals with a period = T0 and Fourier series representations given by ⎨ ⎨ bkejk�0t x(t) = akejk�0t y(t) = k=−� k=−� (a) Show that the Fourier series coefficients of the signal ckejk�0t z(t) = x(t)y(t) = �+� are given by the discrete convolution k=−� ck = �+� −� anbk−n (multiplication property). n= ⎨ ⎨ ⎨ ⎨ bmejm�0t = anbmejn�0t jm�0t z(t) = x(t)y(t) = anejn�0t e n=−� m=−� n=−� −� m= ⎨ ⎨ = anbmej(n+m)�0t n=−� m=−� ⎨ ⎨ = anbk−nejk�0t Let k = n + m ⇒ m = k − n � z(t) n=−� k= � −� � ⎨ ⎨ anbk−n jk�0t Interchange the summations order � z(t) = e k=−� n=−� ⎨ ⎨ � z(t) = x(t)y(t) = ckejk�0t , where ck = anbk−n. k=−� n=−� (c) Suppose that y(t) = x�(t). Express bk in terms of ak, and use the result of part(a) to prove Parseval’s relation for periodic signals. 1
y()=x+(t) ase-jku k=-00 Letk=-k→y(t) a bk=a Now to prove Parseval's relation for period signals we can use the results above as follows: r(t)/2=r(t) *(t)=a(t)y(t) 2celkrot, where c =o bk-n, from part( n=-0 ∑ak-nc=∑ an"n-kekw as proven earlier A common mistake is to using the substitution bk-n=a*+n(i.e. just negating the index n) which would be correct if the relationship was shifting in frequency and then taking the conjugate, while the case at hand is taking the conjugate first and then shifting in frequency to perform the convolution Pave=t/=(t)dt= To Jo k==oo n= and dt yot To Notice that the integration has a value of To only when k=0 and zero for all the other values of k, as seen below. This rule, usually referred to as orthogonality holds for any complex exponential signal, integrated over one period Fork≠0 (1)dt= To Therefore we have: P To ana Db=∑|an2 lr(t)dt
� � +� � +� +� +� +� +� +� +� +� +� +� +� ⎨ jk�0t ⎨ ke−jk�0t y(t) = x� (t) = ake = a� k=−� k=−� ⎨ jk�0t Let k = −k � y(t) = a� −ke k=−� � bk = a� −k Now to prove Parseval’s relation for period signals we can use the results above as follows: |x(t)| 2 = x(t)x� (t) = x(t)y(t) ⎨ ⎨ = ckejk�0t , where ck = anbk−n, from part(a) k=−� n=−� ⎨ ⎨ ⎨ ⎨ = anbk−nejk�0t = ana� jk�0t , as proven earlier. n−ke k=−� n=−� k=−� n=−� A common mistake is to using the substitution bk−n = a� (i.e. just negating k+n the index n) which would be correct if the relationship was shifting in frequency and then taking the conjugate, while the case at hand is taking the conjugate first and then shifting in frequency to perform the convolution. 1 ⎪ T0 1 ⎪ T0 +� T0 0 | T0 0 k= n−kejk�0t dt ⎨ ⎨ Pave = |x(t) 2 dt = ana� −� n=−� ⎨ ⎨ ⎪ T0 1 ana� = T0 n−k ejk�0t dt 0 n=−� k=−� Notice that the integration has a value of T0 only when k = 0 and zero for all the other values of k, as seen below. This rule, usually referred to as orthogonality, holds for any complex exponential signal, integrated over one period. ⎪ T0 1 � � T0 1 For k jk�0t � √= 0 : 0 ejk�0t dt = jk�0 e � = jk�0 (ejk�0T0 − ej0 ) 0 1 = (ejk(2�) − ej0 ) = 0 jk�0 ⎪ T0 ⎪ T0 ⎪ T0 For k = 0 : ejk�0t dt = ej(0)�0t dt = (1)dt = T0. 0 0 0 ⎨ 1 ⎨ +� ⎪ T0 1 +� Therefore we have: Pave = ana�T0 = 2 = x(t)| 2 dt. T0 n |an| T0 0 | n=−� n=−� 2
