The circuit has too many "issues" to be quickly corrected, alas.
It bends the brain somewhat to understand what is intended.
Serious questions:
Where did you get the circuit?
Why do you think it may do what you want?
Resistor values are FAR too small.
eg at full swing I = 9V/50R = 180 mA.
Which the 741 will not do.
Depending on the simulator, simulator OPAMPS can have big muscly arms, lift 100 ton weights, swing rail to rail and accept input rail to rail, and more.
Real world ones wont.
The LED is being voltage driven with no concern for what it thinks of this.
It will die in sorrow.
The circuit critically depends on the mic resistance.
Mic type affects what happens.
Add a say 1 uF to 100 uF capacitor in series with the MIC.
BUT the circuit is still "just wrong".
Best to start again.
Explain what you want to do.
If you have a circuit that somebody else provided please supply a link.
Added:
Had another quick glance.
A fatally fatally fatal problem is that IF the microphone is high impedance then the opamp is trying to place 1.7V across the LED. What this does depends on the LED type.
A white or blue one will do nowt.
A red one may do summat. Or not.
...digs .. LED is red 2V nominal
If the microphone has a DC resistance of close to zero (moving coil) then Vout will be close to zero. Above that the mic DC resistance will affect the DCout setpoint.
Place a cap in series with microphone. If it is an electret type it will stop working. We'll deal with that next.
Increase all resistor values by ABOUT 100 times in value (or more.)
After scaling by 100x make R4 R5 bigger (just remove one) or the other 3 smaller so Vout DC is about 4V.
Add a 100 Ohm to 470 Ohm resistor in series with the LED.
eg if Vout is 4VDC and VLED is 2V and RLED series is 100 Ohm then ILED = (4-2)/100 = 20 mA.
Play
If no mic output place a 100k from MIC+ to V+ 9V (electret drive). (On mic side of the cap you added.)
It works.
(LED may be dead by now)
Report.
For starters, can you tell the dimensions of each matrix ? It will make it easier to think further.
Assuming that u(t) and c are both of same dimensions. If they are, you can just make it :
der(x) = F*x(t)+G*[u(t) + c] , because u(t) is an input ( usually the one we control ) and c is also an input, only it's an external one. u(t) suggests it's time varying input, but that's just a general case, it doesn't have to be.
Edit after your comment:
Now that you have explained that u(t) is a 1x1 scalar, I see a solution, that may be too simple to work. You said you are using lsim command in Matlab, so I suppose you pass state space (ss) system model to it. In that case, since according to your comment u(t) is a scalar, G*u(t) is a constant 6x1 vector. Therefore, you can make a substition and say [G*u + c] is your new G matrix. For now I assume you are familiar with this part of using Matlab, but just in case you need a kickstart:
Matlab documentation on this website:
LINK
Says you can pass systems to lsim in this fashion :
[y,t] = lsim(sys,u,t)
[y,t,x] = lsim(sys,u,t) % for state-space models only
[y,t,x] = lsim(sys,u,t,x0) % with initial state
Best Answer
An integrator has its current state (value) because of the past values.
That is kind of a memory: It memorizes the sum of the "past".
The green sentence follows from the yellow sentence because the yellow sentence defines the integrators as state, and the green sentence just repeats that.