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Given information

I have the following circuit, and frequency graph. The capacitor is connected to ground at the bottom. Unfortunately the graph is in dutch, but it won't matter much. Just understand "Passief laagdoorlaatfilter" means passive low pass filter and \$U_{uit}\$ means \$U_{out}\$.

enter image description here

I understand the following

  • Hz means the amount of periods per second. The frequency spectrum is on the x-axis.
  • The low pass filter attenuates/mitigates the higher frequency's, so without the filter the signal in the graph would be constant (like the dashes indicate)
  • \$f_k\$ is the cutoff frequency

My confusion/question

I am having difficulty understanding how to read this graph. What's happening here? My teacher calls the thing on the y-axis the "transfer (function)". I don't understand what transfer means, and when I try to do some research on it, the pages are filled with way too difficult math and terminology etc. for me to handle.

Things that confuse me:

  • What does transfer or \$V_{out}/V_{in}\$ mean, and why is it on the y-axis? Because I don't understand this, I can't understand the graph either...
  • Why the y-axis also has to do with decibel. I've done some research on decibel, and apparently it's a ratio between an output and input voltage. It's still very confusing to me... I'm going to do some more research on this as soon as I posted this question.
  • How can an analog signal be graphed like this? An analog signal has only one frequency unless it changes periods. So if the frequency of \$V_{in}\$ would be 100 Hz, wouldn't there just be a single discrete value when frequency = 100? But instead this graph looks like a continous curve. So what is it showing the frequency of?

This is very challenging for someone who knows very little about physics...

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    \$\begingroup\$ I don't understand what transfer means, and when I try to do some research on it, the pages are filled with way too difficult math and terminology etc. for me to handle. That's going to be a problem then because "transfer function" is part of the mathematical model of what the circuit does. If the teacher wants you to understand "transfer function", then the teacher wants you to understand the math. \$\endgroup\$ – Solomon Slow Apr 22 '15 at 18:12
  • \$\begingroup\$ @jameslarge I'm sure he doesn't want me doesnt want me to understand that. I'm just a first year robotics student, and hardly know the basics of calculus. He just wants me to understand how the filters work. \$\endgroup\$ – user1534664 Apr 22 '15 at 18:13
  • \$\begingroup\$ How they work? or what they do? @AV23 's answer is pretty good in the what-they-do-department. But if you want to know the theory behind them (i.e., how they work), I'm afraid it's worse than "calculus": You're going to have to use complex analysis. en.wikipedia.org/wiki/Complex_analysis If you're going into any kind of Science or Engineering career, you're going to need to know it anyway, so... \$\endgroup\$ – Solomon Slow Apr 22 '15 at 18:21
  • \$\begingroup\$ @james I meant what they do yeah :P Good catch \$\endgroup\$ – user1534664 Apr 22 '15 at 18:21
  • \$\begingroup\$ I don't think you actually need any calculus to understand transfer functions (phew!), but it certainly comes in handy. \$\endgroup\$ – Greg d'Eon Apr 22 '15 at 19:04
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What does transfer or Vout/Vin mean, and why is it on the y-axis? Because I don't understand this, I can't understand the graph either...

You could make the input signal \$V_{in}\$ anything that you want and solve for \$V_{out}\$. That's fine - it's just a bit of math. However, if you wanted to change \$V_{in}\$ to something else, then you'd have to re-calculate everything. You haven't learned anything; you've just solved the circuit for one signal.

The transfer function of a circuit tells you what \$V_{out}\$ is. It depends on the frequency of your input. If the transfer function is \$H(f)\$, then we can write $$ V_{out} = V_{in} \cdot H(f) $$ and now, if we calculate \$H(f)\$, we know what \$V_{out}\$ looks like for every input frequency.

Why the y-axis also has to do with decibel. I've done some research on decibel, and apparently it's a ratio between an output and input voltage. It's still very confusing to me... I'm going to do some more research on this as soon as I posted this question.

Circuits often deal with a lot of orders of magnitude. Sometimes, you're interested in a signal that's 1 V in amplitude; sometimes, you're looking at 1 uV. That's like multiplying by 0.000001. Yuck - do you like counting zeros?

Decibels are a way of looking at orders of magnitude. Instead of multiplying by 10, you add +20 dB. Now, the difference between 1 V and 1 uV is -120 dB - much easier to read and understand.

How can an analog signal be graphed like this? An analog signal has only one frequency unless it changes periods. So if the frequency of Vin would be 100 Hz, wouldn't there just be a single discrete value when frequency = 100? But instead this graph looks like a continous curve. So what is it showing the frequency of?

This is showing what the output is for any frequency. Let me pretend for a minute that \$f_k\$ = 1000 Hz. Then, look at your graph.

