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
(b)For this part, we plot the individual signals involved and take the sum of products 2 r(-2)6(t-)+(3-t) Figure 2.b: a(t-2)[8(t-3)+u(3
(b) For this part, we plot the individual signals involved and take the sum of products: 1 2 −1 1 2 3 4 5 6 x(t − 2) t −1 1 2 −1 1 2 3 4 5 6 �(t − 1 2 ) 1 2 (1) −1 1 2 3 4 5 6 u(3 − t) 2 −1 −1 1 2 3 4 5 6 x(t − 2)[�(t − 1 2 u(3 − t)] 1 2 ( 1 2 ) 1 2 -1 ) + t t t Figure 2.b: x(t − 2)[�(t − 1 2 ) + u(3 − t)] 6
Problem 3 (a) This problem can be solved in two stages. First we Hip the signal and then we shift by 345 1 234 (b) From the expression, all we have to do is take every odd sample. When you plug in n=0, n=l, n=2... into the expression 2n+l, you end up with the odd samples of the original signal as follows x2n+1
Problem 3 (a) This problem can be solved in two stages. First we flip the signal and then we shift by 2: 1 −1 −4 −3 −2 −1 1 2 3 4 5 x[−n] n 1 2 1 2 −1 2 x[2 − n] 1 −1 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 n 1 2 1 2 −1 2 (b) From the expression, all we have to do is take every odd sample. When you plug in n = 0, n = 1, n = 2 · · · into the expression 2n + 1, you end up with the odd samples of the original signal as follows: x[2n + 1] 1 −3 −2 −1 1 2 3 4 n 1 2 −1 2 7
Problem 4 In order to find the even part me(t)of a signal x(t) we use the relation r(t)+x(-t) 2 Similarly, for the odd part To(t), we use the relation t)-ac-t Using these expressions for the odd and even parts, we end up with the following pictures :a(t) r(-) ce(t)
Problem 4 In order to find the even part xe(t) of a signal x(t), we use the relation: xe(t) = x(t) + x(−t) . 2 Similarly, for the odd part xo(t), we use the relation: xo(t) = x(t) − x(−t) . 2 Using these expressions for the odd and even parts, we end up with the following pictures: x(t) 1 t −4 −3 −2 −1 1 2 3 4 -1 1 2x(−t) 1 2 −4 −2 2 4 t 1 -1 xe(t) 1 −4 −2 2 4 t -1 8
The value of the even part (and the odd part for that matter at t=0 is ambiguous as it depends on how the plot for r (t) is defined at t =0. The plots in this solution assume that the value of a(t)at t=0 is halfway between 0 and 2, i.e. 1. Using a different definition you may get an even part that is discontinuous at t=0. This is also correct provided it is consistent with your assumption of what the value of a(t) is at the discontinuity. For instance, if you assume that a(0)=2, then the plot of the even part will have a"spike"at t=0 of height 2
xo(t) −4 −2 2 4 t 1 -1 The value of the even part (and the odd part for that matter) at t = 0 is ambiguous as it depends on how the plot for x(t) is defined at t = 0. The plots in this solution assume that the value of x(t) at t = 0 is halfway between 0 and 2, i.e. 1. Using a different definition you may get an even part that is discontinuous at t = 0. This is also correct provided it is consistent with your assumption of what the value of x(t) is at the discontinuity. For instance, if you assume that x(0) = 2, then the plot of the even part will have a “spike” at t = 0 of height 2. 9
Problem 5 (a) In order to figure out if a Ct signal a (t)is periodic, we need to find a finite, non-zero value of T such that a(t)=a(t+T) for all t. The smallest T that satisfies this is the fundamental period This function is quite straightforward. We know that the function sin(4t-1)is periodic with period 3. Since the positive and negative cycles of sinusoids have the same shape the square of this function, i.e. a(t)=[sin(4t-1)2 is periodic with fundamental period Also, we can use the relation 11 SIn T 22 which is periodic with period (b)For a DT function an], we need to find a finite, non-zero integer N such that a[nl cn +N for all n. The smallest integer N for which this holds is the fundamental period. If we cannot find such an N, then the function is not periodic We need cos 4(n+N)+ 4 cos「4n+ For the above to hold, the following has to be true for some integer(s)k AN 4n++2k 4 /I Since T is not a rational number, we cannot find an integer n that satisfies this. Thus, the function is not periodic (c) We can use the same steps as we did above but we can start with finding the funda- mental period of the simpler function y[n]=cos(2) We need the following to hold 2T(n +N) 7 7 So we need the following to hold for at least one integer value of k
� � � � � � � � � � � � Problem 5 (a) In order to figure out if a CT signal x(t) is periodic, we need to find a finite, non-zero value of T such that x(t) = x(t + T) for all t. The smallest T that satisfies this is the fundamental period. This function is quite straightforward. We know that the function sin(4t−1) is periodic . Since the positive and negative cycles of sinusoids have the same shape, � with period 2 the square of this function, i.e. x(t) = [sin(4t−1)]2 is periodic with fundamental period . � 4 Also, we can use the relation 1 1 sin2 x = sin 2x, 2 − 2 � which is periodic with period 4 . (b) For a DT function x[n], we need to find a finite, non-zero integer N such that x[n] = x[n + N] for all n. The smallest integer N for which this holds is the fundamental period. If we cannot find such an N, then the function is not periodic. We need cos 4(n + N) + = cos 4n + 4 4 For the above to hold, the following has to be true for some integer(s) k. � � 4n + 4N + = 4n + + 2�k 4 4 � N = k 2 Since � is not a rational number, we cannot find an integer N that satisfies this. Thus, the function is not periodic. (c) We can use the same steps as we did above but we can start with finding the fundamental period of the simpler function y[n] = cos 2�n 7 . We need the following to hold 2�n 2�(n + N) cos = cos 7 7 So we need the following to hold for at least one integer value of k. 10
