Signals & Systems - Non-Periodic wave

[Angello90]

Hi guys,

I have this question in the book whether a signal is Power Signal or Energy Signal.
From what I have read signal is an energy one if its time-averaged power is = 0, or its non periodic.

For non-periodic signal
V(t) \neq V(t+T)
I tried to prove that
V(t) = 230\sqrt{2} Sin (100\pi t)
is non-periodic.

So I picked random values for t=1 and T=1, and I got this eqn:
230\sqrt{2} Sin (100\pi)=230\sqrt{2} Sin (100\pi+1)
Which states that it's non-periodic, ie V(t) is an energy signal.

Now I have an impression that this is not right. Can anyone help me? Book I have is no good with examples.

Plus if anyone could send me a link where there are some examples with Energy, Power, Periodic and Non-Periodic signals?

I'm using "Signals and Systems" by S. Haykin and B. Van Veen.

Thanks guys!

 

[berkeman]

Hi guys,

I have this question in the book whether a signal is Power Signal or Energy Signal.
From what I have read signal is an energy one if its time-averaged power is = 0, or its non periodic.

For non-periodic signal
V(t) \neq V(t+T)
I tried to prove that
V(t) = 230\sqrt{2} Sin (100\pi t)
is non-periodic.

So I picked random values for t=1 and T=1, and I got this eqn:
230\sqrt{2} Sin (100\pi)=230\sqrt{2} Sin (100\pi+1)
Which states that it's non-periodic, ie V(t) is an energy signal.

Now I have an impression that this is not right. Can anyone help me? Book I have is no good with examples.

Plus if anyone could send me a link where there are some examples with Energy, Power, Periodic and Non-Periodic signals?

I'm using "Signals and Systems" by S. Haykin and B. Van Veen.

Thanks guys!

I'm not of great help on your question, but I googled Power Signal or Energy Signal, and got lots of great hits. This PDF is short and to the point:

http://cnx.org/content/m10055/2.10/content_info

You can re-do the full google search if that PDF is not enough help.

 

[bos1234]

Hi,
Remember that T stands for period so its best to work in radians. Its not possible to input an arbitary value (T=-1 or even T=2pi) and say it is not periodic. T=2pi might not be periodc but t=5pi might be periodic, i.e. the curve might repeat every 5pi times.

So, your equationis V(t) = 230\sqrt{2} Sin (100\pi t)
Don't worry about 230... its just the amplitude.
Concentrate on Sin (100\pi t).

Do you know this formula? v(t)=sin(wt+\phi)

w=2\pi f=100\pi, therefore f=50/pi. That is the FUNDAMENTAL FREQUENCY. Hence, T=pi/50

This is the FUNDAMENTAL PERIOD. that is, it repeats every pi/50

-----------
Anyways, you can tell straight away it is periodic because it is similar to sin(t). Only difference is that it has a bigger amplitude and higher freq.

 

[Angello90]

Ok that makes perfect sense! Thanks for that!

But here comes another question, this time for distinct time.

Eqn. x[n] = Cos(2n) is not periodic, as we need x[n+N] = x[n] \forall n \in Z. This follows that x[n+N] = Cos(2n+2N) therefore 2N = \Omega = m2\pi where m \in Z. And there is no integer to solve this question. Simillary if eqn is x[n] = Cos(2n\pi) than signal is periodic with period 1 sample. That also seems clear to me.

But problem arrises with signal in a form x[n] = (-1)^{n}. I tried doing simillar tehnique as in signal above, and I got x[n+N] = (-1)^{n+N}, and than N = \Omega = m2\pi. Again there is no integer to solve this eqn, thus it should be non periodic, but my book tells me that it is periodic, with fundamental period of 2 samples.

That makes me loose the track of what to do with what!

 

[ask_LXXXVI]

But problem arrises with signal in a form x[n] = (-1)^{n}. I tried doing simillar tehnique as in signal above, and I got x[n+N] = (-1)^{n+N}, and than N = \Omega = m2\pi. Again there is no integer to solve this eqn, thus it should be non periodic, but my book tells me that it is periodic, with fundamental period of 2 samples.

That makes me loose the track of what to do with what!

If you compare the 2 sides you can see that (-1)^n = cos(n*pi). As you can see cos(n*pi) has a period of 2 samples.This may help you visualise.

Or, you can see that (-1)^n repeats itself every 2 samples. -1,1,-1,1, ... Thats how the sequence is. So period is again 2.

 

[Angello90]

Ok, but why would you compare (-1)^n with cos(n*pi)?

Also another question about periodic signal:
x(t) = Sin^2(2*pi*f*n);

Do I have to brake Sin^2, into simpler format (i.e. Sin^2(A) = 1/2 (1-Cos (2A) ) )?

 

[ask_LXXXVI]

Ok, but why would you compare (-1)^n with cos(n*pi)?

