Hello again,
Well it is easier to test with DC voltages first then move to AC later. You should see a better result than that, but it is still not perfect yet.
We have a problem with this circuit, and that is that the current mirrors will only work right if they BOTH see the same impedance on their outputs. One solution would be to use a 1k resistor to ground on the output of BOTH upper and lower section current mirror pairs, then for the output stage just construct a differential amplifier with a gain of 10.
What we have now even with a 1k on the output of the lower section is the lower section looks good, but then the upper section does not have a constant 1k impedance on it's output anymore because the output of the lower section changes the bias point for the upper section output, meaning it's no longer a 1k to ground and therefore it will not work as expected, at least not exactly.
With both outputs having a 1k to ground, that means we can at least get the same output from both upper and lower sections. It's then just a matter of using a differential amplifier (four resistors) unless we can find another way to solve this problem.
I am assuming that you are trying to build an instrumentation amplifier, but normally we dont need current mirrors for that. Without the current mirrors we dont have the problem of imbalanced output impedances on the upper and lower sections.
If you want to try this it is not hard to do. Connect a 1k resistor to ground on each current mirror output, then use a regular differential amplifier with the inputs also connected to the outputs of the upper and lower section current mirrors. The output of the diff amp will then be the difference between outputs of the upper and lower sections. The output amp should have a gain of 10 if R2 and R4 are both equal to 10k.
Another idea is to use the two 1k resistors to ground, then use gain of 1 buffers to isolate those outputs from the output stage inputs thus forcing both current mirror pairs to see the same impedance and at the same time supply the right output voltage. This would be a theoretically ideal solution, while the former is an approximation because the output impedances are only very nearly equal in that solution, not exactly equal.