Question 6.2: Periodic excitation and response of a continuous-time system......
Periodic excitation and response of a continuous-time system
A continuous-time system is described by the differential equation
y″(t) + 0.04 y′(t) + 1.58y(t) = x(t).
If the excitation is x(t) = tri(t) ∗ δ_5(t), find the response y(t).
The excitation can be expressed by a CTFS as
\mathrm{x}(t)=\sum_{k=-\infty}^{\infty} \mathrm{c}_{\mathrm{x}}[k] e^{j 2 \pi k t / T_0}where, from Table 6.2,
c_x[k] = (w/T_0 )sinc² (wk /mT_0 ) δ_m[k]with w = 1, T_0 = 5 and m = 1. Then
We know that the CTFS expression for the excitation is a sum of complex sinusoids and the response to each of those sinusoids will be another sinusoid of the same frequency. Therefore, the response can be expressed in the form
\mathrm{y}(t)=\sum_{k=-\infty}^{\infty} \mathrm{c}_{\mathrm{y}}[k] e^{j 2 \pi k t / 5}and each complex sinusoid in y(t) with fundamental cyclic frequency k /5 is caused by the complex sinusoid in x(t) of the same frequency. Subsitituting this form into the differential equation
Gathering terms and simplifying
Therefore, for any particular value of k the excitation and response are related by
[( j2πk /5)² + 0.04( j2πk /5) + 1.58]c_y[k] = c_x[k]and
\frac{c_y[k]}{c_x[k]} = \frac{1}{( j2πk /5)² + 0.04( j2πk /5) + 1.58}The quantity H[k] = \frac{c_y[k]}{c_x[k]} is analogous to frequency response and can logically be called harmonic response. The system response is
\mathrm{y}(t)=(1 / 5) \sum_{k=-\infty}^{\infty} \frac{\operatorname{sinc}^2(k / 5)}{(j 2 \pi k / 5)^2+0.04(j 2 \pi k / 5)+1.58} e^{j 2 \pi k t / 5} .
This rather intimidating-looking expression can be easily programmed on a computer. The signals, their harmonic functions and the harmonic response are illustrated in Figure 6.13 and Figure 6.14.
We can see from the harmonic response that the system responds strongly at harmonic number one, the fundamental. The fundamental period of x(t) is T_0 = 5 s. So y(t) should have a significant response at a frequency of 0.2 Hz. Looking at the response graph, we see a signal that looks like a sinusoid and its fundamental period is 5 s, so its fundamental frequency is 0.2 Hz.
The magnitudes of all the other harmonics, including k = 0, are almost zero. That is why the average value of the response is practically zero and it looks like a sinusoid, a single-frequency signal. Also, notice the phase of the harmonic response at the fundamental. It is 1.5536 radians at k = 1, or almost π/2. That phase shift would convert a cosine into a sine. The excitation is an even function with only cosine components and the response is practically an odd function because of this phase shift.
| Table 6.2 Some CTFS pairs |
| e^{j 2 \pi t / T_0} \xleftrightarrow[m T_0]{\mathcal{F}S}\delta[k-m] |
| \cos \left(2 \pi k / T_0\right) \xleftrightarrow[m T_0]{\mathcal{F}S}(1 / 2)(\delta[k-m]+\delta[k+m]) |
| \sin \left(2 \pi k / T_0\right) \xleftrightarrow[m T_0]{\mathcal{F}S}(j / 2)(\delta[k+m]-\delta[k-m]) |
| 1\xleftrightarrow[T]{\mathcal{F}S}\delta[k], T \text { is arbitrary } |
| \delta_{T_0}(t)\xleftrightarrow[m T_0]{\mathcal{F}S}\left(1 / T_0\right) \delta_m[k] |
| \operatorname{rect}(t / w) * \delta_{T_0}(t) \xleftrightarrow[m T_0]{\mathcal{F}S}\left(w / T_0\right) \operatorname{sinc}\left(w k / m T_0\right) \delta_m[k] |
| \operatorname{tri}(t / w) * \delta_{T_0}(t) \xleftrightarrow[m T_0]{\mathcal{F}S}\left(w / T_0\right) \operatorname{sinc}^2\left(w k / m T_0\right) \delta_m[k] |
| \operatorname{sinc}(t / w) * \delta T_0(t) \xleftrightarrow[m T_0]{\mathcal{F}S} \left(w / T_0\right) \operatorname{rect}\left(w k / m T_0\right) \delta_m[k] |
| t[\mathbf{u}(t)-\mathbf{u}(t-w)] * \delta _{T_0}(t) \xleftrightarrow[m T_0]{\mathcal{F}S} \frac{1}{T_0} \frac{\left[j\left(2 \pi k w / m T_0\right)+1\right] e^{-j\left(2 \pi k w / m T_0\right)}-1}{\left(2 \pi k / m T_0\right)^2} \delta_m[k] |
