Control Systems Test 1

In a feedback control system, the output is determined by a combination of the input signal and the feedback signal. The input signal is the desired output or reference, while the feedback signal is a portion of the actual output that is fed back into the system to compare with the input. This comparison helps the system to adjust its output to minimize the error between the input and the actual output. By considering both the input and feedback signals, the system can continuously adapt and achieve the desired output.

The impulse response is defined as the output of an LTI system due to a unit impulse signal input being applied at time t = 0.

y(t) = h(t) x(t) = h(t) δ(t)

where δ(t) is the unit impulse function and h(t) is the unit impulse response of a continuous-time LTI system.

Calculations:-

Given-

y(t)  = e^-8t 

x(t) = δ(t)

For calculating the transfer function convert the time domain response into Laplace or S domain.

Laplace transform of u(t) is given by

Taking Inverse Laplace transform

y(t) = t.u(t)

• Any feedback path is considered as a valid feedback path if it does not touches any single node twice in its path.
• Feedback paths in the above diagram are af, be, cd, abef, bcde, and abcdef.
• abef is not a valid feedback path because node 'Q' is repeated twice in the path.
• bcde is not a valid feedback path because node 'R' is repeated twice in the path.
• abcdef is not a valid feedback path because node 'Q' and 'R' is repeated twice in the path.

So, af, be and cd are the only valid feedback paths. The sum of the gains of the feedback paths in the signal flow graph is af + be + cd.

TF = H(s) = 1/s,

Input is unit step, R(s) = 1/s

Now output for t > 0 can be calculated as

C(s) = H(s) x R(s)

C(s) = (1/s) x (1/s)

C(s) = 1 / s²

L^-1[C(s)] = t

C(t) = t = ramp function

\(G\left( s \right)H\left( s \right) = \frac{1}{{s\left( {3s + 1} \right)}}{e^{ - 2Ts}}\)

\(G\left( {j\omega } \right)H\left( {j\omega } \right) = \frac{1}{{j\omega \left( {3j\omega + 1} \right)}}{e^{ - 2Tj\omega }}\)

\(\angle G\left( {j\omega } \right)H\left( {j\omega } \right) = - 90 - {\tan ^{ - 1}}\left( {3\omega } \right) - 2T\omega \)

At ω = ω₀, the phase angle is zero.

\( \Rightarrow 0 = - 90 - {\tan ^{ - 1}}\left( {3{\omega _0}} \right) - 2T{\omega _0}\)

\( \Rightarrow {\tan ^{ - 1}}\left( {3{\omega _0}} \right) = - \left( {90 + 2T{\omega _0}} \right)\)

⇒ 3ω₀ = - tan(90 + 2Tω₀)

Concept:

From Nyquist stability criterion, N = P – Z

where, N = number of encirclements of point (–1, 0)

Z = number of zeros of (1 + G(s)·H(s)) or the number of poles of closed-loop system in the right half-plane.

For stability, z = 0,

Calculation:

Open loop transfer function:

\(G\left( s \right)H\left( s \right) = \frac{{2\left( {s + 8} \right)}}{{{s^2} + 7s - 8}} = \frac{{2\left( {s + 8} \right)}}{{\left( {s + 8} \right)\left( {s - 1} \right)}}\)

Number of right side open loop poles (p) = 1

Closed loop transfer function

\(= \frac{{\frac{{2\left( {s + 8} \right)}}{{{s^2} + 7s - 8}}}}{{1 + \frac{{2\left( {s + 8} \right)}}{{{s^2} + 7s - 8}}}}\)

\(= \frac{{2\left( {s + 8} \right)}}{{{s^2} + 7s - 8 + 2s + 16}}\)

\(= \frac{{2\left( {s + 8} \right)}}{{{s^2} + 9s + 8}}\)

\(= \frac{{2\left( {s + 8} \right)}}{{\left( {s + 1} \right)\left( {s + 8} \right)}}\)

Number of closed loop poles (z) = 0

From Nyquist stability

N = P – Z = 1

Therefore Nyquist contour encircle the -1 point once in the counterclockwise direction.

\(G\left( s \right) = \frac{{{e^{ - \frac{s}{{500}}}}}}{{s + 500}}\)

ϕ = ∠G(jω) at ω = 100π

\( = - {\tan ^{ - 1}}\left( {\frac{\omega }{{500}}} \right) - \frac{\omega }{{500}}\)

\( = - {\tan ^{ - 1}}\left( {\frac{{100\pi }}{{500}}} \right) - \frac{{100\pi }}{{500}}\)

\( = - {\tan ^{ - 1}}\left( {\frac{\pi }{5}} \right) - \frac{\pi }{5}\)

= -68.12°

y(t) = A sin (100 πt - 68.12°)

While writing the transfer function of this signal flow graph,

e₂= t_ae₁ + t_be₁ = (t_a+ t_b) e₁ 

Then, signal flow graph will lokk like this:

Consider the block diagram as SFG. There are two feedback loop -G₁G₂H₁ and -G₂G₃H₂ and one forward path G₁G₂ G₃ . So (D) is correct option.

