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Control 3 2023 Past Paper

Sept 1999

Q5:

The block diagram drawn is of a satellite attitude control system. The reference attitude is ${\theta }_R$, while the actual attitude is denoted by ${\theta }_C$. The satellite has a moment of Inertia, $J$, about its axis of rotation, whilst $K$ and $K_v$ are control system gains. The control system also acts to eliminate unwanted rotations due to a disturbance torque $D$.

a) Derive the transfer function relating actual attitude to reference, $\frac{{\theta }_c(s)}{{\theta }_R(s)}$ b) Derive the transfer function relating actual attitude to disturbance, $\frac{{\theta }_c(s)}{D(s)}$ c) A ref input consisting of a step change of $5^o$ is applied at the same instant the satellite experiences a disturbance equiv to a step change of $50$Nm. Given the following values, derive an expression for the satellite’s attitude $\theta (t)$ due to the following inputs: $J=25{\text{kg m}}^{2}K=500\text{Nm/rad}{K}_v=100\text{Nms/rad}$

Solution

a) Taking $D(s)$ to be 0, we get a feedback of $K_{v}s$ over $\frac{1}{J{s}^2}$ , meaning this can be simplified to: $\frac{\frac{1}{J{s}^2}}{1+\frac{K_{v}s}{J{s}^2}}=\frac{1}{J{s}^2+K_{v}s}$, this subsequently cascades with $K$ giving us $\frac{K}{J{s}^2+K_{v}s}$, which finally has a unity gain, meaning that: $\frac{{\theta }_c(s)}{{\theta }_R(s)}=\frac{\frac{K}{J{s}^2+K_{v}s}}{1+\frac{K}{J{s}^2+K_{v}s}}=\frac{K}{J{s}^2+K_{v}s+K}$

b) Taking ${\theta }_R(s)$ to be 0, we get a feedback of $K_{v}s+K$ over $\frac{1}{J{s}^2}$ , which means that: $\frac{{\theta }_c(s)}{D(s)}=\frac{\frac{1}{J{s}^2}}{1+\frac{K_{v}s+K}{J{s}^2}}=\frac{1}{J{s}^2+K_{v}s+K}$

c) If we take ${\theta }_R(s)=\frac{5\ast \pi /180}{s},D(s)=\frac{50}{s},J=25,K=500,K_v=100$

\begin{align} \theta_c(s) &= \frac{500*(\frac{5\pi}{180})+\frac{50}{s}}{s(25s^2+100s+500)} = \frac{\frac{100\pi}{180s} + \frac{2}{s}}{s(s^2+4s+20)}\\ &= \frac{A}{s} + \frac{Bs+C}{s^2+4s+20} \\\\ 1 &= A(s^2+4s+20) + s(Bs+C) \\ 1&=(A+B)s^2 + (4A+C)s + (20)A \\ s^0 &: 1 =20A, A=1/20 \\ s^1 &: 0 =(4A+C), C=-4/20 \\ s^2 &: 0 = (A+B) , B=-1/20 \\ \therefore\theta_c(s) &= \frac{3.7454}{20}[\frac{1}{s} - \frac{s+4}{s^2+4s+20}] \\ & \text{Recognising } s^2+4s+20 = (s+2)^2+4^2 \\ \theta_c(s) &= 0.18727[\frac{1}{s} - \frac{s+2}{(s+2)^2+4^2} - \frac{1}{2}*\frac{4}{(s+2)^2+4^2}] \\ \rightarrow \theta(t)&= 0.18727[1-e^{-2t}(\cos(4t)+\frac{\sin(4t)}{2})] \end{align}

June 1998

Question

Figure Q.5 shows the block diagram of a control system. a) Using block diagram reduction techniques determine the closed loop transfer function of the system shown in Figure Q5. b) Given the individual transfer functions listed below, using Routh’s criteria determine the range of values of K which for the system to remain stable.

