Question 6.17: 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)+1000\mathbf{y}(t)=1000\mathbf{x}(t) is excited by x(t)=4\operatorname{rect}(200t)*\delta_{0.01}(t). Find and graph the response y(t) .
From Example 6.16,
Y(f){\overset{f=\omega/2\pi}{\longrightarrow}}Y(j\omega)={\frac{1000\,\mathrm{X}(j\omega)}{j\omega+1000}}.
The CTFT ( f form) of the excitation is {X}(f)=0.02sinc(f/200)\delta_{100}(f) implying that {\sf X}(j\omega)=0.02sinc(\omega/400\pi)\delta_{100}(\omega/2\pi). Using the scaling property of the periodic impulse,
\mathrm{X}(j\omega)=0.02\mathrm{sinc}(\omega/400\pi)\times2\pi\delta_{200\pi}(\omega)=0.04\pi\mathrm{sinc}(\omega/400\pi)\delta_{200\pi}(\omega)
Therefore the CTFT of the response is
Y(j\omega)={\frac{4000\pi\mathrm{sinc}(\omega/400\pi)\mathrm{\delta}_{200 \pi}(\omega)}{j\omega+1000}}
or, using the defi nition of the periodic impulse,
Y(j\omega)=4000\pi\sum_{k=-\infty}^{\infty}{\frac{\mathrm{sinc}(\omega/400\pi)\delta(\omega-200\pi k)}{j\omega+1000}}.
Now, using the equivalence property of the impulse,
Y(j\omega)=4000\pi\sum_{k=-\infty}^{\infty}\frac{\mathrm{sinc}(k/2)\delta(\omega-200\pi k)}{j200\pi k+1000}
and the inverse CTFT yields the response
y(t)=2000\sum_{k=-\infty}^{\infty}{\frac{\mathrm{sinc}(k/2)}{j200\pi k+100}}e^{j200\pi k t}.
If we separate the k = 0 term and pair each k and −k this result can be written as
\mathrm{y}(t)=2+\sum_{k=1}^{\infty}{\frac{\mathrm{sinc}(k/2)}{{j200\pi k+100}}}e^{j{{2}}00\pi k t}+{\frac{\mathrm{sinc}(-k/2)}{-j200\pi k+1000}}e^{-j200\pi k t}\,.
Using the fact that the sinc function is even and combining the terms over one common denominator,
y(t)=2+\sum_{k=1}^{\infty}\mathrm{sinc}(k/2)\frac{(-j200\pi k+1000)e^{j200\pi k t}+(j200\pi k+1000)e^{-j200\pi k t}}{(200\pi k)^{2}+(1000)^{2}}
y(t)=2+\sum_{k=1}^{\infty}sinc(k/2){\frac{2000\cos(200\pi k t)+400\pi k\sin(200\pi k t)}{(200\pi k)^{2}+(1000)^{2}}}
\mathbf{y}(t)=2+\sum_{k=1}^{\infty}\operatorname{sinc}(k/2){\frac{5\cos(200\pi k t)+\pi k\sin(200\pi k t)}{25+(\pi k)^{2}}}
The response is a constant plus a linear combination of real cosines and sines at integer multiples of 100 Hz (Figure 6.37 and Figure 6.38).
