| (6 intermediate revisions by the same user not shown) | |||
| Line 15: | Line 15: | ||
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] | 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] | ||
</math> | </math> | ||
| − | |||
<math>= a(x_{1}[n]+x_{1}[n+1]+x_{1}[n+2])+b(x_{2}[n]+x_{2}[n+1]+x_{2}[n+2])</math> | <math>= a(x_{1}[n]+x_{1}[n+1]+x_{1}[n+2])+b(x_{2}[n]+x_{2}[n+1]+x_{2}[n+2])</math> | ||
| Line 23: | Line 22: | ||
<math>y_{1}[n-n_{0}] = x_{1}[n-n_{0}]+x_{1}[n-n_{0}+1]+x_{1}[n-n_{0}+2]</math> | <math>y_{1}[n-n_{0}] = x_{1}[n-n_{0}]+x_{1}[n-n_{0}+1]+x_{1}[n-n_{0}+2]</math> | ||
| − | Let <math>x_{2} = x_{1}[n-n_{0}]</math> | + | Let <math>x_{2}[n] = x_{1}[n-n_{0}]</math> |
<math> x_{2}[n] \rightarrow x_{2}[n]+x_{2}[n+1]+x_{2}[n+2]</math> | <math> x_{2}[n] \rightarrow x_{2}[n]+x_{2}[n+1]+x_{2}[n+2]</math> | ||
| − | + | <math>= x_{1}[n-n_{0}]+x_{1}[n-n_{0}+1]+x_{1}[n-n_{0}+2] = y_{1}[n-n_{0}]\therefore</math>System is time-variant | |
| + | |||
| + | (c) Prove that <math>x[n]*\delta[n] = x[n]</math> | ||
| + | |||
| + | <math>x[n]*\delta[n] = \Sigma_{k=-\infty}^\infty x[k]\delta[n-k]</math> | ||
| + | <math>= \Sigma_{k=-\infty}^\infty x[n]\delta[n-k] = x[n]\Sigma_{k=-\infty}^\infty\delta[n-k] = x[n]</math> | ||
| + | <math>\therefore x[n]*\delta[n] = x[n]</math> | ||
| + | |||
| + | |||
| + | Solved by Minwoong Kim | ||
Latest revision as of 16:17, 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.
$ 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
$ 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}[n] = 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
(c) Prove that $ x[n]*\delta[n] = x[n] $
$ x[n]*\delta[n] = \Sigma_{k=-\infty}^\infty x[k]\delta[n-k] $ $ = \Sigma_{k=-\infty}^\infty x[n]\delta[n-k] = x[n]\Sigma_{k=-\infty}^\infty\delta[n-k] = x[n] $ $ \therefore x[n]*\delta[n] = x[n] $
Solved by Minwoong Kim
