The properties of the Fourier transform are summarized below. The properties of the Fourier expansion of periodic functions discussed above are special cases of those listed here. In the following, we assume and .

**Linearity**

**Time shift**

**Proof:**Let , i.e., , we have

**Frequency shift**

**Proof:**Let , i.e., , we have

**Time reversal**

**Proof:**

Replacing by , we get

**Even and Odd Signals and Spectra**If the signal is an even (or odd) function of time, its spectrum is an even (or odd) function of frequency:

and

**Proof:**If is even, then according to the time reversal property, we have

i.e., the spectrum is also even. Similarly, if is odd, we have

i.e., the spectrum is also odd.**Time and frequency scaling**

**Proof:**Let , i.e., , where is a scaling factor, we have

Note that when , time function is stretched, and is compressed; when , is compressed and is stretched. This is a general feature of Fourier transform, i.e., compressing one of the and will stretch the other and vice versa. In particular, when , is stretched to approach a constant, and is compressed with its value increased to approach an impulse; on the other hand, when , is compressed with its value increased to approach an impulse and is stretched to approach a constant.**Complex Conjugation**

**Proof:**Taking the complex conjugate of the inverse Fourier transform, we get

Replacing by we get the desired result:

We further consider two special cases:- If is real, then

i.e., the real part of the spectrum is even (with respect to frequency ), and the imaginary part is odd:

- If is imaginary, then

i.e., the real part of the spectrum is odd, and the imaginary part is even:

If the time signal is one of the four combinations shown in the table (real even, real odd, imaginary even, and imaginary odd), then its spectrum is given in the corresponding table entry:

if is real if is imaginary even, odd odd, even if is Even and even , even , even if is Odd and odd , odd , odd Note that if a real or imaginary part in the table is required to be both even and odd at the same time, it has to be zero.

These properties are summarized below:

1 real even , odd 2 real and even real and even 3 real and odd imaginary and odd 4 imaginary odd , even 5 imaginary and even imaginary and even 6 imaginary and odd real and odd As any signal can be expressed as the sum of its even and odd components, the first three items above indicate that the spectrum of the even part of a real signal is real and even, and the spectrum of the odd part of the signal is imaginary and odd.

- If is real, then
**Symmetry (or Duality)**

Or in a more symmetric form:

**Proof:**As , we have

Letting , we get

Interchanging and we get:

or

In particular, if the signal is even:

then we have

For example, the spectrum of an even square wave is a sinc function, and the spectrum of a sinc function is an even square wave.**Multiplication theorem**

**Proof:**

**Parseval's equation**In the special case when , the above becomes the Parseval's equation (Antoine Parseval 1799):

where

is the energy density function representing how the signal's energy is distributed along the frequency axes. The total energy contained in the signal is obtained by integrating over the entire frequency axes.The Parseval's equation indicates that the

*energy*or*information*contained in the signal is reserved, i.e., the signal is represented equivalently in either the time or frequency domain with no energy gained or lost.**Correlation**The

*cross-correlation*of two real signals and is defined as

Specially, when , the above becomes the*auto-correlation*of signal

Assuming , we have and according to multiplication theorem, can be written as

i.e.,

that is, the auto-correlation and the energy density function of a signal are a Fourier transform pair.**Convolution Theorems**The convolution theorem states that convolution in time domain corresponds to multiplication in frequency domain and vice versa:

**Proof of (a):**

**Proof of (b):**

**Time Derivative**

**Proof:**Differentiating the inverse Fourier transform with respect to we get:

Repeating this process we get

**Time Integration**First consider the Fourier transform of the following two signals:

According to the time derivative property above

we get

and

Why do the two different functions have the same transform?In general, any two function and with a constant difference have the same derivative , and therefore they have the same transform according the above method. This problem is obviously caused by the fact that the constant difference is lost in the derivative operation. To recover this constant difference in time domain, a delta function needs to be added in frequency domain. Specifically, as function does not have DC component, its transform does not contain a delta:

To find the transform of , consider

and

The added impulse term directly reflects the constant in time domain.Now we show that the Fourier transform of a time integration is

**Proof:**First consider the convolution of and :

Due to the convolution theorem, we have

**Frequency Derivative**

**Proof:**We differentiate the Fourier transform of with respect to to get

i.e.,

Multiplying both sides by , we get

Repeating this process we get