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exponential function periodic or not

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neils_arm_strong

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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 .
 

hamidr_karami

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hi
if x is Complex then x=a+bi
where, i=sqrt(-1)
hence, exp(x+2ni)=exp(a+bi+2ni)=exp(a+bi)*exp(2ni)=exp(a+bi)
exp(it)=cos(t)+i sin(t)
hence, the function is periodic.
but if x is real the function is not periodic.
hamid karami
 

v_c

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If you plot \[e^{j t}\] on the real (x) vs. imaginary (y) plane, you will see that it will be a circle with radius 1 centered at the origin. When time \[t=0\] you will be at \[x=1,y=0\]; at \[t=\pi/2\] you will be at \[x=0,y=1\]; at \[t=\pi\] you will be at \[x=-1,y=0\]; etc ... etc

You see that the point in the complex \[x-y\] plane will take \[ 2 \pi\] seconds to get back to where it started at \[t=0\]. So yes, the function is periodic.

Incidentally, as the point is rotating counter-clockwise in the \[x-y\] plane, the instantaneous value of \[x\] represents a cosine wave and the instantaneous value of \[y\] represents a sine wave -- both with period of \[ 2 \pi\].

Best regards,
\[v_c\]
 
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whynot910

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if x=a+jb,j=sqrt(-1), e^x is periodical so long as b is not equal 0
 

electronics_kumar

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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 .
i think that exponential function is a nonperiodic...since it keeping on increasing or decreasing even at infinity...
 

jayc

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Strictly speaking, exp(x) cannot be considered periodic. By definition, if exp(x) is periodic, then there must be some period, T, for which exp(x + T) = exp(x).

Equivalently, exp(x) = exp(x) * exp(T) => exp(T) = 1. This is only true if T = j*n*2*pi, where n is any integer. However, it does not make much sense to define an imaginary period.
 

VirtualThread

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Physics aside, according to mathematics the exponential function can be periodic in complex domain essentially a circle.
 

jraks

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common neils we argued this at college and you still dont get it ,e^x is periodic in mathematical and physical science sense but in reality it is not, if you want more of this lets see you in college
 

brahma

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i dont know how to visualise this physically, can anyone explain me??
 

newmin

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I don't see how e^x can be periodic for any x = a +jb. e^x is periodic if and only if x is a pure imaginary number that is when a= 0.

For a continuous function f(x) to be continuous, f(x) must be equal to f(x+T) where T is a positive real number.
 

claudiocamera

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whynot910 said:
if x=a+jb,j=sqrt(-1), e^x is periodical so long as b is not equal 0
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.
 

eecs4ever

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a doesnt have to equal 0 for e^x to be periodic...

e^x , where x is a element of C (complex numbers)

this function is periodic long the j (imaginary axis) .

but its not periodic along the real axis.

for example,
e^1000 is going ot be much bigger than e^5, and as you increase
along the real axis , e^x will only get bigger.

But along the imaginary axis, e^x is periodic.

for example, e^(7) = e^(7+j*2*pi) = e^(7+j*2*pi*k) where k is any real integer.

if you restrict x to be a real number, then e^x is NOT periodic.
but if x is complex, then its periodic along the imaginary axis.
 

claudiocamera

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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.

e^(a+jb)t can be written as (e^at)*(e^jbt) which is the product of a non periodic funtion with a periodic function, now the ultimate question: can the product of a non periodic function with a periodic function results in a periodic function? I don't think so...
 

newmin

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in complete agreement with camera. :)


And for those who think e^x is periodic function for all x=a+jb , with b not equal to zero, please review the definition of a periodic function. It can be found in any signals and systems book.
 

eecs4ever

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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.
 

newmin

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If you look at your quotation about definition of periodic function, it is obvious that i*2pi is not equal to kT where k is an integer number and T is a real positive number.

Yes, there is no such thing as a period of i*2pi... period is always a real positive number. And you cannot say it is periodic along imaginary axis, unless it is a new definition of periodicity introduced by you.....



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.
 

claudiocamera

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Lets see again the definition of a periodic function:

the definition of periodic function is f(t)=f(t + KT) where T is the period and K is an integer number.

if f(x) = e^x then f(x+i*2*pi) = f(x) since e^(x+i*2pi) = e^x.

I agree but the period woud be i*2*pi and not 2*pi so we would think about having a imaginary period !!! So im my opinion there is a concept bias here. You can analyse e^x considering :

x as a real variable
x as a complex variable

This cases were already analyzed.
 

LouisSheffield

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I've noticed that two seperate problems are being evaluated:
In the change of variables x=a+jb, first consider the trivial case - x is constant.
For this case the exponential is also constant, and at best its period is infinite.
Allowing x to vary linearly for simplicity, the two competing views seem to be:
1) a is constant, but b varies linearly, and
2) both a and b vary linearly (and let's allow separate slopes)

For the first case, exp(a) is of constant amplitude, and exp(jb) is truly periodic.
( remembering that exp(a+jb) = exp(a)exp(jb) )

For the second, we have the case of damped or exponentially growing oscillation.
It is not periodic in the sense that f(a+jb) is never of the same amplitude and phase for some subsequent a and b.

The concept of being "periodic on the imaginary axis" is flawed.
A phasor diagram of the second case would show a spiral in amplitude for all phases - real as well as imaginary.


The problem could certainly be reformulated, though, where true periodicity is achieved. Yet, this is answering a different question (and problem) than the one posed (for example, demodulating the amplitude change by a function of the original function's derivative).
 
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