Question 6 of 6: Sinusoidal steady-state response of a first-order system

Question 6: 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 Figure-3 system is a single first-order block with no feedback, $G(s)=\dfrac{2}{s+0.5}$, i.e. a dc gain of 4 and a pole at $s=-0.5$ (time constant $\tau=2$ s). The input is a cosine of amplitude 3 applied at $t=0$, first at $\omega=0.4$ rad/s and later at $\omega=4$ rad/s.

Find. (a) the steady-state output for $r(t)=3\cos 0.4t$; (b) the time after which the response may be regarded as steady state; (c) the steady-state output for $r(t)=3\cos 4t$.

[Figure not reproduced: Figure 3 redrawn: a single first-order block, 2/(s + 0.5), with no feedback. See the official exam paper.]

Approach. Use the frequency-response property of a stable linear time-invariant system: the steady-state response to a sinusoid is a sinusoid of the same frequency, scaled by $|G(j\omega)|$ and shifted by $\angle G(j\omega)$. The transient is the system's own exponential mode, so its duration is set by the pole, not by the input.

1. Form the frequency response. Substituting $s=j\omega$, $$G(j\omega)=\frac{2}{j\omega+0.5}\quad\Longrightarrow\quad |G(j\omega)|=\frac{2}{\sqrt{\omega^{2}+0.25}},\qquad \angle G(j\omega)=-\tan^{-1}\frac{\omega}{0.5}$$ The magnitude falls and the lag grows with frequency, approaching $-90^\circ$ far above the corner at $\omega=0.5$ rad/s.
2. Part (a): evaluate at $\omega=0.4$ rad/s. Here $$|G(j0.4)|=\frac{2}{\sqrt{0.16+0.25}}=\frac{2}{\sqrt{0.41}}=3.1235,\qquad \angle G(j0.4)=-\tan^{-1}(0.8)=-38.66^\circ$$ Multiplying by the input amplitude of 3, $$\boxed{\;c_{ss}(t)=9.370\cos\!\left(0.4t-38.66^\circ\right)=9.370\cos\!\left(0.4t-0.675\ \text{rad}\right)\;}$$ The drive is just below the corner frequency, so the system still passes most of the available dc gain of 4 and lags by well under $45^\circ$.
3. Part (b): find when the transient has died. With $R(s)=3s/(s^{2}+0.16)$, the output is $C(s)=\dfrac{6s}{(s+0.5)(s^{2}+0.16)}$, whose partial-fraction expansion contains one natural mode from the system pole. Its residue is $$A=\left.\frac{6s}{s^{2}+0.16}\right|_{s=-0.5}=\frac{-3}{0.41}=-7.317$$ so the complete response is $c(t)=-7.317e^{-0.5t}+9.370\cos(0.4t-0.675)$, which correctly gives $c(0)=0$. The transient decays with the system time constant $$\tau=\frac{1}{0.5}=2\ \text{s}$$ and falls below 2 % of the steady-state amplitude after four time constants, so $$\boxed{\;t > 4\tau = 8\ \text{s}\quad(\text{within }1\%\ \text{after }5\tau=10\ \text{s})\;}$$ It is worth stressing that this answer depends only on the pole — changing the input frequency does not change how long the transient lasts.
4. Part (c): evaluate at $\omega=4$ rad/s. Ten times the previous frequency and eight times the corner frequency, $$|G(j4)|=\frac{2}{\sqrt{16+0.25}}=\frac{2}{4.0311}=0.4961,\qquad \angle G(j4)=-\tan^{-1}(8)=-82.87^\circ$$ so the steady-state response is $$\boxed{\;c_{ss}(t)=1.488\cos\!\left(4t-82.87^\circ\right)=1.488\cos\!\left(4t-1.446\ \text{rad}\right)\;}$$
5. Interpret the two results together. Raising the drive frequency by a factor of 10 has cut the output amplitude by a factor of $9.370/1.488=6.3$ and pushed the lag from $-38.7^\circ$ to nearly the $-90^\circ$ asymptote. This is the defining behaviour of a first-order low-pass element: above the corner the magnitude rolls off at $-20$ dB/decade while the phase saturates. The transient duration, however, is unchanged at about 8 s in both cases.