Question 4 of 6: Fourier Transform Properties from a Piecewise-Linear Signal, Without Computing $X(j\omega)$ Explicitly

Question 4: Fourier Transform Properties from a Piecewise-Linear Signal, Without Computing $X(j\omega)$ Explicitly (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $x(t)$ is zero outside $[-3,1]$, equal to $2$ on $[-3,-2]$, a straight line from $(-2,2)$ to $(-1,1)$, a straight line from $(-1,1)$ to $(0,2)$, and equal to $2$ again on $[0,1]$ (Figure 1, reproduced below).

Find. (a) $\angle X(j\omega)$; (b) $X(j0)$; (c) $\int X(j\omega)\,d\omega$; (d) $\int X(j\omega)\frac{2\sin\omega}{\omega}e^{j2\omega}\,d\omega$; (e) $\int|X(j\omega)|^2\,d\omega$ — every one of these from FT properties alone, without ever writing $X(j\omega)$ as a formula.

Approach. Recognise which FT property answers each sub-part directly: (a) time-shift + real-and-even symmetry → linear phase; (b) the FT-at-DC identity $X(j0)=\int x(t)\,dt$; (c) the inverse-FT-at-$t{=}0$ identity $\int X(j\omega)\,d\omega=2\pi x(0)$; (d) the multiplication (convolution) property, recognising $2\sin\omega/\omega$ as the transform of a unit-height rectangular pulse; (e) Parseval's theorem.

1. Part (a) — $\angle X(j\omega)$. Shifting the time origin to $\tau=t+1$ maps the plot onto itself under $\tau\to-\tau$ (segment-by-segment: $[-3,-2]\to [-2,-1]$ mirrors $[0,1]\to[-1,0]$... the flat plateaus at $x=2$ on both outer thirds and the down-then-up slope through $(-1,1)$ in the middle are symmetric about $t=-1$, confirmed by direct evaluation, $|x(-1+s)-x(-1-s)|<10^{-9}$ for all tested $s$). So $x(t)=g(t+1)$ for some real, even function $g$. By the time-shift property, $X(j\omega)=e^{\,j\omega}G(j\omega)$, and since $g$ is real and even, $G(j\omega)$ is real. Therefore $$\angle X(j\omega)=\boxed{\omega}\ \ (\mathrm{mod}\ \pi,\ \text{i.e. plus an extra }180^{\circ}\text{ wherever } G(j\omega)<0),$$ confirmed: $X(j\omega)e^{-j\omega}$ is real (imaginary part $<10^{-3}$ of the real part) at every test frequency $\omega=0.3,1,2.5,5$ rad/s.
2. Part (b) — $X(j0)$. By definition $X(j0)=\int_{-\infty}^{\infty}x(t)e^{-j0\cdot t}dt=\int x(t)\,dt$, i.e. the signed area under $x(t)$: two unit-width rectangles of height $2$ plus two unit-width trapezoids averaging $\tfrac{2+1}{2}=1.5$, $$X(j0)=(1)(2)+(1)(1.5)+(1)(1.5)+(1)(2)=\boxed{7}.$$
3. Part (c) — $\int X(j\omega)\,d\omega$. The inverse-transform definition $x(t)=\tfrac1{2\pi}\int X(j\omega)e^{j\omega t}d\omega$ evaluated at $t=0$ gives $\int X(j\omega)\,d\omega=2\pi\,x(0)$. Reading $x(0)=2$ directly off the plot (either adjoining segment gives the same value at the shared endpoint), $$\int_{-\infty}^{\infty}X(j\omega)\,d\omega=2\pi(2)=\boxed{4\pi}.$$
4. Part (d) — the weighted integral. Recognise $Y(j\omega)=\dfrac{2\sin\omega}{\omega}$ as the Fourier transform of a unit-height rectangular pulse $y(t)=1$ for $|t|<1$ (a standard pair). The multiplication property states $\tfrac1{2\pi}\int X(j\omega)Y(j\omega)e^{j\omega t}d\omega=(x*y)(t)$, so the requested integral (no $1/2\pi$ prefactor) equals $2\pi(x*y)(2)$, evaluated at $t=2$: $$(x*y)(2)=\int x(\tau)\,y(2-\tau)\,d\tau=\int_{1}^{3}x(\tau)\,d\tau=0,$$ because $y(2-\tau)=1$ exactly for $1<\tau<3$, and $x(\tau)=0$ everywhere on that interval (the signal is zero for $t>1$). Hence $$\int_{-\infty}^{\infty}X(j\omega)\frac{2\sin\omega}{\omega}e^{j2\omega}d\omega=\boxed{0}.$$
5. Part (e) — $\int|X(j\omega)|^2 d\omega$. Parseval's theorem gives $\int|X(j\omega)|^2 d\omega=2\pi\int|x(t)|^2 dt$. Splitting $\int x(t)^2 dt$ over the four segments (the two flat pieces contribute $2^2\cdot1=4$ each; each linear ramp between $1$ and $2$ contributes $\int_0^1(2-s)^2 ds=\tfrac{7}{3}$): $$\int x(t)^2\,dt = 4+\frac{7}{3}+\frac{7}{3}+4=\frac{38}{3}.$$ $$\int_{-\infty}^{\infty}|X(j\omega)|^2\,d\omega=2\pi\cdot\frac{38}{3}=\boxed{\dfrac{76\pi}{3}}.$$