Fourier Series (FS)

For periodic signal.

Signals can be constructed by summing sinusoids of different frequencies, amplitudes and phases.

Signals = A0+A1cos(2πf1t+θ1)+A2cos(2πf2t+θ2)+A3cos(2πf3t+θ3)+...A_{0}+A_{1}cos(2\pi f_{1}t + \theta_{1})+A_{2}cos(2\pi f_{2}t + \theta_{2})+A_{3}cos(2\pi f_{3}t + \theta_{3})+...

Periodic signals can be constructed by summing sinusoids that are harmonically related (Fouries Series).

Signals = A0+A1cos(2πf1t+θ1)+A2cos(2πf2t+θ2)+A3cos(2πf3t+θ3)+...A_{0}+A_{1}cos(2\pi f_{1}t + \theta_{1})+A_{2}cos(2\pi f_{2}t + \theta_{2})+A_{3}cos(2\pi f_{3}t + \theta_{3})+...

Where f2=2f1,f3=3f1,f4=4f1,f5=5f1f_{2}=2f_{1},f_{3}=3f_{1},f_{4}=4f_{1},f_{5}=5f_{1}

What is Fourier Series?

definition

It is a way to find out the mathematical representation of the periodic signals.

Periodic waveform = sum of sinusoids that are harmonically related

Harmonically related sinusoids: Sinusoids with frequencies that are integer multiples of a basic (fundamental) frequency

  • fk=kf0f_k = kf_0

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Fourier Series Expansion

definition

x(t)=k=Ckej(2πk0ft)x(t) = \sum_{k=-\infty}^{\infty}C_{k}e^{j(2\pi k_{0}ft)}

oror

x(t)=k=Ckej2πk0tTx(t) = \sum_{k=-\infty}^{\infty}C_{k}e^{j\frac{2\pi k_{0}t}{T}}

where

Ck=1T0T00x(t)ej2πktT0dtC_{k} = \frac{1}{T_{0}}\int_{T_{0}}^{0}x(t)e^{-j2\pi \frac{kt}{T_{0}}}dt

Find Fourier Series Coefficient

how to do

Steps:

  1. Find Frequency (Periodic signal, ff) and Period (TT) of x(t)x(t)

  2. Ck=fT2T2x(t)ej2πktT0dtC_{k} = f\int_{\frac{-T}{2}}^{\frac{T}{2}}x(t)e^{-j2\pi \frac{kt}{T_{0}}}dt

  3. Consider k on both terms

  4. One should be T2T21dt\int_{\frac{-T}{2}}^{\frac{T}{2}}1dt , while another is 0 because of sin(π)sin(\pi) (Use Euler’s Relation!)

  5. Draw the Magnitude and phase spectrum of x(t)x(t)

Note: eatdt=1aeat\int e^{at}dt = \frac{1}{a}e^{at}

examples

Inclass1

Steps :

  1. Find Frequency (Periodic signal, ff) and Period (TT) of x(t)x(t)

sin3πt=sin(2π32t)sin 3\pi t = sin(2\pi\cdot \frac{3}{2}t)

Since the Period is 1f\frac{1}{f} ,

T1=23T_1 = \frac{2}{3}

2cos(5π+0.25π)=2cos(2π52t+0.25π)2cos(5\pi + 0.25\pi) = 2cos(2\pi \cdot \frac{5}{2}t + 0.25\pi)

Since the Period is 1f\frac{1}{f} ,

T2=25T_2 = \frac{2}{5}

Period of sin3πt+2cos(5π+0.25π)sin 3\pi t + 2cos(5\pi + 0.25\pi) :

LCM of T1T_1 and T2T_2 = 2

  1. Ck=fT2T2x(t)ej2πktT0dtC_{k} = f\int_{\frac{-T}{2}}^{\frac{T}{2}}x(t)e^{-j2\pi \frac{kt}{T_{0}}}dt

Ck=1211x(t)ejπktdtC_{k} = \frac{1}{2}\int_{-1}^{1}x(t)e^{-j\pi kt}dt

=1211([sin(3πt)+2cos(5πt+0.25π)]ejπkt)dt= \frac{1}{2}\int_{-1}^{1}([sin(3\pi t) + 2cos(5\pi t + 0.25\pi)]e^{-j\pi kt})dt

  1. Apply Euler’s Formula on both terms

Term 1:

sin(3πt)sin(3\pi t)

