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| + | ==Question 1== | ||
| − | + | This is because real systems have transfer functions with real coefficients. If we write the transfer function H(z) as H(z)=P(z)/Q(z), where P(z) and Q(z) are polynomial, then the poles of the transfer function are the zeros of the polynomial Q(z). But Q(z) has real coefficients (Since the system can be written as a difference equation with real coefficients). | |
| + | |||
| + | ==Question 2== | ||
| + | This implies that the difference equation must has the form | ||
| + | |||
| + | <math> y[n]=\sum_{i=0}^{N-1} b_i x[n-i] -\sum_{k=1}^{M} a_k x[n-k] </math> | ||
| + | |||
| + | ==Question 3== | ||
Revision as of 11:37, 4 November 2013
Hw9_ECE438F13sln
Question 1
This is because real systems have transfer functions with real coefficients. If we write the transfer function H(z) as H(z)=P(z)/Q(z), where P(z) and Q(z) are polynomial, then the poles of the transfer function are the zeros of the polynomial Q(z). But Q(z) has real coefficients (Since the system can be written as a difference equation with real coefficients).
Question 2
This implies that the difference equation must has the form
$ y[n]=\sum_{i=0}^{N-1} b_i x[n-i] -\sum_{k=1}^{M} a_k x[n-k] $
