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## Homework Statement

The impulse response ##h(t)## of a linear time invariant system is real valued. Where * denotes the complex conjugate, the transfer function satisfies:

$$H(- \nu) = H^* (- \nu)$$

Use this result to show that for such a system, given the input ##f(t)= \sin(2 \pi \nu t)##, the output will be:

$$g(t)=(f*h)(t) = |H(\nu)| \sin (2\pi \nu t +arg(H(\nu)))$$

## Homework Equations

Convolution integral is given by: ##g(t)= \int^\infty_{- \infty} f (\tau) h(t-\tau) d \tau##

For input ##f(t)=\exp (j2 \pi \nu t)##:

##g(t)=(f*h)(t)= (\int h(\tau) \exp (- j 2 \pi \nu \tau) d \tau) \ exp (j 2 \pi \nu t) = H (\nu) \exp (j 2 \pi \nu t)##

## The Attempt at a Solution

So, starting from the definition of a convolution:

$$\int^{\infty}_{-\infty} \sin (2\pi \nu \tau) . h(t- \tau) d \tau \ \ (1)$$

We must show that this is equal to:

$$g(t)= |H(\nu)| \sin (2\pi \nu t +arg(H(\nu)))$$

I know that ##H(\nu)= \int^{\infty}_{-\infty} h(t) e^{-2 \pi j \nu t} dt##, so the expression above is equal to:

$$g(t)= (\int^{\infty}_{-\infty} h(t-\tau) e^{-2 \pi j \nu \tau} d \tau) \sin (2\pi \nu t +arg(H(\nu)))$$

So, how can I manipulate

**(1)**further to get to this result?

Any explanation is greatly appreciated.