.. _control_problems-physics: ============================================= Control Package Examples ============================================= Given below, are some comprehensive textbook examples to demonstrate the possible use cases of the Control Module. Example 1 --------- .. image:: Control_Problems_Q1.svg A pole zero plot of an unknown **Transfer Function** is given above. 1. Determine the exact Transfer Function if the continuous time **DC Gain** of the system is **20**. 2. Is the TransferFunction **stable** or **unstable** in nature. 3. Obtain the **unit impulse response** of the system. 4. Find the initial value of the **time-domain response** of system without using the time domain equation. Solution >>> # Imports >>> from sympy import symbols, I, limit, pprint, solve, oo >>> from sympy.physics.control import TransferFunction Subpart 1 >>> s, k = symbols('s k') >>> gain = k # Let unknwon gain be k >>> a = [-3] # Zero at -3 in S plane >>> b = [-1, -2-I, -2+I] # Poles at -1, (-2, j) and (-2, -j) in S plane >>> tf = TransferFunction.from_zpk(a, b, gain, s) >>> pprint(tf) k*(s + 3) ------------------------------- (s + 1)*(s + 2 - I)*(s + 2 + I) >>> gain = tf.dc_gain() >>> print(gain) 3*k*(2 - I)*(2 + I)/25 >>> K = solve(gain - 20, k)[0] # Solve for k >>> tf = tf.subs({k: K}) # Reconstruct the TransferFunction using .subs() >>> pprint(tf.expand()) 100*s ----- + 100 3 ------------------- 3 2 s + 5*s + 9*s + 5 Subpart 2 >>> tf.is_stable() # Expect True, since poles lie in the left half of S plane True Subpart 3 >>> from sympy import inverse_laplace_transform >>> t = symbols('t', positive = True) >>> # Convert from S to T domain for impulse response >>> tf = tf.to_expr() >>> Impulse_Response = inverse_laplace_transform(tf, s, t) >>> pprint(Impulse_Response) -t -2*t 100*e 100*e *cos(t) ------- - ---------------- 3 3 Subpart 4 >>> # Apply the Initial Value Theorem on Equation of S domain >>> # limit(y(t), t, 0) = limit(s*Y(S), s, oo) >>> limit(s*tf, s, oo) 0 Example 2 --------- Find the Transfer Function of the following Spring-Mass dampering system : .. image:: Control_Problems_Q2.svg Solution >>> # Imports >>> from sympy import Function, laplace_transform, laplace_initial_conds, laplace_correspondence, diff, Symbol, solve >>> from sympy.abc import s, t >>> from sympy.physics.control import TransferFunction >>> y = Function('y') >>> Y = Function('Y') >>> u = Function('u') >>> U = Function('U') >>> k = Symbol('k') # Spring Constant >>> c = Symbol('c') # Damper >>> m = Symbol('m') # Mass of block The **DIFFERENTIAL EQUATION** of the system will be as follows: .. math:: \frac{{d^2y(t)}}{{dt^2}} + c\frac{{dy(t)}}{{dt}} + ky(t) = w^2u(t) \\\\ with \ initial \ conditions \\ y(0) = t,\quad\frac{{dy}}{{dt}}\bigg|_{t=0} = 0\\ >>> f = m*diff(y(t), t, t) + c*diff(y(t), t) + k*y(t) - u(t) >>> F = laplace_transform(f, t, s, noconds=True) >>> F = laplace_correspondence(F, {u: U, y: Y}) >>> F = laplace_initial_conds(F, t, {y: [0, 0]}) >>> t = (solve(F, Y(s))[0])/U(s) # To construct Transfer Function from Y(s) and U(s) >>> tf = TransferFunction.from_rational_expression(t, s) >>> pprint(tf) 1 -------------- 2 c*s + k + m*s Example 3 --------- A signal matrix in the time-domain, also known as the *impulse response matrix* **g(t)** is given below. $$g(t) = \begin{bmatrix} (1-t)e^{-t} & e^{-2t} \\ -e^{-t}+5e^{-2t} & \left(-3\sqrt{3}\sin\left(\frac{\sqrt{3}t}{2}\right)+\cos\left(\frac{\sqrt{3}t}{2}\right)\right)e^{-\frac{t}{2}} \end{bmatrix}$$ With Respect to this matrix, find 1. The system matrix (Transfer Function Matrix) in the Laplace domain (**g(t)** → **G(s)**). 2. The number of input and output signals in the system. 3. **Poles** and **Zeros** of the system elements (individual Transfer Functions in Transfer Function Matrix) in the Laplace domain *(Note: The actual poles and zeros of a MIMO system are NOT the poles and zeros of the individual elements of the transfer function matrix)*. Also, visualise the poles and zeros of the individual transfer function corresponding to the **1st input** and **1st output** of the **G(s)** matrix. 4. Plot the **unit step response** of the individual Transfer Function corresponding to the **1st input** and **1st output** of the **G(s)** matrix. 5. Analyse the Bode magnitude and phase plot of the Transfer Function corresponding to **1st input** and **2nd output** of the **G(s)** matrix. Solution >>> # Imports >>> from sympy import Matrix, laplace_transform, inverse_laplace_transform, exp, cos, sqrt, sin, pprint >>> from sympy.abc import s, t >>> from sympy.physics.control import * Subpart 1 >>> g = Matrix([[exp(-t)*(1 - t), exp(-2*t)], [5*exp((-2*t))-exp((-t)), (cos((sqrt(3)*t)/2) - 3*sqrt(3)*sin((sqrt(3)*t)/2))*exp(-t/2)]]) >>> G = g.applyfunc(lambda a: laplace_transform(a, t, s)[0]) >>> pprint(G) [ 1 1 1 ] [----- - -------- ----- ] [s + 1 2 s + 2 ] [ (s + 1) ] [ ] [ 5 1 s + 1/2 9 ] [ ----- - ----- -------------- - ------------------] [ s + 2 s + 1 2 3 / 2 3\] [ (s + 1/2) + - 2*|(s + 1/2) + -|] [ 4 \ 4/] Subpart 2 >>> G = TransferFunctionMatrix.from_Matrix(G, s) >>> type(G) >>> type(G[0]) >>> print(f'Inputs = {G.num_inputs}, Outputs = {G.num_outputs}') Inputs = 2, Outputs = 2 Subpart 3 >>> G.elem_poles() [[[-1, -1, -1], [-2]], [[-2, -1], [-1/2 - sqrt(3)*I/2, -1/2 - sqrt(3)*I/2, -1/2 + sqrt(3)*I/2, -1/2 + sqrt(3)*I/2]]] >>> G.elem_zeros() [[[-1, 0], []], [[-3/4], [4, -1/2 - sqrt(3)*I/2, -1/2 + sqrt(3)*I/2]]] >>> pole_zero_plot(G[0, 0]) # doctest: +SKIP .. plot:: tutorials/physics/control/generate_plots.py q3_3 Subpart 4 >>> tf1 = G[0, 0] >>> pprint(tf1) 2 -s + (s + 1) - 1 ----------------- 3 (s + 1) >>> step_response_plot(tf1) # doctest: +SKIP .. plot:: tutorials/physics/control/generate_plots.py q3_4 Subpart 5 >>> tf2 = G[0, 1] >>> bode_magnitude_plot(tf2) # doctest: +SKIP .. plot:: tutorials/physics/control/generate_plots.py q3_5_1 >>> bode_phase_plot(tf2) # doctest: +SKIP .. plot:: tutorials/physics/control/generate_plots.py q3_5_2 Example 4 --------- 1. A system is designed by arranging **P(s)** and **C(s)** in a series configuration *(Values of P(s) and C(s) are provided below)*. Compute the equivalent system matrix, when the order of blocks is reversed *(i.e. C(s) then P(s))*. $$P(s) = \begin{bmatrix} \frac{1}{s} & \frac{2}{s+2} \\ 0 & 3 \end{bmatrix}$$ $$C(s) = \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix}$$ 2. Also, find the **equivalent closed-loop system** *(or the ratio v/u from the block diagram given below)* for the system (negative-feedback loop) having **C(s)** as the **controller** and **P(s)** as **plant** *(Refer to the block diagram given below)*. .. image:: Control_Problems_Q4.svg Solution >>> # Imports >>> from sympy import Matrix, pprint >>> from sympy.abc import s, t >>> from sympy.physics.control import * Subpart 1 >>> P_mat = Matrix([[1/s, 2/(2+s)], [0, 3]]) >>> C_mat = Matrix([[1, 1], [2, 2]]) >>> P = TransferFunctionMatrix.from_Matrix(P_mat, var=s) >>> C = TransferFunctionMatrix.from_Matrix(C_mat, var=s) >>> # Series equivalent, considering (Input)→[P]→[C]→(Output). Note that order of matrix multiplication is opposite to the order in which the elements are arranged. >>> pprint(C*P) [1 1] [1 2 ] [- -] [- -----] [1 1] [s s + 2] [ ] *[ ] [2 2] [0 3 ] [- -] [- - ] [1 1]{t} [1 1 ]{t} >>> # Series equivalent, considering (Input)→[C]→[P]→(Output). >>> pprint(P*C) [1 2 ] [1 1] [- -----] [- -] [s s + 2] [1 1] [ ] *[ ] [0 3 ] [2 2] [- - ] [- -] [1 1 ]{t} [1 1]{t} >>> pprint((C*P).doit()) [1 3*s + 8 ] [- ------- ] [s s + 2 ] [ ] [2 6*s + 16] [- --------] [s s + 2 ]{t} >>> pprint((P*C).doit()) [ 5*s + 2 5*s + 2 ] [--------- ---------] [s*(s + 2) s*(s + 2)] [ ] [ 6 6 ] [ - - ] [ 1 1 ]{t} Subpart 2 >>> tfm_feedback = MIMOFeedback(P, C, sign=-1) >>> pprint(tfm_feedback.doit()) # ((I + P*C)**-1)*P [ 7*s + 14 -s - 6 ] [--------------- ---------------] [ 2 2 ] [7*s + 19*s + 2 7*s + 19*s + 2] [ ] [ 2 ] [ -6*s - 12 3*s + 9*s + 6 ] [--------------- ---------------] [ 2 2 ] [7*s + 19*s + 2 7*s + 19*s + 2]{t} Example 5 --------- .. image:: Control_Problems_Q5.svg Given, .. math:: G1 &= \frac{1}{10 + s}\\\\ G2 &= \frac{1}{1 + s}\\\\ G3 &= \frac{1 + s^2}{4 + 4s + s^2}\\\\ G4 &= \frac{1 + s}{6 + s}\\\\ H1 &= \frac{1 + s}{2 + s}\\\\ H2 &= \frac{2 \cdot (6 + s)}{1 + s}\\\\ H3 &= 1\\ Where $s$ is the variable of the transfer function (in Laplace Domain). Find 1. The equivalent Transfer Function representing the system given above. 