IronaSona
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- Homework Statement
- .
- Relevant Equations
- 1/1+2piefCR
So am trying to find the Frequency response of a RC LPF between frequencies 100Hz-100kHz ,but i don't know what formula to use .
Frequency Response Between 100Hz and 100kHz
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I don't think you mean $$1+2\pi f CR$$ as you did write, instead of $$1\over 1+2\pi f CR$$
but I could be mistaken.
Frequency response is ##V_{out}/V_{in} ## with a sine ##V_{in}##. Low pass means you are referring to a circuit like here
and the same recommendations apply.
(edit: the link was lost when I went off the page)
##\ ##
but I could be mistaken.
Frequency response is ##V_{out}/V_{in} ## with a sine ##V_{in}##. Low pass means you are referring to a circuit like here
and the same recommendations apply.
(edit: the link was lost when I went off the page)
##\ ##
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IronaSona
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Trying to find the frequency response of this circuitBvU said:
IronaSona
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no, got no ideaberkeman said:
boo
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Sometimes it's hard answering these questions because I have no idea at what level an answer is expected. Essentially the question relates to plotting the magnitude of the transfer function for a simple passive low pass filter. Typically this is done on a logarithmic scale but that wasn't specified. Also it is common to actually plot angular frequency w = 2*pi*f rather than f alone as the independent variable. The transfer function itself is a complex number with both magnitude and phase. In this case it is 1/(1 + 2*pi*f*R*C*j) (as I think you suggested) where j = (-1)^1/2. The magnitude of this transfer function is thus H =1/sqrt(1 + (2*pi*f*R*C)^2). If you like you can can make a simple linear plot of this of magnitude vs f or a typical bode plot of 20*log10(H) vs log10(w).
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Guessed as much. Holler if you don't recognize the equations in the link I gave in #2.IronaSona said:
Can you work with complex numbers ?
That's strange. Why should you have to do an exercise like this if you don't ?IronaSona said:
From where did you (wrongly) quote your relevant equation ?
Jody
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I would do as berkeman said and use KCL. What's the show stopper for you? I'll help you setup the problem and show you an example. To do the plot you'll want to learn about Bode plots.
In case you may be worried about the "direction" of the current arrows: I showed someone in another thread my approach on KCL and how the direction doesn't really matter. I hope it might be helpful for you when you use KCL in the future because I remember feeling very stuck about it too.
edit:
Hey! If you try the problem out may you please scan it or take a picture and share with us your work? Homework section requires you show work. Advice is above is pretty good so far, but we're pushing it with their rules you really ought to give us more of your work even if you come to the wrong conclusion. If we see your work, then it helps us point to us where things do not look right and that will help you for the exams ;)
In case you may be worried about the "direction" of the current arrows: I showed someone in another thread my approach on KCL and how the direction doesn't really matter. I hope it might be helpful for you when you use KCL in the future because I remember feeling very stuck about it too.
edit:
Hey! If you try the problem out may you please scan it or take a picture and share with us your work? Homework section requires you show work. Advice is above is pretty good so far, but we're pushing it with their rules you really ought to give us more of your work even if you come to the wrong conclusion. If we see your work, then it helps us point to us where things do not look right and that will help you for the exams ;)
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boo
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Seriously no need to invoke KCL. This is a simple voltage division problem. I'm sure that you know that if you have a voltage source in series with 2 resistors R and R2 that the voltage drop across R2 (your output voltage) is simply
Vin*R2/(R + R2)
In this case "R2" is the complex impedance of the capacitor namely 1/(j*2*pi*f*C). As you know you can use simple DC techniques to solve steady state sinusoidal problems merely by using complex impedances. Multiplying this out and dividing by Vin on both sides will give you the transfer function representing the ratio of Vout to Vin. Since you are plotting a thousand fold range in frequencies it would be wise to use a log scale. I suspect that you know all of this as you posted a correct version of the transfer function initially absent the factor "j" in the denominator and the disambiguating parenthesis which I assume was simply a typo.
Vin*R2/(R + R2)
In this case "R2" is the complex impedance of the capacitor namely 1/(j*2*pi*f*C). As you know you can use simple DC techniques to solve steady state sinusoidal problems merely by using complex impedances. Multiplying this out and dividing by Vin on both sides will give you the transfer function representing the ratio of Vout to Vin. Since you are plotting a thousand fold range in frequencies it would be wise to use a log scale. I suspect that you know all of this as you posted a correct version of the transfer function initially absent the factor "j" in the denominator and the disambiguating parenthesis which I assume was simply a typo.
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And it seems to me you are not interested either?IronaSona said: