controlcontrol system
In the system whose signal flow graph is shown in figure. \$U_1(s)\$
and \$U_2(s)\$ are inputs. The transfer function
\$\frac{Y(s)}{U_1(s)}\$ is
- \$ \frac{k_1}{JLs^2 + JRs + k_1 k_2}\$
- \$ \frac{k_1}{JLs^2 – JRs – k_1 k_2}\$
- \$ \frac{k_1 – U_2(R + sL)}{JLs^2 + (JR -U_2L)s + k_1 k_2 – U_2R}\$
- \$ \frac{k_1 – U_2(R – sL)}{JLs^2 – (JR -U_2L)s – k_1 k_2 + U_2R}\$
I have a doubt that whether the input \$U_2\$ will come in the transfer function or not? I'm confused between option (1) and (3).
It's eminently possible for an open loop system to have exactly the same TF as a closed loop system: \$ \frac {1}{1+s}\$ could be the TF of a simple series RC circuit with R=1, C=1, or it could be a unity feedback closed loop system with an integrator, \$\frac{1}{s}\$, in the forward path. Or it could be something completely different. It's impossible to tell, from the TF itself, what the primitive structure of the system is.
Equally, putting a Laplace transform into a single box does not mean that the system is open loop. It could be the final abstraction of a primitive closed loop block diagram.
Also, using G for the forward path and H for the feedback path of a CLTF is purely convention. Those letters are not sacrosanct.
Assuming that the order of six the plots is
(i=1) (i=2)
(i=3) (i=4)
(i=5) (i=6)
And that:
Then, also considering the amplitudes, we could sketch a bode plot (like Andy aka suggested - took 5 minutes):
From this, it seems that structure is an derivator (DC gain = 0, positive magnitude slope and phase 90º leading) with two poles (same slope but descending, adds 180º of lag):
$$ Y(s) = \frac{as}{(s+b)(s+c)} U(s) $$
(It could actually be a double/triple/"n" integrator with four/six/"2n" poles, you can discover that by checking if the ascending or descending slope of the gain plot is 20 or 40 or 20*n db/decade).
You can then discover the coefficients by substituting \$Y(s)\$ and \$U(s)\$ in the preceding equation by the inputs and outputs (outputs will be the magnitude gain plus the phase lag), also substituting \$s=jw\$.
Hopefully then you will arrive at your answer.
Best Answer
The transfer function Y/U1 tells you in which way the output of the system (Y) depends on the input signal U1 (only!). Hence, each other input to the system must be set to zero. Another question could be how the output signal Y depends on both inputs - here, we must apply superposition. However, the result would not be a transfer function - just y=f(U1, U2)