MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems-Fall 2003 PROBLEM SET 4 SOLUTION Home Study Exercise -o&e W 3.63 For an Lti system whose frequency response H(w)= 回≤W and which has a continuous-time periodic input signal a(t) with the following Fourier series representation: a(t)=> lesk(m/a), where a is a real number between 0 and 1 How large must W be in order for the output of the system to have at least 90% of the riod of s(t)? Basically, H(w) is an ideal Low Pass Filter(LPF)and we need to find how wide it needs to be, in order to pass at least 90% of its input's average energy per period (i.e. average power) First, let's rewrite the condition above relating the average powers of the input and out put, with Fourier series coefficients ak and bk, respectively P=∑|l2,P Jbrl The required condition, then, would be P≥BP→∑M2≥R∑a2, where R=09(*) k=-0o Then, lets calculate the Fourier series coefficients of the output, bk bk=B(k)=么、(m,1l≤WF=am,H≤W/a0 10.ka>w=10,>W/o
�� �� �� �� �� MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 4 Solution Home Study Exercise - O&W 3.63 For an LTI system whose frequency response is: ⎪ 1, |�| ← W H(j�) = 0, |�| > W. and which has a continuous-time periodic input signal x(t) with the following Fourier series representation: x(t) = �|k| ejk(�/4)t , where � is a real number between 0 and 1 k=−� How large must W be in order for the output of the system to have at least 90% of the average energy per period of x(t)? Basically, H(j�) is an ideal Low Pass Filter (LPF) and we need to find how wide it needs to be, in order to pass at least 90% of its input’s average energy per period (i.e. average power). First, let’s rewrite the condition above relating the average powers of the input and output, with Fourier series coefficients ak and bk, respectively: |bk| 2 Px = |ak| 2 , Py = k=−� k=−� The required condition, then,would be: |bk| 2 Py → RPx � → R |ak| 2 , where R=0.9 (⇒) k=−� k=−� Then, let’s calculate the Fourier series coefficients of the output, bk: ⎪ ⎪ ak, |k�0| ← W ak, |k| ← W/�0 � bk = akH(jk�0) � bk = = 0, |k�0| > W 0, |k| > W/�0 1
And finally, we plug the expressions of ak and bk in the required condition and simplify By matching the the expression of a (t)with the synthesis equation, we can conclude that P=∑ak2=∑2=2∑a2-1(: a is real) (:0R 2-2a 2R R+1 1-a2 2R+(1-)( 2-2 ≥R+1-(1-Ra2( ≥19-0.1a2( pluggin ar2+ log(0.05+0.05a2) 2 log(a) (recall that a <1- log(a)< 0 log(0.05+0.05a2) 2 log(a) After choosing an integer n that satisfies the inequality above, W can be chosen such that
�� �� �� � � � � � � � � � And finally, we plug the expressions of ak and bk in the required condition and simplify. By matching the the expression of x(t) with the synthesis equation, we can conclude that 0 = � and ak = �|k| 4 Px = |ak| 2 = |�|k| | 2 = 2 �2k − 1 (� � is real) k=−� k=−� k=0 2 = − 1 (� 0 0) 2 − 2�2N+2 � → 1.9 − 0.1�2 (plugging in R=0.9) 2N+2 ← 0.05 + 0.05�2 (simplifying a bit) (2N + 2)log(�) ← log(0.05 + 0.05�2 ) log(0.05 + 0.05�2) N + 1 → (recall that � < 1 � log(�) < 0) 2 log(�) log(0.05 + 0.05�2) N → − 1 2 log(�) After choosing an integer N that satisfies the inequality above, W can be chosen such that W → N�0. 2
Problem 1 Consider the Lti system with impulse response given in O&w3.34. Find the Fourier series representation of the output y(t) for the following input (t) 1 From o& W3.34, the impulse response of the Lti system is h(t) From the figure above, we can see that c(t) has a period T=3-w0= 3 First, we calculate the frequency response h(t)e Judt e-jut dt dt+/e-4(e-jutdt e(4-y)d+ 1 4- jw (1-0)+ (0-1)(remember that e-oot3a=0 for any real a 4+ jw+4-jw 16+