=12j(ej3πtej3πt)= \frac{1}{2j}(e^{j3\pi t}-e^{-j3\pi t})

Term 2:

2cos(5πt+0.25π)2cos(5\pi t + 0.25 \pi)

=12(ej(5πt+0.25π)+ej(5πt+0.25π))= \frac{1}{2}(e^{j(5\pi t+0.25\pi)}+e^{-j(5\pi t+0.25\pi)})

=12(ej5πtej0.25π+ej5πt+ej0.25π)= \frac{1}{2}(e^{j5\pi t}\cdot e^{j0.25\pi}+e^{-j5\pi t}+e^{-j0.25\pi})

  1. If the term is a sin function, muiltply jj\frac{j}{j}

sin(3πt)=12j(ej3πtej3πt)jjsin(3\pi t) = \frac{1}{2j}(e^{j3\pi t}-e^{-j3\pi t})\cdot\frac{j}{j}

=j2(ej3πtej3πt)= \frac{-j}{2}(e^{j3\pi t}-e^{-j3\pi t})

  1. For sin term, draw a factor of 1-1 to make - become ++

Since 1=ejπ-1 = e^{j\pi}

sin(3πt)=12j(ej3πtej3πt)jj=j2(ej3πtej3πt)sin(3\pi t) = \frac{1}{2j}(e^{j3\pi t}-e^{-j3\pi t})\cdot\frac{j}{j}= \frac{-j}{2}(e^{j3\pi t}-e^{-j3\pi t})

=j2(ej3πt+(1)ej3πt)=j2(ej3πt+ejπej3πt)= \frac{-j}{2}(e^{j3\pi t}+(-1)\cdot e^{-j3\pi t}) = \frac{-j}{2}(e^{j3\pi t}+e^{j\pi}\cdot e^{-j3\pi t})

Since j=ej0.5π-j = e^{-j0.5\pi}

j2(ej3πt+ejπej3πt)\frac{-j}{2}(e^{j3\pi t}+e^{j\pi}\cdot e^{-j3\pi t})

=12ej0.5π(ej3πt+ejπej3πt)= \frac{1}{2}e^{-j0.5\pi}(e^{j3\pi t}+e^{j\pi}\cdot e^{-j3\pi t})

= 12(ej0.5πej3πt+ej0.5πejπej3πt)\frac{1}{2}(e^{-j0.5\pi}\cdot e^{j3\pi t}+e^{-j0.5\pi}\cdot e^{j\pi}\cdot e^{-j3\pi t})

  1. Now you have the polar form of 2 terms. Find the magnitude and the phase.

Inclass2

2(a)2(a)

Period is 3.

2(b)2(b)

Note x(t)=k=Ckej(2πk0t)x(t) = \sum_{k=-\infty}^{\infty}C_{k}e^{j(2\pi k_{0}t)} where Ck=1T0T00x(t)ej2πktT0dtC_{k} = \frac{1}{T_{0}}\int_{T_{0}}^{0}x(t)e^{-j2\pi \frac{kt}{T_{0}}}dt

Therefore

Ck=1T0[022ej2πktT0dt+23(1)ej2πktT0dt]C_{k} = \frac{1}{T_{0}}[\int_{0}^{2}2e^{-j2\pi \frac{kt}{T_{0}}}dt + \int_{2}^{3}(-1)e^{-j2\pi \frac{kt}{T_{0}}}dt]

A is 2, B is -1.

2(c)2(c)

since k=0k = 0

C0=1T0[022e0dt+23(1)e0dt]C_{0} = \frac{1}{T_{0}}[\int_{0}^{2}2e^{0}dt + \int_{2}^{3}(-1)e^{0}dt]

=13[022dt+23(1)dt]=13[(2220)+((1)3(1)2)]= \frac{1}{3}[\int_{0}^{2}2dt + \int_{2}^{3}(-1)dt] = \frac{1}{3}[(2\cdot2-2\cdot 0) +((-1)\cdot 3 - (-1)\cdot 2 )]

=13(4+(1))=1= \frac{1}{3}(4+(-1)) = 1

2(d)2(d)

Ck=13[022ej2πktT0dt+23(1)ej2πktT0dt]C_{k} = \frac{1}{3}[\int_{0}^{2}2e^{-j2\pi \frac{kt}{T_{0}}}dt + \int_{2}^{3}(-1)e^{-j2\pi \frac{kt}{T_{0}}}dt]

