Question

Question :12 - Signals:

Find whether the given Discrete-time signal is Periodic. If so, then find its Periodicity..

\[\huge \color{green}{ x[n] = e^{\frac{j \cdot \pi \cdot n}{16}} \cos(\frac{n \cdot \pi}{17})}\]

Answer 1

It will definitely be periodic

the exponent is periodic in \(32 \pi \) and the \(\cos\) is periodic in \(34 \pi\).
So \(x[n]\) is periodic in \(\pi\cdot LCM(32,34) = 544 \pi\)

Whether there exists a smaller period, I don't want to find out.

Answer 2

How can exponential has period of \(32 \pi\) ??

It is discrete-time signal and not Continuous-time.. :)

Answer 3

@UnkleRhaukus

Answer 4

@iambatman save me.. :)

Answer 5

The Complex Exponential is Periodic with N = 32 Samples..

And the cosine term is Periodic with N = 34 Samples..

But what will be Overall Period of this overall sequence??

LCM of (32, 34) ??

Answer 6

for a simple signal
\[x(t)=A\cos(\omega t-\varphi)\]
the period is
\[T=2\pi/\omega\]

Answer 7

Uncle Rocks, yes you are right, but this is Discrete Time Signal having sampled values of continuous time signal, n here is an integer..

Answer 8

\(x[n] = Acos(\Omega n - \phi)\)

the period of this sequence is:

\[N = \frac{2 \pi}{\Omega} \times r\]

Answer 9

Where \(r\) is the smallest integer for which N turns to be an Integer.. :)

Answer 10

Like:

\(x[n] = cos(3 \pi n)\)

\[N = \frac{2 \pi}{3 \pi } \times r \implies N = \frac{2}{3} \times r\]

But here N is not integer, so for smallest, r = 3, N = 2 which becomes Integer..

Answer 11

So, for this Discrete Signal, N = 2 Samples is the period..

Answer 12

If you can use any software, then can you show the waveform of my original question?

Answer 13

measuring on my graph
period is = 16.5
which isn't an integer, but it is between 16 and 17

Answer 14

@amistre64

Answer 15

the complex exponent, recollections of trig conversions to e ...

can we convert this into all trig or all exponent?

Answer 16

yes...

Answer 17

By Euler Identity,we can do that..

Answer 18

it might be easier to see when its all in one form ...

Answer 19

\[x[n] = [\cos(\frac{ \pi n}{16}) + j \sin(\frac{\pi n}{16})] \cos(\frac{\pi n}{17})\]

Answer 20

Should I distribute that cosine term into brackets??

Answer 21

and i recall stuff like:

cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

adding we get

cos(a+b) + cos(a-b) = 2cos(a)cos(b)

cos(a)cos(b) = cos(a+b)/2 + cos(a-b)/2

not sure how the i parts pan out tho

Answer 22

i would, to attempt to get this into some linear format

Answer 23

not sure if theres a thrm or something since i havent taken any courses with this stuff

Answer 24

I can distribute and then with trig identities I can make them into addition form too.. But will that be helpful? I don't know when it comes to imaginary..

If it were single imaginary only, then also I can find it,, but extra cos term is bothering me..

Answer 25

sin(a+b) = sin(a)cos(b) + sin(b)cos(a)
sin(a-b) = sin(a)cos(b) - sin(b)cos(a)

works the same, the j part is just the j part .... i believe the combined trig argument is what i would be looking at

Answer 26

Then let me simplify to look more clear form of it..

Answer 27

But it will be painful to do that.. :(

Answer 28

math is often painful, and hardly ever worth it ... but death absolves all absolutes :)

Answer 29

Real part goes like this :

\[\frac{1}{2}\cos( \pi n \frac{33}{272}) + \frac{1}{2}\cos(\pi n \frac{1}{272})\]

Answer 30

Right??

Answer 31

mcd wifi is slow ...

Answer 32

Imaginary part will be:

\[\frac{1}{2}\sin( \pi n \frac{33}{272}) + \frac{1}{2}\sin(\pi n \frac{1}{272})\]

Answer 33

mcd??

Answer 34

mcdonalds

Answer 35

Are you there??

Answer 36

Or it is your wifi connection from there??

Answer 37

daughter is working here, and so today i dropped her off and am using the wifi

Answer 38

but yes, those are fine 33/272 and 1/272

Answer 39

That's great.. :)

Answer 40

so what are the 2 periods for those functions? and can you think of a way to combine them so that they work in 1 period togehter?

Answer 41

I try..

Answer 42

How about finding \(\omega\) by using :

\[\omega = \frac{HCF(Numerator)}{LCM(Denominator)}\]

Answer 43

if that works, give it a shot :)

Answer 44

If I find individual periods, then for Discrete time, their sum will be periodic, it is a proven thing..

Answer 45

For Individual Signals, I got N = 544 Samples..

Answer 46

\[N = \frac{2 \pi}{\frac{\pi}{272}} \implies N = 544\]

Answer 47

thats what im getting

Answer 48

This is Individual periods..

Answer 49

The common \(\omega\) will come out be :

\[\omega = \frac{HCF(\pi , 33 \pi)}{LCM(272, 272)} \implies \omega = \frac{\pi }{272}\]

Answer 50

for the dbl ch

http://www.wolframalpha.com/input/?i=y%3Dcos%28x+pi%2F272%29%2F2+%2B+cos%2833x+pi%2F272%29%2F2+%2B+i%28sin%28x+pi%2F272%29%2F2+%2B+sin%2833x+pi%2F272%29%2F2%29

Answer 51

This will give N = 544 Samples..

Answer 52

I was just about to say to you that please check on wolfram.. :)

Answer 53

theres prolly a simpler way to go about it, but i dont do simple lol

Answer 54

yes, @amistre64 thank you so much.. :)

Answer 55

good luck ;)

Answer 56

What was the simpler method, I don't get yet.. :)

Answer 57

You have more time??

Answer 58

Even this looks a simpler method now, atleast it is giving me an answer. :P

Answer 59

simpler method, find someone smarter than me ;)

someones asking me something in another question, so ill be around

Answer 60

I have one doubt regarding limits of integration, can you clarify?

Answer 61

\[\int\limits_{-\infty}^{\infty} x(\tau) h(t - \tau) d \tau\]

Answer 62

Here, if I put \(t - \tau = \lambda\)

Then how the limits will change?

Answer 63

Still -infinity to infinity or infinity to -infinity??

The initials are the limits of \(\tau\)..

Answer 64

@ganeshie8

Answer 65

Tell me about the limits..

Answer 66

Reading for so long..

Answer 67

\[\int\limits_{-\infty}^{\infty} x(\tau) h(t - \tau) d \tau = \int\limits_{\infty}^{-\infty} x(t-\lambda ) h(\lambda ) (-d\lambda) = \int\limits_{-\infty}^{\infty} x(t-\lambda ) h(\lambda ) d\lambda \]

Answer 68

have to cuz you always ask hard questions :o

Answer 69

Is this hard?

Answer 70

Oh my God, what the hell, why I did not see - with d(lambda) too.. :) I am just a useless guy..

Answer 71

Wait, how this goes if I have summation in place of integration?

Answer 72

\[\sum_{k = -\infty}^{\infty} x[k] h[n-k]\]

Answer 73

If here I put : \(n - k = p\) ??

Answer 74

how do we define \(\large \sum \limits_{k=0}^{-10}f(k)\)
?

Answer 75

Oh!! I think in discrete summation, limits can be interchanged without any worry..

Answer 76

As it is a Sum, it can be written from left to right or right to left..

Answer 77

There -10 will come beneath and 0 will go upwards..

Answer 78

\[\sum_{k = -10}^{0} f(k)\]

Answer 79

it is better to write it in below form :

\[\large \sum \limits_{-10\le k\le 0}f(k)\]

Answer 80

So:

\[\sum_{k = \infty}^{-\infty} = \sum_{k =- \infty}^{\infty} \]

Answer 81

Okay, then??

Answer 82

\[\large \sum_{ -\infty \lt k \lt \infty } x[k] h[n-k] \]

Answer 83

changing variables should not have any effect on limits i guess

Answer 84

We can't write it like this??

\[\large \sum_{ -\infty \lt k \lt \infty } x[k] h[n-k] = \sum_{k = -\infty}^{\infty} x[k]h[n-k]\]

Answer 85

Yes, limits will not change in case of summation..

Answer 86

i don't see anything wrong, but i have never seen a negative infinity as lower limit in a sum

Answer 87

You have not seen??

Answer 88

I am seeing it the time I begin Convolution of Discrete Signals.. :)

Answer 89

it means you're adding up area below the funtion from left end to right end

Answer 90

I am always thankful to you @ganeshie8 , thanks for your help here..

For about 20 minutes, I was thinking, how I can change the limits again to -infty to infty..

Answer 91

Thank you.. :)