Anyone experienced with Qucs(simlar to pspice) ?

[magomago]

Hi all- Just some questions to ask for anyone who knows how to use Qucs

i'm writing my lab and figured it would be nice to maybe do some graphs of this stuff to show what ideal behavior would be like as i get to the experiment i did in class.

So I wanted to produce a corner frequency of about 100. So I'm using a resistor of 9.1kOhms and .18uF

That gives me... fc= 1/(2*pi*RC)= 97.1Hz

close enough for what I want for the high pass filter I wanted to design

So I go in qucs and generate ths, and then later notice the graph is weird. With an input Voltage of 1 to make my life easier, the corner frequency doesn't seem to be about 97.1Hz at all! It looks like it is about between 140 to 150Hz ; at a frequency of ~100Hz I'm seeing an output voltage of maybe .55V

Does anyone have an idea of what is going on? I graphed it and put a table and both show the same thing....

here is a screen shot

http://pics.bbzzdd.com/users/magomago/Screenshot.png


Could anyone explain to me what I'm missing or doing wrong to not have the right corner frequency?

I also noticed I have this problem when I try a low pass filter

 

[DainBramaged]

That screenshot is really cool. :thumbsup:

 

[magomago]

Originally posted by: DainBramaged
That screenshot is really cool. :thumbsup:

Where are all the EEs?

 

[DanFungus]

Not sure if this is your problem, but 9100Ohms != 91kOhms
You say you want 91k, but the screenshot shows 9.1k

 

[jiggahertz]

fc is the cutoff frequency or the frequency at which the output is half (-3 dB) of full scale input. Which is what your graph is showing.

 

[magomago]

Originally posted by: jiggahertz
fc is the cutoff frequency or the frequency at which the output is half (-3 dB) of full scale input. Which is what your graph is showing.

We were taught the Corner Frequency is the frequency at which the output is .707 or 1/sqrt(2)

whats going on

although i'm looking at my notes and I see that a loss of -3dB is supposed to be 1/sqrt(2) ~
halving the signal is -6dB

 

[jiggahertz]

Originally posted by: magomago

Originally posted by: jiggahertz
fc is the cutoff frequency or the frequency at which the output is half (-3 dB) of full scale input. Which is what your graph is showing.

We were taught the Corner Frequency is the frequency at which the output is .707 or 1/sqrt(2)

whats going on

although i'm looking at my notes and I see that a loss of -3dB is supposed to be 1/sqrt(2) ~
halving the signal is -6dB

Yeah, the difference is if you're measuring power or energy. Because you're talking about voltages here you would have to square the numbers to be in units of power. Halving a signal is -3dB... 10*log10(.5) = -3 dB

 

[magomago]

Originally posted by: jiggahertz

Originally posted by: magomago

Originally posted by: jiggahertz
fc is the cutoff frequency or the frequency at which the output is half (-3 dB) of full scale input. Which is what your graph is showing.

We were taught the Corner Frequency is the frequency at which the output is .707 or 1/sqrt(2)

whats going on

although i'm looking at my notes and I see that a loss of -3dB is supposed to be 1/sqrt(2) ~
halving the signal is -6dB

Yeah, the difference is if you're measuring power or energy. Because you're talking about voltages here you would have to square the numbers to be in units of power. Halving a signal is -3dB... 10*log10(.5) = -3 dB

Hmmm okay...I kind of get that When I ran the experiment on a breadboard I did get my the cutoff frequency to result as the frequency at which it was .707
I guess I'm just kind confused why running the same expiriment gives me a different result.

 

[jiggahertz]

Yeah, you're right. I'm not familiar with the software you're using. If you plot the gain manually for an RC circuit, wRC/sqrt(1+(wRC)^2), you'll see that the gain is around .7 for 100 Hz. You may want to put a probe around your voltage source to make sure it's really outputting 1V.

 

[magomago]

Originally posted by: jiggahertz
Yeah, you're right. I'm not familiar with the software you're using. If you plot the gain manually for an RC circuit, wRC/sqrt(1+(wRC)^2), you'll see that the gain is around .7 for 100 Hz. You may want to put a probe around your voltage source to make sure it's really outputting 1V.

pretty sure its outputing 1V - since its a high pass filter it hsould allow high frequencies to pass...and we can see that the high frequecies do allow 1V

i also put a voltage probe around the source and it is putting out 1V ><

awww ><

 

[SaturnX]

Hrmm.. that's strange, I just simulated your circuit in Spice, and I'm getting all the expected values.

For reference, here's the circuit I simulated:

* signal source
v.in 1 0 ac 1 pulse( -1 1 0 0 0 1m 2m )
*
* circuit under test
*
c.1 1 2 .18u ;*C1
r.1 2 0 9100 ;*R1
*
* analysis commands
.ac dec 20 10 1meg
.probe
.end


And here's a screen of the voltage gain, and output voltage, and you can see that at ~97Hz, we have a gain of -3db, and an output voltage of 702 mV, as expected.

Screenshot

Is there a chance that your program somehow simulates parasitics as well? From the probe or signal generator?

 

[SaturnX]

Just a follow up, I downloaded and installed Qucs, just to see what was going on...

I simulated the same circuit you had and output the values through both a probe, and simply using a wire label, and you end up with two different sets of results as seen in my SS:

QUCS

The violet plot is the output directly at the wire label vout, which is the output voltage you want. The red plot is the output of the probe.

What's happening is that the output of the probe is simply the real value of the complex (voltage and phase angle) voltage. That's why you're not getting what you expected to see, and that's why the table for Vout lists a voltage and phase angle.

For example:

0.139 < 82 = 0.0193 + j0.137

and so forth, and the probe simply shows the 0.0193.

Hope that helps!