� � � � Problem 1 Consider the LTI system with impulse response given in O&W 3.34. Find the Fourier series representation of the output y(t) for the following input. x(t) −4 −3 2 3 5 6 · · · −5 −2 −1 2 −1 1 4 · · · t From O & W 3.34, the impulse response of the LTI system is: t h(t) = e−4| |. From the figure above, we can see that x(t) has a period T = 3 � �0 = 2 3 � . First, we calculate the frequency response: � t e−j�tdt � H(j�) = � h(t)e−j�tdt = � e−4| | −� −� 0 e−4(−t) e−j�tdt + � e−4(t) e−j�t = dt � −� � 0 0 (4−j�)t dt + � (−4−j�t)t = e e dt −� 0 � � 0 1 � 1 �� (4−j�)t � + (−4−j�)t = e e 4 − j� � −� −4 − j� 0 1 1 = (1 − 0) + (0 − 1) (remember that e−�+ja = 0 for any real a) 4 − j� −4 − j� 1 1 4 + j� + 4 − j� = + = 4 − j� 4 + j� 16 + �2 8 H(j�) = . 16 + �2 3
Next, we find the Fourier series coefficients of r(t),ak dt [26(t)-6(t-1) dt 3(2c-0 jkwo (1) 2 e-jkwo. for allk Now we are ready to find bk, the Fourier series representation of the output y(t) bk akHGkwo)(o& w, Section 3.8, p 226, and specifically eq (3.124)) 6+(ko)2 3 3/16+(k22, for all k
� � � � � � Next, we find the Fourier series coefficients of x(t), ak: 1 ak = x(t) e−jk�0t dt = 1 � 2 [2α(t) − α(t − 1)] e−jk�0t dt T T 3 −1 = 1 2e−jk�0(0) − e−jk�0(1)⎩ 3 2 1 = − e−jk�0, for allk. 3 3 Now we are ready to find bk, the Fourier series representation of the output y(t): bk = akH(jk�0) (O & W, Section 3.8, p.226, and specifically eq.(3.124) ) 2 1 8 = − e−jk�0 3 3 16 + (k�0)2 � � 8 2 − e−jk 2 3 � = 3 16 + � k 2� ⎩2 , for all k. 3 4
Problem 2 The periodic triangular wave shown below has Fourier series coefficients ak sin(kr/2) ak=13(km)2eP,k≠0 k=0 Consider the LTI system with frequency response HGw) depicted below y() 23-2-91g192g2 Determine values of A1, A2, A3, 01, Q2, and @3 of the LTI filter H(u)such that y(t)=1-cos At the beginning, it is worth noting that the output y(t) contains only a dC component and a single sinusoid with a frequency of 3. H(jw)is a linear system so the output will only have frequency components that exit in the input. Knowing that the input a (t) has a dC component and a fundamental frequency of wo= 3, let's dissect y(t) into a DC component
� � � Problem 2 The periodic triangular wave shown below has Fourier series coefficients ak. x(t) � �2 sin(kω/2)e−jk�/2 ⎧ , k ≤= 0 j(kω)2 1 ak = ⎧� 1 , k = 0. 2 · · · · · · −4 −2 0 2 4 t Consider the LTI system with frequency response H(j�) depicted below: H(j�) x(t) H( ) j� y(t) −�3 −�2 −�1 �1 �2 �3 � A1 A2 A3 Determine values of A1, A2, A3, �1, �2, and �3 of the LTI filter H(j�) such that 3ω y(t) = 1 − cos t . 2 At the beginning, it is worth noting that the output y(t) contains only a DC component and a single sinusoid with a frequency of 3 2 � . H(j�) is a linear system so the output will only have frequency components that exit in the input. Knowing that the input x(t) has a DC component and a fundamental frequency of �0 = 2 , let’s dissect y(t) into a DC component 5
and complex exponentials with a fundamental frequency of wo=2 y(t) 1-e(3)2t1 (-3)t 点3)=∑b be-2,k otherwise b=akH(jk)→H(k)=a, for ak≠0 y(t) has only three non-zero components, therefore H(u) has a non-zero value at those three components, corresponding to A1, A2 and A, as follows H(j(0)o) bo 1 1/2 2=4 H((3)o) b3 (3.)2 4sin(3丌/2 9 j)=7丌2 H(1(-3)o) ÷2 (-3.丌/2) j(9)m2 4sin(-3丌/2) 9 4 HGw) also needs to eliminate the other frequency components of a(t)which do not exist in the output y(t) H(jk0)=0,fork≠0,± To meet the above conditions, the cutoff frequencies 1.2.3 must be chosen to pass the desired components and reject the undesired components. The following inequalities meet that
� � � � � and complex exponentials with a fundamental frequency of �0 = 2 . � � ej 3� t − 1 e−j 3� t 3ω 2 2 y(t) = 1 − cos t = 1 − 2 2 2 1 j(−3) � t = 1 − 2 2 1 ej(3) � t − e 2 2 1 1 j(−3)�0t = bkejk�0t = 1 − ej(3)�0t − e 2 2 k ⎧1, k = 0 − � 1 bk 2 , k = ±3 ⎧0, otherwise � bk = akH(jk�0) � H(jk�0) = bk , for ak ≤= 0. ak y(t) has only three non-zero components, therefore H(j�) has a non-zero value at those three components, corresponding to A1, A2 and A3 as follows: b0 1 H(j(0)�0) = = = 2 = A2. a0 1/2 b3 1 H(j(3)�0) = = − ÷ 2 sin(3 · ω/2) e−j(3)�/2 a3 2 j(3 · ω)2 = − j(9)ω2 ej 3� 2 4 sin(3ω/2) 9 = − j9ω2 (−j) = ω2 = A3. 4(−1) 4 1 sin(−3 · ω/2) e−j(−3)�/2 H(j(−3)�0) = b−3 = − ÷ 2 a−3 2 j(−3 · ω)2 = − 2 j(9)ω2 ej −3� 4 sin(−3ω/2) 9 = −j9ω2 (j) = ω2 = A1. 4(1) 4 H(j�) also needs to eliminate the other frequency components of x(t) which do not exist in the output y(t). � H(jk�0) = 0, for k =≤ 0, ±3 To meet the above conditions, the cutoff frequencies �1,2,3 must be chosen to pass the desired components and reject the undesired components. The following inequalities meet that 6
requirement 0<91<(1)0,(2)0<92<(3)0,(3)0<93<(4)0 (21) Choosing any cutoff frequencies which satisfy (2. 1) would be sufficient. Let's say, for practicality sake, that we want to choose the cutoff frequencies in the midpoints between the desired and undesired frequency components. This gives us the following specific values 2,92,(2+3)u 0+1)u T
requirement: 0 < �1 < (1)�0, (2)�0 < �2 < (3)�0, (3)�0 < �3 < (4)�0 (2.1) Choosing any cutoff frequencies which satisfy (2.1) would be sufficient. Let’s say, for practicality sake, that we want to choose the cutoff frequencies in the midpoints between the desired and undesired frequency components. This gives us the following specific values: �1 = (0 + 1)�0 2 , �2 = (2 + 3)�0 2 , �3 = (3 + 4)�0 2 , where �0 = ω 2 ω 5ω 7ω � �1 = 4 , �2 = 4 , �3 = 2 . 7
Problem 3 Consider a causal discrete-time LTI system whose input an] and output g/nl are related by the following difference equation yn-=yn-1=n+ 2 rn-4 Find the Fourier series representation of the output yIn] when the input is a[n=2+sin(rn/4)-2 cos(Tn/2) First, let's find the frequency response of the system from the difference equation by injecting an input, a[n, that is an eigenfunction of the LTI system r]=en→yn]=H(e)e H(eu) is the frequency response characterizing the system or the eigenvalue of the system By substituting a[n] and yn] in the difference equation y-n-1=m+2nn-4 H(e)e H(e)e-4. H(e)emn e= e+2.eune H(e)el eum 1+2e-Ju4] 1+2e-1u4 Then, we find the Fourier series coefficients, ak, of the given input, possibly by dissecting the input expression into a summation of complex exponentials ]=2+sin(丌n/4)-2cos(m/2)=2+sin(0n)-2cos(2u0m), where wo=- is the greatest Common Factor for the sinusoids frequencies Won won 2e(0×0n+ 2e0k0n⊥1 +c(1)0n_。(-1)on-a(2)nc(-2)un 23 N=8- ak has only eight distinct values and is periodic with a period of N=8