What I meant by comparing was that you can see that the two are one and the same. If you have trouble visualizing periodicity of (-1)^n , you could visualise the periodicity of cos(n*pi).Anyways the two are same.

Also another question about periodic signal:
x(t) = Sin^2(2*pi*f*n);

Do I have to brake Sin^2, into simpler format (i.e. Sin^2(A) = 1/2 (1-Cos (2A) ) )?

Yes by that way you can easily check for periodicity.

 

[jasonjinct]

I always have troubles to determine discrete signal whether periodic or nonperiodic and its fundamental period, even the most basic one x[n] = Cos(2n) and x[n] = Cos(2n\pi)

I know for periodic discrete signal need to follow x[n+N] = x[n]
But I don't know how to apply and relate it like Angello92 do this:

Eqn. x[n] = Cos(2n) is not periodic, as we need x[n+N] = x[n] \\forall n \\in Z . This follows that x[n+N] = Cos(2n+2N) therefore 2N = \\Omega = m2\\pi where m \\in Z . And there is no integer to solve this question. Simillary if eqn is x[n] = Cos(2n\\pi) than signal is periodic with period 1 sample. That also seems clear to me.

And I know this equation N = 2pi / \Omega}

Can help me explain more by solving this discrete signal:
a)x[n] = 5 cos [0.2pi*n]
b)x[n] = 5 cos [6pi*n]
c)x[n] = 5 sin [6pi*n / 35]
d)x[n] = (-1)^{n^2}

Thanks you... some one please help me.

 

[Angello90]

x[n] = 5 cos [0.2\pi n]

Don't worry about 5 - it's amplitude.
Cos[0.2\pi n] = Cos[\Omega n]

thus:

0.2\pi =\Omega and since \Omega = \frac{2\pi}{N}
0.2\pi = \frac{2\pi}{N}
N = \frac{2\pi}{0.2\pi}
N = \frac{2}{0.2}
N = 10
Therefore period is 10.

You do same thing for the rest:

So x[n] = 5 sin [6\pi n / 35]
\frac{6\pi}{35} = \frac{2\pi}{N}
N = \frac{2\pi 35}{6\pi}
N = \frac{70}{6}
6 N = 70
3 N = 35
Therefore period is 35.

With x[n] = (-1)^{n^2} I'm not sure how to do. But it's obvious it's every 2 samples.
@ x[1] = -1
@ x[2] = 1
@ x[3] = -1
@ x[4] = 1
and so on.

Hope that helps

 

[jasonjinct]

The cos part N = 10 I understand.

But:
6N = 70
3N = 35

Why not become N = 35/3?

 

[ask_LXXXVI]

N = 35/3 isn't an integer, it is a rational number. So in such cases period is smallest multiple of the rational number which is an integer.

3N gives 35 which is an integer , thus it is a valid period.

If you recall, the period must be an integer as this is a discrete time signalcoming to the problem of (-1)^{n^2} :-
note that (-1)^n = -1 if n is odd and it is = 1 if n is even.
n^2 is odd when n is odd and it is even when n is even.

thus , (-1)^{n^2} is same as (-1)^n. and it was demonstrated earlier in this thread that (-1)^n has period as 2.

 

[jasonjinct]

Thanks you very much, I understand so far and I will try to work on some exercises 1st

 

[jasonjinct]

x(t) = cos^2(2*pi*t)
=1/2[1+cos(4*pi*t)

so w=2pi/T
4pi = 2pi
T=0.5, is I do the correct way?

But how about
x(t) = sin^3(2t)?
I do like this:
1/4[3sin 2t + sin 6t]

2 = 2pi/T
T = pi

6 = 2pi/T
T = pi/3

The answer is 1/pi. How to do this?

 

[jasonjinct]

Also, beside the question above, here another one
Find the energy for the signal:
x(t) = 5 cos(pi*t) + sin (5pi*t), -infinity<t<infinity.

How to apply the energy formula to this one? is it the X^2(t) in the energy formula substitute with this?
[5 cos(pi*t) + sin (5pi*t)]^2

or the cos and the sin do it separately..

Thank you, please help me for this question and the previous one, help pls..

 

[ask_LXXXVI]

Find the energy for the signal:
x(t) = 5 cos(pi*t) + sin (5pi*t), -infinity<t<infinity.

How to apply the energy formula to this one? is it the X^2(t) in the energy formula substitute with this?
[5 cos(pi*t) + sin (5pi*t)]^2

or the cos and the sin do it separately..

x(t) = 5cos(pi*t) + sin(5pi*t) , is a periodic signal with period pi. This is a power signal, i.e it has a finite power and thus infinite energy over the interval -inf to +inf.

 

[jasonjinct]

ops, sorry is power signal, but how to evaluate the integration
is it take the whole equation as X(t), then square the whole equation? or do integrate it one by one?

Almost, can you help me on the find the fundamental period of x(t) = sin^3(2t)?