=23(02ej23πktdt)13(23ej23πktdt)= \frac{2}{3}(\int_{0}^{2}e^{-j\frac{2}{3}\pi kt} dt) - \frac{1}{3}(\int_{2}^{3}e^{-j\frac{2}{3}\pi kt}dt)

Note eatdt=1aeat\int e^{at}dt = \frac{1}{a}e^{at}

On one side:

2302ej23πktdt\frac{2}{3}\int_{0}^{2}e^{-j\frac{2}{3}\pi kt} dt

=23(1j23πk)[ej23πkt]02= \frac{2}{3}(\frac{1}{-j\frac{2}{3}\pi k})[e^{-j\frac{2}{3}\pi kt}]^2_0

=1jπk[e23πkt]02= \frac{1}{-j\pi k}[e^{\frac{2}{3}\pi kt}]^2_0 Now you found D is 1jπk\frac{1}{-j\pi k}

=1jπk[e23πk2e23πk0]= \frac{1}{-j\pi k}[e^{\frac{2}{3}\pi k2}- e^{\frac{2}{3}\pi k0}]

=1jπk[e43πk1]= \frac{1}{-j\pi k}[e^{\frac{4}{3}\pi k}- 1]

On the other side:

13(23ej23πktdt)\frac{1}{3}(\int_{2}^{3}e^{-j\frac{2}{3}\pi kt}dt)

=13(1j23πkt)[ej23πkt]23= \frac{1}{3}(\frac{1}{-j\frac{2}{3}\pi kt})[e^{-j\frac{2}{3}\pi kt}]^3_2

=1j2πk[ej23πkt]23= \frac{1}{-j2\pi k}[e^{-j\frac{2}{3}\pi kt}]^3_2 Now you found E is 1j2πk\frac{1}{-j2\pi k}

=1j2πk[ej23πk3ej23πk2]= \frac{1}{-j2\pi k} [e^{-j\frac{2}{3}\pi k3}- e^{-j\frac{2}{3}\pi k2}]

=1j2πk[ej2πkej43πk]= \frac{1}{-j2\pi k} [e^{-j2\pi k}- e^{-j\frac{4}{3}\pi k}]

Therefore:

=1jπk[e43πk1]1j2πk[ej2πkej43πk]= \frac{1}{-j\pi k}[e^{\frac{4}{3}\pi k}- 1] - \frac{1}{-j2\pi k} [e^{-j2\pi k}- e^{-j\frac{4}{3}\pi k}]

Inclass3

2(e)2(e)

Sub a 2 to make a common factor.

=1j2πk2[e43πk1]1j2πk[ej2πkej43πk]= \frac{1}{-j2\pi k}2[e^{\frac{4}{3}\pi k}- 1] - \frac{1}{-j2\pi k} [e^{-j2\pi k}- e^{-j\frac{4}{3}\pi k}]

=1j2πk[2e43πk2]+1j2πk[ej2πkej43πk]= \frac{1}{-j2\pi k}[2e^{\frac{4}{3}\pi k}- 2] + \frac{1}{j2\pi k} [e^{-j2\pi k}- e^{-j\frac{4}{3}\pi k}]

=1j2πk(ej2πkej43πk2e43πk+2)= \frac{1}{j2\pi k}(e^{-j2\pi k} - e^{-j\frac{4}{3}\pi k} - 2e^{\frac{4}{3}\pi k} + 2)

=1j2πk(ej2πk3ej43πk+2)= \frac{1}{j2\pi k}(e^{-j2\pi k} - 3e^{-j\frac{4}{3}\pi k} + 2)

Note ej2π=1e^{-j2\pi} = 1

=3j2πk(11ej43πk)= \frac{3}{j2\pi k}(1 - 1e^{-j\frac{4}{3}\pi k})

Make 1 to be e0e^0

=3j2πk(e0ej43πk)= \frac{3}{j2\pi k}(e^0 - e^{-j\frac{4}{3}\pi k})

=3j2πk(ej2πk3ej2πk3ej232πk)= \frac{3}{j2\pi k}(e^{j\frac{2\pi k}{3}}e^{-j\frac{2\pi k}{3}} - e^{-j\frac{2}{3}2\pi k})

Draw ej2πk3e^{-j\frac{2\pi k}{3}}

=3j2πkej2πk3(ej2πk3ej2πk3)= \frac{3}{j2\pi k}e^{-j\frac{2\pi k}{3}}(e^{j\frac{2\pi k}{3}} - e^{-j\frac{2\pi k}{3}})