Solution

a) First we can reorganise the block diagram, moving the summing block in front of $G_2$ behind it: Now we can tackle this standardly: $H_2\text{ in feedback with }{G}_2:\frac{\frac{1}{s+4}}{1+\frac{2s}{s+4}}=\frac{1}{3s+4}$ this then cascades with $G_3$: $\frac{1}{3s^2+4s}$ This block now has $\frac{H_1}{G_2}=s^2(s+4)$ in feedback with it: $\frac{\frac{1}{3s^2+4s}}{1+\frac{s^3+4s^2}{3s^2+4s}}=\frac{1}{s^3+7s^2+4s}$ This now cascades with $G_1$: $\frac{K}{s^3+7s^2+4s}$ This block now has $H_3=1$ in feedback, giving: $\frac{C(s)}{R(s)}=\frac{\frac{K}{s^3+7s^2+4s}}{1+\frac{K}{s^3+7s^2+4s}}=\frac{K}{s^3+7s^2+4s+K}$

b) $b_2=-\frac{1}{7}\left|\begin{array}{cc}1& 4\\ 7& K\end{array}\right|=4-\frac{K}{7}$, $b_0=0$ $c_2=-\frac{1}{4-\frac{K}{7}}\left|\begin{array}{cc}7& K\\ 4-\frac{K}{7}& 0\end{array}\right|=K$ \begin{align} \begin{array}{c|ccccc} s^3 & 1 & 4 \\ s^{2} & 7 & K \\ s^{1} & 4-\frac{K}{7} & 0 \\ s^{0} & K \\ \end{array} \end{align} $4-\frac{K}{7}\ge 0\to K\le 28$ $\therefore 0\le K\le 28$

Tutorials

Block Diagram

Tut 1 Q1:

Question

Figure Q1 shows a block diagram of the positioning control system of the Hubble space telescope. Given that $K=100$ and $K_1=12$, derive the closed loop transfer functions relating angular position $C(s)$ to both the commanded input $R(s)$ and a general disturbance $D(s)$. Determine the response of the telescope $c(t)$ to a unit step input in $R(s)$. Hence deduce the telescope’s response to a unit step disturbance $D(s)$.

Solution

First, for calculating $\frac{C(s)}{R(s)}$, we will take $D(s)=0$, meaning:

We can eliminate the $K_{1}S$ feedback loop relatively simply:

$=\frac{\frac{1}{s^2}}{1+\frac{K_1}{s}}=\frac{1}{s(s+K_1)}$

Then this is in series with $K$ giving us: $\frac{K}{s(s+K_1)}$

Finally this is in a unity feedback loop giving us: $\frac{C(s)}{R(s)}=\frac{\frac{K}{s(s+K_1)}}{1+\frac{K}{s(s+K_1)}}=\frac{K}{s(s+K_1)+K}=\frac{100}{s^2+12s+100}$

For finding $\frac{D(s)}{R(s)}$ we can assume $C(s)$ to be 0, and recenter our block diagram around our input:

This is a simple feedback loop: Giving us $\frac{C(s)}{D(s)}=\frac{\frac{1}{s^2}}{1+\frac{K_{1}s+K}{s^2}}=\frac{1}{s^2+K_{1}s+K}=\frac{1}{s^2+12s+100}$

For a unit step input in $R(s)$, we get $R(s)=\frac{1}{s}$, therefore:

$C(s)=\frac{100}{s(s^2+12s+100)}$ \begin{align} \frac{a}{s} &+ \frac{bs+c}{s^2+12s+100} \\100&= a(s^2+12s+100) + s(bs+c) \\ 100 &= (a+b)s^2 + (12a+c)s + 100a \\ s^0&: 100 = 100a, \quad a =1 \\ s^1 &: 0 = (12+c), \quad c = -12 \\ s^2 &: 0=a+b, \quad b=-1 \\ \therefore C(s)_R &=\frac{100}{s} - \frac{s+12}{s^2+12s+100} \\ &\text{Note that }s^2+12s+100 = (s+6)^2+8^2: \\ C(s)_R &= \frac{100}{s}-\frac{s+6}{(s+6)^2+8} - \frac{6}{8}*\frac{8}{(s+6)^2+8} \\ \therefore c(t)_R &= 100 - e^{-6t}\cos8t - \frac{3e^{-6t}\sin8t}{4} \\ c(t)_R &= 100 - e^{-6t}[\cos8t+\frac{3\sin8t}{4}] \end{align}

$\frac{C(s)}{D(s)}=\frac{1}{100}\ast \frac{C(s)}{R(s)}\Rightarrow c(t{)}_D=\frac{1}{100}c(t{)}_R$ $c(t{)}_D=1-\frac{e^{-6t}}{100}[\cos 8t+\frac{3\sin 8t}{4}]$

Tut 1 Q2:

Question

The ailerons of a large airliner are positioned through an angle ${\delta }_a$ by servo actuators which respond to the pilot’s lateral stick displacement $y$, Figure Q2. Derive the closed loop transfer function of this system relating the aileron displacement to the pilot’s stick motions.

