How to determine if the equation is a periodic signal?

(2What are periodic signals? How to determine if the equation is a periodic signal?

Find the fundamental period:
i) x[n] = sin(2n)
ii) x(t) = cos³(4t)
iii) \[x(t) = 1 + \sin^2(2 \pi t)\]

Re: Periodic Signals

Periodic signals are signals that repeat of after a fixed time.

for example sin(x) is periodic with period 2*pi.

to prove a signal is periodic, you need to show

that f(x) = f(x+T) for some T for all x.

sin(x) = sin(x+2*pi) so let T = 2*pi .
this proves sin(x) is periodic.


x[n] is probably not periodic. why ?

I'll let you work out your own hw problems. good luck.

Re: Periodic Signals

Note that

For a discrete-time signal to be periodic it has to satisfy
\[x[n+N]=x[n]\] where \[N\] is the fundamental period and the condition on it is that it should be an integer.

For a continuous-time signal to be periodic it has to satify
\[x(t+T)=x(t)\] where \[T\] is the fundamental period and there is no restriction on this as in the case of DT signal.


Coming to the given problems

(i) The given signal is \[x[n]=\sin(2n)\]. For it to be periodic it has to satisfy the following equation \[\sin(2n)=\sin(2n+2N)\] for integer values of \[N\]. Note that no value of \[N\] will satisfy the equation. Thus it is aperiodic.

(ii) The given signal is \[x(t)=\cos^3(4t)\]. First let us simplify it in to the form \[x(t)=\frac{1}{4}(\cos(3t)+3\cos(t))\]. So for this to be periodic it has to satisfy the following equation.\[x(t+T)=x(t)\]

Thus \[\cos(3t)\] is periodic with period \[T_1 = 2\frac{\pi}{3}\] and \[\cos(t)\] is periodic with period \[T_2 = 2\pi\]. Thus total period of the signal is (T = LCM(\[2\frac{\pi}{3},2\pi)\]. This turns out to be
\[T = 2\pi\]

(iii) The given signal is \[x(t)=1+\sin^2 (2\pi t)\]. This can be written as \[x(t)=\frac{3}{2}-\frac{1}{1}\cos(4\pi t)\]. The period of \[\cos(4\pi t) \] is \[T_1 = \frac{1}{2}\]. Thus total period is \[T = \frac{1}{2}\].

Hope all this will help you.

thnx

purna!

Re: Periodic Signals

Conclusion:

i) period is \[\frac{1}{\pi }\] , therefore it is aperiodic

ii) period is \[\frac{\pi }{2}\]

iii) period is \[\frac{1}{2}\]