neils_arm_strong
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i think that exponential function is a nonperiodic...since it keeping on increasing or decreasing even at infinity...neils_arm_strong said:My friend argued with me that e^x is periodic with a period of 2πj.I do not agree with him.
Can we define such periodicity where the time period is imaginary? e^x is not even periodic .
If if x=a+jb, e^x is periodic only if a=0 and b≠0. If a <0 the signal is called dumped sine signal, wich means a signal signal whose amplitude decreasy exponetialy. On the other hand, If a>0 the periodic signal e^jb will have its amplitude increasing exponentially until saturation or theoretically until infinite.whynot910 said:if x=a+jb,j=sqrt(-1), e^x is periodical so long as b is not equal 0
claudiocamera said:I mantain that "a" must be zero, the point here is if the function is periodic or not and not its behavior in a particular axis.
the definition of periodic function is f(t)=f(t + KT) where T is the period and K is an integer number.
Now, tell me that this definition stands for e^(a+jb)t if "a" is not zero.
eecs4ever said:claudiocamera said:I mantain that "a" must be zero, the point here is if the function is periodic or not and not its behavior in a particular axis.
the definition of periodic function is f(t)=f(t + KT) where T is the period and K is an integer number.
Now, tell me that this definition stands for e^(a+jb)t if "a" is not zero.
I agree that f(t) = e^(a+jb)t is periodic iff a == 0.
But i thought the question asked about f(x) = e^x .
this is different from f(x) = e^(a+jb)x .
if f(x) = e^x then f(x+i*2*pi) = f(x) since e^(x+i*2pi) = e^x.
so f(x) is periodic long the imaginary axis, with a period of 2*pi.
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