I can understand the point that it is much easier to separate two frequencies that arre widely spaced. However, I don't see how triangular wave and input signal combined to create two frequencies mentioned above.If two frequencies are widely spaced (e.g. one is ten times the other) it is much easier to separate them than if they are close together.
I get it but do you consider the comparator is a modulator here?A simplified radio frequency modulator output is:
Vout=(1+cos(wm*t))*sin(wc*t)
where wm is the message signal angular frequency and wc is the carrier frequency angular frequency, and wm is always much less than wc. This reduces to:
Vout=sin(t*wm+t*wc)/2-sin(t*wm-t*wc)/2+sin(t*wc)
where we see the sum and difference frequencies and the carrier frequency. But this produces waves that are above and below zero at the carrier frequency.
In a PWM system like yours it's a little different because the half cycles are always above zero or below zero and follow the input wave shape, and this is a little difficult to think about it terms of frequencies but we can use a rough approximation to see what happens. The function we can use is:
Vout=(1+cos(wm*t))*(1+sin(wc*t))
Thank you alec_t and MrAl.
I can understand the point that it is much easier to separate two frequencies that arre widely spaced. However, I don't see how triangular wave and input signal combined to create two frequencies mentioned above.
Can I consider comparator is an RF modulator?
Input signal is a baseband and triangular is a carrier. I can draw of output of the comparator but cannot write equation to express them.
I get it but do you consider the comparator is a modulator here?
Could you elaborate about Vout function?
How can I get that?
Thanks Eric,hi anhnha,
Look at this plot of the comparator modulator, its very basic but it shows the modulated output.
The signal in is a50Hz sine wave and the triangular is approx 500Hz.
The modulated signal would be connected to a Res Cap demodulator.
E
EDIT: Changed the image to show the correct sine inp to mod output in same Phase.
I am wondering how can we design a low pass filter that allow only DC component and block all other components.PWM signal is nothing more than a square wave. Also the PWM average voltage is equal to:
Vavg = D *Vpeak Where D is a duty cycle D = Ton/(Ton + Toff). For example if we have 5V square wave with 50% duty cycle the average voltage is equal to 2.5V. To create average voltage from a PWM signal we use a low pass filter.
Yes, in your example input signal is a DC voltage, therefore we can calculate output signal and check it easily.For example if we have a 1V at the input, the PWM output duty cycle will be equal to 20% for 5V supply. If we change Vin to 2.5V the PWM duty cycle will change from 20% to 50%.
And if we back to 1V the PWM duty cycle will back to 20%.
And after the filter we get our original signal.
0.2*5V = 1V
0.5*5V = 2.5V
0.2*5V = 1V
So if input signal is a sine wave the PWM duty cycle will constantly changing in the rhythm of the input sine wave.
We will need ideal low pass filter, but we are unable to build such a filter even if we use ideal components.I am wondering how can we design a low pass filter that allow only DC component and block all other components.
I have checked low pass filter frequency response. To meet the above requirement, the Q (quality factor) of the filter has to be infinite. This means that all component are ideal without parasitic resistance.
Is my understanding correct?
But as you can see in my example I use a DC voltage but this DC voltage is changing with time. So we have a sort of a AC signal in time domain.Yes, in your example input signal is a DC voltage, therefore we can calculate output signal and check it easily.
But how about if input signal is a baseband signal with a range of frequencies. In this case, I failed to calculate output signal (after filter) and check it to make sure that output signal has the same shape with input signal.
Can you tell me why it cannot be built even with ideal components?We will need ideal low pass filter, but we are unable to build such a filter even if we use ideal components.
That seems right! I will have to read more.But as you can see in my example I use a DC voltage but this DC voltage is changing with time. So we have a sort of a AC signal in time domain.
But this dos not change the basic principle of how the circuit is working. What is a different between my DC signal that change with the time and sinusoidal signal? Or yours baseband signal ? In time domain not a big difference. Because we can take a "snapshots" of a input signal and treat it as a DC signal.
The input signal don't change the principles of operation of a circuit.
I have reread your post and now I see that output signal will constantly change in the rhythm of the input signal.
How about the case that duty cycle change constantly?
For example, in this picture D changes constantly, say D = 20% only for one cycle, next cycle D =30% and next cycle D =40% ... then how can I calculate output voltage?
I think we should compute average voltage over each cycle, right?
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