MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems-Fall 2003 PROBLEM SET 1 SOLUTIONS (E1)(O&W154) (a) For the r=l case, we have 1+1+12 n=0 For the r+ 1 case, by carrying out the long division, we can see that 1-r 1 N b) Using the formula we just derived for the r+ 1 case, we have 1 If r lim n=0
� � � � MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 1 Solutions (E1) (O&W 1.54) (a) For the r = 1 case, we have: n=N−1 1 = 1 + 11 + 12 + + 1N−1 · · · n=0 = N For the r =→ 1 case, by carrying out the long division, we can see that 1 r = 1 + r + r1 + r2 + + rN−1 + N 1 − r · · · 1 − r N �−1 Nr n = r + 1 − r n=0 N �−1 1 − rN nr = 1 − r n=0 (b) Using the formula we just derived for the r =→ 1 case, we have � N−1 nr = lim rn N�� n=0 n=0 1 − rN = lim N�� 1 − r 1 rN = 1 − r − lim N�� 1 − r If | | r < 1 Nr lim = 0 N�� 1 − r So, � 1 nr = 1 − r n=0 1
+2a2+3a3 a(1+2a+3a Now we can seperate the contents of the paranthesis on the right-hand-side(rHs) of the equation above as follows a(1+a+a2+…+a+2a2+3a3+…) (1+a+ )+a(a+2a2+3a3+…) Note that the contents of the second paranthesis on the RhS is the very expression we are trying to evaluate (1-a)∑ n=0 Using the result from part(b)for a<1 a
�� � �� � � � � �� �� � � �� � = � �� �� � �� � (c) n�n = � + 2�2 + 3�3 + · · · n=0 = � 1 + 2� + 3�2 + � · · · Now we can seperate the contents of the paranthesis on the right-hand-side (RHS) of the equation above as follows: n�n = � 1 + � + �2 + + � + 2�2 + 3�3 + � · · · · · · n=0 = � 1 + � + �2 + � + � � + 2�2 + 3�3 · · · + · · · Note that the contents of the second paranthesis on the RHS is the very expression we are trying to evaluate: n�n = � 1 + � + �2 + + n�n · · · n=0 n=0 (1 − �) n�n = � 1 + � + �2 + � · · · n=0 �n n=0 Using the result from part (b) for | | � < 1, (1 − �) n�n = 1 − � n=0 n�n = (1 − �)2 n=0 2
a n=0 1 一a 1-a
(d) �� �� � k−1 n=k �n = n=0 �n − n=0 �n = 1 1 − � − 1 − �k 1 − � �k = . 1 − � 3
Problem 1 (a) For this, we convert the part that is in cartesian form to polar form and proceed from tan 2 Plugging this into the expression we want to evaluate, we have √3+ 32e In graphical form, this can be represented as follows Figure 1.la: Magnitude and Phase plot
Problem 1 (a) For this, we convert the part that is in cartesian form to polar form and proceed from there: � � 1 � � � 3 + j = tan−1 �3 = �/6 � �� 2 3 + j = � 3 + 12 | | = 2 � 3 + j = 2e�/6 . Plugging this into the expression we want to evaluate, we have: � 5 −j�/3 � 2ej�/6 �5 −j�/3 � 3 + j e = e j5�/6 −j�/3 = 25 e e· = 32ej�/2 In graphical form, this can be represented as follows: ≥e 32 ∞m Figure 1.1a: Magnitude and Phase plot 4
Problem 2 (a) This problem can be solved in stages. First we flip the signal 2 345678 -5-4 2.a.1 Next we scale the time axis by 2 Figure 2.a 2: x(-3 Now we shift by 3 because the time axis has been scaled down by three (1-5) Figure 2.a3: a (1-3)
Problem 2 (a) This problem can be solved in stages. First we flip the signal: 2 3 4 5 6 7 8 9 −6 −5 −4 −3 −2 −1 x(−t) t 1 1 −1 2 Figure 2.a.1: x(−t) : 3 1 Next we scale the time axis by −6 −3 6 9 x(−t 3 ) t 3 1 −1 2 Figure 2.a.2: x(−3 )t Now we shift by 3 because the time axis has been scaled down by three: −6 −3 3 6 x(1 − t 3 ) t 9 1 2 −1 Figure 2.a.3: x(1 − 3 )t 5