Question

What is period T of x(t)

x(t) = cos(3t) + cos(10t)

I was thinking that T = 2*pi, is this correct?

Answer 1

The period of the function is the smallest number T > 0 such that
x(t) = x(t+T)

Remember that

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

Answer 2

13pi/10

Answer 3

hmm if i type it in wolframalpha i also got 2pi :)

http://www.wolframalpha.com/input/?i=period+cos%283t%29+%2B+cos%2810t%29

Answer 4

but jamesj, what about Cn?

i can write x(t) as:

\[1/2e ^{j3t}+1/2e ^{-j3t}+1/2e ^{j10t}+1/2e ^{-j10t}\]

Is \[C1=1/2 + 1/2\]

can i just add those 1 terms together?

Answer 5

2pi is not the answer. 2pi is the period of cos x and sin x.

Let me ask you, what the period of cos(3x) all be itself?

Answer 6

thats

\[2\pi/3 = T\]

Answer 7

period of (cos3t )=2pi/3,n (cos10t)=2pi/10
use the trick take LCM of periods

Answer 8

right. Now, as
x(t) = cos(3t) + cos(10t)
= 2 . cos (13t/2) . cos(7t/2)

what period works for both expressions?

Answer 9

but i dont know that answer of:

2 . cos (13t/2) . cos(7t/2)

Answer 10

cos(13t/2) has period 2pi/(13/2) = 4pi/13
cos(7t/2) has period 2pi/(7/2) = 4pi/7

Now find their lcm.

Or, as Sam suggests, find the lcm of
period of (cos3t ) = 2pi/3
and period(cos10t) = 2pi/10

Answer 11

If you're having trouble with the lcm concept for periods, work some examples:
what is period of
- cos x + cos 2x
- cos 2x + cos 3x

Answer 12

but what is lcm? and how can i calculate it :(

Answer 13

LCM = lowest common multiple. Take integer multiples of both periods until you find the the smallest pair which are equal.

For example, for
cos x + cos 2x

the periods are 2pi and pi. Clearly the LCM of those two periods is 2pi. Hence the period of cos x + cos 2x is 2pi.

For cos 2x + cos 3x
the periods are pi and 2pi/3

Now the integer multiples of pi are: pi, 2pi, 3pi, ...
intger multiples of 2pi/3 are: 2pi/3, 4pi/3, 2pi, ...

hence their LCM is 2pi and the period of cos 2x + cos 3x is 2pi.

Use the same method here.

Answer 14

oke, so

T1 = 2pi/3-> 2pi/3, 4pi/3, 6pi/3....

T2 = 2pi/10 -> 2pi/10, 4pi/10, 6pi/10....

Answer 15

until i find the match?

Answer 16

yes

Answer 17

there must be a faster way

Answer 18

divide T1 with T2?

Answer 19

not necessarily. you're in university now; don't expect the problems to be so elementary.

Answer 20

hmm.. i was thinking about dividing the 2 periods

Answer 21

That doesn't work, most emphatically not.

Answer 22

Just do it, it's not hard.

Answer 23

ahhh alright.. and my other question...

about Cn? have you read that one?

Answer 24

What is your other question exactly?

Answer 25

i have to find Cn of x(t) (fourier)

You can write x(t) as form of Euler, as i typed in above...

Cn for n=1, is 1/2 + 1/2, but can i add there 2 terms together...

Answer 26

I don't understand your question. You're trying to find the Fourier series of something?

Your function x(t) = cos(3t) + cos(10t) is a Fourier series in its own right.
[ x(t) = 0 cos t + 0 cos(2t) + 1 cos(3t) + 0 cos(4t) + ... etc. ]

But what is it exactly you're asking?

Answer 27

a formula to find cn is: \[cn*e ^{jn \omega t}\]

like cos(t) = (e^j3t + e^-j3t)/2
for n=+/-1 cn = 1/2
for other values its zero

Answer 28

the same is for cos(10t)

for n=+/- 1 Cn = 1/2
other n Cn = 0

Answer 29

But what is C1 for x(t)

Answer 30

Wait. Are you trying to find the coefficients c_n such that
\[ x(t) = \sum_{i=-\infty}^\infty c_n e^{njx} \]
?

Answer 31

yes :)

Answer 32

i just gave you the answers, but can i add my 2 answers?

Answer 33

Then notice that if x(t) = cos(3x) and nothing else, then c_n = 0 except for n = +3, -3.

Likewise for cos(10x), except for n = +10, -10

Hence for x(t) = cos(3x) + cos(10x),
the only non-zero coefficients c_n are those for which c_n = -10, -3, 3 and 10

Answer 34

ahh oke...

but what is C_n for -10, -3, 3 and 10?

Answer 35

c'mon! What is it if x(t) = cos(3x) and nothing else? What is
\[ c_3, c_{-3} \]

Answer 36

thats 1/2

Answer 37

yes.

Answer 38

but what is x(t) = cos(3t) + cos(10t)

OWHHH pellet!

>.<

its 1/2 for n=-10,-3, 3 and 10

Answer 39

yes

Answer 40

i was just confused.. about adding 1/2 with the other 1/2

Answer 41

buttt.. all clear now :)

thank you