{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

zTransform - ECE 3250 THE z-TRANSFORM Fall 2008 1 Power...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 3250 THE z -TRANSFORM Fall 2008 1. Power series We’ll have occasion to consider infinite series of the form X n = -∞ c n z - n , where z is a complex variable and the c n are given complex numbers. We’ll want to know whether the series converges for at least some z C and, if so, for exactly what values of z the series converges. The special form of power series makes for a tidy theory of convergence. Fact 1: If R o > 0 is such that the sequence { c n R - n o : n 0 } is bounded, then the series X n =0 c n z - n converges for every z C satisfying | z | > R o . Similarly, if R o > 0 is such that the sequence { c n R - n o : n < 0 } is bounded, then the series - 1 X n = -∞ c n z - n converges for every z C satisfying | z | < R o . Proof: Consider the first assertion. Suppose R o > 0 is such that the sequence { c n R - n o : n 0 } is bounded from above in magnitude, say by M > 0. If | z | > R o , then X n =0 | c n z - n | = X n =0 | c n R - n o | ( R o / | z | ) n M X n =0 | ( R o / | z | ) n = M/ (1 - R o / | z | ) , where the last equality holds by geometric-series reasoning since R o / | z | < 1. It follows that the infinite sequence { c n z - n } is absolutely summable and is therefore summable by Facts 3 and 7 from the handout on sequences and series. The proof of the second assertion is similar. Fact 2: There exists some z C for which the series X n =0 c n z - n converges if and only if there exists some R o > 0 for which the sequence { c n R - n o : n 0 } is bounded. Similarly, there exists some z C for which the series - 1 X n = -∞ c n z - n converges if and only if there exists some R o > 0 for which the sequence { c n R - n o : n < 0 } is bounded. Proof: The “if” part is just Fact 1. As for the “only if,” suppose z o C is such that P n =0 c n z - n o (respectively, P - 1 n = -∞ c n z - n ) converges. Then the sequence {| c n || z o | - n : n 0 } (respectively, {| c n || z o | - n : n < 0 } ) must be bounded, so taking R o = | z o | proves the assertion(s). 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 Suppose now that { c n : n Z } are given complex numbers, and suppose there exists R o > 0 such that the sequence { c n R - n o } is bounded. Define R a as follows: R a = inf { R o > 0 : { c n R - n o : n 0 } is bounded } . Observe that R a = 0 is possible. Similarly, define R b via R b = sup { R o > 0 : { c n R - n o : n < 0 } is bounded } . Observe that R b = is possible. Taken together, Facts 1 and 2 reveal that the series X n =0 c n z - n converges for every z C satisfying | z | > R a and diverges for every z satisfying | z | < R a . To see why it diverges for such z , note that if the series converged for some z o with | z o | = R o < R a , then { c n R - n o : n 0 } would be bounded, and Fact 1 along with the definition of R a would imply that R o R a , a contradiction. Similarly, the series - 1 X n = -∞ c n z - n converges if | z | < R b and diverges if | z | > R b . The infinite series P n = -∞ c n z - n therefore converges at least for z -values in the region R a < | z | < R b . Such z -values exist if and only if R a < R b , in which case the indicated set of z -values constitutes an annular region centered on z = 0 in the complex plane. The series may or may not converge for some z -values satisfying | z | = R a or | z | = R b , but we pay little attention to those borderline values. One can show that the series diverges for
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}