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So I need to find what will be the average of a flipped square wave. I know that the high amplitude is 15V and the low amplitude is -9V and the average of the original signal is 2V. What will be the average of the inverted square wave?

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    \$\begingroup\$ Since this is homework you need to show your work. Otherwise your question is likely to be closed. Add a diagram. Hit the edit link below your question ... \$\endgroup\$ – Transistor Oct 15 '19 at 21:02
  • \$\begingroup\$ I don't have a diagram and it's not a homework question but I have tried calculating it like so: 15*x-9*y=2 then I set x to any number and calculated y from the expression and I then used this 15*y-9*x=z and just plugged in the x and y from the previous expression but that is not the correct ansver. \$\endgroup\$ – user229923 Oct 15 '19 at 21:09
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The waveform.

$$ V_{avg} = \frac {15x -9y}{x+y} = 2 \ \text V $$

Since \$ x + y = 1 \$

$$ \frac {15x -9(1-x)}{1} = 2 \ \text V $$ $$ 24x = 11 $$ $$ x = 11/24 $$

So the duty cycle is a little below 50%.

I call it flipped but what I mean is that the on and off time is flipped not the voltages.

Then set \$x = \frac {13}{24}\$ and \$y = \frac {11}{24}\$ and calculate the average using the first equation.

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  • \$\begingroup\$ Thanks a lot this is exactly right. \$\endgroup\$ – user229923 Oct 16 '19 at 6:17
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If you have a square wave (50% duty cycle) with a positive amplitude of 15 volts and a negative amplitude of -9 volts, then the average is (15 +(-9))/2 = 3 volts not 2 volts. Its not clear what you mean by flipping. I assume it means that the 15 volts becomes -15 volts and the -9 volts become +9 volts. In that case, the average changes sign and becomes -3 volts.

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  • \$\begingroup\$ I guess the duty cycle needs to be calculated from the given numbers. \$\endgroup\$ – Eugene Sh. Oct 15 '19 at 21:22
  • \$\begingroup\$ I'm sorry that I was not clear let me explain. I don't know what the duty cycle is I only know that the average of the signal is 2V so that means that the duty cycle can't be 50%. And by flipping it I mean invert the signal so that the on-time or high state becomes the new off time low time state. \$\endgroup\$ – user229923 Oct 15 '19 at 21:28
  • \$\begingroup\$ @TihomirRaicevic: when the signal is "flipped", is the high voltage still +15, and the low voltage -9? \$\endgroup\$ – Peter Bennett Oct 15 '19 at 21:31
  • \$\begingroup\$ I call it flipped but thats what I mean the on and off time is flipped not the voltages. \$\endgroup\$ – user229923 Oct 15 '19 at 21:37
  • \$\begingroup\$ You really need to add it to the question. So You are given the wave with DC=d, and want to find the average for the wave with DC=100%-d. Do you know how to calculate d? \$\endgroup\$ – Eugene Sh. Oct 15 '19 at 21:39
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Vpp = +15 - (-9) = 24Vpp from +15 to -9V.

15V-24/2 = 3V is the average Vdc Half swing =+/-12 V

The result is 3V+/-12 = 15V,-9V for a 24V swing.

Where does the input Vavg=2V come from?
It's not shown.

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A good way to think of it is as a weighted average:

D in 0..1 represents fraction of the time the signal is at V_h.

...then (1 - D) represents fraction of the time the signal is at V_l

...and V_avg = D*V_h + (1 - D)*V_l

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