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<math> | <math> | ||
| − | \begin{align}ax_{1}[n]+bx_{2}[n] \rightarrow ax_{1}[n]+bx_{2}[n]+ax_{1}[n+1]+bx_{2}[n+1]+ax_{1}[n+2]+bx_{2}[n+2] | + | \begin{align}ax_{1}[n]+bx_{2}[n] &\rightarrow ax_{1}[n]+bx_{2}[n]+ax_{1}[n+1]+bx_{2}[n+1]+ax_{1}[n+2]+bx_{2}[n+2] |
| − | = a(x_{1}[n]+x_{1}[n+1]+x_{1}[n+2])+b(x_{2}[n]+x_{2}[n+1]+x_{2}[n+2]) | + | &= a(x_{1}[n]+x_{1}[n+1]+x_{1}[n+2])+b(x_{2}[n]+x_{2}[n+1]+x_{2}[n+2]) |
| − | = ay_{1}[n]+by_{2}[n] \therefore System is linear | + | &= ay_{1}[n]+by_{2}[n] \therefore System is linear |
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 16:54, 30 June 2008
(a) Derive the condition for which the discrete time complex exponetial signal x[n] is periodic.
$ x[n] = e^{jw_{o}n} $ $ x[n] = x[n+N] = e^{jw_{o}(n+N)} = e^{jw_{o}n}e^{jw_{o}N} $ to be periodic $ e^{jw_{o}N} = 1 = e^{j2\pi k} $ $ \therefore w_{o}N = 2\pi k $ $ \Rightarrow \frac{w_{o}}{2\pi} = \frac{K}{N} \Rightarrow $Rational number $ \therefore \frac{w_{o}}{2\pi} $ shold be a rational number
(b) Show that the system described by
$ y[n] = x[n] + x[n+1] + x[n+2] $ is a LTI system.
$ \begin{align}ax_{1}[n]+bx_{2}[n] &\rightarrow ax_{1}[n]+bx_{2}[n]+ax_{1}[n+1]+bx_{2}[n+1]+ax_{1}[n+2]+bx_{2}[n+2] &= a(x_{1}[n]+x_{1}[n+1]+x_{1}[n+2])+b(x_{2}[n]+x_{2}[n+1]+x_{2}[n+2]) &= ay_{1}[n]+by_{2}[n] \therefore System is linear \end{align} $
y_{1}[n-n_{0}] = x_{1}[n-n_{0}]+x_{1}[n-n_{0}+1]+x_{1}[n-n_{0}+2]
Let x_{2} = x_{1}[n-n_{0}]
x_{2}[n] \rightarrow x_{2}[n]+x_{2}[n+1]+x_{2}[n+2]
= x_{1}[n-n_{0}]+x_{1}[n-n_{0}+1]+x_{1}[n-n_{0}+2] = y_{1}[n-n_{0}]
\therefore System is time-variant
