Step Functions — Question 3

Question 3 – Solution

Strategy. Re-express the phase at each switch separately. The second switch occurs at a different oscillator phase.

Step 1: Identify the window and endpoint values. The function equals $\sin t$ on $[\pi/2,3\pi/2)$ and zero elsewhere. In particular $f(\pi/2)=1$, while $f(3\pi/2)=0$. At the latter time the left limit is $-1$; the graph distinguishes it from the assigned value.

Step 2: Shift the two phases correctly. With $a=\pi/2$ and $b=3\pi/2$, we have $\sin(v+a)=\cos v$ and $\sin(v+b)=-\cos v$. Therefore $$f=u_a(t)\cos(t-a)+u_b(t)\cos(t-b),$$ $$\boxed{F(s)=\frac{s(e^{-\pi s/2}+e^{-3\pi s/2})}{s^2+1}.}$$ The plus sign in the transform reflects the phase reversal at shutoff. After $b$, the two shifted cosine terms cancel exactly, as the original window requires.

Step 3: Use finite support to evaluate zero. The defining integral has finite support, so it converges for every real $s$; finite-interval integration gives the displayed expression on the whole real line. Directly, $$F(0)=\int_a^b\sin t\,dt=0,$$ $$F'(0)=-\int_a^b t\sin t\,dt =-[-t\cos t+\sin t]_a^b=\boxed 2.$$ The formula also gives $F(0)=0$ and expands as $2s+O(s^2)$, verifying the derivative.

Step 4: Pair the equal and opposite lobes. Write $t=\pi+x$ with $-\pi/2\le x\le\pi/2$. Pair $x$ with $-x$. For $0<x<\pi/2$, the earlier value is $\sin x>0$ and the later value is $-\sin x$. Their weighted contribution is $$e^{-\pi s}\sin x\,[e^{sx}-e^{-sx}]>0\qquad(s>0).$$ Integrating these pairs proves strict positivity, despite cancellation at $s=0$. This also agrees with the positive numerator and denominator of the formula for $s>0$.