# Steady-State Output from Transfer Function 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.

• 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
– Ben
Nov 3 '19 at 1:25
• @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 Nov 3 '19 at 2:28
• After you have your exponential on the numerator and on the denominator, you simply have to evaluate the gain and phase shift
– Ben
Nov 3 '19 at 2:48
• Now, you've got the gain and phase shit. So apply the gain to the signal and phase shift I think you made a mistake with the phase shift sign btw. x(t) = sin(t) y(t) = 3.21 sin(t - 0.224)
– Ben
Nov 3 '19 at 3:50

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.