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The progress I have made is as follows:

$\sin(t)$ is our signal therefore $\omega = 1 = 2\pi f$ and $f$ = $\frac{1}{2 \pi}$

Also, $f_s$ = 10Hz therefore T = $\frac{1}{f_s}$ = 0.1s

$H(z) = \frac{z}{z-0.7}$ so let $z = e^{j\omega T}$ where $\omega T = 1 \times0.1 = 0.1$

$H(\omega) = \frac{e^{j0.1}}{e^{j0.1} - 0.7}$

Taking the denominator:

$\cos(0.1) + j\sin(0.1) = 0.995 + 0.099j$

therefore,

$\cos(0.1) -0.7 + j\sin(0.1) = 0.311∠0.324$

$H(\omega) = \frac{1∠0.1}{0.311∠0.324}$

$H(\omega) = 3.21∠0.224$

From here I am out of ideas on how to continue. Any advice appreciated.

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  • $\begingroup$ hint : e^jx = cos(x) + j sin(x) So your denominator is : cos(0.1) - 0.7 +j sin(0.1). You can convert it back to an exponential $\endgroup$
    – Ben
    Nov 3, 2019 at 1:25
  • $\begingroup$ @Ben thanks for the hint, I think I have the bulk of the problem worked out now, I'm just not sure how to use that answer to answer the rest of the question $\endgroup$
    – Mikey
    Nov 3, 2019 at 2:28
  • $\begingroup$ After you have your exponential on the numerator and on the denominator, you simply have to evaluate the gain and phase shift $\endgroup$
    – Ben
    Nov 3, 2019 at 2:48

1 Answer 1

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You may know that one important property of linear time-invariant (LTI) systems is that the complex exponential $e^{j\omega_0}$ is an eigenfunction, and the corresponding eigenvalue is given by the system's frequency response evaluated at $\omega_0$.

So the response to an input signal $$x[n]=e^{jn\omega_0}$$ is given by $$y[n]=H(e^{j\omega_0})e^{jn\omega_0}$$

Now try to show that from this it follows that the response to a sinusoidal input signal $$x[n]=\sin(\omega_0n)$$ is given by $$y[n]=\big|H(e^{j\omega_0})\big|\sin\left[\omega_0n+\angle H(e^{j\omega_0})\right]$$ (if the system is real-valued). Now should you have all the information to compute the reconstructed output signal.

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