=32πkej2πk3(ej2πk3ej2πk3)j= \frac{3}{2\pi k}e^{-j\frac{2\pi k}{3}}\frac{(e^{j\frac{2\pi k}{3}} - e^{-j\frac{2\pi k}{3}})}{j}

Use Euler’s Formula : sinθ=ejθejθ2jsin\theta = \cfrac{e^{j\theta}-e^{-j\theta}}{2j}

=3πkej2πk3(ej2πk3ej2πk3)2j= \frac{3}{\pi k}e^{-j\frac{2\pi k}{3}}\frac{(e^{j\frac{2\pi k}{3}} - e^{-j\frac{2\pi k}{3}})}{2j}

=3πkej2πk3sin(2πk3)= \frac{3}{\pi k}e^{-j\frac{2\pi k}{3}} sin(\frac{2\pi k}{3})

G is 3πk\frac{3}{\pi k} and F is 2πk3\frac{2\pi k}{3}

2(f)2(f)

Ck=3πkej2πk3sin(2πk3)C_k = \frac{3}{\pi k}e^{-j\frac{2\pi k}{3}} sin(\frac{2\pi k}{3})

C1=3πej2π3sin(2π3)C_1 = \frac{3}{\pi}e^{-j\frac{2\pi}{3}}sin(\frac{2\pi}{3})

=3πsin(2π3)ej2π3= \frac{3}{\pi}sin(\frac{2\pi}{3})e^{-j\frac{2\pi}{3}}

H=3πsin(2π3)=334πH = \frac{3}{\pi}sin(\frac{2\pi}{3}) = \frac{3\sqrt3}{4\pi}

θ=2π3\theta = \frac{2\pi}{3}

Magnitude and Phase Spectrum

Magnitude Spectrum

Note it must be even function. (Symmetric)

y axis :

Ck=12A\left |C_k\right | = \frac{1}{2}A

Note: In Polar form,

z=ρejθz = \rho e^{-j\theta}

where ρ\rho is the y axis of Magnitude spectrum.

In Rectangular form z=a+jbz = a + jb,

Ck=a2+b2\left |C_k\right | = \sqrt{a^2+b^2}

Phase Spectrum

Note it must be odd function. (Anti-symmetric)

y axis :

Ck=tan1imaginaryreal\angle C_k = tan^{-1}\frac{imaginary}{real}

Note: In Polar form,

z=ρejθz = \rho e^{-j\theta}

where θ\theta is the y axis of phase spectrum.

In Rectangular form z=a+jbz = a + jb,

Ck=tan1ba\angle C_k = tan^{-1}\frac{b}{a}

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Fourier Transform (FT)

For non-periodic signal (No period)

Describe a continuous-time, aperiodic signal in in terms of frequency content.

The frequency components are defined for all real values of the frequency.

What is Fourier Transform?

Definition

  • Fourier transform: aperiodic signal, TT\rightarrow \infty

When T is increased,

the density of the frequency components increases

the envelope of the magnitude of the scaled spectral components remains the same.

TT\rightarrow \infty, converge into a continuous spectrum

ckT=2kω0sinkω02=sinkω02kω02=sinckω02πc_kT = \frac{2}{k\omega_0}sin\frac{k\omega_0}{2}=\frac{sin\frac{k\omega_0}{2}}{\frac{k\omega_0}{2}}=sinc\frac{k\omega_0}{2\pi}

limTckT=sincω2π\lim_{T \rightarrow \infty} c_kT = sinc\frac{\omega}{2\pi}

sincλ=sinπλπλsinc\lambda = \frac{sin\pi\lambda}{\pi\lambda}

sinc0=1sinc 0 = 1

Fourier transform - Definition:

When carry out Fourier Transform, the Time domain changes from Time tt to Frequency ω\omega or changes from Frequency ω\omega to Time tt.

Forward transform, i.e., time to frequency:

X(ω)=x(t)ejωtdtX(\omega) = \int_{-\infty }^{\infty }x(t)e^{-j\omega t} dt

Inverse transform, i.e., frequency to time:

x(t)=12πX(ω)ejωtdωx(t) = \frac{1}{2\pi}\int_{-\infty }^{\infty }X(\omega)e^{j\omega t}d\omega

The function is basically the same with Fourier Series but T become a variable.

