Ampitude of a wave: f(t)= cos(t)+sin(t+(pi/4))

[J.T]

Evening all

I have a function
f(t)= cos(t)+sin(t+(pi/4))

To find the amp of this wave do i just use the general constant formula (not sure its called that)
asinx+bcosv= Sqrt(a^2+b^2)sin(x+v)

so the amp is sqrt(1^2+1^2)

so sqrt2 or 1.4142?

thanks

JT

 

[Deleted member 4993]

Evening all

I have a function
f(t)= cos(t)+sin(t+(pi/4))

To find the amp of this wave do i just use the general constant formula (not sure its called that)
asinx+bcosv= Sqrt(a^2+b^2)sin(x+v) ...... Incorrect

so the amp is sqrt(1^2+1^2)

so sqrt2 or 1.4142? ..... No

thanks

JT

A * cos(Θ) + B * sin(Θ) = √(A² + B²) * sin (Θ + Φ)

where

sin(Φ) = A/√(A² + B²)

 

[Deleted member 4993]

cos(t)+sin(t+(π/4))

=cos(t) + sin(t)*cos(π/4) + cos(t) * sin(π/4)

= 1/√2 * sin(t) + (1 + 1/√2) * cos(t)

Now apply the form I have shown you in the post above....

 

[J.T]

thanks for that. I get.

√((1/√2)² + (1 + 1/√2)²)

=1.8478

regards

JT