Solution

We can eliminate the unity feedback over $\frac{1}{s(0.1s+1)}$ giving us: $\frac{1}{0.1s^2+s+1}$ , which then feeds into another $\frac{1}{s}$, as well as a $4$ term giving us $\frac{4}{s(0.1s^2+s+1)}$ , finally this has a unity feedback meaning the final $\frac{{\delta }_a(s)}{y(s)}$ is: $\frac{{\delta }_a(s)}{y(s)}=\frac{4}{0.1s^3+s^2+s+1}=\frac{40}{s^3+10s^2+s+10}$

Tut 1 Q3:

Question

Figure Q3 shows the block diagram of an electric locomotive motor where ωd is the desired angular velocity of motor as set by the driver whilst ω is the actual angular velocity achieved. The block $G_2(s)$ represents the load transfer function (i.e. the locomotive) whilst $G_1(s)$ is the armature controller transfer function. The locomotive motor can experience torque disturbances due to braking, given by $T_d(s)$. Derive the transfer functions relating output angular velocity to desired angular velocity, $\frac{\omega (s)}{{\omega }_d(s)}$ and output angular velocity to torque disturbance, $\frac{\omega (s)}{Td(s)}$ , given:

Solution

First to find $\frac{\omega (s)}{{\omega }_d(s)}$, we can set $T_d(s)=0$, meaning that we have $G_1{G}_2=\frac{10}{(s+1)(2s+0.5)}$ in feedback with 0.1, this becomes $\frac{\frac{10}{(s+1)(2s+0.5)}}{1+\frac{0.1}{(s+1)(2s+0.5)}}=\frac{100G_1{G}_2}{100+G_1{G}_2}=\frac{10}{0.1+(s+1)(2s+0.5)}$ , which would now be in series with $540$, this becomes $\frac{5400}{0.1+(s+1)(2s+0.5)}=\frac{5400}{2s^2+2.5s+0.6}$, which finally has unity feedback giving us: $\frac{\omega (s)}{{\omega }_d(s)}=\frac{\frac{5400}{2s^2+2.5s+0.6}}{1+\frac{5400}{2s^2+2.5s+0.6}}=\frac{2700}{s^2+1.25s+2700.75}$

Finding $\frac{\omega (s)}{T_d(s)}$, we set ${\omega }_d(s)$ to 0: So our $\frac{\omega (s)}{T_d(s)}$ is just a feedback loop of $G_2=\frac{1}{2s+0.5}$ with $540.1G_1=\frac{5401}{s+1}$ \begin{align} \frac{\omega(s)}{T_d(s)} &= \frac{\frac{1}{2s+0.5}}{1+\frac{5401}{(s+1)(2s+0.5)}} = \frac{1}{2s+0.5+\frac{5401}{s+1}} \\&= \frac{s+1}{(2s^2+2s)+(0.5s+0.5)+5401} \\ &=\frac{s+1}{2s^2+2.5s+5401.5} \\ &= \frac{0.5(s+1)}{s^2+1.25s+2,700.75} \end{align}

Tut 1, Q8, Pt1

Tut 1, Q8, Pt2

Question

Determine the closed loop transfer function of the system shown in Figure Q8. Investigate the stability of this system when:

Solution

Feedback Loop $g_1,h_1$ = $\frac{\frac{1}{s+1}}{1+\frac{0.5s}{(s+1)}}=\frac{0.5s}{1+1.5s}$, this then cascades with $g_2$ , making $\frac{s}{(s+2)(1.5s+1)}=\frac{s}{1.5s^2+4s+2}$ , This collective then has a feedback of $h_2$ which makes the full thing: \begin{align} \frac{C(s)}{R(s)} &= \frac{\frac{s}{1.5s^2+4s+2}}{1+\frac{s}{(s+1)(1.5s^2+4s+2)}} = \frac{s}{1.5s^2+4s+2+\frac{s}{s+1}} \\ &= \frac{s(s+1)}{(1.5s^3+4s^2+2s) + (1.5s^2+4s+2) + s} \\ &= \frac{s(s+1)}{1.5s^3+5.5s^2+7s+2} \\ &= \frac{\frac{2}{3}s(s+1)}{s^3 + (11/3 )s^2 +(14/3)s+(4/3)} \end{align}

TF & Stability

Tut 1 Q4:

Question

Calculate the poles and zeros of the systems with the following transfer functions. Comment on the stability of each and the expected response characteristics.

