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
Before exploring the other methods, let's first find the Fourier series for y(t), shown below, which is a periodic triangular function with a period of T. y(t) will be useful for the Method(c) y(t) (t) ()+dk We will find the Fourier series for a(t) and from it, we will find the Fourier series for y() z(t)e-Jkwo dt swot T Ti(-ikwo) (En-c6)=7n(=k AwoL kuo1⊥a-jku0oT1 Kwoi)I Thus dk=eljko 2-2 cos(hwoTi) TTk2w2
� � � Before exploring the other methods, let’s first find the Fourier series for y(t), shown below, which is a periodic triangular function with a period of T. y(t) will be useful for the Method(c): y(t) t 1 T −T1 T1 T · · · · · · − 2 2 z(t) = dy(t) dt 1 T1 · · · − T1 · · · T −T1 T t 2 2 −1 T1 F Let z(t) = dy(t) , z(t) �⇒ ek , and y(t) F = jk�0 ek dt �⇒ dk 1 We will find the Fourier series for z(t) and from it, we will find the Fourier series for y(t), as follows: ⎪ ⎪ T1 1 z(t) e−jk�0t dt = 1 ⎩⎪ 0 1 )e−jk�0t dt + (−1 )e−jk�0t ek = ( dt T T T −T1 T1 0 T1 1 1 � 1 � T T1 (−jk�0) e−jk�0t 0 − e−jk�0t T1 � 1 jk�0T1 + 1 − e−jk�0T1 � = |−T1 |0 = T T1 (−jk�0) 1 − e � −1 = −1 2 − (ejk�0T1 + e−jk�0T1 ) � = T T1jk�0 [2 − 2 cos(jk�0T1)] . T T1jk�0 d Thus, k = ek( 1 ) = 2 − 2 cos(k�0T1) . jk�0 T T1k2�2 0 6
Method(c): By dissecting the signal into simpler components Here, we will dissect (t)into ni(t)and 2(t)which we know their Fourier Series(using the result of dk above and the time-shifting property) r(t)=a1(t)+r2(t), and let c1(t) bk and a2(t) ak= bk +Ck=(2) 9=30 2-2cos(ku0(1) (3)(1)k26 32(2+) Although this result looks different from those found in the previous methods, further simplification will show that they are identical a=2-208ko+ck)=3k4a(2-2csk(2+c) 32(、ckEo)(2+ek0 3k2(4+2ek0-2ek0-k2-2e~k0-e) 1 (4-ck02-2e-1k0-1) 3k26 (3 k2(1-c0 (remember that e-3kuwo ekwo2 for T=3) which is the same answer found in previous methods
� � � • Method (c): By dissecting the signal into simpler components: Here, we will dissect x(t) into x1(t) and x2(t) which we know their Fourier Series (using the result of dk above and the time-shifting property). x1(t) 2 −4 −3 −2 −1 0 1 2 3 4 t x2(t) −4 −3 −2 −1 0 1 2 3 4 t 1 x(t) = x1(t) + x2(t) , and let x1(t) F and x2(t) F �⇒ bk �⇒ ck � ak = bk + ck = (2) 2 − 2 cos(k�0(1)) + (1)2 − 2 cos(k�0(1)) e−jk�0(−1) (3)(1)k2�2 (3)(1)k2�2 0 0 = 2 − 2 cos k�0 (2 + ejk�0 ). 3k2�0 2 Although this result looks different from those found in the previous methods, further simplification will show that they are identical: 1 ak = 2 − 2 cos k�0 (2 + ejk�0 ) = 3k2�2 (2 − 2 cos k�0)(2 + ejk�0 ) 3k2�2 0 0 1 = 3k2�0 2 (2 − ejk�0 − e−jk�0)(2 + ejk�0 ) 1 � 4 + 2ejk�0 − 2ejk�0 − ejk�02 − 2e−jk�0 0 � = 3k2�2 − e 0 1 � 1 � = jk�02 − 2e−jk�0 � = jk�02 − 2e−jk�0 � − 1 3k2�0 2 4 − e 3k2�0 2 3 − e = 1 1 − ejk�02� (remember that e−jk�0 = ejk�02 for T = 3) k2�2 0 = 1 1 − ejk 4 3 � , which is the same answer found in previous methods. k2�2 0 7
Method (d): using the analysis equation In the process of evaluating the analysis equation, the following integral will save us a lot of derivation steps te dt a≠0 2+t) dt dt+te"-3kwot dt-2t dr 21 wo jkwo kawa two -jkwo(1) 3 hwo , 6 jho 2 1(1) kwo jho e-jkwo(1) k2w2 k2w 2 RwG. 1 e-jhuo 3jhw kwo k2w2 kwa kwa 1 3(k2 12, 2ekup2 3k2∞6 k2w2 (remember that e- o ko2 for T=3) -eika), which is the same answer found in previous methods k2w
