You're railing your op-amp.

Basically, the output can't swing to the power rails. As such, even though your circuit has a gain of 100, the *maximum* voltage you could ever reasonably expect to get out of the output would be +-15V.

Note that this maximum is *independent of the gain*.

There are basically two things you can do. You can increase your power supply voltages, or decrease your input signal magnitude.

Note that no real op-amp can ever swing *completely* to the rails. A modern rail-rail output op-amp may swing to a few millivolts of the rails.

However, you're using the ancient (and extremely crappy) 741 op amp. This is **not** a rail-rail op-amp. From the 741 datasheet:

VS = ±15V

RL ≥ 10 kΩ

Min: ±12v

Typ: ±14v

You can see with ±15V rails, and a 10 kΩ load on the output (what you have), the output swing is typically +14V - -14V, with a worst-case situation being ±12v.

So basically, your circuit is doing exactly what you would expect, given the op-amp you are using.

The input to this system is denoted by \$x(t)\$ and its output by \$y(t)\$.

To prove that the system is *homogeneous*, what you need to show is
that

if input \$x(t)\$ produces output \$y(t)\$,
then the output produced by input \$\alpha\cdot x(t)\$ is \$\alpha\cdot y(t)\$.

Here, \$\alpha\$ is an arbitrarily chosen real number, that is, the highlighted
statement must hold regardless of the value of \$\alpha\$. Similarly,
the requirement must be satisfied regardless of the choice of input \$x(t)\$.
In other words, finding one pair of input and output signals \$x(t)\$ and \$y(t)\$
and one real number \$\alpha\$ for which the highlighted statement is true
is not sufficient; it has to hold for *all* such choices.

To prove that the system is additive, what you need to show is that

if inputs \$x_1(t)\$ and \$x_2(t)\$ produce outputs \$y_1(t)\$
and \$y_2(t)\$ respectively,
then the output produced by input \$x_1(t)+x_2(t)\$ is
\$y_1(t)+y_2(t)\$.

Once again, cherry-picking is not allowed; the statement has to hold
for all choices of input signals and corresponding output signals.

It is not too hard to verify that your system is both homogeneous
and additive.

## Best Answer

An example for a system with memory could be this counter with \$a_0=0\$

$$a_{n} = a_{n-1} + 1$$

Calculating \$a_{n}\$ requires knowledge about the previous state \$a_{n-1}\$ (except for the case of \$a_{0}\$)

A counter without a memory could look like this

$$a_n(n) = n$$

\$a_{n}\$ can be calculated directly from the input n. Knowledge about a previous state is not required. Both examples are doing the same thing and you might be tempted to say they are the same thing, but under the consideration of memory-less-ness

^{1}they are not the same.A filter may or may not be easy to implement in one way or the other. At first glance, requiring memory appears to be entirely disadvantageous. However, being able to access previous state directly can make an algorithm very quick for example.

A nice example are video codecs. To save space, some video codes only look at the differences between one frame and the next one. Of course this makes it hard to go to an arbitrary frame and jump around in the video, because in theory, every previous frame needs to be calculated. But if you are just watching the video continuously in one go, you will not experience that problem. Encoding a video this way can make its file size smaller.

Knowing that something is stateless

^{2}gives you an advantage, because you know that you should always get the same result for the same input. If you don't, there's either something wrong with that part of your system or if you can guarantee that its working properly then there must be something else that has a memory and causing the differences.A counter that breaks causality could look like this

$$a_{n} = a_{n+1} - 1$$

That's almost ridiculous. I cannot tell you the number without knowing the next number, but that in turn requires knowledge about its next number...where does this end...or rather start? \$\infty -1\$? What value does that have? Such filters cannot be built.

^{1}is that a word?^{2}I guess that's a better word