1- Suppose that you are recording sound in a room with an echo. so your microphone picks up both the original signal and a delayed, attenuated version of the signal. You decide to model this process as an LTI system with impulse response h(t)=delta (t)+beta delta (t-alpha ) a)(5p) Determine the frequency response, H(w) , of the system. b)(20p) For alpha =2 and beta =1 , find and plot the magnitude response and phase response of the system.

a) To determine the frequency response, $H(w)$, of the system, we need to take the Fourier transform of the impulse response, $h(t)$.<br /><br />The impulse response is given by:<br />$h(t) = \delta(t) + \beta \delta(t - \alpha)$<br /><br />Taking the Fourier transform of $h(t)$, we get:<br />$H(w) = \mathcal{F}\{h(t)\} = \mathcal{F}\{\delta(t)\} + \beta \mathcal{F}\{\delta(t - \alpha)\}$<br /><br />Using the properties of the Fourier transform, we have:<br />$\mathcal{F}\{\delta(t)\} = 1$ and $\mathcal{F}\{\delta(t - \alpha)\} = e^{-jw\alpha}$<br /><br />Substituting these values, we get:<br />$H(w) = 1 + \beta e^{-jw\alpha}$<br /><br />b) For $\alpha = 2$ and $\beta = 1$, the magnitude response and phase response of the system can be found by substituting these values into the expression for $H(w)$.<br /><br />Magnitude response:<br />$|H(w)| = \sqrt{1 + \beta^2 - 2\beta\cos(w\alpha)}$<br /><br />Phase response:<br />$\angle H(w) = \arctan\left(\frac{\beta\sin(w\alpha)}{1 + \beta\cos(w\alpha)}\right)$<br /><br />Substituting $\alpha = 2$ and $\beta = 1$, we get:<br />$|H(w)| = \sqrt{1 + 1 - 2\cos(2w)}$<br />$\angle H(w) = \arctan\left(\frac{\sin(2w)}{2\cos(2w)}\right)$<br /><br />To plot the magnitude response and phase response, we can use a graphing tool or software. The magnitude response will show the variation of the magnitude of $H(w)$ with respect to the frequency, while the phase response will show the variation of the phase of $H(w)$ with respect to the frequency.