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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

  1. \$ \frac{k_1}{JLs^2 + JRs + k_1 k_2}\$
  2. \$ \frac{k_1}{JLs^2 - JRs - k_1 k_2}\$
  3. \$ \frac{k_1 - U_2(R + sL)}{JLs^2 + (JR -U_2L)s + k_1 k_2 - U_2R}\$
  4. \$ \frac{k_1 - U_2(R - sL)}{JLs^2 - (JR -U_2L)s - k_1 k_2 + U_2R}\$

graph

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).

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2 Answers 2

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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)

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  • \$\begingroup\$ so, option (1) is right ? \$\endgroup\$
    – Vedanshu
    Feb 15, 2017 at 16:05
  • \$\begingroup\$ Yes, correct. The transfer function Y/U1 is according to (1). \$\endgroup\$
    – LvW
    Feb 15, 2017 at 18:13
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A block diagram solution (complementing the answer from LvW):

Dirceu Rodrigues Jr.

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