Math Problem Statement

Given x(t) shown below, determine its CTFT X(ω). DO NOT sketch X(ω). Hint: Decompose x(t) into the sum of basic signals. Then use CTFT pairs and properties to obtain X(ω).

Solution

To determine the Continuous-Time Fourier Transform (CTFT) X(ω)X(\omega) of the given function x(t)x(t), we will first decompose x(t)x(t) into simpler components. From the plot, x(t)x(t) consists of a series of rectangular pulses of varying heights and durations.

Decomposition of x(t)x(t)

  1. Identify segments:
    • From 4t<2-4 \leq t < -2: x(t)=0x(t) = 0
    • From 2t<1-2 \leq t < -1: x(t)=2x(t) = 2
    • From 1t<2-1 \leq t < 2: x(t)=3x(t) = 3
    • From 2t<52 \leq t < 5: x(t)=2x(t) = 2
    • From 5t<75 \leq t < 7: x(t)=0x(t) = 0

Represent x(t)x(t) using rectangular functions:

  • Rectangular function rect(tab)\text{rect}\left(\frac{t-a}{b}\right) is 1 for ta<b2|t-a| < \frac{b}{2} and 0 otherwise.

Using the step function, u(t)u(t), we can express x(t)x(t) as follows:

[ x(t) = 2 \left[u(t+2) - u(t+1)\right] + 3 \left[u(t+1) - u(t-2)\right] + 2 \left[u(t-2) - u(t-5)\right] ]

CTFT of each component:

  • CTFT of u(t)u(t): F{u(t)}=πδ(ω)+1jω\mathcal{F}\{u(t)\} = \pi \delta(\omega) + \frac{1}{j\omega}

  • CTFT of shifted u(ta)u(t - a): F{u(ta)}=πδ(ω)+1jωejωa\mathcal{F}\{u(t - a)\} = \pi \delta(\omega) + \frac{1}{j\omega} e^{-j\omega a}

  • CTFT of u(ta)u(tb)u(t-a) - u(t-b): F{u(ta)u(tb)}=1jω(ejωaejωb)\mathcal{F}\{u(t-a) - u(t-b)\} = \frac{1}{j\omega} \left(e^{-j\omega a} - e^{-j\omega b}\right)

Applying the CTFT to each term:

  1. For 2[u(t+2)u(t+1)]2 \left[u(t+2) - u(t+1)\right]: X1(ω)=21jω(ejω1ejω2)X_1(\omega) = 2 \cdot \frac{1}{j\omega} \left(e^{j\omega 1} - e^{j\omega 2}\right)

  2. For 3[u(t+1)u(t2)]3 \left[u(t+1) - u(t-2)\right]: X2(ω)=31jω(ejω2ejω1)X_2(\omega) = 3 \cdot \frac{1}{j\omega} \left(e^{j\omega 2} - e^{-j\omega 1}\right)

  3. For 2[u(t2)u(t5)]2 \left[u(t-2) - u(t-5)\right]: X3(ω)=21jω(ejω2ejω5)X_3(\omega) = 2 \cdot \frac{1}{j\omega} \left(e^{-j\omega 2} - e^{-j\omega 5}\right)

Final CTFT X(ω)X(\omega):

X(ω)=X1(ω)+X2(ω)+X3(ω)X(\omega) = X_1(\omega) + X_2(\omega) + X_3(\omega)

Substitute the results: X(ω)=2jω(ejω1ejω2)+3jω(ejω2ejω1)+2jω(ejω2ejω5)X(\omega) = \frac{2}{j\omega} \left(e^{j\omega 1} - e^{j\omega 2}\right) + \frac{3}{j\omega} \left(e^{j\omega 2} - e^{-j\omega 1}\right) + \frac{2}{j\omega} \left(e^{-j\omega 2} - e^{-j\omega 5}\right)

This is the CTFT of the given signal x(t)x(t).

Would you like a detailed step-by-step solution for each part, or do you have any specific questions?

Here are some related questions you might find helpful:

  1. How is the Fourier transform of a rectangular pulse derived?
  2. What properties of the Fourier transform can be used to simplify expressions?
  3. How does the shifting property of the Fourier transform work?
  4. What is the significance of the impulse function δ(ω)\delta(\omega) in the Fourier domain?
  5. How can we use symmetry properties of x(t)x(t) to simplify the CTFT?

Tip: Always check the limits and scaling when dealing with piecewise functions to avoid mistakes in the decomposition process.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Continuous-Time Fourier Transform (CTFT)
Signal Decomposition
Rectangular Function
Step Function

Formulas

CTFT of u(t): F{u(t)} = πδ(ω) + 1/(jω)
CTFT of shifted u(t - a): F{u(t - a)} = πδ(ω) + 1/(jω) * e^(-jωa)
CTFT of u(t-a) - u(t-b): F{u(t-a) - u(t-b)} = 1/(jω) * (e^(-jωa) - e^(-jωb))

Theorems

Fourier Transform Properties
Shifting Property
Scaling Property

Suitable Grade Level

University Level (Signal Processing/Engineering)