Problem 1(O&w 3. 22(a)-only the signal in Figure p3.22(c)) Determine the Fourier series representation for the signal a (t) c(t) 2+t, for -2<t<O c(t for 0<t<1 r(t) periodic with period T=3→==等 a goal of this problem solution is to show different ways to reaching the same answer Finding the Fourier series coefficients of a signal using the analysis equation usually requires the most effort, but can be reverted to if everything else fails. Oftentimes, a signal can be dissected into simpler signals that are easier to analyze or can be derived from a simpler signal by integration, differentiation, time shifting, or any combination of the properties of the Fourier series (see Table 3.1, O&w, p. 206) We will start with finding ao, which is usually straight-forward and doesn't require much effort, and then explore the different methods for finding ak+0 ao T/(t)dt=-(the total area under the curve for one period)I (2+1)=1 The following are four possible methods to calculate ak+0, the Fourier series coefficients of r(t)fork≠0: Method (a): Using the integration property Let g(t)=do -r(t)=g(t)dt+, where p is the value of r(t)at the beginning of the period, and it equals to zero for the period we selected that starts at t=-2. Note that, since we are trying to find ak+, the value of p is not important because it only ffects the DC level of a(t)and we have already calculated it by finding ao g(t) 234 1
� � 0 Problem 1 (O&W 3.22 (a) - only the signal in Figure p3.22 (c)) Determine the Fourier series representation for the signal x(t). x(t) 2 · · · · · · x(t) = 2 + t, for − 2 ← ←t 2 − 2t, for 0 ← ←t 1. −5 −4 −3 −2 −1 0 1 2 3 4 t T 2� x(t) periodic with period T = 3 �0 = 2 T � ⇒ = 3 A goal of this problem solution is to show different ways to reaching the same answer. Finding the Fourier series coefficients of a signal using the analysis equation usually requires the most effort, but can be reverted to if everything else fails. Oftentimes, a signal can be dissected into simpler signals that are easier to analyze or can be derived from a simpler signal by integration, differentiation, time shifting, or any combination of the properties of the Fourier series (see Table 3.1, O&W, p.206). We will start with finding a0, which is usually straight-forward and doesn’t require much effort, and then explore the different methods for finding ak=0 √ : ⎪ 1 1 1 a0 = x(t)dt = (the total area under the curve for one period) = (2 + 1) = 1. T T 3 3 The following are four possible methods to calculate ak√=0, the Fourier series coefficients of x(t) for k =√ 0: • Method (a): Using the integration property: Let g(t) = dx(t) x(t) = g(t)dt + p, where p is the value of x(t) at the beginning of dt ⇒ the period, and it equals to zero for the period we selected that starts at t = −2. Note that, since we are trying to find ak=0 √ , the value of p is not important because it only affects the DC level of x(t) and we have already calculated it by finding a0. g(t) −5 −4 −3 2 3 4 −1 t −2 1 · · · · · · 3
Note that g(t) must have a zero DC level, otherwise a ramping signal will be included in s(t) making it non-periodic, and unbounded. By definition, g(t) should have a zero DC level because the derivative operation eliminates it, so this can be used as a double-check After finding bk, the Fourier series coefficients for g(t), we can use the Fourier series properties to find ak, the Fourier series coefficients for a(t) (1)e- 3kwotdt 1 0 - swot JAe-jkwot (1-ck02-2e-k0+2) kwo (e2+2e-1k0-3) bk(from the Integration property, Table 3.1, o &W, p. 206) jkwo 3jkwo (3-2e-k0-e2) (1-efkwo2) (remember that e -kuo =eikw02 for T=3) (1-c)=2(1-c) Method(b): Using the integration property twice Let's define v(t)as the following 1)≈2r(t) dt =a(t)=/ u(t)dt dt+p= g(t)dt+p Similar to the discussion in Method(a) of the DC level of g(t), v(t) must have a zero dC level. In addition, its limited integration over one period must also have a zero DC level We can find v(t) by differentiating g(t). However, in our case, but not always, we can find u(t)directly from r(t)in one step, by placing an impulse at each point of time where the slope of x(t) changes abruptly. The value of that impulse (i.e its area)is the change in slope of a(t) at that point
