Transfer Function and Block Diagrams Questions and Answers

Answer: A

When two blocks are in parallel, they receive the same input and their outputs are added at a summing point, so the overall transfer function is the sum.

\[ G(s) = G_1(s) + G_2(s) \]

The product \(G_1(s)G_2(s)\) is the result for blocks in cascade (series). The ratio \(G_1/G_2\) does not correspond to any simple connection, and \(\dfrac{G_1 G_2}{G_1 + G_2}\) is the form of two resistors in parallel, not of transfer functions in parallel. The answer is option A.

Answer: A

The circuit is a series resistor with a capacitor across the output, the first order RC low pass circuit. With zero initial conditions, in the Laplace domain the capacitor has the impedance \(1/(Cs)\), so the output is a potential divider.

\[ \begin{aligned} \frac{E_o(s)}{E_i(s)} = \frac{1/(Cs)}{R + 1/(Cs)} \\ &= \frac{1}{RCs + 1} \end{aligned} \]

The pole is at \(s = -1/RC\) and the dc gain is 1. The option \(1/(RCs)\) is the transfer function of an integrator, and \(RCs + 1\) and \(RCs - 1\) are not transfer functions of this circuit, since they would increase with frequency. The answer is option A.

Answer: A

Take the Laplace transform of the step response, then divide by the transform of the unit step, \(1/s\).

\[ \begin{aligned} Y(s) \\ &= \frac{1}{s} - \frac{7/3}{s+1} + \frac{3/2}{s+2} - \frac{1/6}{s+4} \end{aligned} \]

\[ \begin{aligned} H(s) = sY(s) \\ &= 1 - \frac{7s}{3(s+1)} + \frac{3s}{2(s+2)} - \frac{s}{6(s+4)} \end{aligned} \]

Putting this over a common denominator gives \(H(s) = \dfrac{s + 8}{(s+1)(s+2)(s+4)}\). A check at \(s = 0\): the dc gain is \(8/8 = 1\), which matches the final value of the step response, \(y(\infty) = 1\). Also the residue of \(Y(s)\) at \(s = -1\) is \(H(-1)/(-1) = -7/3\), which agrees with the coefficient of \(e^{-t}\). The answer is option A.

Answer: B

The original diagram has the block \(P\) before the summing point: \(Z = PX \pm Y\). To move the summing point ahead of (before) the block, the signal \(Y\) that is added after \(P\) must be divided by \(P\) so that it has the same effect once it goes through the block.

\[ Z = P\left(X \pm \frac{Y}{P}\right) = PX \pm Y \]

So the correct replacement is to put the block \(1/P\) in the path of \(Y\), then sum with \(X\), and then pass the sum through \(P\), which is figure (B). In figure (A) the second \(P\) multiplies \(Y\) by \(P\), giving \(PX \pm PY\), which is wrong. In (C) the output is \(P(X \pm Y) \ne PX \pm Y\). Figure (D) is the rule for moving a take-off point, not a summing point. The answer is option B.

Answer: D

Reduce the diagram step by step. The blocks \(G_2\) and \(G_3\) are in parallel, so they combine to \(G_2 + G_3\), and the blocks \(G_1\) and \(G_4\) in series combine to \(G_1G_4\).

The feedback path \(H_1\) is taken from a point before the block \(G_2 + G_3\). Moving that take-off point after the block means that \(H_1\) must be divided by \(G_2 + G_3\), giving the inner loop with forward gain \(G_1G_4(G_2 + G_3)\) and feedback \(H_1/(G_2 + G_3)\), with positive feedback.

\[ \begin{aligned} \frac{G_1G_4(G_2 + G_3)}{1 - G_1G_4(G_2 + G_3)\cdot\dfrac{H_1}{G_2 + G_3}} \\ &= \frac{G_1G_4(G_2 + G_3)}{1 - G_1G_4H_1} \end{aligned} \]

This is a single block with the outer feedback \(H_2\) around it, which is figure (D). Figures (A) and (B) are intermediate stages, not the complete reduction, and (C) leaves out \(G_2 + G_3\) and \(H_2\). The answer is option D.

Answer: C

Take the Laplace transform of the step response and divide by the transform of the unit step, \(1/s\).

\[ \begin{aligned} C(s) \\ &= \frac{1}{s} + \frac{0.2}{s + 60} - \frac{1.2}{s + 10} \end{aligned} \]

The common denominator is \(s(s + 10)(s + 60)\), and the numerator works out to \(600\):

\[ \begin{aligned} \frac{(s+10)(s+60) + 0.2s(s+10) - 1.2s(s+60)}{s(s+10)(s+60)} \\ &= \frac{600}{s(s+10)(s+60)} \end{aligned} \]

\[ \begin{aligned} \frac{C(s)}{R(s)} = sC(s) \\ &= \frac{600}{(s + 10)(s + 60)} \\ &= \frac{600}{s^2 + 70s + 600} \end{aligned} \]

The dc gain is \(600/600 = 1\), matching the final value 1. The answer is option C.