Match the following : Codes :List - I List - II (spectrum |G(w)| in the original figure) (a) Rectangular Pulse (i) 
(b) Double-sided Exponential (ii) 
(c) Cosine Pulse (iii) 
(d) Damped Sine (iv) 
a-(iv), b-(ii), c-(i), d-(iii)
Two questions sort all four spectra: does it have ripples, and is it centred at the origin or split in two?
Ripples mean a rectangular time window. A sharp-edged pulse transforms to a sinc, whose side lobes are the ripples:
\(\text{rect}(t/\tau)\ \longleftrightarrow\ \tau\,\text{sinc}\!\left(\dfrac{\omega\tau}{2}\right)\)
A smooth, gradually decaying time function transforms to a smooth, ripple-free spectrum. So (i) and (iv) belong to the rectangular family, and (ii) and (iii) to the exponential family.
Two lobes mean multiplication by a sinusoid. Multiplying any signal by \(\cos\omega_0t\) shifts its spectrum to \(\pm\omega_0\):
\(x(t)\cos\omega_0t\ \longleftrightarrow\ \tfrac{1}{2}\left[X(\omega-\omega_0)+X(\omega+\omega_0)\right]\)
So a single central lobe means no carrier; a split pair means the signal oscillates.
Now assign all four.
| Signal | Ripples? | Centred or split? | Spectrum |
|---|---|---|---|
| (a) Rectangular pulse | yes — sharp edges | centred | (iv) |
| (b) Double-sided exponential | no — smooth \(e^{-a|t|}\) | centred | (ii) |
| (c) Cosine pulse | yes — a rect times a cosine | split | (i) |
| (d) Damped sine | no — smooth envelope | split | (iii) |
which is option 4.
The two ripple-free spectra confirmed. The double-sided exponential gives a Lorentzian centred at zero,
\(e^{-a|t|}\ \longleftrightarrow\ \dfrac{2a}{a^{2}+\omega^{2}}\)
and the damped sine is that same Lorentzian shape shifted to \(\pm\omega_0\):
\(e^{-at}\sin\omega_0t\,u(t)\ \longleftrightarrow\ \dfrac{\omega_0}{(a+j\omega)^{2}+\omega_0^{2}}\)
The underlying principle. Smoothness in one domain buys compactness in the other. A discontinuity in time — the vertical edge of a rectangle — forces the spectrum to decay only as 1/ω, which is what the ripples are; a signal with no discontinuity decays far faster. This is exactly why window functions such as Hamming and Hanning are used in spectral analysis: rounding the edges of the observation window suppresses the side lobes.
Hence, the correct matching is a-(iv), b-(ii), c-(i), d-(iii).
Similar Questions
Fourier transform or Fourier integral of a function f(t) is given by
Match the following lists :
List – I List – II a. 1 i. \(\pi\delta(\omega)+\dfrac{1}{j\omega}\) b. u(t) ii. 1 c. δ(t) iii. \(\dfrac{-2}{\omega^{2}}\) d. | t | iv. \(2\pi\delta(\omega)\) The Fourier transform of the following time domain function is :
Important Questions from Fourier Transform
- Match the LIST-I with LIST-IIChoose the correct answer from the options given below:
LIST-I LIST-II A. Continuous and periodic signal I. continuous time Fourier series B. Continuous and aperiodic signal II. discrete time Fourier series C. Discrete and periodic signal III. continuous time Fourier transform D. Discrete and aperiodic signal IV. discrete time Fourier transform - The Fourier transform of the signal $x(t)=e^{at}u(-t)$, $a0$ is
The Fourier transform for the given waveform.
Fourier transform or Fourier integral of a function f(t) is given by
A continuous real-valued signal $x(t)$ has finite positive energy and $x(t) = 0, \forall t < 0$. From the list given below, select ALL the signals whose continuous-time Fourier transform is purely imaginary.
