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The amplitude of any harmonic can be calculated as 2/( p x n).

For the special case of a 50% duty-cycle ideal square wave, the even harmonics have an amplitude of zero.

The collection of sine-wave-frequency components and their amplitudes is called the spectrum.

In the case of the 50% duty-cycle square wave, the zeroth harmonic is 0.5 v. Its amplitude is equal to the average value of the signal. This is sometimes called the zeroth harmonic. To allow a DC offset, or a nonzero average value, the DC component is stored in the zero-frequency value.

Since sine waves are all centered about zero, any combination of sine waves can only describe waveforms in the time domain that are centered about zero. There is one other special frequency value, 0 Hz. If the transition-voltage range of the ideal square wave were to double to 0 v to 2 v, the amplitudes of each harmonic would double as well. It is important to note that the amplitudes of higher sine-wave-frequency components decrease with 1/f. We can even calculate the amplitude of the 1001 st harmonic.

The amplitude of the third harmonic is 0.21 v.

n = the harmonic number, only odd allowedДЇor example, an ideal square wave with 50% duty-cycle and 0 v to 1 v transition has a first harmonic amplitude of 0.63 v.

A n = the amplitude of the n th harmonic.

The amplitudes, A n, of the odd harmonics are given by: Equation 2-3 It is only odd harmonics that have values. The amplitudes of all the even harmonics (e.g., 2 GHz, 4 GHz, 6 GHz) are all zero. Luckily, it is possible to calculate the DFT exactly for this special case of an ideal square wave. But what are the amplitudes of each sine wave? The only way to determine this is to perform a DFT on the ideal square wave. We expect to see components at f = 1 GHz, 2 GHz, 3 GHz, and so on. If the ideal square-wave repeat frequency is 1 GHz, the sine-wave frequency values in its spectrum will be multiples of 1 GHz.

Figure 2-6 Time and frequency domain views of an ideal square wave.

Bipolar square wave harmonics