Consider the block diagram as a SFG. Two forward path G₁G₂ and G₃ and three loops -G₁G₂ H₂, -G₂H₁, -G₃ H₂

There are no nontouching loop. So (B) is correct.

P₁ = 5 x 3 x 2 = 30, Δ = 1 - (3x - 3) = 10

Note: the provided diagram is too small/low-resolution to unambiguously read the direction and connectivity of every branch. I cannot reliably recompute the transfer function without a clearer diagram. Please upload a higher-resolution image or give the directed-graph gains in text (node-to-node gains and node numbering). Once the exact directed edges and their polarities are confirmed, I will recompute the transfer function using Mason's gain formula step by step.

Assumption checklist for re-evaluation (please confirm):

• Which node is the first node after R (its label/number).
• Exact directed edges and their gains (for example: R → 2 : 1, 2 → 3 : -1, etc.).
• Direction and gain of the long feedback arc (for example: C → 2 : 5).

After you provide a clear image or the above adjacency/gain list I will:

• identify all forward paths and their gains,
• list all individual loops and their gains,
• compute Δ and each Δ_k (Mason's formula),
• give the final transfer function C/R and select the correct option.

s³ – s² + 3s + 1 = 0

put s = z + 1

⇒ (z + 1)³ – (z + 1)² + 3(z + 1) + 1 = 0

⇒ z³ + 1 + 3z² + 3z – z² – 2z – 1 + 3z + 3 + 1 = 0

⇒ z³ + 2z² + 4z + 4 = 0

Taking Routh criteria for the given characteristic equation

There are no sign changes in the first column of the Routh array.

X₁ = 1 - H₁

X₂ = (G₁ + G₃) X₁ - X₃H₁

X₃ = X₂G₂ + G₄

The negative feedback closed-loop system, subjected to 15V, has a forward gain of 2 and a feedback gain of 0.5.

• The output voltage is calculated as the product of the forward gain and the effective input voltage.
• The effective input voltage is the difference between the input voltage and the feedback voltage.

To find the output voltage:

• Calculate the feedback voltage: Output Voltage multiplied by Feedback Gain.
• Effective input = Input voltage - Feedback voltage.
• Output voltage = Forward gain x Effective input voltage.

After solving, the output voltage is 15V and the error voltage is 7.5V.

Four loops -G₁G₄, -G₁G₂G₅, -G₁,G₂G₅G₇ and -G₁G₂G₃G₃G₇.

There is no nontouching loop. So (B) is correct.

Given \(\rm G\left( s \right)H\left( s \right) = \frac{K}{{s\left( {s + 4} \right)\left( {{s^2} + 4s + 12} \right)}}\)

\(\rm 1 + G\left( s \right)H\left( s \right) = 0\)

\(\rm 1 + \frac{K}{{s\left( {s + 4} \right)\left( {{s^2} + 4s + 12} \right)}} = 0\)

\(\therefore K = - s\left( {s + 4} \right)\left( {{s^2} + 4s + 12} \right)\)

for break point calculation we should have \(\rm \frac{{dK}}{{ds}} = 0\)

\(\rm \therefore \frac{d}{{ds}}\left[ {\left( {{s^4} + 8{s^3} + 28{s^2} + 48s} \right)\left( { - 1} \right)} \right] = 0\)

\(\rm 4{s^3} + 24{s^2} + 56s + 48 = 0\)

\(\rm {s^3} + 6{s^2} + 14s + 12 = 0\)

Break away point are, \(\rm -2, -2+j√2, -2-j√2\)

∴ Number of complex Break away points are = 2

Concept:

Velocity error coefficient is given

\({K_V} = \mathop {\lim }\limits_{s \to 0} s(G\left( s \right)H\left( s \right)\)

Calculations:

\({K_V} = \mathop {\lim }\limits_{s \to 0} s(G\left( s \right)H\left( s \right)\)

\(1000 = \mathop {\lim }\limits_{s \to 0} \;\frac{{s\left( {{K_P} + {K_D}s} \right)100}}{{s\left( {s + 10} \right)}}\)

1000 = 10 kp

Kp = 100

Characteristic Equation:

1 + (G(s) H(s) = 0

\(1 + \frac{{\left( {100 + {K_D}s} \right)100}}{{s\left( {s + 10} \right)}} = 0\)

⇒ s² + (10+100k_D)s+10⁴ = 0

Comparing with standard second order equation:

2ξw_n = 100 k_D + 10

\(2\left( {\frac{1}{2}} \right)\left( {100} \right) = {K_D} + 10\) 

90 = 100 k_D

K­_D = 0.9

\(\dot x\left( s \right) = \left[ {\begin{array}{*{20}{c}} 1&1\\ 0&1 \end{array}} \right]x\left( t \right) + \left[ {\begin{array}{*{20}{c}} {{b_1}}\\ {{b_2}} \end{array}} \right]u\left( t \right)\)

\(C\left( t \right) = \left[ {\begin{array}{*{20}{c}} {{d_1}}&{{d_2}} \end{array}} \right]x\left( t \right)\)

