MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems--Fall 2003 PROBLEM SET 2 SOLUTIONS (E1)O&W1.38(a) (a) From Figure 1. 34 in O&W, we have the following 6△(t) Figure 2.o1.1 Narrow Pulse Now, consider SA(2t)which is a time compressed version of Figure 1. 34 in Oppenheim and willsky Figure 2.O1.2: Time Compressed Narrow Pulse The area under this new pulse is of course 1. If we take the limit as 4-0,we end up with
� � � � MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 2 Solutions (E1) O&W 1.38(a) (a) From Figure 1.34 in O&W, we have the following ��(t) 1 � t Figure 2.O1.1: Narrow Pulse Now, consider ��(2t) which is a time compressed version of Figure 1.34 in Oppenheim and Willsky. ��(2t) 1 2 1 � t Figure 2.O1.2: Time Compressed Narrow Pulse 1 The area under this new pulse is of course 2 . If we take the limit as � � 0, we end up with: 1
6(2t)=limb△(2t) The area under which remains 3. From the result in Section 1.4.2 of the text we 6(2)=亓6(t) We can also show this using the relationship between the unit step function and the unit impulse (i.e. the unit impulse is the time derivative of the unit step function). For a given A, the approximation of both unit steps t) and uA(2t)are shown to the right. Note that u,△(2t) reaches unity at t=会 Making a change of variable s= 2t where ds 6△(2)6△(t) 2dt and using Eqn(1. 73)in O& w, gives 2dt 1du△(2)12 2 dt 2△△ The area enclosed by a(2t) is half of that of Sa(t) as shown to the right. Since &(t)is defined by it's area, we get d(2t)=38(t). In general 6(a)=一0(t) holds for any nonzero real number a. The above is a proof of the time scaling property, which tells us that we've simply squeezed da(t) by a
� � �(2t) = lim ��(2t) ��0 The area under which remains 1 . From the result in Section 1.4.2 of the text we 2 have: 1 �(2t) = �(t) 2 u�(2t) We can also show this using the relationship 1 between the unit step function and the unit �� u�(t) impulse (i.e. the unit impulse is the time � derivative of the unit step function). For a given �, t the approximation of both unit steps � 0 � u�(t) and u�(2t) are shown to the right. Note 2 � that u�(2t) reaches unity at t = �. 2 Making a change of variable s = 2t where ds = ��(2t) ��(t) 1 2dt and using Eqn(1.73) in O&W, gives � du�(s) du�(2t) ��(s) = = ds 2dt 1 du�(2t) 1 2 1 = = = . 2 dt 2 � � t 0 � The area enclosed by ��(2t) is half of that of 2 � ��(t) as shown to the right. Since �(t) is defined by it’s area, we get �(2t) = 1 �(t). In general 2 1 �(at) = �(t) |a| holds for any nonzero real number a. The above is a proof of the time scaling property, which tells us that we’ve simply squeezed ��(t) by a factor of two. 2
(E2)O&W2.33(a-(i) (a)()We need to find the homogeneous and particular solutions with the final solution being the sum of the two. Let's start with the particular solution which we will denote yp(t). Let the particular solution take the form of the input for t> 0, (t)=Aet. Substituting into(P2 33-1)for t>0 we ha 3.Aet +2Ae Which a g noting the homogeneous solution as yn(t), we have yn(t)= Best where B and s are constants to be determined. Substituting into(P2 33-1)with the input set to o we have Bse+ 2Be 0 Thus the homogeneous solution takes the form Be-2t. The output is then given y(t)= yn(t)+ yp(t) We know that the system is initially at rest, so at t=0, the output has to be zero B Thus, the output, y(t) + Note that we had to solve for B with the additional information(in addition to the differential equation) that the system was at initial rest. This is because LCCDES are Not complete characterizations of systems. In general, we need more information to compute the output when given and input signal
(E2) O&W 2.33 (a-(i)) (a) (i) We need to find the homogeneous and particular solutions with the final solution being the sum of the two. Let’s start with the particular solution which we will denote yp(t). Let the particular solution take the form of the input for t > 0, yp(t) = Ae3t . Substituting into (P2.33 − 1) for t > 0 we have, 3t 3t 3Ae3t + 2Ae = e 1 A = 5 Which 3t gives us the particular solution, yp(t) = 1 e . 5 For the homogeneous part, let’s try a general exponential, again for t > 0. Denoting the homogeneous solution as yh(t), we have yh(t) = Best where B and s are constants to be determined. Substituting into (P2.33 − 1) with the input set to 0 we have, 3t Bse3t + 2Be = 0 s + 2 = 0 s = −2 Thus the homogeneous solution takes the form Be−2t . The output is then given by y(t) = yh(t) + yp(t) = Be−2t + 1 3t e 5 We know that the system is initially at rest, so at t = 0, the output has to be zero: y(t) = Be−2(0) + 1 3t e 5 1 B = −5 1 3t Thus, the output, y(t) = −1 e−2t + 5 e . 5 Note that we had to solve for B with the additional information (in addition to the differential equation) that the system was at initial rest. This is because LCCDEs are NOT complete characterizations of systems. In general, we need more information to compute the output when given and input signal. 3