Solution

i) Zeros at $-1$, Poles at $s=-2,s=-3$, stable, no oscillations ii) Zeros at $-3$, Poles at $s=\frac{-8\pm \sqrt{64-80}}{2}=-4\pm 2j,$ Stable System, Oscillatory iii) Zeros at $-3,-1$, Poles at: $s=-3,s=2,s=-1\pm 3j$. Unstable System, Oscillatory iv) Zeros at …, Poles at: $s=-4,s=-1,s=1.5\pm 4j$ , Unstable System, Oscillatory.

Tut 1 Q5:

Question

Figure Q5 shows a block diagram for a feedback control system. Determine the values of the closed loop poles and hence the stability of the system with the following combinations of controller and plant transfer functions.

Solution

i) $\frac{C(s)}{R(s)}=\frac{\frac{(s+1)}{s^2+s-2}}{1+\frac{(s+1)}{s^2+s-2}}=\frac{s+1}{s^2+2s-1}$ Poles at $s=-1\pm \sqrt{2}j$ . Stable System with oscillations.

ii) $\frac{C(s)}{R(s)}=\frac{\frac{3s+3}{s^2+s-2}}{1+\frac{3s+3}{s^2+s-2}}=\frac{3s+3}{s^2+4s+1}$ Poles at $s=-2\pm \sqrt{3}j$ . Stable system with oscillations

iii) $\frac{C(s)}{R(s)}=\frac{\frac{2s}{s^2-3s+2}}{1+\frac{2s}{s^2-3s+2}}=\frac{2s}{s^2-s+2}$ Poles at $s=0.5(1\pm \sqrt{3}j)$ , Unstable System, Oscillatory

iv) $\frac{C(s)}{R(s)}=\frac{\frac{5s}{s^2-3s+2}}{1+\frac{5s}{s^2-3s+2}}=\frac{5s}{s^2+2s+2}$ Poles at $s=-1\pm j$ . Stable System. Oscillatory

Tut 2, Q1:

Question

A controller with proportional gain $K_p$ is in series with a plant with transfer function $G(s)=\frac{1}{s(0.05s+1)}$ and employs a unity negative feedback loop.

a) Determine the response of the system to a unit step input if the feedback loop is open.

b) If the feedback loop is closed, determine the response to a unit step input given: i) $K_p$ = 0.5, ii) $K_p$ = 10.

c) What is the lowest value of $K_p$ which gives a non-oscillatory response?

Solution

a) $K_p$ cascades giving us: $\frac{K_p}{s(0.05s+1)}$ Assuming no feedback (open loop), our response to a unit step would be: $\frac{K_p}{s^2(0.05s+1)}=\frac{20K_p}{s^2(s+20)}$

\begin{align} \frac{1}{s^2(s+20)} &= \frac{A}{s^2}+ \frac{B}{s} + \frac{C}{s+20} \\ 1 &= A(s+20) + B(s^2+20s) + Cs^2 \\ &= (B+C)s^2 + (A+20B)s + 20A \\ s^0 &: 1 = 20A,\quad A=\frac{1}{20} \\ s^1 &: 0=(A+20B), \quad B =-\frac{1}{400} \\ s^2 &: 0=(B+C), \quad C=\frac{1}{400} \\ \therefore C(s) &= \frac{1}{s^2(s+20)} \\&= 20K_p[\frac{0.05}{s^2} - \frac{0.0025}{s} + \frac{0.0025}{s+20}] \\ \therefore c(t) &= \frac{K_p}{20}[20t-1+e^{-20t}] \end{align}

