PI

Thread Starter

lomax1

Joined Feb 6, 2014
27
I HAVE HAD THIS THROWN, KICKED AND EVEN SPAT AT ME BUT DOES NOT SEEM TO GO IN....

∏ (pi) if I have an equation 2∏t does this mean 2 of that number or 2 halves? so confused....
 

MrChips

Joined Oct 2, 2009
30,802
It usually means two times pi times t

2 x ∏ x t

where ∏ = 3.141592653589793

Give us the context so we can better help you.
 

MrChips

Joined Oct 2, 2009
30,802
Nice.

Usually T refers to the period of a repetitive signal
and t is a time variable.

Hence

t/T x 2∏

refers to the phase angle at time t where the phase angle is calculated in radians.

(There are 2∏ radians in 360 degrees or one period.)
 

Thread Starter

lomax1

Joined Feb 6, 2014
27
So then, t= 0.002 Seconds, T(Tau)= 1.6 Seconds, T = 50 milliseconds and the equation is : y(t)=e-t/t(tau)sin(2∏t/T) It would be written as: e-0.002/1.6sin(2x50/∏)?
 

shteii01

Joined Feb 19, 2010
4,644
o...k, and the first bit before the sin is ok?
Yes, it looks alright, the t is in seconds, the tau is in seconds. So you have: e to the power of -t/Tau
\(e^{\frac{-t}{Tau}}=2.71^{\frac{-0.002}{1.6}}=0.998\)
I think you can round it up to 1 or just keep 0.998 or 0.99
 

WBahn

Joined Mar 31, 2012
30,052
So then, t= 0.002 Seconds, T(Tau)= 1.6 Seconds, T = 50 milliseconds and the equation is : y(t)=e-t/t(tau)sin(2∏t/T) It would be written as: e-0.002/1.6sin(2x50/∏)?
Notice that you have

y(t)=e-t/t(tau)sin(2∏t/T)

This is very ill-formed. Strictly speaking, what you have written is

\(
y(t) \; = \; e\,-\, \( \frac{t}{t} \) \tau \, \sin \( \frac{2 \pi t}{T} \)
\)

What I'm pretty sure you meant to write is

\(
y(t) \; = \; e^{- \frac{t}{\tau}} \, \sin \( \frac{2 \pi t}{T} \)
\)

which, in text, would be written as

y(t) = e^(- t/tau ) * sin(2∏t/T)
 
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