Problems

Part a

Convert each linear factor to standard form \((s/\omega_b + 1)\):

• \((s + 10) = 10\,(s/10 + 1)\), a zero with \(\omega_b = 10\) rad/s.
• \((s + 100) = 100\,(s/100 + 1)\), a pole with \(\omega_b = 100\) rad/s.
• \((s + 1000) = 1000\,(s/1000 + 1)\), a pole with \(\omega_b = 1000\) rad/s.

Collecting the constants, \[\begin{aligned} H(s) &= \frac{1000 \cdot 10\,(s/10 + 1)} {100\,(s/100 + 1) \cdot 1000\,(s/1000 + 1)} = \underbrace{\frac{1}{10}}_{K} \cdot \frac{s/10 + 1}{(s/100 + 1)(s/1000 + 1)}. \end{aligned}\]

Part b

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

Part c

Following the procedure of subsection 13.1.3, sketch each factor’s Bode plot individually and graphically sum them.

DC gain and break frequencies. \(K = 1/10 = -20\) dB. Break frequencies (ascending): \(10\), \(100\), \(1000\) rad/s.

Sketch individual factor magnitudes and sum. The building blocks are:

• \(K = 0.1\): flat at \(-20\) dB.
• Zero at \(\omega_b = 10\): flat at \(0\) dB below \(10\), then \(+20\) dB/dec above.
• Pole at \(\omega_b = 100\): flat at \(0\) dB below \(100\), then \(-20\) dB/dec above.
• Pole at \(\omega_b = 1000\): flat at \(0\) dB below \(1000\), then \(-20\) dB/dec above.

Adding these in dB produces the combined asymptote in : flat at \(-20\) dB, rising at \(+20\) dB/dec from \(\omega = 10\) to \(0\) dB at \(\omega = 100\), flat again, then falling at \(-20\) dB/dec above \(\omega = 1000\).

Sketch individual factor phases and sum. Each real factor transitions \(\pm 90°\) over two decades centered on its break frequency. Adding the three contributions produces the combined phase asymptote in : starting at \(0°\), peaking near \(+45°\) around \(\omega = 10\)–\(100\), and settling to \(-90°\) at high frequency.

Refine. The break frequencies are each a decade apart, so the \(3\) dB corrections at each break do not overlap significantly. compares the asymptotic sketch to the actual Bode plot.

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.

Part a

Substituting \(s \mapsto j\omega\), \[\begin{aligned} H(j\omega) &= \frac{10}{j\omega + 5}. \end{aligned}\]

Part b

Multiplying numerator and denominator by the complex conjugate of the denominator, \[\begin{aligned} H(j\omega) &= \frac{10}{j\omega + 5} \cdot \frac{-j\omega + 5}{-j\omega + 5} = \frac{10(5 - j\omega)}{\omega^2 + 25} = \frac{50}{\omega^2 + 25} - j\frac{10\omega}{\omega^2 + 25}. \end{aligned}\]

Evaluating at each frequency:

• \(\omega = 0\): \(H(j0) = 2 + j0\), so \(|H| = 2\) and \(\angle H = 0°\).
• \(\omega = 1\): \(H(j1) = \frac{50}{26} - j\frac{10}{26} \approx 1.923 - j0.385\), so \(|H| = \frac{10}{\sqrt{26}} \approx 1.961\) and \(\angle H = -\arctan(1/5) \approx -11.3°\).
• \(\omega = 5\): \(H(j5) = \frac{50}{50} - j\frac{50}{50} = 1 - j1\), so \(|H| = \sqrt{2} \approx 1.414\) and \(\angle H = -45°\).
• \(\omega = 50\): \(H(j50) = \frac{50}{2525} - j\frac{500}{2525} \approx 0.0198 - j0.198\), so \(|H| = \frac{10}{\sqrt{2525}} \approx 0.199\) and \(\angle H \approx -84.3°\).

Part c

The points trace a semicircle in the lower-half of the complex plane, from \(H(j0) = 2\) on the positive real axis to the origin as \(\omega \to \infty\). This is characteristic of a first-order lag system.

Part d

As \(\omega \to 0\): \(|H| \to 10/5 = 2\) (the DC gain) and \(\angle H \to 0°\).

As \(\omega \to \infty\): \(|H| \to 10/\omega \to 0\) and \(\angle H \to -90°\).

Part e

From part b, at \(\omega = 5\): \(|H(j5)| = \sqrt{2}\) and \(\angle H(j5) = -45°\). From , \[\begin{aligned} y(t) &= 4\sqrt{2} \cos(5t - \pi/4) \approx 5.66 \cos(5t - \pi/4). \end{aligned}\]

Part a

Taking the Fourier transform of the ODE with zero initial conditions, \[\begin{aligned} j\omega\, Y(\omega) + 4\, Y(\omega) &= 8\, U(\omega). \end{aligned}\] Solving for \(H(j\omega) = Y(\omega)/U(\omega)\), \[\begin{aligned} H(j\omega) &= \frac{8}{j\omega + 4}. \end{aligned}\]

Part b

Taking the Laplace transform with zero initial conditions, \[\begin{aligned} s\,Y(s) + 4\,Y(s) &= 8\,U(s), \end{aligned}\] so \(H(s) = 8/(s + 4)\). Substituting \(s \mapsto j\omega\), \[\begin{aligned} H(j\omega) = \frac{8}{j\omega + 4}, \end{aligned}\] which matches part (a). The two approaches must agree because the Fourier transform of the ODE with zero initial conditions is equivalent to the Laplace transform with zero initial conditions evaluated on the imaginary axis \(s = j\omega\).

Part c

At \(\omega = 0\), \[\begin{aligned} |H(j0)| = \left|\frac{8}{4}\right| = 2. \end{aligned}\]

For a constant input \(u = 1\), the steady state satisfies \(\dot{y} = 0\), so \(4y = 8\), giving \(y = 2\). The ratio \(y/u = 2\) matches the DC gain.

Part d

The magnitude is \(|H(j\omega)| = 8/\sqrt{\omega^2 + 16}\). As \(\omega \to \infty\), the denominator grows without bound, so \(|H| \to 0\).

Physically, the system has a time constant \(\tau = 1/4\) s. When the input oscillates much faster than \(1/\tau = 4\) rad/s, the system’s state cannot keep up — it begins to change in one direction before the input reverses, so the output amplitude shrinks toward zero.

Part e

Setting \(|H(j\omega)| = 2/\sqrt{2}\), \[\begin{aligned} \frac{8}{\sqrt{\omega^2 + 16}} &= \frac{2}{\sqrt{2}} = \sqrt{2}. \end{aligned}\] Solving, \(\omega^2 + 16 = 32\), so \(\omega = 4\) rad/s. This is the break frequency, equal to the pole location \(a = 4\). The phase at the break frequency is \[\begin{aligned} \angle H(j4) = -\arctan(4/4) = -45°. \end{aligned}\]

Part a

\[\begin{aligned} |H_1| &= \sqrt{9 + 16} = 5, &\quad \angle H_1 &= \arctan(4/3) \approx 53.1°, \\ |H_2| &= \sqrt{1 + 4} = \sqrt{5}, &\quad \angle H_2 &= \arctan(-2/1) \approx -63.4°. \end{aligned}\]

Part b

\[\begin{aligned} H_1 H_2 &= (3 + j4)(1 - j2) = 3 - j6 + j4 - j^2 8 = 11 - j2. \end{aligned}\] Checking magnitudes: \(|H_1 H_2| = \sqrt{121 + 4} = \sqrt{125} = 5\sqrt{5}\) and \(|H_1|\,|H_2| = 5\sqrt{5}\). They agree.

Checking phases: \(\angle(H_1 H_2) = \arctan(-2/11) \approx -10.3°\) and \(\angle H_1 + \angle H_2 \approx 53.1° - 63.4° = -10.3°\). They agree.

Part c

\[\begin{aligned} \frac{H_1}{H_2} &= \frac{3 + j4}{1 - j2} \cdot \frac{1 + j2}{1 + j2} = \frac{3 + j6 + j4 - 8}{1 + 4} = \frac{-5 + j10}{5} = -1 + j2. \end{aligned}\] Checking magnitudes: \(|H_1/H_2| = \sqrt{1 + 4} = \sqrt{5}\) and \(|H_1|/|H_2| = 5/\sqrt{5} = \sqrt{5}\). They agree.

