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Problem 13.1 (GAUCHE)

Consider a system with i/o ODE \[\begin{aligned} \label{eq:gauche_ode} \ddot{y} + a\,\dot{y} + b\,y &= b\,u \end{aligned}\] for constants \(a, b \in \mathbb{R}\).

  1. Derive the frequency response function \(H(j\omega)\) and the transfer function \(H(s)\). Hint: either can be found from the other.

  2. Let \(u(t) = 7 \cos(5 t + 3)\). What is the steady state forced response \(y(t)\) in terms of \(a, b\)? Hint: this shouldn’t require much computation.

  3. Now let \(u(t) = 3\,\delta(t)\), an impulse. What is the impulse response \(y(t)\) in terms of the inverse Fourier transform \(\mathcal{F}^{-1}\) and \(H(j\omega)\)? Do not substitute in for \(H(j\omega)\) or inverse transform.

  4. Use computer software to plot the Bode plot of \(H(j\omega)\) for \(a = b = 1\).

  5. For \(b = 1\), for what range of \(a\) will there be a complex conjugate pair of poles?1 Hint: consider comparing the transfer function derived in part (a) to the standard form of the second-order transfer function in figure 13.4a.

Part a

We use the Fourier transform to derive the frequency response function \(H(j\omega)\), although you may use inspection. Transforming with initial conditions zero, \[\begin{aligned} (j\omega)^2 Y(\omega) + a\, (j\omega) Y(\omega) + b\, Y(\omega) &= b\, U(\omega). \end{aligned}\] Letting \(H(j\omega) = Y(\omega)/U(\omega)\), \[\begin{aligned} H(j\omega) &= \frac{b} {(j\omega)^2 + a (j\omega) + b}. \end{aligned}\] The transfer function is \[\begin{aligned} H(s) &= H(j\omega)|_{j\omega\mapsto s} \\ &= \frac{b} {s^2 + a s + b}. \end{aligned}\]

Part b

From , the sinusoidal frequency response is \[\begin{aligned} y(t) &= A |H(j\omega_0)| \cos(\omega_0 t + \psi + \angle H(j\omega_0)) \\ &= 7 |H(j\omega_0)| \cos(5 t + 3 + \angle H(j\omega_0)). \end{aligned}\] So all we need are the magnitude and phase of the frequency response function \(H(j\omega)\) at \(\omega = \omega_0 = 5\) rad/s. From above, \[\begin{aligned} H(j5) &= \frac{b} {(j5)^2 + a (j5) + b} \\ &= \frac{b} {b - 25 + j 5 a}. \\ \end{aligned}\] The magnitude is \[\begin{aligned} |H(j5)| &= \frac{|b|} {|b - 25 + j 5 a|} \\ &= \frac{|b|} {\sqrt{(b - 25)^2 + 25 a^2}} \end{aligned}\] and the phase is \[\begin{aligned} \angle H(j5) &= \angle(b) - \angle(b - 25 + j 5 a) \\ &= 0 - \arctantwo(5 a,b - 25) \\ &= - \arctantwo(5 a,b - 25). \end{aligned}\]

Part c

From , we know the unit impulse response \(h(t)\) is \[\begin{aligned} h(t) &= \mathcal{F}^{-1}(H(j\omega)). \end{aligned}\] In this problem, the impulse is \(u = 3\delta(t)\). From superposition,2 \[\begin{aligned} y(t) &= 3 \mathcal{F}^{-1}(H(j\omega)). \end{aligned}\]

Part d

First, define the transfer function.

a = 1;
b = 1;
H = tf([b],[1,a,b])
H =
 
       1
  -----------
  s^2 + s + 1
 
Continuous-time transfer function.

Now, the Bode plot. The results are shown in .

figure
bode(H)
grid on

Part e

The standard form of the denominator yields3 the equality, in terms of damping ratio \(\zeta\) and natural frequency \(\omega_n\), \[\begin{aligned} s^2 + 2\zeta\omega_n + \omega_n^2 = s^2 + a s + b. \end{aligned}\] Equating \(s\)-coefficients, \[\begin{aligned} \omega_n = \sqrt{b} \end{aligned}\] and \[\begin{aligned} \zeta = \frac{a} {2\omega_n} = \frac{a} {2\sqrt{b}}. \end{aligned}\] If \(b = 1\), \(a = 2 \zeta\). Recall that \(\zeta \in [0,1)\) is underdamped and corresponds to complex roots. Therefore, complex roots emerge when \[\begin{aligned} a \in [0,2). \end{aligned}\]

For stable systems, that is all. For unstables systems, complex roots occur when \(\zeta \in (-1,0]\). So, although the exercise is mum on stability, technically complex roots occur when \[\begin{aligned} a \in (-2,2). \end{aligned}\]

Problem 13.2 (TICKLE)

Let a transfer function \(H\) be \[\begin{aligned} \frac{10(s + 100)}{s^2 + 2\, s + 100}. \end{aligned}\] Use \(H\) to respond to the following questions and imperatives.

  1. Write \(H\) as a product of standard-form transfer functions.

  2. Find the frequency response function \(H(j\omega)\) without simplifying.

  3. Use the axes below to sketch the Bode plot of \(H\).

The following was generated from a Jupyter notebook with the following filename and kernel.

disp(['notebook filename: ',the_notebook,'.ipynb'])
disp(['notebook kernel: ',the_kernel])
notebook filename: bode_exercise_01.ipynb
notebook kernel: matlab
sys = tf(10*[1,100],[1,2,100])
sys =
 
    10 s + 1000
  ---------------
  s^2 + 2 s + 100
 
Continuous-time transfer function.
sys_a = tf_factor(sys)
sys_a(:,:,1,1) =
 
        100
  ---------------
  s^2 + 2 s + 100
 

sys_a(:,:,2,1) =
 
  0.01 s + 1
 

sys_a(:,:,3,1) =
 
  10
 
3x1 array of continuous-time transfer functions.
% [f,ax_mag,ax_phase] = bode_multi(G); % get axis handles
f = bode_multi(stack(1,sys,sys_a));

hgsave(f,'figures/temp');

Problem 13.3 (ME)

Let a transfer function \(H\) be \[\begin{aligned} H(s) = \frac{1000 (s+10)} {(s+100)(s+1000)}. \end{aligned}\] Use \(H\) to respond to the following questions and imperatives.

  1. Write \(H\) as a product of standard-form transfer functions.

  2. Find the frequency response function \(H(j\omega)\) without simplifying.

  3. Use the axes below to sketch the Bode plot of \(H\).


  1. Here is \(H\) written as a product of standard-form transfer functions: \[\begin{aligned} H(s) &= \frac{1} {10} \left(\frac{1} {10} s + 1\right) \frac{1} {\frac{1} {100}s + 1} \cdot \frac{1} {\frac{1} {1000}s + 1}. \end{aligned}\]

  2. Applying the usual substitution, \[\begin{aligned} H(j\omega) &= H(s)|_{s=j\omega} \\ &= \frac{1} {10} \left(\frac{1} {10} j \omega + 1\right) \frac{1} {\frac{1} {100} j \omega + 1} \cdot \frac{1} {\frac{1} {1000} j \omega + 1}. \end{aligned}\]

  3. The plot is below.

Problem 13.4 (ELMO)

Consider a system with transfer function \[\begin{aligned} H(s) = \frac{100 (s+9)} {(s+5)(s+6)(s^2 + 8 s + 32)}. \end{aligned}\]

  1. Identify the poles and zeros of \(H\).

  2. Derive the frequency response function \(H(j\omega)\). Do not simplify the expression.

  3. Create a Bode plot of \(H\).

  4. Let the system have sinusoidal input \(u(t) = 2 \cos(3 t)\). What is the steady-state system output \(y(t)\)?

  5. Let the system have the same sinusoidal input as previously. Simulate its forced response for nine seconds and plot it.

clear; close all

The general approach we take is to use functions from Matlab’s Control Systems toolbox to make our lives easier. In that spirit, we begin by defining a transfer function model. When we have a partially factored transfer function, it is often easiest to use the following trick: let’s define s to be a transfer function object and build our full \(H\) transfer function from that.

s = tf([1,0],[1])
s =
 
  s
 
Continuous-time transfer function.