2. Pole-Zero plot of the system. Solution >>> from sympy.abc import s >>> from sympy.physics.control import * >>> G1 = TransferFunction(1, 10 + s, s) >>> G2 = TransferFunction(1, 1 + s, s) >>> G3 = TransferFunction(1 + s**2, 4 + 4*s + s**2, s) >>> G4 = TransferFunction(1 + s, 6 + s, s) >>> H1 = TransferFunction(1 + s, 2 + s, s) >>> H2 = TransferFunction(2*(6 + s), 1 + s, s) >>> H3 = TransferFunction(1, 1, s) >>> sys1 = Series(G3, G4) >>> sys2 = Feedback(sys1, H1, 1).doit() >>> sys3 = Series(G2, sys2) >>> sys4 = Feedback(sys3, H2).doit() >>> sys5 = Series(G1, sys4) >>> sys6 = Feedback(sys5, H3) >>> sys6 # Final unevaluated Feedback object Feedback(Series(TransferFunction(1, s + 10, s), TransferFunction((s + 1)**3*(s + 2)*(s + 6)**2*(s**2 + 1)*(-(s + 1)**2*(s**2 + 1) + (s + 2)*(s + 6)*(s**2 + 4*s + 4))*(s**2 + 4*s + 4)**2, (s + 1)*(s + 6)*(-(s + 1)**2*(s**2 + 1) + (s + 2)*(s + 6)*(s**2 + 4*s + 4))*((s + 1)**2*(s + 6)*(-(s + 1)**2*(s**2 + 1) + (s + 2)*(s + 6)*(s**2 + 4*s + 4))*(s**2 + 4*s + 4) + (s + 1)*(s + 2)*(s + 6)*(2*s + 12)*(s**2 + 1)*(s**2 + 4*s + 4))*(s**2 + 4*s + 4), s)), TransferFunction(1, 1, s), -1) >>> sys6.doit() # Reducing to TransferFunction form without simplification TransferFunction((s + 1)**4*(s + 2)*(s + 6)**3*(s + 10)*(s**2 + 1)*(-(s + 1)**2*(s**2 + 1) + (s + 2)*(s + 6)*(s**2 + 4*s + 4))**2*((s + 1)**2*(s + 6)*(-(s + 1)**2*(s**2 + 1) + (s + 2)*(s + 6)*(s**2 + 4*s + 4))*(s**2 + 4*s + 4) + (s + 1)*(s + 2)*(s + 6)*(2*s + 12)*(s**2 + 1)*(s**2 + 4*s + 4))*(s**2 + 4*s + 4)**3, (s + 1)*(s + 6)*(s + 10)*(-(s + 1)**2*(s**2 + 1) + (s + 2)*(s + 6)*(s**2 + 4*s + 4))*((s + 1)**2*(s + 6)*(-(s + 1)**2*(s**2 + 1) + (s + 2)*(s + 6)*(s**2 + 4*s + 4))*(s**2 + 4*s + 4) + (s + 1)*(s + 2)*(s + 6)*(2*s + 12)*(s**2 + 1)*(s**2 + 4*s + 4))*((s + 1)**3*(s + 2)*(s + 6)**2*(s**2 + 1)*(-(s + 1)**2*(s**2 + 1) + (s + 2)*(s + 6)*(s**2 + 4*s + 4))*(s**2 + 4*s + 4)**2 + (s + 1)*(s + 6)*(s + 10)*(-(s + 1)**2*(s**2 + 1) + (s + 2)*(s + 6)*(s**2 + 4*s + 4))*((s + 1)**2*(s + 6)*(-(s + 1)**2*(s**2 + 1) + (s + 2)*(s + 6)*(s**2 + 4*s + 4))*(s**2 + 4*s + 4) + (s + 1)*(s + 2)*(s + 6)*(2*s + 12)*(s**2 + 1)*(s**2 + 4*s + 4))*(s**2 + 4*s + 4))*(s**2 + 4*s + 4), s) >>> sys6 = sys6.doit(cancel=True, expand=True) # Simplified TransferFunction form >>> sys6 TransferFunction(s**4 + 3*s**3 + 3*s**2 + 3*s + 2, 12*s**5 + 193*s**4 + 873*s**3 + 1644*s**2 + 1484*s + 712, s) >>> pole_zero_plot(sys6) # doctest: +SKIP .. plot:: tutorials/physics/control/generate_plots.py q5 References ^^^^^^^^^^ 1. `testbook.com `_ 2. `www.vssut.ac.in `_