  • 100 Hz is to the left of \$f_k\$. That means when you put a 100 Hz signal in, you get out the exact same signal.
  • 10 kHz is to the right of \$f_k\$. Now, the output has dropped quite a bit: down to -40 dB. That means the output is 100 times smaller than the input.

Continue this for any other frequency and you get the continuous transfer function.


Replying to some comments,

Why does the influence of the capacitor get smaller as the frequency gets higher?

The charge on a capacitor is $$ Q = CV $$ so the current through the capacitor is $$ I = C\frac{dV}{dt} $$ Think about what happens when you change the input frequency.

  • At low frequencies, a sine wave doesn't change very fast, so \$\frac{dV}{dt}\$ is small, and the capacitor doesn't let much current through it.
  • At high frequencies, \$\frac{dV}{dt}\$ is big, so \$I\$ can be big, too. Now, the capacitor lets a lot of current through, and the output voltage gets lower (the \$V = IR\$ drop across the resistor gets big).
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  • \$\begingroup\$ You're very good at explaining things simply! I understand how to read and interpret the graph now. I have one last confusion, that I already mentioned on another comment... How is \$f_{out}\$ dependent on the frequency of \$f_{in}\$? I still don't get this last detail I'm afraid. I understand it formula wise, since \$H\$ is a function of \$f\$, but not when I think about the signal itsself, moving about in the circuit. \$\endgroup\$ – user1534664 Apr 22 '15 at 19:08
  • \$\begingroup\$ Thanks! The cool thing is that f_out = f_in. You've got a good point that comes up in phasor analysis: when you're looking at circuits in AC (like here), you just need to keep track of the amplitude and phase of all of your voltages - all of the frequencies at the end will be the same as what you put in. \$\endgroup\$ – Greg d'Eon Apr 22 '15 at 19:16
  • \$\begingroup\$ I think this confusion might have to do with this: Why does the influence of the capacitor get smaller as the frequency gets higher? The frequency is just how fast the period passes, but that doesn't affect the amplitude right? \$\endgroup\$ – user1534664 Apr 22 '15 at 19:16
  • \$\begingroup\$ Actually, the capacitor does change the output amplitude - I'll edit some info about that into my answer. \$\endgroup\$ – Greg d'Eon Apr 22 '15 at 19:22
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    \$\begingroup\$ @user1534664 Do you see how the U_in and U_out markers are arrows? They show you that the voltages are measured from ground to those two nodes. \$\endgroup\$ – Greg d'Eon Apr 22 '15 at 21:49
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This is a common plot type to show the frequency response of a system with a single input and a single output. Audio amplifiers, for example, are often characterized in this way.

Since the purpose of the graph is to show what the system does as a function of frequency, it naturally has frequency on the X axis. However, what is not obvious unless you are used to looking at these graphs, is that the frequency is plotted logarithmically on the X axis. Each tick on the X axis represents a frequency multiple, not a frequency increment. It is irresponsible to not clearly say so, but unfortunately that happens too often. For example, each tick on the X axis may be one octave (multiple of 2) or one decade (multiple of 10), or some other number. At least in this case the slope of the dotted line is explicitly labeled as -6 dB/octave.

The Y axis is the gain of the system. However, just like with the X axis in this example, the gain is also shown logarithmically. This time though, that is clearly indicated with the legend "dB", meaning "deci-Bels". dB is used to express power ratios. Specifically dB = 10*Log10(Power2/Power1). Since power is proportional to the square of the voltage, we often use dB to express voltages ratios: dB = 20*Log10(Voltage2/Voltage1).

Note that 0 dB always specifies a ratio of 1:1, which means the output voltage just follows the input voltage. In this case the system is a single-pole R-C filter. At low frequencies it leaves the signal alone, and at high frequencies it attenuates the signal ever more with frequency. That's what R-C low pass filters do, and also what the graph is showing you.

