(New page: CT LTI system: <math>y(t)=6x(t)+4x(t-3)\!</math> == Part A == The unit impulse response of this system is: <math>h(t)=6\delta(t)\!+4\delta(t-3)</math> Taking the laplace transform of ...) |
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<math>x(t) = \frac{3}{2}e^{i7t}+ \frac{3}{2}e^{-i7t} + \frac{11}{2i}e^{i4t}- \frac{11}{2i}e^{i(4t-3)}</math> and we know that <math>w=1\!</math> | <math>x(t) = \frac{3}{2}e^{i7t}+ \frac{3}{2}e^{-i7t} + \frac{11}{2i}e^{i4t}- \frac{11}{2i}e^{i(4t-3)}</math> and we know that <math>w=1\!</math> | ||
| − | So, the response | + | So, the response of this system to the signal I defined in Q1 is: |
<math>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)</math> | <math>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)</math> | ||
<math>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)}</math> | <math>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)}</math> | ||
Latest revision as of 13:36, 26 September 2008
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)} $
