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) (:0<la|< P=∑|2=∑向ak2, where N is the largest integer, such that N≤W/o k=-0 2 k=0 1 1--, for any complex B≠0 Plugging in(* N+1 1>R 2-2a 2R R+1 1-a2 2R+(1-)( 2-2 ≥R+1-(1-Ra2( ≥19-0.1a2( pluggin ar2+< 0.05+0.05a2(simplifying a bit) (2N+2)log(a)≤log(0.05+0.05a2) N+1> 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 < |�| < 1) 1 − �2 � N Py = |bk| 2 = |ak| 2 , where N is the largest integer, such that N ← W/�0 k=−� k=−N N N = |�|k| | 2 = 2 �2k − 1 (� � is real) k=−N k=0 1 − (�2) N+1 M � � 1 − �M+1 n = 2 − 1 � = , for any complex � ≤= 0 1 − �2 1 − � n=0 Plugging in (⇒): 1 − (�2) N+1 � � 2 2 − 1 → R − 1 1 − �2 1 − �2 2 − 2�2N+2 2R → − R + 1 1 − �2 1 − �2 2 − 2�2N+2 2R + (1 − R)(1 − �2) → 1 − �2 1 − �2 2 − 2�2N+2 → R + 1 − (1 − R)�2 (� 1 − �2 > 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