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  • The ratio \$V_{out}/V_{in}\$ may be interpreted as the fraction of the input voltage \$V_{in}\$ that the low pass filter allows to 'pass' to \$V_{out}\$. This is after all the function of a low (or any other) pass filter.
  • The decibel (dB) is (in this case) a logarithmic unit of the ratio of voltages. The value in dB is given by \$20\log_{10}(V_{out}/V_{in})\$. Logarithmic scales such as this are convenient as they allow you to express large changes in a small region of a graph.
  • Well, in this case, we are effectively looking at the decibel value of the voltage ratio for different analog signals of different frequencies, and plotting our observations on the graph. There is no time variation of a single analog signal, only different cases of different frequencies. Each case contributes one point to the graph, like you mentioned, and together, they form a continuous curve across all frequencies.
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  • \$\begingroup\$ Thank you. Does the y-axis represent $V_{out}/V_{in}$ or the logarithm formula you mentioned? Also, the values that the filter allows to pass are all the values below the curve right? \$\endgroup\$ – user1534664 Apr 22 '15 at 17:55
  • \$\begingroup\$ The y-axis is marked 'dB' and is therefore in the logarithmic scale. The filter works by reducing an input voltage to a fraction of itself - for each input voltage, there is a unique output voltage. The ratio must always lie on the curve, not below or above. In a low pass filter, this fraction is 1 (i.e. 0 db) for zero frequency, and rapidly decreases when moving to larger frequencies - the 'transfer' is less at higher frequencies. \$\endgroup\$ – AV23 Apr 22 '15 at 18:01
  • \$\begingroup\$ Thanks! I finally understand what the ratio does now. I'm slowly getting there... In your answer, what did you mean by "different analog signals"? I can only see two: the voltage input and voltage output. In other words, I don't quite understand how the analog signals is changing with respect to the different frequencys. \$\endgroup\$ – user1534664 Apr 22 '15 at 18:06
  • \$\begingroup\$ We supply an input voltage at some frequency, look at the output voltage, and mark the corresponding point on the graph. Then we switch off the input and completely remove all its traces on the circuit. Now, we supply input at a different frequency to get another point on the graph and so on. The output voltage for the same input signal will in general also depend on the frequency of the input. \$\endgroup\$ – AV23 Apr 22 '15 at 18:12
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    \$\begingroup\$ Notice that the curve in the graph approaches an asymptote that looks like a straight line when plotted on the logarithmic scale. That's the other reason why decibels are used. \$\endgroup\$ – Solomon Slow Apr 22 '15 at 18:16
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An analog signal has only one frequency unless it changes periods.

That is absolutely not true, except for the special case of a sine wave signal. See Fourier series.

Take a square wave, for instance, of period p. Its fundamental frequency (which is what you're thinking of) is f = 1/p, but it also has frequency components called harmonics of decreasing amplitudes at frequencies 3f, 5f, 7f...

The lowpass filter graph shows that the gain (Vout / Vin) decreases as the frequency increases, so if you were to pass a square wave through the filter, the fundamental would remain strong, but the upper harmonics would be reduced in strength; as it turns out this tends to "round off" the corners of a square wave and make it more similar to a sine wave.

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  • \$\begingroup\$ I have very basic knowledge about electronics and signals, so I appreciate your feedback :) I don't have time to research your answer right now, but I'll look into harmonics tomorrow! \$\endgroup\$ – user1534664 Apr 22 '15 at 20:01
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    \$\begingroup\$ I added a link to get you started. Fourier analysis is pretty central to audio and RF signal processing. \$\endgroup\$ – Russell Borogove Apr 22 '15 at 20:14
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Transfer(function) means the relationship between what you put into it to what you get out of it.

You for instance eat food and well, poop from time to time.

input -> output

In mathematical terms this is expressed as a ratio $$\frac{output}{input}$$

Both can be whatever values you want. Usually you pick values that are of interest to you. In your case two voltages are of interest. Which voltage is input and which one is output is totally arbitrary, but it happens to be the way it is $$\frac{V_{out}}{V_{in}}$$

Why is it on the y-axis?

Stuff on the y axis is often some value that changes depending on another value. The mathematical term for that is that it is a function. Well, things slowly come together now: that's why it's called a transferfunction, because it is not a constant value but changing depending on some other value.

Just like your transfer function depends on several things, like if you suffered from constipation, you would eat as normal, but poop less.

What value does it depend on?

To answer that question, you have to find the formula of the transfer function. In your case, that's just a voltage divider

$$\frac{V_{out}}{V_{in}}=\frac{\frac{1}{j\omega C}}{R + \frac{1}{j \omega C}}$$

Recap that the capacitor has a complex resistance (or impedance) \$\frac{1}{j\omega C}\$

As you can see, it the depends on \$\omega\$, which is why the frequency is the x axis.

Why dB?

Basically speaking it stretches the plot of this function into a nicer form.

An analog signal has only one frequency unless it changes periods.

That's plain wrong. A signal is analog if it is continuous, not just 1 and 0 (digital) but all values in between are possible, to give a very simplified example.

So if the frequency of \$V_{in}\$ would be 100 Hz, wouldn't there just be a single discrete value when frequency = 100?

Yes, with the formula \$\omega = 2\pi f\$ plug 100 Hz into the formula above you you know the ratio between both voltages or \$V_{out}\$ for that matter.

The point is that the transfer function allows you to calculate the the result of any signal with any frequency. It even works with signals that have many frequencies.

The plot tells you that higher frequencies result in a lower ratio, ie. the value of the transfer function for high frequencies is low. That's why it is called a low pass filter, because low frequencies will be transferred almost unchanged (the ratio is close to 1), but higher frequencies come out with lowered values.

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