From Example 6.16,
Y(f)⟶f=ω/2πY(jω)=jω+10001000X(jω).
The CTFT ( f form) of the excitation is X(f)=0.02sinc(f/200)δ100(f) implying that X(jω)=0.02sinc(ω/400π)δ100(ω/2π). Using the scaling property of the periodic impulse,
X(jω)=0.02sinc(ω/400π)×2πδ200π(ω)=0.04πsinc(ω/400π)δ200π(ω)
Therefore the CTFT of the response is
Y(jω)=jω+10004000πsinc(ω/400π)δ200π(ω)
or, using the defi nition of the periodic impulse,
Y(jω)=4000π∑k=−∞∞jω+1000sinc(ω/400π)δ(ω−200πk).
Now, using the equivalence property of the impulse,
Y(jω)=4000π∑k=−∞∞j200πk+1000sinc(k/2)δ(ω−200πk)
and the inverse CTFT yields the response
y(t)=2000∑k=−∞∞j200πk+100sinc(k/2)ej200πkt.
If we separate the k = 0 term and pair each k and −k this result can be written as
y(t)=2+∑k=1∞j200πk+100sinc(k/2)ej200πkt+−j200πk+1000sinc(−k/2)e−j200πkt.
Using the fact that the sinc function is even and combining the terms over one common denominator,
y(t)=2+∑k=1∞sinc(k/2)(200πk)2+(1000)2(−j200πk+1000)ej200πkt+(j200πk+1000)e−j200πkt
y(t)=2+∑k=1∞sinc(k/2)(200πk)2+(1000)22000cos(200πkt)+400πksin(200πkt)
y(t)=2+∑k=1∞sinc(k/2)25+(πk)25cos(200πkt)+πksin(200πkt)
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).