# simplify this...

Discussion in 'Math' started by jut, Mar 27, 2009.

1. ### jut Thread Starter AAC Fanatic!

Aug 25, 2007
224
2
$e^{(-1+2j)t}$+$e^{(-1-2j)t}$

my attempt:
$e^{-t}e^{j2t}$ + $e^{-t}e^{-j2t}$

$e^{-t}(e^{j2t}$ + $e^{-j2t})$

euler's identity says...

cos(2t)=$e^{j2t}$+$e^{-j2t}$/2

so...

$e^{(-1+2j)t}$+$e^{(-1-2j)t}$ = $e^{-t}(2cos(2t))$

it seems right, but it's wrong! I checked the proof by substituting numbers into the original equation and the final equation and they evaluate to different numbers.

Any help would be nice.

Last edited: Mar 27, 2009
2. ### Ratch New Member

Mar 20, 2007
1,068
4
jut,

Your solution is correct, but your substitution is wrong. I plotted both the given expression and your solution. They were identical.

Ratch

3. ### jut Thread Starter AAC Fanatic!

Aug 25, 2007
224
2
Thanks. OK the solution was correct. I must have goofed up the substitution last night.

Strangely enough, MATLAB simplified the expression to: 2*exp(-1)*cos(2*t)
which was very close.

Last edited: Mar 27, 2009
4. ### Ratch New Member

Mar 20, 2007
1,068
4
jut,

I disagree with both MATLAB and your assessment. 2*exp(-1)*cos(2*t) = 2*0.37*cos(2t) is a constant amplitude sinusoidal wave. The correct solution is an exponentially damped sinusoidal wave.

Ratch