# Signals and Systems coursework help

1. Jan 15, 2007

### Kbob08

1. The problem statement, all variables and given/known data
Detmerine the Power and rms value for each of the following signals:

a)5 + 10cos(100t+pi/3)

... more signals

3. The attempt at a solution

It's hard to right out integral signs or I dont know the way to do so I'll use I(lower bound, upper bound) to denote an integral

a)P= lim t->inf. (1/T) [I(-T/2,T/2) [5 + 10cos(100t+pi/3)^2dt]

=lim t->inf. (1/T) [25T + I(-T/2,T/2) 100 cos^2(100t+pi/3)dt + I(-T/2,T/2)50cos(100t + pi/3)dt->0

= lim t->inf. (1/T) [25T + 50T {I(-T/2,T/2)cos(200t+2pi/3)dt}->0

= 25 +50 = 75, Rms 75^ (-1/2)

does it look like i'm following the right idea?

2. Jan 15, 2007

### Staff: Mentor

Welcome to the PF, Kbob08. To learn how to insert math symbols in your posts, check out the LaTex tutorial here:

https://www.physicsforums.com/forumdisplay.php?f=151

And on your question, what is the function you've written? Is is a voltage waveform? You can certainly calculate the RMS value if it is a voltage, but you need to know something about the current or load impedance to get a power out of it....

3. Jan 15, 2007

### Corneo

Usually in signal books they don't denote any units for these types of problems. Otherwise you would have to do an additional step to make sure the units are matched.

4. Jan 15, 2007

### Staff: Mentor

Well, IMO checking units is important. It sure is important in the real world. So you just assume that the function shows is a voltage waveform into 50 Ohms?

5. Jan 16, 2007

### Corneo

Yes I agree your right that checking units is important. I was just commenting that most textbook problems leave out units of the function in question. The meat of the problem for this specific exercise is just performing the integration.

To the OP: You have the right idea and your work is correct. It may help to remember for signals in the form

$$C \cos (\omega_0 t + \theta)$$, the power is $$P_g = C^2/2$$. You can prove this yourself pretty straight forward by performing the same integration you have done and using some trig substitution. Notice that the result depends solely on the amplitude and nothing else.