Is $e^{(1+j2t)}$ a periodic function?

Question

Is $e^{(1+j2t)}$ a periodic function? I think yes,because $e^{(1+j2t)}=e^1 \times e^{j2t}$,

$e^{j2t}=cos(2t)+jsin(2t)$,and it is periodic,and $e^1$ is a constant, so i think a constant value multiply a periodic function is still a periodic function,is my thinking right?

score 2 · Accepted Answer · edited Sep 22 '19 at 06:27

2

Yes ! , $f(t)=e^{1+2jt}$ is periodic with period $\pi$, $$f(t+\pi)=e^{1+2j(t+\pi)}= e^{1+2jt} e^{2j \pi}=f(t), \forall t\in R$$ As $j=\sqrt{-1},e^{2j\pi}=cos(2\pi)+jsin(2\pi)=1. $

edited Sep 22 '19 at 06:27

XM551

219

answered Sep 22 '19 at 04:16

Z Ahmed

43,235

1

2]pi? i think the period of $e^{(1+j2t)}$ is $\frac{2\pi}{2}=\pi$ – XM551 Sep 22 '19 at 04:31
@XM551 yes you are right, see my edit. – Z Ahmed Sep 22 '19 at 05:31

score 1 · Answer 2 · answered Sep 22 '19 at 06:20

Yes, $f(t)=e^{1+j2t}$ is a periodic function because

$$f(t)=e^{1+j2t}=e^1e^{j2t}$$ which is a constant times a periodic function. The period is $\pi$ since $f(t+\pi)=f(t)$. We also know that for a constant $c$

$cf(t)$ is periodic with period $\pi$.
Every constant function is periodic and has no fundamental period.

therefore you are correct to deduce that a constant times a periodic function remains a periodic function.

Is $e^{(1+j2t)}$ a periodic function?

2 Answers2