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Symmetry Properties of the Fourier Transform

Module by: Carlos E. Davila. E-mail the author

Summary: Looks at some of the consequences of conjugate symmetry of the Fourier transform of real signals.

When x(t)x(t) is real, the Fourier transform has conjugate symmetry, X(-jΩ)=X(jΩ)*X(-jΩ)=X(jΩ)*. It is not hard to see this:

X ( j Ω ) * = ∫ - ∞ ∞ x ( t ) e - j Ω t d t * = ∫ - ∞ ∞ x ( t ) e - j Ω t * d t = ∫ - ∞ ∞ x ( t ) e j Ω t d t = X ( - j Ω ) X ( j Ω ) * = ∫ - ∞ ∞ x ( t ) e - j Ω t d t * = ∫ - ∞ ∞ x ( t ) e - j Ω t * d t = ∫ - ∞ ∞ x ( t ) e j Ω t d t = X ( - j Ω )
(1)

where the second equality uses the definition of a Riemann integral as the limiting case of a summation, and the fact that the complex conjugate of a sum is equal to the sum of the complex conjugates. The third equality used the fact that the complex conjugate of a product is equal to the product of complex conjugates.

Letting X(jΩ)=a(jΩ)+jb(jΩ)X(jΩ)=a(jΩ)+jb(jΩ), it follows that

X ( - j Ω ) = a ( - j Ω ) + j b ( - j Ω ) X ( - j Ω ) = a ( - j Ω ) + j b ( - j Ω )
(2)

and

X ( j Ω ) * = a ( j Ω ) - j b ( j Ω ) X ( j Ω ) * = a ( j Ω ) - j b ( j Ω )
(3)

Equating Equation 2 and Equation 3 gives a(jΩ)=a(-jΩ)a(jΩ)=a(-jΩ) and b(-jΩ)=-b(jΩ)b(-jΩ)=-b(jΩ), which implies that the real and imaginary parts of X(jΩ)X(jΩ) have even and odd symmetry, respectively. A consequence of this is that X(jΩ)=X(jΩ)*=X(-jΩ)X(jΩ)=X(jΩ)*=X(-jΩ), that is, the magnitude of the Fourier transform has even symmetry. It can similarly be shown that the phase of the Fourier transform has odd symmetry.

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