22-Elec-A3 Signals and Communications · December 2013

Question 4 of 6: Fourier-Transform Properties Without Computing the Transform

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2013 — 07-Elec-A3, Signals and Communications. Three hours, closed book (one two-sided aid sheet and an approved calculator permitted). Six questions, all of equal value at 20 marks; five constitute a complete paper.

All six questions are solved here. The examination marks only the first five answers submitted, but this document is a study resource, so every question — and every sub-part — is worked in full.

Reference texts.

B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. — Ch. 3 (signal transmission), Ch. 4 (amplitude modulation), Ch. 5 (angle modulation), Ch. 6 (sampling).
B. P. Lathi, Linear Systems and Signals, 2nd ed. — Ch. 6 (Fourier series), Ch. 7 (Fourier transform).
A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. — Ch. 4 (Fourier transform properties), Ch. 7 (sampling), Ch. 10 (z-transform).
S. Haykin, Communication Systems, 5th ed. — Ch. 3 (amplitude modulation), Ch. 4 (angle modulation).
J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. — Ch. 3 (z-transform and inversion).

Question 4: Fourier-Transform Properties Without Computing the Transform (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. From Figure 1, $x(t)$ is zero outside $-3 \le t \le 1$ and inside that window consists of four straight segments:

Interval	$x(t)$
$-3 \le t \lt -2$	constant, $x = 2$
$-2 \le t \lt -1$	linear, $2 \to 1$
$-1 \le t \lt 0$	linear, $1 \to 2$
$0 \le t \le 1$	constant, $x = 2$

Find. The phase of $X(j\omega)$, the DC value, the area under $X(j\omega)$, a weighted integral of $X(j\omega)$, and the energy integral — all without evaluating the transform in closed form.

Approach. Recognise the symmetry of $x(t)$, which fixes the phase; then use the DC-value, inverse-transform-at-a-point, multiplication/convolution and Parseval properties in turn, evaluating each in the time domain where the work is trivial.

[Figure not reproduced: The signal x(t), read from the exam figure. It is EVEN about t = -1, which is the single observation that unlocks every part of the question. See the official exam paper.]

Identify the symmetry and hence the phase (part a). Checking pairs of points about $t = -1$: $x(0)=x(-2)=2$, $x(-3)=x(1)=2$, $x(-0.5)=x(-1.5)=1.5$. The signal is therefore even about $t=-1$. Writing $x(t) = y(t+1)$ with $y$ real and even, the time-shift property gives $$X(j\omega) = e^{\,j\omega}\,Y(j\omega), \qquad Y(j\omega)\ \text{real (and even)}$$ Since a real quantity contributes a phase of 0 or $\pi$, $$\boxed{\,\angle X(j\omega) = \omega \quad (\text{plus } \pi \text{ wherever } Y(j\omega) \lt 0)\,}$$ i.e. the phase is purely linear with slope $+1$ rad per rad/s, apart from $\pi$ jumps at the sign changes of $Y$.
Evaluate the DC value (part b). Setting $\omega=0$ in the analysis equation gives $X(j0) = \int_{-\infty}^{\infty}x(t)\,dt$, the net area. Adding the two rectangles and the two trapezoids: $$X(j0) = \underbrace{2(1)}_{[-3,-2]} + \underbrace{\tfrac{2+1}{2}(1)}_{[-2,-1]} + \underbrace{\tfrac{1+2}{2}(1)}_{[-1,0]} + \underbrace{2(1)}_{[0,1]} = 2 + 1.5 + 1.5 + 2$$ $$\boxed{\,X(j0) = 7\,}$$
Evaluate the area under the transform (part c). The synthesis equation at $t=0$ reads $x(0) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X(j\omega)\,d\omega$, so the integral is $2\pi$ times the signal's value at the origin. From the figure $x(0) = 2$, giving $$\boxed{\,\int_{-\infty}^{\infty}X(j\omega)\,d\omega = 2\pi\,x(0) = 4\pi \approx 12.57\,}$$ Note the deliberate duality between parts (b) and (c): one is an area in time read as a value in frequency, the other an area in frequency read as a value in time.
Recognise the weighting function in part (d). The factor $\dfrac{2\sin\omega}{\omega}$ is the Fourier transform of the unit-height rectangular pulse $h(t) = 1$ for $|t| \lt 1$ and 0 otherwise. The integral therefore has exactly the form of a synthesis equation evaluated at $t=2$: $$\int_{-\infty}^{\infty}X(j\omega)H(j\omega)e^{j2\omega}\,d\omega = 2\pi\left[\frac{1}{2\pi}\int X(j\omega)H(j\omega)e^{j\omega t}d\omega\right]_{t=2} = 2\pi\,(x*h)(2)$$ using the convolution property $X(j\omega)H(j\omega) \leftrightarrow (x*h)(t)$.
Evaluate the convolution at $t = 2$ (part d). By definition $(x*h)(2) = \int x(\tau)h(2-\tau)\,d\tau$, and $h(2-\tau)$ is 1 only for $1 \lt \tau \lt 3$. But $x(\tau) = 0$ for $\tau \gt 1$, so the two windows do not overlap at all: $$(x*h)(2) = 0 \quad\Longrightarrow\quad \boxed{\,\int_{-\infty}^{\infty}X(j\omega)\frac{2\sin\omega}{\omega}e^{j2\omega}d\omega = 0\,}$$ The result is zero purely on support grounds — $x*h$ is non-zero only on $[-4,2)$, and $t=2$ is exactly its right-hand edge.
Apply Parseval's theorem (part e). For energy signals $\int|X(j\omega)|^2 d\omega = 2\pi\int |x(t)|^2 dt$. The two rectangles contribute $4$ each and each ramp contributes $\int_0^1 (2-u)^2du = \tfrac{7}{3}$: $$\int_{-\infty}^{\infty}|x(t)|^2 dt = 4 + \tfrac{7}{3} + \tfrac{7}{3} + 4 = \tfrac{38}{3}$$ $$\boxed{\,\int_{-\infty}^{\infty}|X(j\omega)|^2 d\omega = 2\pi\left(\tfrac{38}{3}\right) = \frac{76\pi}{3} \approx 79.59\,}$$

