Absolutely Summable

Definition

A signal x[n] is absolutely summable iff $ \sum_{n=-\infty}^\infty |x[n]| < \infty $.

Context

O/W #10.6:

x[n] is absolutely summable.

Then, $ \sum_{n=-\infty}^\infty |x[n]| < \infty $.

However, notice that since $ 1^{-n} = 1 $, we can also say that $ \sum_{n=-\infty}^\infty |x[n]| 1^{-n} = \sum_{n=-\infty}^\infty |x[n] 1^{-n}| < \infty $.

Since the ROC (region of convergence) of a z-transform is $ \{z | x(t) z^{-n} \text{ is absolutely summable}\} $, $ \{z:|z|=1\} $ (ie, the unit circle) is in the ROC.


Absolutely Integrable

Definition

A signal x(t) is absolutely integrable iff $ \int_{-\infty}^\infty |x(t)| dt < \infty $.

Context

O/W #9.6:

x(t) is absolutely integrable.

Then, $ \int_{-\infty}^\infty |x(t)| dt < \infty $.

However, notice that since $ e^0 = e^{-0t} = 1 $, we can also say that $ \int_{-\infty}^\infty |x(t)| e^{-0t} dt = \int_{-\infty}^\infty |x(t) e^{-0t}| dt < \infty $.

Since the ROC (region of convergence) of a Laplace transform is $ \{s | {x(t) e^{-st}} \text{ is absolutely integrable}\} $, s = 0 (and consequenty the $ j\omega $-axis) is in the ROC.

Alumni Liaison

Meet a recent graduate heading to Sweden for a Postdoctorate.

Christine Berkesch