� � Problem 3 Consider a causal discrete-time LTI system whose input x[n] and output y[n] are related by the following difference equation: 1 y[n] − y[n − 1] = x[n] + 2x[n − 4] 4 Find the Fourier series representation of the output y[n] when the input is x[n] = 2 + sin(ωn/4) − 2 cos(ωn/2). First, let’s find the frequency response of the system from the difference equation by injecting an input, x[n], that is an eigenfunction of the LTI system: j�n � y j�) ej�n x[n] = e [n] = H(e H(ej�) is the frequency response characterizing the system or the eigenvalue of the system. By substituting x[n] and y[n] in the difference equation: 1 y[n] − y[n − 1] = x[n] + 2x[n − 4] 4 j�) ej�(n−1) j�(n−4) H(ej�) ej�n − 1 · H(e = ej�n + 2 · e 4 1 j�) ej�n − j�) ej�n ej�(−1) j�n ej�(−4) H(e · H(e = ej�n + 2 · e 4 H(ej�) ej�n 1 j�n � 1 + 2 e−j�4 � 1 − e−j� = e 4 � H(ej�) = 1 + 2 e−j�4 . 1 − 1 e−j� 4 Then, we find the Fourier series coefficients, ak, of the given input, possibly by dissecting the input expression into a summation of complex exponentials: x[n] = 2 + sin(ωn/4) − 2 cos(ωn/2) = 2 + sin(�0n) − 2 cos(2�0n), ω where �0 = is the Greatest Common Factor for the sinusoids frequencies 4 ej�0n − e−j�0n ej2�0n + e−j2�0n = 2 ej(0)�0n + − 2 · 2j 2 1 1 = 2 ej(0)�0n + ej(1)�0n − j(−1)�0n − ej(2)�0n − ej(−2)�0n e 2j 2j � N = 8 � ak has only eight distinct values and is periodic with a period of N = 8. 8
kkk,kk 0 3,4,5 And finally, we find the Fourier series coefficients, bk, of the output y[n 1+2e-jk04 1+2e-k4 bk= akH(eko)=ap ek We have only few non-zero coefficients, so we can go ahead and evaluate them. In doing so it is sometimes useful while computing the value of the complex exponential, to visualize the the complex vector elko going around the unit circle. As the integer k increases by one, the complex vector's angle increases by an angle of g 1+2e-(0x 1-1e-j() by aH(e(wo) 1)1+2e-)z 1+2(-1 2/1- 0.1247+j0.5806 H(e (-1)uo 1+2e--1)r 1+2(-1) 0.1247-j0.5806 b2=a2H(e120)=(-1)1+2e-1e)x eF=(-1) 1+2(1) 28235+j0.7059 1+2e- 1+2(1) e-(-2) 28235-j0.7059 4, As a double-check for our answer, notice that b-k=b* which indicates that the output yn is real. We expect this because the input is real and because the LTi system applies a real operation, i.e. the difference equation. We also expect this from a mathematical point of
� � � � � � � ⎧⎧−1, k = −2 ⎧⎧⎧−2 1 j ⎧⎧ , k = −1 ⎧2, k = 0 � ak = ⎧ 1 ⎧2j , k = 1 ⎧⎧⎧⎧⎧−1, k = 2 ⎧ 0, k = 3, 4, 5 And finally, we find the Fourier series coefficients, bk, of the output y[n]: 4 1 + 2 e− 4 jk � bk = akH(ejk�0) = ak · 1 + 2 e−jk�04 = ak · 1 − 1 e−jk � 1 − 1 e−jk�0 4 4 4 1 + 2 e−jk� = ak · 1 − 1 e−jk � 4 4 We have only few non-zero coefficients, so we can go ahead and evaluate them. In doing so, it is sometimes useful while computing the value of the complex exponential, to visualize the the complex vector ejkω going around the unit circle. As the integer k increases by one, the complex vector’s angle increases by an angle of β. � 1 + 2 6 b0 = a0H ej(0)�0 ⎩ = (2) 1 + 2 e−j(0)� = (2) = 1 − 1 e−j(0) � 1 − 1 3 4 4 4 4 = 8 b1 = a1H � ej(1)�0 ⎩ = 1 1 + 2 e−j(1)� 1 1 + 2(−1) = 1 − 1 e−j(1) � 1 2j 4 2j 4 ( 1 − j �2 ) 4 1 − 1 �2 = 0.1247 + j 0.5806 b−1 = a−1H � ej(−1)�0 ⎩ = �−1 2j � 1 + 2 e−j(−1)� 1 − 1 4 e−j(−1) � 4 = �−1 2j � 1 + 2(−1) 1 − 1 4 ( � 1 2 + j � 1 2 ) = 0.1247 − j 0.5806 b2 = a2H � ej(2)�0 ⎩ = (−1) 1 + 2 e−j(2)� 1 − 1 4 e−j(2) � 4 = (−1) 1 + 2(1) 1 − 1 4 (−j) = −2.8235 + j 0.7059 b−2 = a−2H � ej(−2)�0 ⎩ = (−1) 1 + 2 e−j(−2)� 1 − 1 4 e−j(−2) � 4 = (−1) 1 + 2(1) 1 − 1 4 (j) = −2.8235 − j 0.7059 b3,4,5 = 0. As a double-check for our answer, notice that b−k = b� k which indicates that the output y[n] is real. We expect this because the input is real and because the LTI system applies a real operation, i.e. the difference equation. We also expect this from a mathematical point of 9
view because the input is real, i.e. a-k=ak, and H(e-u)=h*(eu), from the calculated expression
view because the input is real, i.e. a−k = a� k, and H(e−j�) = H�(ej�), from the calculated expression. 10