Checking phases: \(\angle(H_1/H_2) = \arctan(2/(-1)) = 180° - \arctan 2 \approx 116.6°\) and \(\angle H_1 - \angle H_2 \approx 53.1° + 63.4° = 116.5°\). They agree (up to rounding).

Part d

In decibels, \(20\log_{10}|H_1 H_2| = 20\log_{10}|H_1| + 20\log_{10}|H_2|\). That is, the dB magnitude of a product is the sum of the individual dB magnitudes. Since any transfer function can be factored into simple terms, its Bode magnitude plot (in dB) is obtained by adding the Bode plots of the individual factors. The same additive property holds for phases directly (without the dB conversion).

Part a

Substituting \(s \mapsto j\omega\), \[\begin{aligned} H(j\omega) = \frac{a - j\omega}{a + j\omega}. \end{aligned}\] The magnitude is \[\begin{aligned} |H(j\omega)| = \frac{|a - j\omega|}{|a + j\omega|} = \frac{\sqrt{a^2 + \omega^2}}{\sqrt{a^2 + \omega^2}} = 1. \end{aligned}\]

Part b

Using the angle-of-quotient rule, \[\begin{aligned} \angle H(j\omega) &= \angle(a - j\omega) - \angle(a + j\omega). \end{aligned}\] The numerator \(a - j\omega\) lies in the fourth quadrant (positive real, negative imaginary for \(\omega > 0\)), so \(\angle(a - j\omega) = -\arctan(\omega/a)\). The denominator \(a + j\omega\) lies in the first quadrant, so \(\angle(a + j\omega) = \arctan(\omega/a)\). Therefore, \[\begin{aligned} \angle H(j\omega) &= -\arctan(\omega/a) - \arctan(\omega/a) \\ &= -2\arctan(\omega/a). \end{aligned}\] Note the limiting behavior: at \(\omega = 0\), \(\angle H = 0°\), and as \(\omega \to \infty\), \(\angle H \to -2(\pi/2) = -180°\). The phase decreases monotonically from \(0°\) to \(-180°\).

Part c

For \(a = 10\):

Part d

Since \(|H| = 1\), the output is \(y(t) = \cos(\omega_0 t + \phi)\). The output peak occurs when \(\omega_0 t + \phi = 0\), i.e., at \(t = -\phi/\omega_0\). Since the input peak is at \(t = 0\), the delay is \[\begin{aligned} \tau = \frac{-\phi}{\omega_0} = \frac{2\arctan(\omega_0/a)}{\omega_0}. \end{aligned}\]

Part e

The time delay is not constant across frequencies — it decreases with \(\omega\). This means that different frequency components of a composite signal arrive at the output at different times, causing phase distortion. The waveform shape is altered even though every frequency component has the same amplitude. A system with frequency-dependent delay is said to be dispersive.

Part a

The zero is at \(s = -2\). The poles are at \(s = -1\) and \(s = -5\). All lie on the negative real axis.

Part b

At \(\omega = 3\), the point on the imaginary axis is \(j3\). The three vectors are:

• Zero vector: from \(-2\) to \(j3\), i.e., \(j3 + 2 = 2 + j3\), with length \(\sqrt{4 + 9} = \sqrt{13}\).
• Pole 1 vector: from \(-1\) to \(j3\), i.e., \(j3 + 1 = 1 + j3\), with length \(\sqrt{1 + 9} = \sqrt{10}\).
• Pole 2 vector: from \(-5\) to \(j3\), i.e., \(j3 + 5 = 5 + j3\), with length \(\sqrt{25 + 9} = \sqrt{34}\).

Part c

As \(\omega \to \infty\), all three vectors point nearly straight up with length \(\approx \omega\). The magnitude is \[\begin{aligned} |H(j\omega)| = \frac{|j\omega + 2|}{|j\omega + 1|\,|j\omega + 5|} \approx \frac{\omega}{\omega \cdot \omega} = \frac{1}{\omega} \to 0. \end{aligned}\] There are more poles (2) than zeros (1), so the denominator vectors collectively grow faster than the numerator vector.