� � � � � � � � � � � � � � � � � � � • Method (d): using the analysis equation: In the process of evaluating the analysis equation, the following integral will save us a lot of derivation steps: ⎪ t 1 at teatdt = a − a e , for any a = 0 2 √ ⎪ 1 ⎪ 1 1 � x(t)e−jk�0t dt = x(t)e−jk�0t ak = dt T 3 −2 ⎩⎪ −� ⎪ 1 � 1 0 (2 − 2t)e−jk�0t = (2 + t)e dt −jk�0t dt + 3 −2 0 1 ⎩ ⎪ 1 ⎪ 1 ⎪ � 0 e−jk�0t dt + te−jk�0t dt − 2 te−jk�0t = 2 dr � 3 −2 −2 0 � � � � 0 � e−jk�0t �1 � �1 1 � 1 t � 1 t � = 2 � + � e−jk�0t 3 −jk�0 � k2�2 − jk�0 e−jk�0t � − 2 k2�2 − −2 0 −2 0 jk�0 0 1 −2 e−jk�0(1) − e−jk�0(−2)� 1 1 (−2) e−jk�0(−2) = + 3 jk�0 k2�0 2 − k2�2 − 0 jk�0 ⎩� � �� 1 (1) 1 e−jk�0(1) −2 k2�2 − jk�0 − k2�0 2 0 1 −2 e−jk�0 2 k2 1 �0 2 − k2 1 �0 2 jk�02 2 − jk�0 jk�02 = + ejk�02 + e e 3 jk�0 jk�0 2 2 2 −k2�0 2 e−jk�0 + e−jk�0 + jk�0 k2�2 0 1 −2 1 2 e−jk�0 2 e−jk�0 2 e = + −jk�0 + + 3 jk�0 − k2�0 2 jk�0 k2�2 k2�2 0 0 2 1 2 + ejk�02 jk�02 jk�02 e e jk�0 − k2�0 2 − jk�0 1 2 3 1 + ejk�02 = 3 −k2�0 2 e−jk�0 k2�0 2 − k2�0 2 1 � = 3 − 2e−jk�0 jk�02� 3k2�0 2 − e = 1 1 − ejk�02� (remember that e−jk�0 = ejk�02 for T = 3) k2�2 0 = 1 1 − ejk 4 3 � , which is the same answer found in previous methods. k2�2 0 = 9 1 − ejk 4 3 � . 4k2�2 8
Problem 2 O w323(a) Given ak, the Fourier series coefficients of a periodic continuous time signal with period 4 determine the signal a(t) The Fourier series coefficients ak are given as follows in k/4 T, otherwise k Here are some of the facts we know about a(t) a0=0- no DC component in a(t) ·T=4→o=2x/4=丌/2 Gr* sin(-k/ 4) sin(k /4) (-)smkx/4= Thus r(t) is a real signal (O&W, Section 3.5.6, p. 204) Noting that j=ej/2-G)=(ei/2)=ekm/2= ejkwo =e-jkuwo( -1), we can consider r(t) to be a time-shifted version of another signal y(t)such that x()=y/(t +1), where v(0)-, bo =0, ba 0- sin k/4 and ak =bge/ko() By backtracking the derivation equation of bk, we can find the signal i(t)which has the same bk but can have a different DC level (i.e. bo) Sin b≠0=b≠0 k丌/41/c1kx/4-c- k 2 (1)e?kott The integration above suggests that <t< 0(t lo, elsewhere in the same period T=4
� � � � � � Problem 2 O & W 3.23 (a) Given ak , the Fourier series coefficients of a periodic continuous time signal with period 4, determine the signal x(t). The Fourier series coefficients ak are given as follows: a �0, k = 0 k = �(j)k sin k�/4 , otherwise. k� Here are some of the facts we know about x(t): • a0 = 0 ⇒ no DC component in x(t) • ⇒ T = 4 �0 = 2�/4 = �/2 • � �k (j) −k sin(−k�/4) 1 − sin(k�/4) a−k = = −k� j −k� (−j) k sin(k�/4) = = a� k. k� Thus x(t) is a real signal (O&W, Section 3.5.6, p.204). e−jk�0(−1) Noting that j = ej�/2 ⇒ (j)k = ej�/2 �k = ejk�/2 = ejk�0 = , we can consider x(t) to be a time-shifted version of another signal y(t) such that: F x(t) = y(t + 1), where y(t) �⇒ b0 = 0, bk=0 = sin k�/4 and ak = bkejk�0(1) √ k� By backtracking the derivation equation of bk, we can find the signal ˆ b y(t) which has the same k but can have a different DC level (i.e. b0) : ˆbk=0 √ = bk=0 = sin k�/4 = 1 ejk�/4 − e−jk�/4 √ k� k� 2j 1 1 � = (4) jk(� ejk�/4 − e−jk�/4 � 2 ) ⎪ 1 2 ) 1 1 1 2 jk�0( 1 − ejk�0(− 2 ) � 1 (1)ejk�0t = e = dt. 1 2 T jk�0 T − The integration above suggests that 1 yˆ(t) = 1, −2 < t < 1 2 0, elsewhere in the same period T=4. 9
0(t) Note that the same conclusion can be reached by noticing that i(t) is the same signal in Example 3.5(O& W, p 193)with Ti=and T=4 To find y(t), which has bo=0, we first calculate bo and then subtract it from g(t) 1/2 bo T 0()dt (1)dt <t< →()=()-→y(O)= 号<|<2 x(t)=y(t+1) 1.5<t<-0.5 0.5<t<2.5 ketches of y(t)and r t) are shown below y(t) 3/4 3/4 1/4
� � yˆ(t) −3 −2 −1 0 1 2 3 t 1 −T 2 T 2 · · · · · · Note that the same conclusion can be reached by noticing that yˆ(t) is the same signal in Example 3.5 (O& W, p.193) with T 1 1 = 2 and T = 4. To find y(t), which has b0 = 0, we first calculate ˆb0 and then subtract it from yˆ(t) : ⎪ 1 1 ⎪ 1/2 1 ˆb0 = yˆ(t)dt = (1)dt = T T 4 −1/2 4 1 2 2 3 1 1 < t < ⇒ y(t) = yˆ(t) − 4 � y(t) = 4 , − 1 1 4 , < |t| < 2. − 2 3 −1.5 < t < −0.5 x(t) = y(t + 1) = 4 , ⇒ 1 4 − , −0.5 < t < 2.5 Sketches of y(t) and x(t) are shown below: y(t) −3 −2 −1 1 2 3 t 3/4 −1/4 · · · · · · x(t) −3 −2 1 2 3 t 3/4 −1/4 −1 · · · · · · 10
