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

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The discussion centers on determining if the signals x(t) and y(t) are the same, given specific parameters for frequency, time shift, and phase. Both signals have the same frequency, but one is a cosine function and the other is a sine function. The initial conclusion that x(t) equals y(t) is challenged, as the relationship between cosine and sine requires an additional phase shift of -π/2 for equivalence. Participants clarify that without this adjustment, the two signals cannot be considered the same. The conversation highlights the importance of recognizing the differences between cosine and sine functions in signal analysis.
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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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