Now we can define \(H\) straightforwardly.

H = 100*(s+9)/((s+5)*(s+6)*(s^2+8*s+32))
H =
 
              100 s + 900
  ------------------------------------
  s^4 + 19 s^3 + 150 s^2 + 592 s + 960
 
Continuous-time transfer function.

Part a

The poles and zeros can be found easily as follows.

pole(H)
ans =
           -4 +          4i
           -4 -          4i
           -6 +          0i
           -5 +          0i
zero(H)
ans =
    -9

Part b

This is probably just as easily done by-hand: \[\begin{aligned} H(j\omega) &= H(s)_{s\mapsto j\omega} \\ &= \frac{100(j\omega + 9)}{(j\omega + 5)(j\omega + 6)((j\omega)^2 + j\omega 8 + 32)}. \end{aligned}\]

Clearly, this could be simplified, but the exercise implores us not to.

Part c

We take the easy way. 

figure
bode(H)
grid on

Part d

From Equation sin.10, the output amplitude is \[\begin{aligned} 2 |H(j3)|. \end{aligned}\] Evaluating \(H(j3)\) is easy with Matlab.

Hj3 = freqresp(H,3)
Hj3 =
     0.059622 -    0.72718i

We’re only interested in the magnitude for this problem.

Hj3_mag = abs(Hj3)
output_amplitude = 2*Hj3_mag
Hj3_mag =
      0.72962
output_amplitude =
       1.4592

Part e

The simulation will be conducted with lsim.

t_a = linspace(0,9,300); % time array
u_a = 2*cos(3*t_a); % input array
y_a = lsim(H,u_a,t_a); % forced response

Let’s plot it with the input for fun. The result is shown in . 

figure
plot(t_a,y_a,t_a,u_a);
xlabel('time (s)')
legend('forced response')

Just as we expected from part d, it looks like the amplitude approaches 1.46.

Problem 13.5 (HUM)

In many measurement systems, an interference signal (i.e., an uncontrolled, undesirable signal that appears in a signal) that often appears in measurements is the mains hum, which arises from the mains power grid. Its fundamental frequency is \(f_1 = 60\) Hz, and much smaller-amplitude components appear at higher harmonics.

Consider the following measurement system. A pressure sensor has transfer function \[\begin{aligned} G(s) = \frac{1\cdot 10^{3}}{s^2 + 1\cdot 10^3 s + 1\cdot10^{6}}. \end{aligned}\] with units V/Pa. The desired measurement signal has maximum angular frequency \(\omega_\text{sig} = 100\) rad/s. Mains hum is observed in the measurement signal with magnitude \(m = 0.5\) V.

Design a first-order low-pass filter \[\begin{aligned} H(s) = \frac{1} {s/\omega_b + 1} \end{aligned}\] for the output of the sensor with the following steps:

  1. Derive the filter frequency response function \(H(j\omega)\).

  2. Compute the magnitude \(|H(j\omega)|\) of the frequency response function.

  3. Solve for the break frequency \(\omega_b\) such that \(|H(j\omega_\text{sig})| = 0.97\). This design will leave the desired signal only lightly attenuated but will further attenuate the mains hum.

  4. Compute the attenuation of the mains hum amplitude \(|H(j60\cdot2\pi)|\) and the corresponding filter output \(|H(j60\cdot2\pi)| m\).

  5. Compute the steady-state filtered output amplitude for a sensor input \(1 \sin(\omega_\text{sig} t)\) kPa.

  6. Suppose a greater attenuation of the mains hum interference is desired. How could the filter \(H(s)\) be altered to reduce it further without significantly attenuating the desired signal?

