Fourier transform or Fourier integral of a function f(t) is given by
\(\displaystyle\int_{-\infty}^{+\infty}f(t)\,e^{-j\omega t}\,dt\)
Two things define the Fourier transform — limits of −∞ to +∞, and a kernel of \(e^{-j\omega t}\) — and only option 4 has both.
\(F(\omega)=\int_{-\infty}^{+\infty}f(t)\,e^{-j\omega t}\,dt\)
Why the limits must be two-sided. The transform decomposes a signal into everlasting sinusoids, which have no beginning and no end, so the analysis must cover all time. Truncating the lower limit to zero, as options 1 and 2 do, silently assumes the signal is causal — and discards whatever happens for \(t\lt0\). That one-sided integral with the \(e^{-j\omega t}\) kernel is the unilateral transform, a different object.
Why the kernel carries a minus sign. The forward transform correlates \(f(t)\) against \(e^{+j\omega t}\), and correlation takes the complex conjugate — hence \(e^{-j\omega t}\). The positive exponent of option 3 is the kernel of the inverse transform:
\(f(t)=\dfrac{1}{2\pi}\int_{-\infty}^{+\infty}F(\omega)\,e^{+j\omega t}\,d\omega\)
Using the wrong sign yields \(F(-\omega)\) instead of \(F(\omega)\) — harmless for real even signals, wrong in general.
| Option | What it actually is |
|---|---|
| 1 | Laplace transform — kernel \(e^{-st}\) |
| 2 | Unilateral Fourier transform — causal signals only |
| 3 | Inverse transform kernel |
| 4 | Fourier transform |
Option 1 is worth a second look because the two transforms are close relatives. Put \(s=\sigma+j\omega\) and set \(\sigma=0\): the Laplace kernel becomes the Fourier kernel. So the Fourier transform is the Laplace transform evaluated on the imaginary axis — valid whenever that axis lies inside the region of convergence. The extra \(e^{-\sigma t}\) is what lets Laplace handle growing signals that have no Fourier transform at all.
Convergence. The integral exists when \(f(t)\) is absolutely integrable; sinusoids and steps are not, and appear in the transform as impulses, admitted through the generalised-function extension of the theory.
Hence, the correct expression is the two-sided integral with kernel e−jωt.
Similar Questions
Match the following :
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) 
Codes :
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.
Match the following :
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) Codes :
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.
