# obtain the Fourier transform of a rectangular pulse (gate function) shown in figure

find the fourier transform of the rectangular pulse obtain the Fourier transform of a rectangular pulse (gate function) shown in figure ?

**Fourier Transform (FT)**

**Fourier Transform (FT) **fourier transform provides effective reversible link frequency domain and time domain representation of the signal. non-periodic signals can be represented with the help of fourier transform.

**Periodic Signals**

(i) non-periodic signals can be represented with the help of fourier transform.

(ii) fourier transform provides effective reversible link between frequency domain and time domain representation of the signal.

(iii) for non-periodic signals, t_{o} hence = 0. therefore, spacing between the spectral components becomes infinitesimal and hence the spectrum appears to be continuous.

**Fourier Transform**

The fourier transform of x(t) is defined as

x(0) = x(t)e^{-jt} dt

x(f) = x(t) e^{-j2ft} dt

here,x(t) = time domain representation of the signal and x(0) or x(f) = frequency domain representation of the signal.

is the frequency.

Sometimes x(0) is also written as x(jo).

Similarly, x(t) can be obtained from x by inverse fourier transform, i.e.,

x(t) = 1/2 x(0) e^{-jt} d = x(t) e^{-2ft} dt

A fourier transform pair is represented as

x(t) = x(0)

x(t) = x(f)

**Transform Existence of Fourier Dirchlet Conditions**

**(a) Single-valued property **x(t) must have only finite values at any instant over a finite time interval t.

**(d) Finite-discontinuities** x(t) should have at the most finite number of discontinuities over a finite time interval T.

**(c) Finite peaks **the signal x(t) should have finite number of maxima and minima over a finite time interval t.

**(d) Absolute integration**

|x(t)| dt < 0

these conditions are sufficient but not necessary for the signal to be fourier transformable.

**Properties of Fourier Transform**

**Linearity**

if x(t) = x(0)

and y(t) = y(0)

then, z(t) = ax(t) + by(t) ………. z(0) = a x(0) + by(0)

**Time Shift**

y(t) = x(t – t_{o}) y(0) = e^{-jto} x(0)

**Frequency Shift**

x(t) = e^{jot} x(t) = y(0) = x(0 – 0_{0})

**Time Scaling**

y(t) = x(at) y(0) = 1/|a| x(0/a)

**Frequency-differentiation**

-jt x(t) = d/d x(0)

**Time differentiation**

dx/dt x(t) j x

**Convolution**

z(t) = x(t)* y(t) = z(0) = x(0). y(0)

**Integration**

x d 1/j x + x (0)

**Modulation**

z(t) = x(t) y(t) = z

= 1/2 [x [0]* y(0)]

**Duality**

x(t) = 2 x(-0)

**Symmetry**

let x(t) by real signal and

x(0) = x_{r} (0) + jx_{1}(0)

then, x_{e} (t) = x_{r} (0)

x_{0} (t) = j x _{f}(0)

**Parseval’s Theorem or Rayleigh’s Theorem**

E = |x(t)^{2}| dt = 1/2 |x(0)^{2}| d0

= |x(f)^{2}| df

**Example 1. **obtain the Fourier transform of the signal e^{-at} u(t) and plot its magnitude and phase spectrum.

**Sol.** x(t) = e^{-at} u(t)

x(0) = e^{-at} u(t) e^{-jt} dt

x(0) = e^{-t(a + j)}u(t) dt

x(0) = e^{-t(at + j)} dt

x(0) = 1/a + j

e^{-at }u(t) 1/a + j

**to obtain magnitude and phase spectrum**

x(0) = 1/a + j

x(0) = 1/a + j x a – j/a – j

x(0) = a – j/a^{2} + 0^{2}

x(0) = a/a^{2} + 0^{2} – j 0/a^{2} + 0^{2}

|x (0)| = [a/a^{2} + 0^{2} ]^{2} + [0/a^{2} + 0^{2}]^{2}

= [a^{2} + 0^{2}/(a^{2} + 0^{2})^{2} = 1/a^{2} + 0^{2}

< x(0) = tan ^{-1} [0/a^{2} + 0^{2}/a/a^{2} + 0^{2}] = – tan^{-1} (0/a)

**Example 2.** obtain Fourier transforms of following functions

(i) x(t) = (t)

(ii) x(t) = 1

(iii) x(t) = sgn (t)

(iv) x(t) = u(t)

**Sol.** (i) x(t) = (t)

x (0) = x(t) e^{-jt} dt

x(0) = (t)e^{-jt} dt

by shifting property,

x(t) (t – t_{o}) dt = x(t_{o})

in above equation x(t) = e^{-jt} and t_{o} = 0

hence, x(0) = [e^{-jt}]_{to} = 1

thus, (t) = 1

(ii) x(t) = 1

here, |x(t)| dt = dt this means dirichlet condition is not satisfied. but its fourier transform can be calculated with the help of duality property. we know that

(t) = 1

x(0) = 1

(t) = x(0)

the duality property states that

x(t) = 2x(-0)

here, x(t) = 1, then x(-0) will be, (-0), then we have,

1 = 2(-0)

(0) = (-0)

1 = 2 (0)

(iii) x(t) = sgn (t)

the sgn (t) function can be expressed as

x(t) = 2u(t) – 1

Differentiating both the sides,

d/dt x(t) = 2d/dt u(t) = 2(t)

taking fourier transform on both sides,

f[(d/dt x(t)] = 2f [(t)]

j x(0) = 2

x(0) = 2/j

thus, sgn(t) 2/j

(iv) x(t) = u(t)

sgn (t) = 2u(t) – 1

2u (t) = 1 + (sgn) (t)n

f [2u(t)] = f[1] + f[(sgn)(t)]

2u(t) = 2 (0) + 2/j

u(t) = (0) + 1/j

**Example 3. obtain the Fourier transform of a rectangular pulse (gate function) shown in figure.**

**Sol.** x(0) = (1) e^{-jt} dt

x(0) = [e^{-jt}/-j]^{t}

x(0) = – 1/j [e^{jt} – e^{jt}]

x(0) = 2/0. [e^{jt} – e^{-jt}/2j]

x(0) = 2/0 sin (t)

we know that

sin c (0) = sin

x(0) = 2t sin

x(0) = 2t sin c(t/2)

thus,

[rectangular pulse amplitude, period 2t or rect (t/2t)] 2t sin c(t)

if A rect (t/t) AT sin c(t/2)

**Magnitude and phase plot**

since, x(0) is real, x(0) |= 2t sin c(t)

and <x (0) = 0

|x (0)| = 2t sin c(t) = 2 sin (t)

t = + n

t = + n

0 = + t + 2/t

using l’ hospital’s rule, lim 2 sin (t) = 2t

**Example 4.** find the inverse fourier transform of the rectangular spectrum shown in fignre.

**Sol.** x(t) = 1/2 [x (0) e^{jt} d

x(t) = 1/2 1.e^{jt} d = 1/2 [e^{jt}/jt]

x(t) = 1 sin t

x(t) sin (t)/(t)

x(t) = sin c(t) …………………(1)

- (1) goes to zero at t = + 2

using l’ hosplital rule lim 1/t sin t

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**Class 6 **

Hindi social science science maths English

**Class 7**

Hindi social science science maths English

**Class 8**

Hindi social science science maths English

**Class 9 **

Hindi social science science Maths English

**Class 10**

Hindi Social science science Maths English

**Class 11 **

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chemistry business studies biology accountancy

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