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y(t)= h(t)*x(t) where h(t) is a decaying exponential and x(t)= sin(5t) u(t). Find y(t) using convolution theorem. I'm confused about the sine wave. If i write sinusoid in exponential form then I get imaginary parts as well. can someone please help

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    \$\begingroup\$ y(t) must be real, so the imaginary parts of the convolution integral must cancel. Can you show your working so far. \$\endgroup\$ – Chu Mar 17 '15 at 18:12
  • \$\begingroup\$ Thanks Chu. I gave my solution below. Hopefully its correct \$\endgroup\$ – Fox Knue Mar 19 '15 at 15:05
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Hint: You have to combine the resulting complex exponentials into sine and cosine terms:

$$\sin x=\frac{e^{jx}-e^{-jx}}{2j}\\ \cos x=\frac{e^{jx}+e^{-jx}}{2}$$

If you use \$h(t)=e^{-at}u(t)\$, you should end up with the expression

$$y(t)=\frac{1}{a^2+25}\left[a\sin(5t)-5\cos(5t)+5e^{-at}\right]u(t)$$

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  • \$\begingroup\$ Thanks Matt. Yes, i totally missed the integral formula \$\endgroup\$ – Fox Knue Mar 19 '15 at 15:06
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I've solved it like this:

Let's say h(t)= exp(-5t) u(t).

Because we have u(t), y(t)= 0 for t<0 since unit step is zero for values less than '0'

For t>0 we get lower limit=0 and upper limit=t

y(t)= x(t)*h(t)

=  ∫ x(Ω) h(t-Ω) dΩ

=  ∫ sin(Ω) exp(-5t+5Ω) dΩ  *convolution expression*

=   exp(-5t) ∫sin(Ω) exp(5Ω) dΩ

using integral formula:

exp(ax) sin bx dx = (exp(ax) / (a 2 + b 2) (asin bx - bcos bx) + c

And then apply limits (lower limit=0 ; upper limit=t)

P.S i don't have much experience writing mathematical expressions like this so excuse my notations

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