b) If the feedback loop is closed, with a unity feedback, we get: $\frac{C(s)}{R(s)}=\frac{K_p}{0.05s^2+s+K_p}$ With a unit step response: $C(s)=\frac{K_p}{s(0.05s^2+s+K_p)}$ i) $K_p=0.5$ $C(s)=\frac{20K_p}{s(s^2+20s+10)}$ Solving $(s^2+20s+10)$ for factors: $\frac{-20\pm \sqrt{400-40}}{2}=-10\pm 3\sqrt{1}0=10\pm 9.4868=-0.5132,-19.4868$ $C(s)=\frac{20K_p}{s(s+0.51)(s+19.49)}$ \begin{align} C(s)&=\frac{1}{s(s+0.51)(s+19.49)}\\ &= \frac{A}{s}+\frac{B}{s+0.51}+\frac{C}{s+19.49} \\ \text{So:}\\ 1 &=A(s+0.51)(s+19.49) + Bs(s+19.49) + Cs(s+0.51) \\ &= A(s^2+20s+9.94) + B(s^2+19.49s) + C(s^2+0.51s) \\ &= s^2(A+B+C) +s(20A+19.49B+0.51C) + 9.94A \\ \text{Given} & \text{ this:} \\ s^0 &: 1= 9.94A,\quad A = 0.101 \\ s^1 &: 0= 20A+19.49B+0.51C,\quad 19.49B+0.51C = -2.02 \\ s^2 &: 0 = A+B+C, \quad B+C=-0.101 \\ \therefore &\quad 18.98B = -2.02 - (-0.05151) = -1.96849 \\ &\quad B= -0.1037, \quad C=-0.0027 \\ C(s) &= \frac{0.101}{s} -\frac{0.1037}{s+0.51} -\frac{0.0027}{s+19.49} \\ \therefore c(t) &=0.101-0.1037e^{-0.51t}-0.0027e^{-19.49t} \end{align}

c) $s^2+20s+20K\to s^2+2\xi {\omega }_{n}s+{\omega }_n^2$ $\to {\omega }_n^2=20K,2\xi {\omega }_n=20$ $\to 4{\xi }^2{\omega }_n^2=400\to (2\xi {)}^2=\frac{400}{20K}$ $\to 2\xi =\sqrt{\frac{20}{K}}\to \xi =\sqrt{\frac{5}{K}}$

System is oscillatory for $\xi <1$: Oscillatory for $\sqrt{\frac{5}{K}}<1$ Oscillatory for $K>5$

Routh’s Criterion

Tut 1, Q6

Question

Use Routh’s Criterion to determine the stability characteristics of the following systems. In each case determine the number of positive roots.

Solution

i) $b_1=-\frac{1}{4}\left|\begin{array}{cc}1& 1\\ 4& 0\end{array}\right|=1$ $\begin{array}{ccc}{s}^2& 1& 1\\ s^1& 4& 0\\ s^0& 1& \end{array}$ The system is stable

ii) $b_2=-\frac{1}{4}\left|\begin{array}{cc}1& 5\\ 4& 6\end{array}\right|=14/4$ $b_0=0$ $c_2=-\frac{4}{14}\left|\begin{array}{cc}4& 6\\ 14/4& 0\end{array}\right|=6$ $\begin{array}{ccc}{s}^3& 1& 5\\ s^2& 4& 6\\ s^1& 14/4& 0\\ s^0& 6& \end{array}$ The system is stable

iii) $b_2=-\frac{1}{3}\left|\begin{array}{cc}1& -6\\ 3& 10\end{array}\right|=-28/3$ $b_0=0$ $c_2=\frac{3}{28}\left|\begin{array}{cc}3& 10\\ -28/3& 0\end{array}\right|=10$ $\begin{array}{ccc}{s}^3& 1& -6\\ s^2& 3& 10\\ s^1& -28/3& 0\\ s^0& 10& \end{array}$ Two sign changes, Two positive roots, unstable system

iv) $b_3=-\left|\begin{array}{cc}1& 2\\ 1& 4\end{array}\right|=-2$ $b_1=-\left|\begin{array}{cc}1& 8\\ 3& 0\end{array}\right|=24$ $c_3=\frac{1}{2}\left|\begin{array}{cc}1& 4\\ -2& 24\end{array}\right|=16$ $c_1=0$ $d_3=-\frac{1}{16}\left|\begin{array}{cc}1& 4\\ 16& 0\end{array}\right|=4$ $\begin{array}{cccc}{s}^4& 1& 2& 8\\ s^3& 1& 4& \\ s^2& -2& 24& \\ s^1& 16& 0& \\ s^0& 4& & \end{array}$ Two sign changes, Two Positive Roots, unstable system

Tut 1, Q7

Question

Use Routh’s Criterion to determine the range of values of K which give stability for the systems with closed loop characteristic equations:

