CTFT of the signum and unit-step functions

Find the CTFT of x(t) = sgn(t) and then use that result to find the CTFT of x(t) = u(t).

Question

CTFT of the signum and unit-step functions

Find the CTFT of x(t) = sgn(t) and then use that result to find the CTFT of x(t) = u(t).

Accepted Answer

Applying the integral formula directly we get

[latex]\mathrm{X}(f)=\int_{-\infty}^{\infty} \operatorname{sgn}(t) e^{-j 2 \pi f t} d t=-\int_{-\infty}^0 e^{-j 2 \pi f t} d t+\int_0^{\infty} e^{-j 2 \pi f t} \boldsymbol{d} t [/latex]

and these integrals do not converge. We can use a convergence factor to find the generalized CTFT. Let [latex]x_\sigma(t)=\operatorname{sgn}(t) e^{-\sigma|t|}[/latex] with σ> 0. Then

[latex]\begin{aligned} & \mathrm{X}_\sigma(f)=\int_{-\infty}^{\infty} \operatorname{sgn}(t) e^{-\sigma \mid t \mid} e^{-j 2 \pi f t} d t=-\int_{-\infty}^0 e^{(\sigma-j 2 \pi f) t} d t+\int_0^{\infty} e^{-(\sigma+j 2 \pi f) t} d t, \ & \mathrm{X}_\sigma(f)=-\left.\frac{e^{(\sigma-j 2 \pi f) t}}{\sigma-j 2 \pi f} ight|_{-\infty} ^0-\left.\frac{e^{-(\sigma+j 2 \pi f) t}}{\sigma+j 2 \pi f} ight|_0 ^{\infty}=-\frac{1}{\sigma-j 2 \pi f}+\frac{1}{\sigma+j 2 \pi f} \end{aligned}[/latex]

and

[latex]\mathrm{X}(f)=\underset{{\sigma \longrightarrow 0}}{ \lim } \mathrm{X}_\sigma(f)=1 / j \pi f[/latex]

or in the radian-frequency form
X( jω ) = 2/jω.

To find the CTFT of x(t) = u(t), we observe that
u(t) = (1/2)[sgn(t) + 1]

So the CTFT is

[latex]\begin{aligned} & \mathrm{U}(f)=\int_{-\infty}^{\infty}(1 / 2)[\operatorname{sgn}(t)+1] e^{-j 2 \pi f t} d t=(1 / 2)[\underbrace{\int_{-\infty}^{\infty} \operatorname{sgn}(t) e^{-j 2 \pi f t} d t}_{=\mathcal{F}(\operatorname{sgn}(t))=1 / j \pi f}+\underbrace{\int_{-\infty}^{\infty} e^{-j 2 \pi f t} d t}_{=\mathcal{F}(1)=\delta(f)}] \ & \mathrm{U}(f)=(1 / 2)[1 / j \pi f+\delta(f)]=1 / j 2 \pi f+(1 / 2) \delta(f) \end{aligned}[/latex]

or in the radian-frequency form
U( jω ) = 1/jω + πδ(ω) .

Question 6.5: CTFT of the signum and unit-step functions Find the CTFT of ......

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