  7. Find the sensor’s natural frequency \(\omega_n\) and damping ratio \(\zeta\). Its peak output magnitude occurs at a frequency of \(\omega_p = \omega_n \sqrt{1 - 2 \zeta^2}\). Compute \(\omega_p\) for the sensor \(G\).

  8. Compute the steady-state filtered output amplitude for a sensor input \(1 \sin(\omega_p t)\) kPa. Explain why it is greater than the filtered output magnitude at \(\omega_\text{sig}\) from part e.

  1. The filter frequency response function \(H(j\omega)\) is \[\begin{aligned} H(j\omega) &= H(s)|_{s\mapsto j \omega} \\ &= \frac{1} {j \omega/\omega_b + 1 \end{aligned}\]

  2. The magnitude \(|H(j\omega)|\) of the frequency response function is \[\begin{aligned} |H(j\omega)| &= \left| \frac{1} {j \omega/\omega_b + 1} \right| \\ &= \frac{1} {\left| j \omega/\omega_b + 1 \right|} \\ &= \frac{1} {\sqrt{\omega^2/\omega_b^2 + 1} \end{aligned}\]

  3. The frequency at which \(|H(j\omega)| = 0.97\) is \[\begin{aligned} |H(j\omega_\text{sig})| &= 0.97 = \frac{1} {\sqrt{\omega_\text{sig}^2/\omega_b^2 + 1}} \Rightarrow \\ \omega_b &= \frac{\omega_\text{sig}}{\sqrt{1/0.97^2 - 1}} \\ &\approx 4 \omega_\text{sig} = 400~\text{rad/s \end{aligned}\]

  4. The attenuation of the mains hum amplitude is \[\begin{aligned} |H(j60\cdot2\pi)| \approx 0.7277 \text{ V/V \end{aligned}\] The corresponding filter output is \[\begin{aligned} |H(j60\cdot2\pi)| m &= |H(j60\cdot2\pi)| (0.5) \approx 0.3639 \text{ \end{aligned}\]

  5. The steady-state signal output amplitude for a sensor input \(1 \sin(\omega_\text{sig} t)\) kPa is \[\begin{aligned} 1000 |G(j\omega_\text{sig})| |H(j\omega_\text{sig})| &= 970 |G(j\omega_\text{sig})| \\ &\approx 970 \cdot 0.001004 \\ &\approx 0.9748\text{ V \end{aligned}\]

  6. For a greater attenuation of the mains hum interference without significantly attenuating the desired signal, use a higher-order filter (e.g., second order) with a similar break frequency.

  7. The natural frequency is, from the standard form of the second-order transfer function, \(\omega_n = \sqrt{1\cdot 10^6} = 1000\) rad/s. Similarly, the damping ratio is \(\zeta = 1\cdot10^3/(2 \omega_n) = 1/2\). Therefore, the peak amplitude frequency is \[\begin{aligned} \omega_p &= \omega_n \sqrt{1 - 2 \zeta^2} \\ &\approx 707.1 \text {rad/s \end{aligned}\]

  8. The steady-state filtered output amplitude for a sensor input \(1 \sin(1\cdot10^3 t)\) kPa at the peak frequency of the sensor is \[\begin{aligned} 1000 |G(j\omega_p)| |H(j\omega_p)| &\approx 1000 (0.001155) (0.4924) \\ &\approx 0.3713 \text{ V \end{aligned}\] It is not greater than the filtered output magnitude at \(\omega_\text{sig}\) from part e (\(0.9748\) V) because the filter significantly attenuated at that frequency (i.e., \(|H(j\omega_p)| \approx 0.4924\)).

  1. Superposition applies to linear systems and this system is linear; in fact, any system with a frequency response function is linear.↩︎

  2. Superposition applies to linear systems and this system is linear; in fact, any system with a frequency response function is linear.↩︎

  3. We could also use the quadratic formula. Tomato tomato.↩︎

Online Resources for Section 13.7

No online resources.