Part d

\[\begin{aligned} |H(j0)| &= \frac{2}{1 \cdot 5} = 0.4, \\ |H(j1)| &= \frac{\sqrt{5}}{\sqrt{2}\cdot\sqrt{26}} = \frac{\sqrt{5}}{\sqrt{52}} \approx 0.310, \\ |H(j3)| &= \frac{\sqrt{13}}{\sqrt{10}\cdot\sqrt{34}} = \frac{\sqrt{13}}{\sqrt{340}} \approx 0.196, \\ |H(j10)| &= \frac{\sqrt{104}}{\sqrt{101}\cdot\sqrt{125}} \approx \frac{10.20}{112.4} \approx 0.0908. \end{aligned}\]

Part e

The zero at \(s = -2\) has the most pronounced effect near \(\omega \approx 2\) rad/s. At that frequency, the zero vector (from \(-2\) to \(j2\)) has its smallest “relative” length compared to the pole vectors, in the sense that the zero’s break frequency \(\omega = 2\) is where the zero factor transitions from flat to \(+20\) dB/decade on the Bode plot. Geometrically, at low frequencies (\(\omega \ll 2\)), the zero vector has nearly constant length \(\approx 2\), contributing little variation. Near \(\omega = 2\), the zero vector begins to grow significantly, temporarily slowing the magnitude rolloff caused by the poles. This produces a visible “shelf” or flattening in the Bode plot near \(\omega = 2\).

Part a

The fundamental frequency is \(\omega_n = 5\) rad/s. The components are:

• DC component (\(n = 0\)): constant \(2\).
• First harmonic (\(n = 1\)): \(3\cos(5t)\) at \(\omega = 5\) rad/s.
• Third harmonic (\(n = 3\)): \(\cos(15t)\) at \(\omega = 15\) rad/s.

Note the second harmonic (\(n = 2\)) is absent.

Part b

In standard form, \[\begin{aligned} H(s) = \frac{1}{\frac{s}{5} + 1}. \end{aligned}\] The frequency response function is \[\begin{aligned} H(j\omega) = H(s)\big|_{s \to j\omega} = \frac{1}{\frac{j\omega}{5} + 1} = \frac{1}{1 + j\frac{\omega}{5}}. \end{aligned}\]

Part c

At \(\omega = 0\) (DC): \[\begin{aligned} H(j0) = \frac{1}{1 + 0} = 1 \quad\Rightarrow\quad |H| = 1,\quad \angle H = 0°. \end{aligned}\]

At \(\omega = 5\) rad/s (first harmonic): \[\begin{aligned} H(j5) = \frac{1}{1 + j} \quad\Rightarrow\quad |H| = \frac{1}{\sqrt{2}},\quad \angle H = -45°. \end{aligned}\]

At \(\omega = 15\) rad/s (third harmonic): \[\begin{aligned} H(j15) = \frac{1}{1 + j3} \quad\Rightarrow\quad |H| = \frac{1}{\sqrt{10}},\quad \angle H = -\arctan(3) \approx -71.6°. \end{aligned}\]

Part d

Each component of the input is scaled in amplitude by \(|H|\) and shifted in phase by \(\angle H\) at that component’s frequency: \[\begin{aligned} y(t) = 2 + \frac{3}{\sqrt{2}}\cos(5t - 45°) + \frac{1}{\sqrt{10}}\cos\!\big(15t - \arctan(3)\big). \end{aligned}\]

Part e

This is a low-pass filter (break frequency \(\omega_b = 5\) rad/s). The system passes the DC component unchanged, moderately attenuates the first harmonic (by \(1/\sqrt{2} \approx -3\) dB at the break frequency), and significantly attenuates the third harmonic (by \(1/\sqrt{10} \approx -10\) dB), making the output smoother than the input.

Part a

• Zero: \(s = -10\).
• Poles: \(s = -5\) and \(s = -100\).

All poles have strictly negative real parts, so the system is asymptotically stable.

Part b

Substituting \(s \to j\omega\), \[\begin{aligned} H(j\omega) = \frac{50(j\omega + 10)}{(j\omega + 5)(j\omega + 100)}. \end{aligned}\]

Part c

Factor the DC value out of each first-order term: \[\begin{aligned} H(s) &= \frac{50 \cdot 10 \cdot \left(\frac{s}{10} + 1\right)} {5 \cdot 100 \cdot \left(\frac{s}{5} + 1\right) \left(\frac{s}{100} + 1\right)} = \frac{\frac{s}{10} + 1} {\left(\frac{s}{5} + 1\right)\left(\frac{s}{100} + 1\right)}. \end{aligned}\] The DC gain is \(H(0) = 1\) (i.e., \(0\) dB). The building blocks are: a constant gain of \(1\), a first-order zero with break frequency \(10\) rad/s, a first-order pole with break frequency \(5\) rad/s, and a first-order pole with break frequency \(100\) rad/s.