\(A = \left[ {\begin{array}{*{20}{c}} 1&1\\ 0&0 \end{array}} \right]B = \left[ {\begin{array}{*{20}{c}} {{b_1}}\\ {{b_2}} \end{array}} \right]C = \left[ {\begin{array}{*{20}{c}} {{d_1}}&{{d_2}} \end{array}} \right]\)

Controlling matrix, \({Q_C} = \left[ {\begin{array}{*{20}{c}} B&{AB} \end{array}} \right]\)

\({Q_C} = \left[ {\begin{array}{*{20}{c}} {{b_1}}&{{b_1} + {b_2}}\\ {{b_2}}&{{b_2}} \end{array}} \right]\)

\(\left| {{Q_C}} \right| = {b_1}{b_2} - {b_2}\left( {{b_1} + {b_2}} \right) = - b_2^2 \ne 0\)

Given the system is controllable for all values of b₂ except b₂ = 0

Observability matrix, \({Q_A} = \left[ {\begin{array}{*{20}{c}} C\\ {CA} \end{array}} \right]\)

\({Q_A}\left[ {\begin{array}{*{20}{c}} {{d_1}}&{{d_2}}\\ {{d_1}}&{{d_1} + {d_2}} \end{array}} \right]\)

\(\left| {{Q_A}} \right| = {d_1}\left( {{d_1} + {d_2}} \right) - {d_1}{d_2} = d_1^2 \ne 0\)

Given system is homogeneous whose solution is X(t) = e^At X(0)

Now, e^At = L^-1 {(sI - A)^-1}

\(\left[ {sI - A} \right] = s\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}0&1\\{ - 2}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}s&{ - 1}\\2&s\end{array}} \right]\)

\({\left( {sI - A} \right)^{ - 1}} = \frac{{Adj\left( {sI - A} \right)}}{{\left| {sI - A} \right|}}\)

\(Adj\left( {sI - A} \right) = {\left[ {\begin{array}{*{20}{c}}{{C_{11}}}&{{C_{12}}}\\{{C_{21}}}&{{C_{22}}}\end{array}} \right]^T} = {\left[ {\begin{array}{*{20}{c}}s&{ - 2}\\1&s\end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}}s&1\\{ - 2}&s\end{array}} \right]\)

|sI - A| = s² + 2

\({\left( {sI - A} \right)^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\frac{s}{{{s^2} + 2}}}&{\frac{1}{{{s^2} + 2}}}\\{\frac{{ - 2}}{{{s^2} + 2}}}&{\frac{s}{{{s^2} + 2}}}\end{array}} \right]\)

\({e^{At}} = {L^{ - 1}}\left[ {\begin{array}{*{20}{c}}{\frac{s}{{{s^2} + 2}}}&{\frac{1}{{{s^2} + 2}}}\\{\frac{{ - 2}}{{{s^2} + 2}}}&{\frac{s}{{{s^2} + 2}}}\end{array}} \right]\)

\({L^{ - 1}}\left\{ {\frac{s}{{{s^2} + 2}}} \right\} = {L^{ - 1}}\left\{ {\frac{s}{{{s^2} + {{\left( {\sqrt 2 } \right)}^2}}}} \right\} = \cos \sqrt {2t} \)

\({L^{ - 1}}\left\{ {\frac{1}{{{s^2} + 2}}} \right\} = {L^{ - 1}}\left\{ {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 2 }}}&{\frac{{\sqrt 2 }}{{{s^2} + {{\left( {\sqrt 2 } \right)}^2}}}}\end{array}} \right\} = \frac{1}{{\sqrt 2 }}\sin \sqrt 2 t\)

\({L^{ - 1}}\left\{ {\frac{{ - 2}}{{{s^2} + 2}}} \right\} = {L^{ - 1}}\left\{ { - \begin{array}{*{20}{c}}{\sqrt 2 }&{\frac{{\sqrt 2 }}{{{s^2} + {{\left( {\sqrt 2 } \right)}^2}}}}\end{array}} \right\} = - \sqrt 2 \sin \sqrt 2 t\)

\(\therefore {e^{At}} = \left[ {\begin{array}{*{20}{c}}{\cos \sqrt {2\;t} }&{\frac{1}{{\sqrt 2 }}\sin \sqrt 2 \;t}\\{ - \sqrt 2 \sin \sqrt 2 \;t}&{\cos \sqrt 2 \;t}\end{array}} \right]\)

\(\therefore X\left( t \right) = {e^{At}}X\left( 0 \right) = {e^{At}}\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\cos \sqrt 2 \;t + \frac{1}{{\sqrt 2 }}\sin \sqrt 2 \;t}\\{\cos \sqrt 2 \;t - \sqrt 2 \sin \sqrt 2 \;t}\end{array}} \right]\)

∴ Output response Y(t) = [1 - 1] X(t)

\(= \left[ {1 - 1} \right]\left[ {\begin{array}{*{20}{c}}{\cos \sqrt 2 \;t + \frac{1}{{\sqrt 2 }}\sin \sqrt 2 \;t}\\{\cos \sqrt 2 \;t - \sqrt 2 \sin \sqrt 2 \;t}\end{array}} \right]\)

\(= \frac{3}{{\sqrt 2 }}\sin \sqrt 2 \;t\)