(New page: Category:2010 Fall ECE 438 Boutin ---- From the first question, we knew that <math> -a^{n}u[-n-1] = \mathcal{Z}^{-1}\bigg\{\frac{1}{1-az^{-1}}\bigg\} \text{ where } |z|<|a|. \,\!</...)
 
 
(4 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
[[Category:2010 Fall ECE 438 Boutin]]
 
[[Category:2010 Fall ECE 438 Boutin]]
 
+
----
 +
== Solution to Q2 of Week 5 Quiz Pool ==
 
----
 
----
  
Line 24: Line 25:
 
\,\!</math>
 
\,\!</math>
 
----
 
----
Back to  
+
<span style="color:green"> I saw many students made a mistake about this kind of question, such as </span>
 +
 
 +
<math> -a^{n}u[-(n-3)-1] = \mathcal{Z}^{-1}\bigg\{\frac{z^{-3}}{1-az^{-1}}\bigg\} \text{ where } |z|<|a|. \;\;\; \text{This is not correct!!} \,\!</math>
 +
 
 +
<span style="color:green"> which the time-shifting is applied only to the unit step and not to the power of <math>a</math>. </span>
 +
<span style="color:green"> Thus, you should be very careful when you use the time-shifting property. You have to apply time-shifting to all <math>n</math>, when you find the inverse Z-transform. </span> -[[User:han83|Jaemin]]
 +
----
 +
Back to [[ECE438_Week5_Quiz|Lab Week 5 Quiz Pool]]
  
 
Back to [[ECE438_Lab_Fall_2010|ECE 438 Fall 2010 Lab Wiki Page]]
 
Back to [[ECE438_Lab_Fall_2010|ECE 438 Fall 2010 Lab Wiki Page]]
  
 
Back to [[2010_Fall_ECE_438_Boutin|ECE 438 Fall 2010]]
 
Back to [[2010_Fall_ECE_438_Boutin|ECE 438 Fall 2010]]

Latest revision as of 17:39, 19 September 2010


Solution to Q2 of Week 5 Quiz Pool


From the first question, we knew that

$ -a^{n}u[-n-1] = \mathcal{Z}^{-1}\bigg\{\frac{1}{1-az^{-1}}\bigg\} \text{ where } |z|<|a|. \,\! $

And the time-shifting property of Z-transform is defined as

$ x[n-k] = \mathcal{Z}^{-1}\bigg\{z^{-k}X(z)\bigg\} \text{ when } x[n] = \mathcal{Z}^{-1}\bigg\{X(z)\bigg\}\,\! $

Therefore, if we use the time-shifting property of Z-transform, then

$ -a^{n-3}u[-(n-3)-1] = \mathcal{Z}^{-1}\bigg\{\frac{z^{-3}}{1-az^{-1}}\bigg\} \text{ where } |z|<|a|. \,\! $

Combined with the result from the linearity of Z-transform, then

$ \begin{align} \mathcal{Z}^{-1}\bigg\{\frac{2z^{-3}}{1-az^{-1}}\bigg\} \text{ for } |z|<|a| &= -2a^{n-3}u[-(n-3)-1], \\ &= -2a^{n-3}u[-n+2] \end{align} \,\! $


I saw many students made a mistake about this kind of question, such as

$ -a^{n}u[-(n-3)-1] = \mathcal{Z}^{-1}\bigg\{\frac{z^{-3}}{1-az^{-1}}\bigg\} \text{ where } |z|<|a|. \;\;\; \text{This is not correct!!} \,\! $

which the time-shifting is applied only to the unit step and not to the power of $ a $. Thus, you should be very careful when you use the time-shifting property. You have to apply time-shifting to all $ n $, when you find the inverse Z-transform. -Jaemin


Back to Lab Week 5 Quiz Pool

Back to ECE 438 Fall 2010 Lab Wiki Page

Back to ECE 438 Fall 2010

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010