1. ## Power and energy

Hi,
I would like to know if there is a quick analysis that I can do to determine if a signal is a power signal or energy signal?

I have those signals:

a- u(t)+5*u(t-1)-2*u(t-2)

b- e^(-5t)*u(t)

c- (1-e^(-5t))*u(t)

d- r(t) (ramp function)

e- r(t)-r(t-1)

f- t^(1/4)*u(t-3)

We are asked to determine if the signals are power , energy signals or neither.
Thank you
B

Hi,
I would like to know if there is a quick analysis that I can do to determine if a signal is a power signal or energy signal?

I have those signals:

a- u(t)+5*u(t-1)-2*u(t-2)

b- e^(-5t)*u(t)

c- (1-e^(-5t))*u(t)

d- r(t) (ramp function)

e- r(t)-r(t-1)

f- t^(1/4)*u(t-3)

We are asked to determine if the signals are power , energy signals or neither.
Thank you
B
I looks to me (roughly) like if the signal decays as time goes to infinity then you use the energy technique (integrate the square of the curve) and if the signal doesn't decay then you use the power technique (average of the signal over a time period.) See here. Graphing is probably the fastest check, though it will not always be obvious if the integral of the square of the signal is infinite, you ought to be able to tell in most cases by simply looking.

-Dan

3. Thank you

However I still need some help to compute the energy of the following signals:

cos(10pi*t)*u(t)*u(2-t)

and

u(t)-u(t-15)

For the second signal , I found -infinity . Am I right?

For the first, I dont really know how to do it.

Thank you
B

Thank you

However I still need some help to compute the energy of the following signals:

cos(10pi*t)*u(t)*u(2-t)

and

u(t)-u(t-15)

For the second signal , I found -infinity . Am I right?

For the first, I dont really know how to do it.

Thank you
B
The first graph is for the energy signal of the cos signal. It looks to me like the integral is non-infinite, so I would use that one, and not the power signal technique.

The second graph is for the second signal, the u(t) - u(t - 15). It looks again like the integral of the energy is nice and finite, so again I would use the energy signal technique.

-Dan