To find whether x(t) and y(t) signals are same?

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SUMMARY

The discussion centers on determining the equivalence of two signals, x(t) and y(t), defined as x(t)=cos(ω1(t+τ1) +δ1) and y(t)=sin(ω2(t+τ2) +δ2) with specific parameters: ω1=π/3, ω2=π/3, τ1=0, τ2=1, δ1=2π, and δ2=-π/3. The conclusion drawn by the initial poster, that x(t) equals y(t) due to ω1 equaling ω2, is challenged by another participant who points out the fundamental difference between cosine and sine functions. The correct relationship should account for the phase difference between the two functions, specifically incorporating a term of -π/2.

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Homework Statement


let x(t)=cos(ω1(t+τ1) +δ1) and sin (ω2(t+τ2) +δ2)
determine whether both the signal are same for given data
ω1=∏/3 , ω2=∏/3
τ1=0 , τ1=1
δ1=2∏ , δ1=-∏/3

Homework Equations


The Attempt at a Solution


solution is as given-:
as ω1=ω2,
so
ω1τ1+ δ1=ω2τ2+ δ2+2∏k where k is integer
so x(t)=y(t)

i did not understand how they can conclude like that?
 
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hi amaresh92! :smile:

i assume you mean …
amaresh92 said:
let x(t)=cos(ω1(t+τ1) +δ1) and sin (ω2(t+τ2) +δ2)
determine whether both the signal are same for given data
ω1=∏/3 , ω2=∏/3
τ1=0 , τ2=1
δ1=2∏ , δ2=-∏/3
as ω1=ω2,
so
ω1τ1+ δ1=ω2τ2+ δ2+2∏k where k is integer
so x(t)=y(t)

i did not understand how they can conclude like that?

(try using the X2 button just above the Reply box :wink:)

if they were both cos, that would be correct, but since one is cos and the other is sin, it would have to be

ω1τ1+ δ12τ2+ δ2+2∏k - ∏/2 :confused:
 
:redface:
tiny-tim said:
hi amaresh92! :smile:

if they were both cos, that would be correct, but since one is cos and the other is sin, it would have to be

ω1τ1+ δ12τ2+ δ2+2∏k - ∏/2 :confused:

Oops.
I agree. Now. :redface:
 
Last edited:

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