(E3)O&W2.44(a) (a) Consider the general signals that satisfy the given restrictions as depicted below. Note, that the graph is of the time-flipped version of h(T) h(-r) 12 h(-r{+T1+T2) T1-T2 h(-TI-Ti-T2) T1↑T1+212 Figure 2.03: x(t)and h(t)at Break Points in Flip and Slide
(E3) O&W 2.44 (a) (a) Consider the general signals that satisfy the given restrictions as depicted below. Note, that the graph is of the time-flipped version of h(� ). h(−� ) � −T2 T2 x(� ) � −T1 T1 h(−� + T1 + T2) � −T1 − T2 −T1 − 2T1 −T1 h(−� − T1 − T2) � T1 + T2 T1 T1 + 2T2 Figure 2.03: x(t) and h(t) at Break Points in Flip and Slide 4
Using the Hip and slide method to perform the convolution, we see that the product of the two signals(based on overlap)is first non-zero at t=-T1-T2 and it stops being non-zero at t=1+T2. Thus, the output, y(t)is zero for t>T1+ T2 Generally, when convolving two CT signals of finite duration, the result starts being non-zero at the sum of the time indices when the two original signals start being non- zero. Similarly, the result of the convolution ends being non-zero at the sum of the time indices when the signals to be convolved end being non-zero
Using the flip and slide method to perform the convolution, we see that the product of the two signals (based on overlap) is first non-zero at t = −T1 − T2 and it stops being non-zero at t = T1 + T2. Thus, the output, y(t) is zero for |t| > T1 + T2. Generally, when convolving two CT signals of finite duration, the result starts being non-zero at the sum of the time indices when the two original signals start being nonzero. Similarly, the result of the convolution ends being non-zero at the sum of the time indices when the signals to be convolved end being non-zero. 5
Problem 1 (a) With short duration DT sequences, it is often simplest to find their convolution by centering copies of one of the signals about each of the non-zero samples of the other signal and scaled by the value of the sample at that location. The result is the sum of all the shifted and scaled signals. Thus, yIn is given by the sum of the following -6-5
Problem 1 (a) With short duration DT sequences, it is often simplest to find their convolution by centering copies of one of the signals about each of the non-zero samples of the other signal and scaled by the value of the sample at that location. The result is the sum of all the shifted and scaled signals. Thus, y[n] is given by the sum of the following signals. x[n + 2] 1 2 −1 −2 −4 −3 −2 −1 1 2 3 4 5 2 3 4 5 −6 −5 n −x[n + 1] 1 2 −1 −2 −6 −5 −4 −3 −2 −1 1 2 3 4 5 2 3 4 5 n 6
n 2 12345 Figure 2.1.a1:x[n] Scaled and shifted The sum of these yields the following sequence for yIn yIn 2 Figure 2.1.a2: gn
x[n − 2] 1 2 −1 −6 −5 −4 −3 −2 1 2 3 4 5 2 3 4 5 n x[n − 3] 1 2 −1 −6 −5 −4 −3 −2 −1 1 2 3 4 5 2 3 4 5 n Figure 2.1.a.1: x[n] Scaled and shifted The sum of these yields the following sequence for y[n]: 1 2 3 −1 −2 −6 −5 −4 −3 −2 −1 1 2 3 4 5 2 3 4 5 y[n] n Figure 2.1.a.2: y[n] 7
(b)For this part, we can again use the shift and scale method since the sequence x[n of a short duration as given below cure 2.1. b: nl Thus, we can write the output as a sum of scaled shifted inputs as follows: y]=2u2-n]+2+h(1-n]+2+2u-n]+2+3u-m-1]+2n+u-n-2 ∑2+u2-n-
(b) For this part, we can again use the shift and scale method since the sequence x[n] is of a short duration as given below: 1 −1 −6 −5 −4 −3 −2 −1 1 2 3 4 5 2 3 4 5 x[n] n Figure 2.1.b: x[n] Thus, we can write the output as a sum of scaled shifted inputs as follows: y[n] = 2nu[2 − n] + 2n+1u[1 − n] + 2n+2u[−n] + 2n+3u[−n − 1] + 2n+4u[−n − 2] � 4 = 2n+ku[2 − n − k] k=0 8
Problem 2 (a)From the definition of the convolution, we have the following expression for the output h(t-Tar(r) illustrated in the diagram. The ranges are t<-1 and- n up into 2 regions as Based on the given a(t) and h(t), we can break the integrati 123456 h(t-T) 4-3-2-1 123456 1<t h(t-T) (t-7)
� � Problem 2 (a) From the definition of the convolution, we have the following expression for the output y(t): y(t) = h(t − � )x(� )d� −� Based on the given x(t) and h(t), we can break the integration up into 2 regions as illustrated in the diagram. The ranges are t < −1 and t → −1. x(� ) −4 −3 −2 −1 1 2 3 4 5 6 1 � t −4 −3 −2 −1 1 2 3 4 5 6 1 h(t − � ) −1 � e2(t−� ) t −4 −3 −2 −1 1 2 3 4 5 6 1 h(t − � ) −1 � t e2(t−� ) � t t < 9
For the range t-1, the (r)h(t-r)is non-zero for r>t. So the expression for y(t) is given by y(t)
� � � � � � � � � � � � For the range t t. So the expression for y(t) is given by: y(t) = h(t − � )x(� )d� = e2(t−� ) e−�d� t t � � 1 �� 2t −3� = e e d� = e2t − e−3� t 3 t 1 2t = e − e−3t 3 1 −t = e 3 10
