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Question

Inverse Fourier Transform of δ(ω - ω 0) is ______.

The correct answer is \(\frac {e^{jω_0t}}{2\pi}\)

Understanding the Inverse Fourier Transform Problem

The question asks us to find the Inverse Fourier Transform of a specific function in the frequency domain, which is \(\delta(\omega - \omega_0)\). This function is a shifted version of the Dirac delta function in the frequency domain. Finding its Inverse Fourier Transform means converting this frequency representation back into a time-domain signal.

Defining the Inverse Fourier Transform

The Inverse Fourier Transform \(x(t)\) of a signal \(X(\omega)\) in the frequency domain is given by the integral formula:

\(x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega\)

This formula takes a function of frequency, \(X(\omega)\), and produces a function of time, \(x(t)\).

Applying the Formula

In this particular problem, our frequency-domain signal is given as \(X(\omega) = \delta(\omega - \omega_0)\). We substitute this into the Inverse Fourier Transform formula:

\(x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega - \omega_0) e^{j\omega t} d\omega\)

Now, we need to evaluate this integral involving the Dirac delta function.

Utilizing the Dirac Delta Function Property

The key property of the Dirac delta function \(\delta(x - a)\) is its sifting property. The sifting property states that for any function \(f(x)\) that is continuous at \(x = a\):

\(\int_{-\infty}^{\infty} f(x) \delta(x - a) dx = f(a)\)

In our integral, the variable of integration is \(\omega\). The function \(f(\omega)\) is \(e^{j\omega t}\), and the shift in the delta function is \(a = \omega_0\). According to the sifting property, the integral \(\int_{-\infty}^{\infty} \delta(\omega - \omega_0) e^{j\omega t} d\omega\) evaluates to the value of \(e^{j\omega t}\) at \(\omega = \omega_0\).

So, \(\int_{-\infty}^{\infty} \delta(\omega - \omega_0) e^{j\omega t} d\omega = e^{j\omega_0 t}\).

Calculating the Final Result

Now we substitute the result of the integral back into the Inverse Fourier Transform formula:

\(x(t) = \frac{1}{2\pi} \cdot e^{j\omega_0 t}\)

Thus, the Inverse Fourier Transform of \(\delta(\omega - \omega_0)\) is \(\frac{e^{j\omega_0 t}}{2\pi}\).

Revision Table: Common Fourier Transform Pairs

Understanding common Fourier Transform pairs can be very helpful.

Time Domain \(x(t)\) Frequency Domain \(X(\omega)\) Notes
\(\delta(t)\) \(1\) Delta function at origin
\(1\) \(2\pi \delta(\omega)\) Constant function
\(e^{j\omega_0 t}\) \(2\pi \delta(\omega - \omega_0)\) Complex exponential
\(\cos(\omega_0 t)\) \(\pi [\delta(\omega - \omega_0) + \delta(\omega + \omega_0)]\) Cosine function
\(\sin(\omega_0 t)\) \(j\pi [\delta(\omega - \omega_0) - \delta(\omega + \omega_0)]\) Sine function

Additional Information: The Significance of \(\delta(\omega - \omega_0)\)

The frequency-domain function \(\delta(\omega - \omega_0)\) represents a signal that contains energy only at a single frequency \(\omega = \omega_0\). This is characteristic of a pure sinusoidal signal or, more generally, a complex exponential signal in the time domain. The result we found, \(\frac{e^{j\omega_0 t}}{2\pi}\), is indeed a complex exponential signal in the time domain, confirming this interpretation. The factor of \(2\pi\) is part of the definition of the Fourier Transform pair convention used.

Understanding the Fourier Transform and Inverse Fourier Transform of basic signals like the Dirac delta function and complex exponentials is fundamental in signal processing and system analysis.

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Important Questions from Standard Signals

  1. The following statements relate to sampling distributions. Choose the correct code for the statements being correct or incorrect.

    Statement I: Sampling distribution of mean is normally distributed irrespective of the type of population distribution and size of samples.

    Statement II : The standard deviation of the sampling distribution of mean is less than the standard deviation of the population distribution.

  2. The value of \(\mathop \smallint \limits_{ - \infty }^{ + \infty } {e^{ - t}}\delta \left( {2t - 2} \right)dt\), where \(\delta \left( t \right)\) is the Dirac delta function, is

  3. \(\mathop \smallint \nolimits_{ - 7}^2 \left( {{t^2} + {t^3} + 1} \right)\delta \left( {t - 3} \right)dt = \_\_\_\_\)
  4. Which of the following points CANNOT be observed about a unit impulse function if it is assumed in the form of a pulse?

  5. Which of the following is NOT one of the sampling techniques?

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