The transform pair notation:

x(t)X(ω)x(t) \leftrightarrow X(\omega)

Look at the Fourier Transform Table. (bottom)

Note ω=2πf=2π1T\omega = 2\pi f = 2\pi \frac{1}{T}

Functions that you need to know

Retangular pulse

Delta Pulse Function

Pulse Function / Rectangular Function

pτ(t)p_{\tau}(t)

Exponential function

Exponential signal:

x(t)=ebtu(t)x(t)=\mathrm{e}^{-b t} u(t)

  • Time-domain representation
  • If b>0, exp(-bt) -> 0

What is u(t)? - Unit Step Function

The Unit Step Function - u(t)

Materials support

More about The Laplace Transform

Fourier Transform Properties

FT Properties

FT duality

Difference between FS and FT

Difference betwene FS and FT

Fourier Transform Table


## How to use the FT table?

examples

(a)(a)

Note the formula ebtu(t)1jω+be^{-bt}u(t) \leftrightarrow \frac{1}{j\omega + b} from the table.

In this case, obviously, b=5b = 5.

x(t)=2e5tu(t)X(ω)=21jω+5=2jω+5x(t) = 2e^{-5t}u(t) \leftrightarrow X(\omega) = 2\cdot \frac{1}{j\omega + 5} = \frac{2}{j\omega + 5}

(b)(b)

Note the property time scaling. The formula is x(at)1aX(ωa)x(at) \leftrightarrow \frac{1}{a}X(\frac{\omega}{a})

v1(t)=x(2t)=2e5t2u(2t)V1(ω)=12X(ω2)=122jω2+5=2jω+10v_1(t) = x(2t) = 2e^{-5t\cdot 2}u(2t) \leftrightarrow V_1(\omega) = \frac{1}{2}X(\frac{\omega}{2}) = \frac{1}{2}\cdot \frac{2}{j\frac{\omega}{2}+5} = \frac{2}{j\omega+10}

(c)(c)

Note the property of both time scaling and time shift,

The formula is x(at)1aX(ωa)x(at) \leftrightarrow \frac{1}{a}X(\frac{\omega}{a}) and x(tc)X(ω)ejωcx(t-c) \leftrightarrow X(\omega)e^{-j\omega c}

v2(t)=x(2t1)=x(2(t12))V2(ω)=12X(ω2)ejω2=2ejω2jω+10v_2(t) = x(2t-1) = x(2(t-\frac{1}{2})) \leftrightarrow V_2(\omega) = \frac{1}{2}X(\frac{\omega}{2})e^{\frac{-j\omega}{2}} = \frac{2e^{\frac{-j\omega}{2}}}{j\omega+10}

(d)(d)

v3(t)=2e(5+j)tu(t)=2e5tu(t)ejt=x(t)ejtv_3(t) = 2e^{(-5+j)t}u(t) = 2e^{-5t}u(t)\cdot e^{jt} = x(t)e^{jt}

Note the property is mulitplication by a complex exponential. The formula is x(t)ejω0tX(ωω0)x(t)e^{j\omega_0 t} \leftrightarrow X(\omega-\omega_0)

Therefore obviously ω0=1\omega_0 = 1.

v3(t)=x(t)ejtV3(ω)=X(ωω0)=2j(ω1)+5v_3(t)= x(t)e^{jt} \leftrightarrow V_3(\omega) = X(\omega - \omega_0) = \frac{2}{j(\omega-1)+5}

Questions of p(t)​

(a)(a)

Period(T)=40=4Period (T) = 4 - 0 = 4

τ=4\tau = 4

x(t)=2p4(t2)x(t) = 2\cdot p_4(t-2)

(b)(b)

Note the formula : pτ(t)τsincτω2πp_\tau(t) \leftrightarrow \tau sinc\frac{\tau\omega}{2\pi}

x(t)=2p4(t2)X(ω)=24sinc4ω2π=8sinc2ωπx(t) = 2\cdot p_4(t-2) \leftrightarrow X(\omega) = 2\cdot 4sinc\frac{4\omega}{2\pi} = 8sinc\frac{2\omega}{\pi}

(c)(c)

y(t)=x(t)+(x(t))=x(t)x(t)y(t) = x(t) + (- x(-t)) = x(t)-x(-t)

(d)(d)

y(t)=x(t)x(t)Y(ω)=X(ω)X(ω)y(t) = x(t) - x(-t) \leftrightarrow Y(\omega) = X(\omega) - X(-\omega)

Some Tricky Question

Let’s say you have this question.

