Signals & System Lecture 14

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Fourier Transform Overview

• Objective: In this chapter, we analyze continuous time signals to obtain their frequency spectrum.
  • The spectrum consists of two graphs:
    1. Magnitude Response
    2. Phase Response

Key Definitions

1. Fourier Transform:

   • For a given function $x(t)$, the Fourier Transform is given by: $$F\{x(t)\}=X(j\omega )={\int }_{-\infty }^{\infty }x(t)e^{-j\omega t}dt$$
     • There is no need for the Region of Convergence (ROC) in this case.

2. Inverse Fourier Transform:

   • For a given function $X(\omega )$, the inverse Fourier Transform is defined as: $$x(t)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }X(\omega )e^{j\omega t}d\omega$$
     • Again, there is no need for ROC here.

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Examples of Fourier Transforms

1. Case 1:

   • Given $x(t)=e^{-at}u(t)$, where $u(t)$ is the unit step function: $${\int }_{-\infty }^{\infty }{e}^{-at}u(t)dt={\int }_0^{\infty }{e}^{-at}dt=\frac{1}{a}$$
     • Fourier Transform of $x(t)$ is given by: $$X(j\omega )=\frac{1}{a+j\omega }$$

2. Case 2:

   • Given $x(t)=e^{at}u(-t)$, where $u(-t)$ is the reverse unit step function: $${\int }_{-\infty }^{\infty }{e}^{at}u(-t)e^{-j\omega t}dt=\frac{1}{a-j\omega }$$

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Combining the Two Cases

• For $x(t)=u(t)e^{-at}+e^{at}u(-t)$, the Fourier Transform is found by adding the results of the above two cases: $$X(j\omega )=\frac{1}{a+j\omega }+\frac{1}{a-j\omega }$$ Simplifying, we get: $$X(j\omega )=\frac{2a}{a^2+{\omega }^2}$$

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Frequency Response

• For a given signal $x(t)=e^{-at}u(t)$, the frequency response is obtained as follows: $$X(\omega )=\frac{1}{a+j\omega }$$
  • Magnitude Response: $$|X(\omega )|=\frac{1}{\sqrt{a^2+{\omega }^2}}$$
  • Phase Response: $$\angle X(\omega )=-{\tan }^{-1}\left(\frac{\omega }{a}\right)$$

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Phase and Magnitude Response of $X(\omega )$

• Phase Response: The phase response is antisymmetric.
• Magnitude Response: The magnitude response is symmetric.

Phase and Magnitude Calculations:

For the signal $X(\omega )=\frac{1}{a+j\omega }$, we calculate the magnitude and phase for specific values of $\omega$.

$\omega$	$A$	$X(\omega )$	$	$\angle X(\omega )$
$0$	$\frac{1}{a}$	${\tan }^{-1}(0)=0^{\circ }$
$a$	$\frac{1}{\sqrt{2a}}$	${\tan }^{-1}(1)=45^{\circ }$
$2a$	$\frac{1}{\sqrt{5a}}$	${\tan }^{-1}(2)=63.43^{\circ }$
$3a$	$\frac{1}{\sqrt{10a}}$	${\tan }^{-1}(3)=71.56^{\circ }$
$\infty$	$0$	${\tan }^{-1}(\infty )=90^{\circ }$

Graphical Representation:

• Magnitude Response: $|X(\omega )|$

  • The graph illustrates how the magnitude decreases as $\omega$ increases.
  • It starts at $\frac{1}{a}$ for $\omega =0$ and gradually decreases towards 0 as $\omega \to \infty$.
  • Key points on the graph:
    • At $\omega =a$: $|X(\omega )|=\frac{1}{\sqrt{2a}}$
    • At $\omega =2a$: $|X(\omega )|=\frac{1}{\sqrt{5a}}$

• Phase Response: $\angle X(\omega )$

  • The phase angle $\angle X(\omega )$ shifts gradually from $0^{\circ }$ at $\omega =0$ to $-90^{\circ }$ as $\omega \to \infty$.

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Constant Signal Fourier Transform

• The Fourier transform of a constant signal $x(t)=A$ over all time $-\infty <t<\infty$: $$X(\omega )=2\pi A\delta (\omega )$$
  • This represents an impulse function at $\omega =0$, indicating that a constant signal has all its frequency concentrated at zero frequency.

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Fourier Transform of the Delta Function

• Non-Integrable Nature of the Delta Function: Since the delta function, $\delta (\omega )$, is not integrable in the conventional sense, we apply the Fourier Transform properties. The Fourier Transform of $\delta (\omega )$ is given as:

  $$F\{\delta (\omega )\}=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\delta (\omega )e^{j\omega t}d\omega$$
  • At $\omega =0$, the exponential term becomes $e^0=1$.

• Therefore, we have:

  $$F\{\delta (\omega )\}=\frac{1}{2\pi }\cdot 1$$

• Inverse Fourier Transform of $\delta (\omega )$:

  $$F^{-1}\{\delta (\omega )\}=\frac{1}{2\pi }$$

  This property shows that the inverse Fourier Transform of the delta function is a constant factor of $\frac{1}{2\pi }$.

• Multiplying by $2\pi$ on Both Sides:

  $$2\pi \cdot F^{-1}\{\delta (\omega )\}=1$$

• Fourier Transform of a Constant Signal: Taking the Fourier Transform of both sides for a constant $A$:

  $$2\pi F\{A\}=F\{1\}$$
  • The result gives us:
  $$F\{A\}=2\pi A\delta (\omega )$$

This shows that the Fourier Transform of a constant signal results in a scaled delta function in the frequency domain.

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