1. ## sin(t)*cos(t) convolution

Hello everyone, what is the convolution of sin(t) with cos(t). The provided answer seems to be merely t/2.sin(wt)... However, the integral im getting includes this expression along with other sin and cos terms.

Another question, whats the integral of sin(X) FROM -inf to +inf....

Any help is appreciated...

2. You can compute the convolution $\sin t * \cos t$ or from definition...

$\sin t * \cos t = \int_{0}^{t} \sin u \cdot \cos (t-u)\cdot du$ (1)

... or using the Laplace Transform. In that case is $\mathcal{L} \{\sin t\} = \frac{1}{1+s^{2}}$ , $\mathcal{L} \{\cos t\} = \frac{s}{1+s^{2}}$ so that...

$\mathcal {L} \{\sin t * \cos t \}= \frac{s}{(1+s^{2})^{2}}$ (2)

... and ...

$\sin t * \cos t = \mathcal{L}^{-1} \{\frac {s}{(1+s^{2})^{2}}\} = \frac{t\cdot \sin t}{2}$ (3)

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
You can compute the convolution $\sin t * \cos t$ or from definition...

$\sin t * \cos t = \int_{0}^{t} \sin u \cdot \cos (t-u)\cdot du$ (1)
Convolution is defined as:

$(f*g)(t)=\int_{-\infty}^{\infty} f(\tau)g(t-\tau)\; d\tau$

So in this case the convolution cannot be defined by this integral.

What you have defined is the convolution of $u(t)\sin(t)$ and $u(t)\cos(t)$, which may be what the OP wants, but is not what they asked for.

CB

4. Thanks guy for the help, well turns out that i've done a horrible calculation mistake, the result worked out fine. Well i would have loved to use laplace transform but we (as students) are obliged to resolve to "convolution" for the time being. And concerning the 0 --> t or the -inf ---> +inf dispute, the result i got was using the 1st set of limits...

And i still dont get why it can't be defined.

(thanks once more)

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# convolution sin x cos x

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