Determining output of a LTI system

Question

Consider following LTI system $$y[n] - 2y[n-1] = x[n]$$ where $x[n]$ is the input to the system and $y[n]$ is the output. Let $x[n] = \cos[n\frac{\pi}{3}] + 2\cos[n\frac{\pi}{2} + \frac{\pi}{4}]$, determine the Fourier series coefficients for $y[n]$.

My try: First of all we should determine impulse response $h[n]$. So we have $h[n] - 2h[n-1] = \delta[n]$ and assuming initial rest condition leads to $h[n] = 2^nu[n]$. Then the frequency response could be computed as follows: $$H(e^{j\omega}) = \sum_{n = -\infty}^{+\infty}h[n]e^{-jwn} = \sum_{n = 0}^{+\infty}2^ne^{-jwn} = \sum_{n = 0}^{+\infty}(2e^{-jw})^n$$ It's a geometric series and divergent because $|2e^{-j\omega}| = 2 >1$. So we should conclude that response to $x[n]$ doesn't exist? Also is initial rest a valid assumption here?

Edit: Let $a_k$ be the Fourier series coefficients for the input signal $x[n]$. Then the Fourier series coefficients for output is $b_k = a_kH(jk\omega_0)$ where $\omega_0$ is the Fundamental frequency of the input. The problem is that for each $k$ the geometric series diverges so $b_k$ doesn't exist.

Answer 1

You are right that the system $h[n] = 2^n.u[n]$ is unstable in absolute summability sense. And absolute summability of impulse response of an LTI system is the required condition for BIBO stability. Meaning Bounded Input bounded output stability. Here the input is bounded and it is only corresponding to 4 frequencies in digital frequency domain $\omega = -\pi /3, \pi /3, -\pi/2, \pi/2$. The magnitude of input cannot be greater than 3 at any value of $n$. But, the output will not be bounded because the system is not BIBO stable.

Even if you start the system at 0, meaning $y[n] = 0, \forall n < 0$, $y[n]$ will blow up as $n$ grows. $$y[n] = \sum^{\infty}_{m=0}2^m.x[n-m], \forall n \ge 0$$ Putting $x[n] = cos[\frac{\pi n}{3}]$ as input: $$y[n] = \sum^{\infty}_{m=0}2^m. \frac{e^{j\pi (n-m)/3} + e^{-j\pi (n-m)/3}}{2}$$ $$y[n] = \frac{e^{j\pi n/3}}{2}.\sum^{\infty}_{m=0}(2.e^{-j\pi/3})^m + \frac{e^{-j\pi n/3}}{2}.\sum^{\infty}_{m=0}(2.e^{j\pi/3})^m$$ Since both the sums are going to blow up as $n$ grows, we can say that $y[n]$ is not bounded.

Another way of saying this is that $H(e^{j\omega})$ does not exist for this system ,because DTFT sum does not converge. And so we cannot get $|H(e^{j\omega})|$ and $\angle{H(e^{j\omega})}$ for any $\omega \in [-\pi, \pi]$.

If we considered the system to be anti-causal, then the unit circle will fall inside ROC because ROC will become $|z|<2$. And the anti-causal but stable system's impulse response will be given by : $h[n] = 2^n.u[-n]$.

But this will be wrong thing to do, because the difference equation represents a Causal System, since the current output sample depends only on current input sample and a past output sample.

Answer 2

The region of Convergence of the system is $ \vert z \vert >2$. Sinusoids are eigen functions of LTI systems over infinite time, since the ROC does not include the unit circle hence system is not stable and hence an infintely running periodic signal will not converge.

More info after comments

The BIBO property of the system, if the input is unbounded even a BIBO stable system can diverge, the boundedness of an input is defined differently to a system, for a bounded signal it's maximum amplitudes should be bounded, absolute summability is required for systems and not signals, if this system had an ROC that included unit circle, we will get a sinusoid output

Initial conditions have nothing to do with stability itslef, if a system is stable than a finite initial condition will act as a transient and will die out after sometime. Initial conditions impact causality and linearity (to some extent) of a system.

Link to discussion on initial condition: LTI system and initial conditions

Answer 3

I would think that your exercise if more related to the concept of Fourier series than to system identification. In an engineering fashion, you would like to suggest the following line of thinking (honestly, I did not do it yet):

1. Find the Fourier series for the input (which has periodicity): you can do this for the two terms separately, since the system is linear
2. Since each input term is a pure tone, its output will be a pure tone, potentially delayed and attenuated, and can be represented by a Fourier series, with unknown coefficients
3. Equate the (unknown) LHS with the RHS, using for instance the shift in time property
4. Solve the system