Energy of a CT signal

$ E = \int_{t1}^{t2} |x(t)|^2\ dt \! $

Power of a CT signal

$ P = \frac{1}{t2-t1} \int_{t2}^{t1} |f(t)|^2\ dt \! $

Energy of a DT signal

$ E = \sum_{n=n1}^{n2} |x[n]|^2 $

Power of a DT signal

$ P = \frac{1}{n2-n1+1} \sum_{n=n1}^{n2} |x[n]|^2 $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang