Analyze the frequency domain envelope of a square wave having variable frequency, duty cycle and rise time.
The frequency domain envelope rolls off at a rate of 6 dB per octave or 20 dB per decade. The initial roll off starts at a frequency defined at 1 / (π *
pulse width). Slowing the rise and fall time of the square wave will cause an additional 20 dB of attenuation to start at a frequency defined at 1 / (π *
rise time). All even numbered n-Th. terms will be in the null for a 50 % duty cycle, but slight changes away from 50 % causes the minimum level
and frequency of the null to shift dramatically. Note: The vertical blue marker is the location of the n-th user defined harmonic of the fundamental
frequency ( fo ).
Change the "Fundamental Frequency" variable to observe the start point of the 20 dB per decade roll. A higher fundamental causes the roll off to
start at a higher frequency. Set the frequency back to 10 MHz and adjust the duty cycle from 50 % to 51%. Notice how this decreases the width
and depth of the null for just a 1 % change in duty cycle and how it shifts the frequency of the null for a larger change. This is one reason why some
signal amplitudes change greatly from sample to sample. Set the duty cycle to 50 % again and adjust the "Rise Time" variable to some portion of
the pulse width. Notice how this change adds additional attenuation (20 dB / decade) at the frequency defined at 1 /(π * rise time). The null will be
less dominant and also changed in frequency. The pulse width variable, τ base, is equal to the period times the duty cycle. τ base assumes zero
rise and fall time. Keep the user defined rise time, τ r, less than τ base.
The example assumes a fall time equal to the rise time. Finally, try a very narrow pulse to simulate a unit impulse function δ . Notice how the
frequency domain envelope remains flat further out in frequency. This should be expected since the 20 dB roll off doesn't start until 1 /(π * pulse
width. Narrow pulse widths can be the result of reflections do to an improperly terminated transmission line.
---------------------------User defined Independent Variables------------------Fundamental Frequency (Hz)
Duty Cycle (%)
6
fo := 10⋅ 10
Vertical blue marker: n-th harmonic
Duty_Cycle := 50
Rise Time (sec)
− 100
τ r := 2.5⋅ 10
n := 3
--------------------------MathCad Analysis------------------Scan Range & Step Size
6
Calculate Period
6
7
f := 1⋅ 10 , 2⋅ 10 .. 500⋅ 10
Marker Frequency
T :=
1
τ base :=
fo
Magnitude Of Pulse
i := n⋅ fo
Marker
100
−8
⋅T
τ base = 5 × 10
−8
τ := τ base − τ r
⎛ 1 ⎞
⎟ Mark := n⋅ fo
⎝ i ⎠
Marker := ⎜
i
Duty_Cycle
Pulse Width Determined At The 50 %
Points Of The Rise And Fall Transition Times
A := 1
i := 1, 100.. 10000
Pulse Width For A Zero Rise And Fall Time)
The Amplitude Of The Spectral Components
sin(π ⋅ f ⋅ τ r)
τ sin( π ⋅ f ⋅ τ )
fDomain( f ) := 2⋅ A ⋅ ⋅
⋅
T
(π ⋅f⋅τ )
π ⋅f⋅τ r
Start Of 1 / (π * pulse width)
The 20dB/Decade Roll Off.
f1 :=
τ = 5 × 10
Start Of 1 / (π * rise time) The additional
20 dB/Decade Roll Off.
1
π ⋅τ
f2 :=
6
1
π ⋅τ r
99
f1 = 6.366× 10
f2 = 1.273× 10
Frequency
1
Amplitude
0.1
Marker i
fDomain( f )
0.01
1×10
1×10
−3
−4
6
1×10
7
1×10
8
1×10
1×10
9
10
1×10
Mark , f
Frequency
Figure 1. Frequency domain of a square wave with frequency, duty cycle and rise time as user defined variables.