Part d

Break frequencies and their contributions:

• Pole at \(\omega = 5\) rad/s: magnitude slope \(-20\) dB/dec, phase \(0° \to -90°\).
• Zero at \(\omega = 10\) rad/s: magnitude slope \(+20\) dB/dec, phase \(0° \to +90°\).
• Pole at \(\omega = 100\) rad/s: magnitude slope \(-20\) dB/dec, phase \(0° \to -90°\).

The net asymptotic magnitude is \(0\) dB for \(\omega < 5\), then falls at \(-20\) dB/dec between \(5\) and \(10\) rad/s, is flat (the shelf formed by the pole–zero pair) between \(10\) and \(100\) rad/s, and rolls off at \(-20\) dB/dec above \(100\) rad/s. The high-frequency asymptotic phase is \(0° + (-90°) + 90° + (-90°) = -90°\).

Part e

At \(\omega = 50\): \[\begin{aligned} H(j50) &= \frac{j5 + 1}{(j10 + 1)(j0.5 + 1)}. \end{aligned}\] The magnitude is \[\begin{aligned} |H(j50)| &= \frac{\sqrt{1^2 + 5^2}} {\sqrt{1^2 + 10^2}\;\sqrt{1^2 + 0.5^2}} = \frac{\sqrt{26}}{\sqrt{101}\;\sqrt{1.25}} = \sqrt{\frac{26}{126.25}}. \end{aligned}\] In decibels, \[\begin{aligned} |H(j50)|_\text{dB} = 10\log_{10}\!\frac{26}{126.25} \approx -6.9 \text{ dB}. \end{aligned}\] The phase is \[\begin{aligned} \angle H(j50) &= \arctan(5) - \arctan(10) - \arctan(0.5) \approx 78.7° - 84.3° - 26.6° = -32.2°. \end{aligned}\]

Part a

The poles \(s = -3 \pm j4\) both have negative real part (\(-3 < 0\)), so the system is asymptotically stable. The natural frequency is the magnitude of either pole: \[\begin{aligned} \omega_n = |{-3 + j4}| = \sqrt{9 + 16} = 5 \text{ rad/s}. \end{aligned}\] The damping ratio is \[\begin{aligned} \zeta = \frac{-\operatorname{Re}(p)}{\omega_n} = \frac{3}{5} = 0.6. \end{aligned}\]

Part b

From the sinusoidal response property, the steady-state output is \[\begin{aligned} y(t) &= A\,|H(j\omega_0)|\cos\!\big(\omega_0 t + \psi + \angle H(j\omega_0)\big) \\ &= 4 \cdot \tfrac{1}{2}\cos(6t + 0° - 90°) \\ &= 2\cos(6t - 90°) = 2\sin(6t). \end{aligned}\]

Part c

The unit step has Laplace transform \(U(s) = 1/s\), so \[\begin{aligned} Y(s) = H(s)\,U(s) = \frac{20}{s\,(s + 2)(s + 5)}. \end{aligned}\] By the Final Value Theorem, \[\begin{aligned} y(\infty) = \lim_{s \to 0} s\,Y(s) = \lim_{s \to 0} \frac{20}{(s + 2)(s + 5)} = \frac{20}{2 \cdot 5} = 2. \end{aligned}\] Note this equals the DC gain \(H(0)\), as expected for a stable system driven by a unit step.

Part d

Applying the through-variable divider rule from with two parallel impedances, \[\begin{aligned} \frac{\mathcal{F}_1}{\mathcal{F}_S} = \frac{1/Z_1}{1/Z_1 + 1/Z_2}. \end{aligned}\]

Part e

Asymptotic slope contributions on the Bode magnitude plot:

1. First-order pole: \(-20\) dB/decade.
2. First-order zero: \(+20\) dB/decade.
3. Complex conjugate pole pair: \(-40\) dB/decade.

Total phase contributions on the Bode phase plot:

1. First-order pole: \(-90°\).
2. First-order zero: \(+90°\).
3. Complex conjugate pole pair: \(-180°\).