Basics Of Control Systems Block Diagram And Sfgs Practice Questions

We know phase crossover frequency is that frequency at which phase of the open loop transfer function is $-180^{\circ}$ .

$Y(s)=\underbrace{G_{1}(s) R(s)}_{Y_{1}(s)}+\underbrace{G_{2}(s) D(s)}_{Y_{2}(s)}$ Considering first $R(s)$ only, then $Y(s)$ is $Y_{1}(s)$

Now considering $D(s)$ only, then $Y(s)$ is $Y_{2}(s)$

Hence, $G_{1}(s)$ and $G_{2}(s)$ both are equal.

$\frac{Y}{R}=\frac{G_{1}(1-0)+G_{2}(1-0)}{1-\left[-G_{1} H\right]}=\frac{G_{1}+G_{2}}{1+G_{1} H}$

Using block diagram reduction, we get, $\frac{Y(s)}{X(s)}=H(s)=\frac{s^{2}+1}{s^{3}+2 s^{2}+s+1}$

The transfer function due to the disturbance torque $T_{d}(s) \text { is }$

The steady value of response for unit impulse input

Substituting the given values into above equation, we get $\theta(0)=-\frac{1}{2}=-0.5 \mathrm{rad}$

Transfer function

Given blocks has transfer function $G_{1}(s)$ and $G_{2}(s)$ In cascade connection $G_{1}(s) \cdot G_{2}(s)$ In parallel connection $G_{1}(s)+G_{2}(s)$ From the given options: (For cascading/ summing/differncing) (A) $G_{1}(s) G_{2}(s) \Rightarrow$ Realization is possible. (B) $\frac{G_{1}(s)}{G_{2}(s)} \Rightarrow$ Realization is not possible because the gain of blocks given as $G_{1}(s)$ and $G_{2}(s)$ (C) $G_{1}(s)\left(\frac{1}{G_{1}(s)}+G_{2}(s)\right)=1+G_{1}(s) G_{2}(s)$ Realization is possible. (D) $G_{1}(s)\left(\frac{1}{G_{1}(s)}-G_{2}(s)\right)=1-G_{1}(s) G_{2}(s)$ Realization is possible.

Currently no explanation available

Redrawing the block diagram with $X_{1}(s)=0$ The transfer function $T(s)=\frac{Y(s)}{X_{2}(s)}=\frac{G(s)}{1+G(s) H(s)} \quad\ldots(i)$ Here, $G(s)=\frac{1}{s}$ and $H(s)=\frac{s}{s+1}$ $\frac{Y(s)}{X_{2}(s)}=\frac{1 / s}{1+\frac{1}{s} \times \frac{s}{s+1}}=\frac{(s+1)}{s(s+2)}$

Converting the block d iagram into signal flow graph on Forward paths, $P_{1}=G_{1} G_{2} ; P_{2}=G_{2} \cdot 1 ; P_{3}=1 \cdot 1=1$ So, the transfer function is $\frac{C(s)}{R(s)}=G_{1} G_{2}+G_{2}+1$

Using Mason's gain formula,

For option (d), $\frac{Y(s)}{U(s)}=\frac{100 / s^{3}}{1+\frac{100}{s^{2}}}$ Which is not transfer function of H(s) .

$L_{1}$ and $L_{3}$ are non-touching.

Take off points is moved $G_{2}$ so $H/G_{2}$

From KVL in both loops In first loop, $V_{i}(s)=I_{1}(s) Z_{1}(s)+\left[I_{1}(s)-I_{2}(s)\right] Z_{3}(s)$ $V_{i}(s)=I_{1}(s)\left[Z_{1}(s)+Z_{3}(s)\right]-I_{2}(s) Z_{3}(s)$ $\frac{V_{i}(s)}{Z_{1}(s)+Z_{3}(s)}=I_{1}(s)-\frac{I_{2}(s) \cdot Z_{3}(s)}{Z_{1}(s)+Z_{3}(s)} \; \; ...(i)$ In second loop, $\left[I_{2}(s)-I_{1}(s)\right] Z_{3}(s)+I_{2}(s) Z_{2}(s)+I_{2}(s) Z_{4}(s)=0$ $I_{2}(s)\left(Z_{2}(s)+Z_{3}(s)+Z_{4}(s)\right)=I_{1}(s) \cdot Z_{3}(s)$ $G_{2}=\frac{I_{2}(s)}{I_{1}(s)}=\frac{Z_{3}(s)}{Z_{2}(s)+Z_{3}(s)+Z_{4}(s)}$ From SFG, $I_{1}(s)=V_{i} G_{1}(s)+I_{2}(s) H(s)$ $I_{1}(s)=V_{i}(s) \frac{1}{Z_{1}(s)+Z_{3}(s)}+I_{2}(s) \frac{Z_{3}(s)}{Z_{1}(s)+Z_{3}(s)}$ $\therefore \quad H(s)=\frac{Z_{3}(s)}{Z_{1}(s)+Z_{3}(s)}$ (comparing above two equations)