Simple Impulse Response

Question

I am trying to find the impulse response to this system:

At first it looks easy enough, $y(n)=x(n)(-1)^n$, set $x(n)=\delta(n)$, which makes $h(n)=(-1)^n$. But something seems wrong.

What am I missing?

(considering that $(-1)^n=\cos(\pi n)=e^{j\pi n}$, this system seems more complicated than it looks. It seems like $y(n)$ is $x(n)$ with a 180$^\circ$ phase shift.)

Answer 1

as the comments to your question mention you have not done the first step: system classification

1. Is the system linear? We have to input three different signal in the system and compare the outputs: the two signals $x_1$ and $x_2$ and the sum signal $x_{1+2} = x_1 + x_2$

\begin{align} y_1 &= (-1)^t x_1 \\ y_2 &= (-1)^t x_2 \\ y_{1+2} &= (-1)^t (x_1 + x_2) \\ \end{align}

and even at this point it is easy to see that $y_1 + y_2 = y_{1+2}$ that means the system is linear

2. Is the system time variant? it's usually very easy to say yes here when your system is a function of the time (or n in your case) $z(t) = (-1)^t $ Mathematically you have to input a single signal $x(t)$ at two points in time and check if the system gives you different outputs at different times:

\begin{align} y_1 &= (-1)^t x(t) \\ y_2 &= (-1)^t x(t-\Delta t) \\ \end{align}

to compare 1 and 2 you have to time shift the output of the first system to make sure they overlap the way you want.

\begin{align} y_1 &= (-1)^t x(t) \\ y_{1-\Delta t} &= (-1)^{t - \Delta t} x(t - \Delta t) \\ y_2 &= (-1)^t x(t-\Delta t) \\ \end{align}

the comparison shows that those two are not the same, which means your system changes over time

\begin{align} (-1)^{t - \Delta t} x(t - \Delta t) &= (-1)^t x(t-\Delta t) \\ (-1)^{t - \Delta t} &= (-1)^t \\ (-1)^{ - \Delta t} &= 1 \\ \end{align}

this is not true in general, that means your system is linear and time variant and you can not use the description of a LTI System. But there's a definition of the impulse response of a time variant system that you could use:

Time-varying "impulse response"

\begin{align} y(t) = \int h(\tau, t) x(t - \tau) d\tau \end{align}

since your system is memory less and kausal (i don't think there's a trivial mathematical proof for those two properties but it is easy to see), your time variant impulse response is rather simple:

\begin{align} h(\tau, t) = \delta(\tau) (-1)^t \end{align}

which leads to the collapse of the integral in the time variant convolution (just check the rules for integration over the dirac function if you don't remember them):

\begin{align} y(t) = \int \delta(\tau) (-1)^t x(t - \tau) d\tau \\ y(t) = (-1)^t x(t) \end{align}

And the last line ( and basically your first line) is the best description for linear memory less time variant systems - a simple multiplication with a time variant function or in general just:

\begin{align} y(t) = x(t) z(t) \\ \end{align}

bye bye stackexchange you managed to annoy me enough in my first two answers for me to decide not to come back.