� � � � � � � Note that g(t) must have a zero DC level, otherwise a ramping signal will be included in x(t) making it non-periodic, and unbounded. By definition, g(t) should have a zero DC level because the derivative operation eliminates it, so this can be used as a double-check. After finding bk, the Fourier series coefficients for g(t), we can use the Fourier series properties to find ak, the Fourier series coefficients for x(t) 1 ⎪ ⎪ 1 g(t)e−jk�0t dt = 1 �⎪ 0 (1)e−jk�0t dt + (−2)e−jk�0t bk = dt T T 3 −2 0 �0 �1 1 1 � 1 � � = e−jk�0t − 2 −jk�0 e−jk�0t � −1 jk�02 − 2e−jk�0 � � = + 2 3 −jk�0 � −2 � 3jk�0 1 − e 0 1 � jk�02 + 2e−jk�0 − 3 � = e . 3jk�0 1 ak = bk (from the Integration property, Table 3.1, O &W, p.206 ) jk�0 1 1 � jk�02 + 2e−jk�0 − 3 � 1 � 3 − 2e−jk�0 jk�02 � = e = 3k2�2 − e jk�0 3jk�0 0 = 1 1 − ejk�02� (remember that e−jk�0 = ejk�02 for T = 3) k2�2 0 1 1 − ejk 4 3 � � 1 1 − e−jk 2� = = 3 . k2�2 k2�2 0 0 • Method (b): Using the integration property twice: Let’s define v(t) as the following: ⎪ ⎪ ⎪ d2x(t) dg(t) v(t) = dt2 = � x(t) = v(t) dt dt + p = g(t) dt + p dt Similar to the discussion in Method(a) of the DC level of g(t), v(t) must have a zero DC level. In addition, its limited integration over one period must also have a zero DC level. We can find v(t) by differentiating g(t). However, in our case, but not always, we can find v(t) directly from x(t) in one step, by placing an impulse at each point of time where the slope of x(t) changes abruptly. The value of that impulse (i.e its area) is the change in slope of x(t) at that point. 4
10 To find ck, the Fourier series coefficients of v(t), let's take the period between -1 and 2, which contains two impulses Note that we can also take the period between -2 and 1, but we have to be careful not include the impulses at both -2 and 1. In other words, we can take the period between 2+0 and 1+d or the period between -2-d and 1-0 7/(-koat=1(/(-35()+35(-1)e-ud) [-6(t)+6(t-1) e-gkwot Now to find ak, we just need to use the integration property two times (from the Integration property, Table 3.1, O &W, p. 206) Kuo 2(e-jkwo_1) (1 e-k), which is the same answer found in Method(a)
� � � � � −5 −4 −3 −2 −1 0 1 2 3 4 t v(t) −3 · · · · · · 3 3 To find ck, the Fourier series coefficients of v(t), let’s take the period between -1 and 2, which contains two impulses. Note that we can also take the period between -2 and 1, but we have to be careful not include the impulses at both -2 and 1. In other words, we can take the period between −2 + � and 1 + � or the period between −2 − � and 1 − �. ⎪ ck = 1 v(t)e−jk�0t dt = 1 �⎪ 2 [−3�(t) + 3�(t − 1)] e−jk�0t dt T T 3 −1 ⎪ 2 = [−�(t) + �(t − 1)] e−jk�0t dt −1 = −e−jk�0(0) + e−jk�0(1) = e−jk�0 − 1. Now to find ak, we just need to use the integration property two times: 1 1 ak = ck (from the Integration property, Table 3.1, O &W, p.206 ) jk�0 jk�0 = 1 e−jk�0 � (jk�0)2 − 1 = 1 1 − e−jk�0 � k2�2 0 1 = 1 − e− 3 jk 2� , which is the same answer found in Method(a). k2�2 0 5