It is obvious, this function contains time shift and can be transformed into pτ(t)p_\tau (t) using the common form pairs.

Step 1: Check the available functions

Notice there is a e3jωe^{-3j\omega}.

x(tc)X(ω)ejωcx(t-c) \leftrightarrow X(\omega)e^{j\omega c}

Notice there is a sinsin function.

sincλ=sinπλπλsinc\lambda = \frac{sin\pi\lambda}{\pi\lambda}

pτ(t)τsincτω2πp_\tau(t) \leftrightarrow \tau sinc \frac{\tau\omega}{2\pi}

Step 2: Divide the parts

X(ω)=4sin(3ω)e3jωω=4sin(3ω)ωe3jωX(\omega) = \frac{4sin(3\omega)e^{-3j\omega}}{\omega} = 4 \frac{sin(3\omega)}{\omega} \cdot e^{-3j\omega}

Step 3: Find the λ\lambda variable

If you look closely, the formula sincλ=sinπλπλsinc\lambda = \frac{sin\pi\lambda}{\pi\lambda} requires a π\pi.

Therefore, plug a ππ\frac{\pi}{\pi} inside.

X(ω)=4sin(3ω)ωe3jω=4sin(3ωππ)ωe3jω=4sin(π3ωπ)ωe3jωX(\omega) = 4 \frac{sin(3\omega)}{\omega} \cdot e^{-3j\omega} = 4\frac{sin(3\omega\frac{\pi}{\pi})}{\omega}\cdot e^{-3j\omega} = 4\frac{sin(\pi\frac{3\omega}{\pi})}{\omega}\cdot e^{-3j\omega}

the λ=3ωπ\lambda = \frac{3\omega}{\pi}.

Step 4: Turn it to sincλsinc\lambda

To make the bottom part match the formula sincλ=sinπλπλsinc\lambda = \frac{sin\pi\lambda}{\pi\lambda}, plug another ππ\frac{\pi}{\pi}.

X(ω)=4sin(π3ωπ)ωe3jω=4sin(π3ωπ)ω3ππ3ππe3jωX(\omega)= 4\frac{sin(\pi\frac{3\omega}{\pi})}{\omega}\cdot e^{-3j\omega} = 4\frac{sin(\pi\frac{3\omega}{\pi})}{\omega}\cdot\frac{3\frac{\pi}{\pi}}{3\frac{\pi}{\pi}}\cdot e^{-3j\omega}

X(ω)=43ππsin(π3ωπ)ω3ππe3jω=43ππsin(π3ωπ)π3ωπe3jωX(\omega) = 4\cdot 3\frac{\pi}{\pi}\cdot \frac{sin(\pi\frac{3\omega}{\pi})}{\omega \cdot 3\frac{\pi}{\pi}}\cdot e^{-3j\omega} = 4\cdot 3\frac{\pi}{\pi}\cdot \frac{sin(\pi\frac{3\omega}{\pi})}{\pi \cdot \frac{3\omega}{\pi}}\cdot e^{-3j\omega}

Now you have the term sinπλπλ\frac{sin\pi\lambda}{\pi\lambda}. Make it sincλsinc\lambda .

X(ω)=12sinc3ωπe3jωX(\omega) = 12 \cdot sinc\frac{3\omega}{\pi}\cdot e^{-3j\omega}

Step 5: Find the τ\tau variable

To carry Fourier transform pτ(t)τsincτω2πp_\tau(t) \leftrightarrow \tau sinc \frac{\tau\omega}{2\pi}, you need to make π\pi become 2π2\pi.

X(ω)=12sinc6ω2πe3jωX(\omega) = 12 \cdot sinc\frac{6\omega}{2\pi}\cdot e^{-3j\omega}

Now you get the variable τ\tau. τ=6\tau = 6.

X(ω)=26sinc6ω2πe3jωX(\omega) = 2 \cdot 6sinc\frac{6\omega}{2\pi}\cdot e^{-3j\omega}

Step 6: Carry Fourier transform

X(t)=2p6(t3)X(ω)=26sinc6ω2πe3jωX(t) = 2\cdot p_6(t-3) \leftrightarrow X(\omega) = 2 \cdot 6sinc\frac{6\omega}{2\pi}\cdot e^{-3j\omega}