x [n] = > x [k]8 [n - k] (4) K=-00 1 for n = -1 For example if x [n] = 2 for n = 0 -1 for n = 1 (5), then based on Eqn. 4, 0 otherwise x[n] = [n + 1] + 28[n] - 8[n - 1] (6) There are two ways to look at the convolution equation. First we can decompose the input signal into delta functions. Based on signal decomposition (Eqn. 4) and convolution Eqn. 2 and 3, we can represent output as a sum of a series of shifted and scaled (based on each of the input signal value) versions of impulse response functions: 00 y[n] = h[n] * x[n] = h[n] * x [k]8 [n - k] = > x [k]h[n - k] (7) K=-00 K = -00 For x[n] defined in Eqn. 5, we have y[n] = x[n] * h[n] = h[n] * (8[n + 1] + 28[n] - 8[n - 1]) = h[n + 1] + 2h[n] - h[n - 1] (8) Eqn. 7 and 8 represent convolution output from the viewpoint of the input signal, i.e., how the input signal is decomposed into impulses, with each impulse passing through the LTI system, and then the individual outputs are added to generate the final output. (Think about how the LTI property can be used to generate the output in Eqn. 8.) The contribution of each sample/impulse in the input signal to many samples in the output signal is evident from Eqn. 7 and 8. For example, in Eqn. 8, x[-1]=1 contributes h[n+1] to the output, x[0]=2 contributes 2h[n] to the output, and x[1]=-1 contributes -h[n-1] to the output.Task: 1. Calculate the convolution of the following two signals y[n]| = x[n] * h[n] 2 1.0 1 0s 0 0.0 05 1.0 15 0 25 U 05 10 15 25 3l -1 05 2 -1.0 h[n] x[n] Figure 2. h[n] and x[n] a. Write x[n] as a sum of delta functions similar to Egn. 6. b. Compute convolution using the method shown in Egn. 8. Plot your y[n] in a figure similar to Fig. 2