Answer Happy

Show this graph in a 3D Form in Matlab, label correctly the axes.
(a) Write down the function g(t) that has the shape of a sine wave that increases linearly in frequency from 0 Hz at t=0 s to 5 Hz at t = 10 s. Answer: The instantaneous frequency of function g(t) at time t is t 282 5 Hz 10 s and since the phase of a sine wave is 2 times the integrated frequency so far, we get Answer: for (b) and (c) 1 0.8 0.6 (b) Plot the graph of this function using MATLAB's plot command. (c) Add to the same figure (this can be achieved using the hold command) in a different colour a graph of the same function sampled at 5 Hz, using the stem command. 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 g(t) = sin · (2 [² ƒ(²²) a²²). 0 f(t) = t 2 o 4 - sin 2n o 6 -sin e की o 00 + o o 10
t 0:0.01:10; f = sin(pi*t. "2/2): plot(t, f); hold; t2 = 0:1/5: 10; stem(t2, sin(pi t2.-2/2), 'I'); (d) [Extra credit] Plot the graph from (c) separately. Can you explain its symmetry? [Hints: sampling theorem, aliasing). Answer: A sine wave with a frequency f larger than half the sampling frequency f. cannot be distinguished based on the sample values from a sine wave of frequency f.- f. In other words, the sample values would have looked the same had we replaced the instantaneous frequency f(t) with f./2-1f./2-f(t), and the latter is symmetric around f/2, which is in this graph 2.5 Hz and occurs at t = 5 s. [The above is of course just a hand-waving argument, but shall be sufficient for this exercise. There are actually a few more conditions fulfilled here that lead to the exact symmetry of the plot. Firstly, since we started sampling at t = 0s with f. = 5 Hz, the positions of the sample values end up being symmetric around t = 5 s. Secondly, at the symmetry point t = 5 s, the sine wave was at a symmetric peak from where increasing or decreasing the phase has the same result.] 1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 G U T T T G G G- 3 0 Q 10