Question 6.16: System analysis using the CTFT A system described by the dif......
System analysis using the CTFT
A system described by the differential equation \mathbf{y}^{\prime}(t)+\mathrm{{1000y}}(t)=\mathrm{{1000x}}(t) is excited by x(t)=4{\mathrm{rect}}(200t). Find and graph the response y(t).
If we Fourier transform the differential equation we get
{j2}\pi f{{Y}}(f)+1000\,\mathrm{Y}(f)=1000\,\mathrm{X}(f)
which can be rearranged into
\mathbf{Y}(f)={\frac{1000\,{X}(f)}{j2\pi f+1000}}.
The CTFT of the excitation is \mathrm{X}(f)=0.02sinc(f/200). Therefore the CTFT of the response is
Y(f)=\frac{20sinc( f/200)}{j2\pi f+1000}
or, using the definition of the sinc function and the exponential definition of the sine function,
Y(f)=20\frac{\sin(\pi f/200)}{(\pi f/200)\left(j2\pi f+1000\right)}=4000\frac{e^{j2\pi f/400}-e^{-j2\pi f/400}}{j2\pi f(j2\pi f+1000)}.
To find the inverse CTFT, start with the CTFT pair e^{-\alpha t}\,{\mathbf{u}}(t){\boldsymbol{}}\overset{\mathcal F }{\longleftrightarrow } 1/(j2\pi f+\alpha),\ \alpha\gt 0,
e^{-1000t}\,{\bf u}(t)\overset{\mathcal F}{\longleftrightarrow } \frac{1}{j2\pi f+1000}.
Next use the integration property,
\int\limits_{-\infty}^{t}{g}(\lambda)d\lambda\overset{\mathcal F}{\longleftrightarrow } {\frac{G(f)}{j2\pi f}}+(1/2){\mathrm{G}}(0)\delta(f).
\int\limits_{-\infty }^{t}{e^{-1000\lambda } } u(\lambda )d(\lambda )\overset{\mathcal F}{\longleftrightarrow } \frac{1}{j2\pi f} \frac{1}{j2\pi f +1000}+\frac{1}{2000} \delta (f)
Then apply the time-shifting property,
{g}(t-t_{0})\overset{\mathcal F}{\longleftrightarrow } {G}(f)e^{-j2\pi ft_{0}}.
\int\limits_{0}^{t+1/400}{e^{-1000\lambda } }d\lambda \overset{\mathcal F}{\longleftrightarrow } \frac{1}{j2\pi f} \frac{e^{j2\pi f/400} }{j2\pi f+1000} +\underbrace{\frac{e^{j2\pi f/400} }{2000} \delta (f)}_{=\delta(f)/2000}
\int\limits_{0}^{t-1/400}{e^{-1000\lambda } }d\lambda \overset{\mathcal F}{\longleftrightarrow } \frac{1}{j2\pi f} \frac{e^{-j2\pi f/400} }{j2\pi f+1000} +\underbrace{\frac{e^{-j2\pi f/400} }{2000} \delta (f)}_{=\delta(f)/2000}
Subtracting the second result from the first and multiplying through by 4000
4000\int\limits_{-\infty }^{t+1/400}{e^{-1000\lambda } } u(\lambda )d\lambda -4000\int\limits_{-\infty }^{t-1/400}{x^{-1000\lambda } } u(\lambda )d\lambda \overset{\mathcal F}{\longleftrightarrow } \frac{4000}{j2\pi f} \frac{e^{j2\pi f/400}-e^{-j2\pi f/400} }{j2\pi f+1000}
4000\left[\int\limits_{-\infty }^{t+1/400}{e^{-1000\lambda } } u(\lambda )d\lambda -\int\limits_{-\infty }^{t-1/400}{x^{-1000\lambda } } u(\lambda )d\lambda\right] \overset{\mathcal F}{\longleftrightarrow } \frac{4000}{j2\pi f} \frac{e^{j2\pi f/400}-e^{-j2\pi f/400} }{j2\pi f+1000}
The two integral expressions can be simplified as follows.
\int\limits_{-\infty }^{t+1/400}{e^{-1000\lambda } } u(\lambda )d\lambda =\left \{ \begin{matrix} (1/1000)(1-e^{-1000(t+1/400)}), &t\geq-1/400 \\ 0, & t\lt -1/400\end{matrix} \right \} ={\frac{1}{1000}}(1-e^{-1000(t+1/400)})u(t+1/400)
\int\limits_{-\infty }^{t-1/400}{e^{-1000\lambda } } u(\lambda )d\lambda =\left \{ \begin{matrix} (1/1000)(1-e^{-1000(t-1/400)}), &t\geq 1/400 \\ 0, & t\lt 1/400\end{matrix} \right \} ={\frac{1}{1000}}(1-e^{-1000(t-1/400)})u(t-1/400)
Then
4\Bigl[(1-e^{-1000(t+1/400)})\mathbf{u}(t+1/400)-(1-e^{-1000(t-1/400)})\mathbf{u}(t-1/400)\Bigr] \overset{\mathcal F}{\longleftrightarrow } {\frac{4000}{j2\pi f}}{\frac{e^{j2\pi f/400}-e^{-j2\pi f/400}}{j2\pi f+1000}}
Therefore the response is
\begin{array}{c l c r}{{y(t)=4[(1-e^{-1000(t+1/400)})\,\mathrm{u}(t+1/400)-(1-e^{-1000(t-1/400)})\,\mathrm{u}(t-1/400)]}}\end{array}
(Figure 6.35 and Figure 6.36).
