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

  • 2
    \$\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

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


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.