CT LTI system: $ y(t)=6x(t)+4x(t-3)\! $


Part A

The unit impulse response of this system is:

$ h(t)=6\delta(t)\!+4\delta(t-3) $

Taking the laplace transform of the unit impulse response of this system gives us:

$ H(s)=6+e^{-3s}\! $


Part B

The signal used in Q1 is:

$ x(t) = 3cos(7t) + 11sin(4t)\! $

which is also equal to:

$ x(t) = \frac{3}{2}e^{i7t}+ \frac{3}{2}e^{-i7t} + \frac{11}{2i}e^{i4t}- \frac{11}{2i}e^{i(4t-3)} $ and we know that $ w=1\! $

So, the response of this system to the signal I defined in Q1 is:

$ y(t)=\bigg(6+e^{-3i}\bigg)\bigg(\frac{3}{2}e^{i7t}+ \frac{3}{2}e^{-i7t} + \frac{11}{2i}e^{i4t}- \frac{11}{2i}e^{-i4t}\bigg) $

$ y(t)=9e^{i7t}+\frac{3}{2}e^{i(7t-3)}+9e^{-i7t}+\frac{3}{2}e^{-i(7t+3)}+\frac{33}{i}e^{i4t}+\frac{11}{2i}e^{i(4t-3)}-\frac{33}{i}e^{-i4t}-\frac{11}{2i}e^{-i(4t+3)} $

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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva