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# The Periodic Pulse Signal

Module by: Don Johnson. E-mail the author

Summary: This module examines the spectrum of the periodic pulse signal.

## Example

Let's calculate the spectrum of the periodic pulse signal shown here. The pulse width is Δ Δ , the period T T , and the amplitude A A. The complex Fourier spectrum of this signal is given by c k =AT0Δe(i)2πkΔTdt=(Ai2πk)(e(i)2πkΔT1) c k A T t 0 Δ 2 k Δ T A 2 k 2 k Δ T 1 At this point, simplifying this expression requires knowing an interesting property.

1e(iθ)=eiθ2(eiθ2eiθ2)=eiθ22isinθ2 1 θ θ 2 θ 2 θ 2 θ 2 2 θ 2
(1)
Armed with this result, we can simply express the Fourier series coefficients for our pulse sequence.
c k =Ae(i)πkΔTsinπkΔTπk c k A k Δ T k Δ T k
(2)
Because this signal is real-valued, we find that the coefficients do indeed have conjugate symmetry: c k = c k ¯ c k c k . The periodic pulse signal has neither even nor odd symmetry; consequently, no additional symmetry exists in the spectrum. Because the spectrum is complex valued, to plot it we need to calculate its magnitude and phase.
| c k |=A|sinπkΔTπk| c k A k Δ T k
(3)
c k =πkΔT+πnegsinπkΔTπksignk c k k Δ T neg k Δ T k sign k
(4)

### Exercise 1

What is the value of c 0 c 0 ? Recalling that this spectral coefficient corresponds to the signal's average value, does your answer make sense?

#### Solution

c 0 =AΔT c 0 A Δ T . This quantity clearly corresponds to the periodic pulse signal's average value.

The somewhat complicated expression for the phase results because the sine term can be negative; magnitudes must be positive, leaving the occasional negative values to be accounted for as a phase shift of π . The neg· neg · equals -1 if its argument is negative and zero otherwise.

Also note the presence of a linear phase term (the first term in c k c k is proportional to frequency kT k T . Comparing this term with that predicted from delaying a signal, a delay of Δ2 Δ 2 is present in our signal. Advancing the signal by this amount centers the pulse about the origin, leaving an even signal, which in turn means that its spectrum is real-valued. Thus, our calculated spectrum is consistent with the properties of the Fourier spectrum.

## Exercise 2

Investigate the half-wave rectified sine wave's spectrum for a linear phase term. If one is present, show how to delay or advance the signal to create an even or odd signal. If one is not present, convince yourself that no delay would yield a signal having even or odd symmetry.

### Solution

A half-wave rectified sine wave occurs when Δ=T2 Δ T 2 . Thus, c k ¯=(πk2)πnegsinπk2πk c k k 2 neg k 2 k , which corresponds to a linear function of k k. The linear phase can be removed by delaying the signal by 14 1 4 of a period.

The phase plot shown in Figure 2 requires some explanation as it does not seem to agree with what Equation 4 suggests. There, the phase has a linear component, with a jump of π every time the sinusoidal term changes sign. We must realize that any integer multiple of 2π 2 can be added to a phase at each frequency without affecting the value of the complex spectrum. We see that at frequency index 4 the phase is nearly π The phase at index 5 is undefined because the magnitude is zero in this example. At index 6, the formula suggests that the phase of the linear term should be less than (more negative) than π . In addition, we expect a shift of π in the phase between indices 4 and 6. Thus, the phase value predicted by the formula is a little less than (2π) 2 . Because we can add 2π 2 without affecting the value of the spectrum at index 6, the result is a slightly negative number as shown. Thus, the formula and the plot do agree. In phase calculations like those made in MATLAB, values are usually confined to the range π π by adding some (possibly negative) multiple of 2π 2 to each phase value.

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