(New page: Category:ECE301Spring2011Boutin Category:Problem_solving ---- = Practice Question on Causal LTI systems defined by a linear, constant coefficient difference equation = Consider th...) |
|||
| Line 20: | Line 20: | ||
---- | ---- | ||
=== Answer 1 === | === Answer 1 === | ||
| − | + | ||
| + | a) | ||
| + | |||
| + | <math>\mathfrak F (y[n]-\frac{1}{2}y[n-1]) = \mathfrak F (x[n])</math> | ||
| + | |||
| + | <math>\mathcal Y (\omega) - \frac{1}{2}\mathfrak F (y[n-1]) = \mathcal X (\omega)</math> | ||
| + | |||
| + | <math>\mathcal Y (\omega) - \frac{1}{2}e^{-j\omega} \mathcal Y (\omega) = \mathcal X (\omega)</math> | ||
| + | |||
| + | <math>\mathcal Y (\omega) = \frac{1}{1-\frac{1}{2}e^{-j\omega}}\mathcal X (\omega)</math> | ||
| + | |||
| + | <math>\mathcal H (\omega) = \frac{1}{1-\frac{1}{2}e^{-j\omega}}</math> | ||
| + | |||
| + | b) | ||
| + | |||
| + | <math>h[n]=\mathfrak F ^{-1} (\mathcal H (\omega))= \mathfrak F ^{-1} \Big( \frac{1}{1-\frac{1}{2}e^{-j\omega}} \Big)</math> | ||
| + | |||
| + | <math>use \mathfrak F (a^n u[n]) = \frac{1}{1-ae^{-j\omega}}</math> | ||
| + | |||
| + | <math>h[n] = \Big(\frac{1}{2}\Big)^n u[n]</math> | ||
| + | |||
=== Answer 2 === | === Answer 2 === | ||
Write it here. | Write it here. | ||
Revision as of 15:51, 8 March 2011
Contents
Practice Question on Causal LTI systems defined by a linear, constant coefficient difference equation
Consider the LTI system defined by the difference equation
$ y[n]-\frac{1}{2}y[n-1]=x[n]\ $
a) What is the frequency response of this system?
b) What is the unit impulse response of this system?
c) What is the system's response to the input $ x[n] = \left( \frac{1}{5}\right)^n u[n] \ $?
c) What is the system's response to the input $ x[n] =\cos \left(\frac{\pi}{2} n \right) \ $?
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
a)
$ \mathfrak F (y[n]-\frac{1}{2}y[n-1]) = \mathfrak F (x[n]) $
$ \mathcal Y (\omega) - \frac{1}{2}\mathfrak F (y[n-1]) = \mathcal X (\omega) $
$ \mathcal Y (\omega) - \frac{1}{2}e^{-j\omega} \mathcal Y (\omega) = \mathcal X (\omega) $
$ \mathcal Y (\omega) = \frac{1}{1-\frac{1}{2}e^{-j\omega}}\mathcal X (\omega) $
$ \mathcal H (\omega) = \frac{1}{1-\frac{1}{2}e^{-j\omega}} $
b)
$ h[n]=\mathfrak F ^{-1} (\mathcal H (\omega))= \mathfrak F ^{-1} \Big( \frac{1}{1-\frac{1}{2}e^{-j\omega}} \Big) $
$ use \mathfrak F (a^n u[n]) = \frac{1}{1-ae^{-j\omega}} $
$ h[n] = \Big(\frac{1}{2}\Big)^n u[n] $
Answer 2
Write it here.
Answer 3
Write it here.