Solution

i) $b_3=-\left|\begin{array}{cc}1& 1\\ 3& 2\end{array}\right|=1$ $b_1=-\left|\begin{array}{cc}1& K\\ 3& 0\end{array}\right|=3K$ $c_3=-\left|\begin{array}{cc}3& 2\\ 1& 3K\end{array}\right|=2-9K$ $c_1=0$ $d_3=-\frac{1}{2-9k}\left|\begin{array}{cc}3& 2\\ 2-9K& 0\end{array}\right|=2$ $\begin{array}{cccc}{s}^4& 1& 1& 8\\ s^3& 3& 2& \\ s^2& 1& 3K& \\ s^1& 2-9K& 0& \\ s^0& 2& & \end{array}$ $\therefore 2-9K\ge 0\to 9K\le 2$ $\to K\le \frac{2}{9}$

ii) $b_3=-\left|\begin{array}{cc}1& 3\\ 1& 2\end{array}\right|=1$ $b_1=-\left|\begin{array}{cc}1& K\\ 1& 0\end{array}\right|=K$ $c_3=-\left|\begin{array}{cc}1& 2\\ 1& K\end{array}\right|=2-K$ $c_1=0$ $d_3=-\frac{1}{2-k}\left|\begin{array}{cc}1& 2\\ 2-K& 0\end{array}\right|=2$ $\begin{array}{cccc}{s}^4& 1& 3& 8\\ s^3& 1& 2& \\ s^2& 1& K& \\ s^1& 2-K& 0& \\ s^0& 2& & \end{array}$ $\therefore 2-K\ge 0\to K\le 2$

Tut 1, Q8, Pt2

Tut 1, Q8, Pt1

Question

Determine the closed loop transfer function of the system shown in Figure Q8. Investigate the stability of this system when:

Solution

From Part 1 $\frac{C(s)}{R(s)}=\frac{\frac{2}{3}s(s+1)}{s^3+(11/3)s^2+(14/3)s+(4/3)}$ $b_2=-\frac{3}{11}\left|\begin{array}{cc}1& 14/3\\ 11/3& 4/3\end{array}\right|=-3/11\ast (12/9-154/9)=426/11$ $b_0=0$ $c_2=-\frac{11}{426}\left|\begin{array}{cc}11/3& 4/3\\ 426/11& 0\end{array}\right|=4/3$ $\begin{array}{ccc}{s}^3& 1& 14/3\\ s^2& 11/3& 4/3\\ s^1& 426/11& 0\\ s^0& 4/3& \end{array}$ No sign changes indicates that this is a stable system.

Root Locus

Tut 3, Q2:

Using the standard rules, draw the approximate root-locus diagrams for the systems in question Q1.

Tut 3, Q6:

$$y/d=\frac{\frac{K(1+s)}{s(s^2+4s+4)}}{1+\frac{K(1+s)}{(1+0.5s)s(s^2+4s+4)}}$$

$\frac{K(1+s)}{s(s^2+4s+4)+\frac{K(1+s)}{1+0.5s}}=\frac{K(1+s)(1+0.5s)}{(1+0.5s)(s^3+4s^2+4s)+K(1+s)}$ $\frac{K(0.5s^2+1.5s+0.5)}{(0.5s^4+2s^3+2s^2)+(s^3+4s^2+4s)+Ks+K}=\frac{K(0.5s^2+1.5s+0.5)}{0.5s^4+3s^3+6s^2+(K+4)s+K}$

$\frac{K(s^2+3s+1)}{s^4+6s^3+12s^2+2(K+4)s+2K}$

Worked Examples

Block Diagrams

Ex 1:

$$\begin{array}{rl}{G}_2{H}_2\text{ feedback loop:}& =(\frac{G_2}{1+G_2{H}_2})\\ \text{In series with }{G}_1& \text{, Top Block becomes:}\\ & =\frac{G_2{G}_1}{1+G_2{H}_2}\\ \text{Feedback loop with }{H}_1& \\ \frac{\frac{G_2{G}_1}{1+G_2{H}_2}}{1+\frac{G_2{G}_1}{1+G_2{H}_2}{H}_1}& =\frac{G_1{G}_2}{1+G_2{H}_2+H_1}\\ \therefore \frac{C(s)}{R(s)}& =\frac{G_1{G}_2}{1+G_2{H}_2+H_1}\end{array}$$

Block_Diagram_Worked_Examples.excalidraw