Question 3 of 6: Sinusoidal Steady-State Response of a First-Order System

Question 3: Sinusoidal Steady-State Response of a First-Order System (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The open-loop first-order plant of Figure 2,

driven from rest ($c(0)=0$) by $r(t)=3\cos\omega t$ applied at $t=0$, first with $\omega=0.4$ rad/s and then with $\omega=4$ rad/s.

Find. (a) the steady-state output for $\omega=0.4$ rad/s; (b) the elapsed time after which the transient has died away and that steady state is actually established; (c) the steady-state output for $\omega=4$ rad/s.

Approach. For a stable linear system the steady-state response to a sinusoid is the input sinusoid scaled by $|G(j\omega)|$ and shifted by $\angle G(j\omega)$; the transient is governed solely by the system pole, so the "time to reach steady state" is a fixed multiple of the time constant and is completely independent of the drive frequency.

(a) Steady-state response at $\omega=0.4$ rad/s

1. Evaluate the frequency response. Substituting $s=j\omega$ with $\omega=0.4$ rad/s, $$G(j0.4) \;=\; \frac{2}{0.5+j0.4}.$$ The magnitude is the ratio of the magnitudes and the phase is the difference of the phases: $$|G(j0.4)| = \frac{2}{\sqrt{0.5^{2}+0.4^{2}}} = \frac{2}{0.6403} = 3.1235,$$ $$\angle G(j0.4) = -\tan^{-1}\!\left(\frac{0.4}{0.5}\right) = -\tan^{-1}(0.8) = -38.66^{\circ} = -0.6747\ \text{rad}.$$
2. Scale and shift the input. Multiplying the input amplitude by $|G|$ and adding $\angle G$ to its argument, $$\boxed{\;c_{ss}(t) \;=\; 3(3.1235)\cos\!\left(0.4t-38.66^{\circ}\right) \;=\; 9.370\,\cos\!\left(0.4t-0.6747\right). \;}$$ The output is amplified almost 3.1-fold and lags the input by just under $39^{\circ}$, which is expected: $0.4$ rad/s sits below the plant's corner at $0.5$ rad/s, so the plant is still near its DC gain of $2/0.5 = 4$.

(b) When is the system actually in steady state?

1. Extract the time constant. The complete response is the sum of a forced sinusoid and a natural term set by the single pole at $s=-0.5$: $$c(t) = \underbrace{A\,e^{-0.5t}}_{\text{transient}} \;+\; \underbrace{9.370\cos(0.4t-0.6747)}_{\text{steady state}},$$ with $A$ fixed by the zero initial condition. The pole gives $$\tau = \frac{1}{0.5} = 2\ \text{s}.$$
2. Apply the settling criterion. The transient decays as $e^{-t/\tau}$, reaching $e^{-4}=1.83\%$ of its initial value after four time constants and $e^{-5}=0.67\%$ after five. On the usual engineering criteria, $$\boxed{\;t \gtrsim 4\tau = 8\ \text{s (2\% criterion)},\qquad t \gtrsim 5\tau = 10\ \text{s (1\% criterion)}. \;}$$ So the system is in steady state for roughly $t>8$ s onward — strictly, for all $t\to\infty$, but for practical measurement after about four to five time constants.

(c) Steady-state response at $\omega=4$ rad/s

1. Re-evaluate the frequency response one decade higher. With $\omega=4$ rad/s, $$|G(j4)| = \frac{2}{\sqrt{0.5^{2}+4^{2}}} = \frac{2}{4.0311} = 0.4961,\qquad \angle G(j4) = -\tan^{-1}(8) = -82.88^{\circ}.$$ The drive is now well above the corner frequency, so the plant behaves almost as a pure integrator: the gain has collapsed by a factor of $6.3$ and the lag has grown to nearly the $-90^{\circ}$ asymptote.
2. Write the response. $$\boxed{\;c_{ss}(t) \;=\; 3(0.4961)\cos\!\left(4t-82.88^{\circ}\right) \;=\; 1.4884\,\cos\!\left(4t-1.4465\right). \;}$$
3. Note what did not change. Raising the drive frequency by a factor of ten changes the amplitude and the phase of the forced term but leaves the pole — and therefore the time constant — untouched. The answer to part (b) is unchanged at $8$–$10$ s: the transient still needs four to five time constants to disappear regardless of how fast the input is oscillating. That invariance is the point the question is testing.