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 region where a(r)h(t-T)is non-zero is from -1-00. So, the expression for y(t)is given by T)(7 2(t-7)a-7 -rdr ezt 3 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 < −1, the region where x(� )h(t − � ) is non-zero is from −1 � ∗. So, the expression for y(t) is given by: y(t) = h(t − � )x(� )d� = e2(t−� ) e−�d� −1 −1 � � 1 �� 2t −3� = e e d� = e2t − e−3� −1 3 −1 1 2t+3 = e 3 For the range t → −1, the x(� )h(t − � ) is non-zero for � > 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