Every part has been answered by inspection of the time-domain figure; at no point was the closed-form expression for $X(j\omega)$ needed, which is precisely the skill the question is testing.

Part	Quantity	Result
(a)	$\angle X(j\omega)$	$\omega$ (linear phase; $+\pi$ jumps where $e^{-j\omega}X$ is negative)
(b)	$X(j0)$	7
(c)	$\int X(j\omega)\,d\omega$	$4\pi \approx 12.57$
(d)	$\int X(j\omega)\frac{2\sin\omega}{\omega}e^{j2\omega}d\omega$	0
(e)	$\int \|X(j\omega)\|^2 d\omega$	$76\pi/3 \approx 79.59$

Topic. Using Fourier-transform properties — symmetry, time shift, DC value, duality, convolution and Parseval — in place of direct integration.

Key relations. $x(t-t_0) \leftrightarrow e^{-j\omega t_0}X(j\omega)$; real and even $\Rightarrow$ transform real and even; $X(j0)=\int x\,dt$; $x(0) = \frac{1}{2\pi}\int X\,d\omega$; $x*h \leftrightarrow XH$; $\int|x|^2dt = \frac{1}{2\pi}\int|X|^2d\omega$.

Why this works. Each requested quantity is a single number, and every one of them is the value of some time-domain functional that the properties expose directly. The linear-phase result is the deepest: any signal that is symmetric about some point $t_0$ has a transform whose phase is exactly $-\omega t_0$ modulo $\pi$, because shifting the symmetry point to the origin makes the transform real. Recognising $2\sin\omega/\omega$ as a transformed rectangle converts an intimidating frequency integral into a one-line support argument in time — and the answer is zero simply because the two supports fail to overlap.

Common pitfall. Assuming the signal is even about $t=0$ because it “looks symmetric”, which gives zero phase and loses part (a). A second pitfall in part (d) is dropping the $2\pi$ factor when converting the integral to a synthesis equation, and a third is trying to compute $X(j\omega)$ explicitly for part (e) — a long and error-prone route to a number Parseval delivers in two lines.

Source. Oppenheim & Willsky, Signals and Systems, 2nd ed., §4.3–4.6 (property tables); Lathi, Linear Systems and Signals, 2nd ed., §7.3.

← Question 3 Next: Question 5 →

Interval	\(x(t)\)
\(-3 \le t \lt -2\)	constant, \(x = 2\)
\(-2 \le t \lt -1\)	linear, \(2 \to 1\)
\(-1 \le t \lt 0\)	linear, \(1 \to 2\)
\(0 \le t \le 1\)	constant, \(x = 2\)

Part	Quantity	Result
(a)	\(\angle X(j\omega)\)	\(\omega\) (linear phase; \(+\pi\) jumps where \(e^{-j\omega}X\) is negative)
(b)	\(X(j0)\)	7
(c)	\(\int X(j\omega)\,d\omega\)	\(4\pi \approx 12.57\)
(d)	\(\int X(j\omega)\frac{2\sin\omega}{\omega}e^{j2\omega}d\omega\)	0
(e)	\(\int \|X(j\omega)\|^2 d\omega\)	\(76\pi/3 \approx 79.59\)