How can I read the graph of a low pass filter - Electrical Engineering Stack Exchange most recent 30 from electronics.stackexchange.com 2019-09-20T18:26:56Z https://electronics.stackexchange.com/feeds/question/166143 https://creativecommons.org/licenses/by-sa/4.0/rdf https://electronics.stackexchange.com/q/166143 5 How can I read the graph of a low pass filter user1534664 https://electronics.stackexchange.com/users/72918 2015-04-22T17:38:00Z 2015-04-23T13:28:42Z <h1>Given information</h1> <p>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}\$. </p> <p><img src="https://i.stack.imgur.com/yTDbk.png" alt="enter image description here"></p> <h1>I understand the following</h1> <ul> <li>Hz means the amount of periods per second. The frequency spectrum is on the x-axis.</li> <li>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)</li> <li>\$f_k\$ is the cutoff frequency</li> </ul> <h1>My confusion/question</h1> <p>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.</p> <p>Things that confuse me:</p> <ul> <li>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...</li> <li>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.</li> <li>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?</li> </ul> <p>This is very challenging for someone who knows very little about physics... </p> https://electronics.stackexchange.com/questions/166143/-/166144#166144 3 Answer by AV23 for How can I read the graph of a low pass filter AV23 https://electronics.stackexchange.com/users/72881 2015-04-22T17:50:43Z 2015-04-22T18:43:35Z <ul> <li>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.</li> <li>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.</li> <li>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.</li> </ul> https://electronics.stackexchange.com/questions/166143/-/166149#166149 4 Answer by Olin Lathrop for How can I read the graph of a low pass filter Olin Lathrop https://electronics.stackexchange.com/users/4512 2015-04-22T18:30:29Z 2015-04-22T18:30:29Z <p>This is a common plot type to show the <i>frequency response</i> of a system with a single input and a single output. Audio amplifiers, for example, are often characterized in this way.</p> <p>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.</p> <p>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*Log<sub>10</sub>(Power2/Power1). Since power is proportional to the square of the voltage, we often use dB to express voltages ratios: dB = 20*Log<sub>10</sub>(Voltage2/Voltage1).</p> <p>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.</p> https://electronics.stackexchange.com/questions/166143/-/166157#166157 4 Answer by Greg d'Eon for How can I read the graph of a low pass filter Greg d'Eon https://electronics.stackexchange.com/users/49251 2015-04-22T18:56:25Z 2015-04-22T21:49:43Z <blockquote> <p>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...</p> </blockquote> <p>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.</p> <p>The <strong>transfer function</strong> 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 <em>every</em> input frequency.</p> <blockquote> <p>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.</p> </blockquote> <p>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?</p> <p>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.</p> <blockquote> <p>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?</p> </blockquote> <p>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.</p> <ul> <li>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.</li> <li>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.</li> </ul> <p>Continue this for any other frequency and you get the continuous transfer function.</p> <hr> <p>Replying to some comments,</p> <blockquote> <p>Why does the influence of the capacitor get smaller as the frequency gets higher?</p> </blockquote> <p>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. </p> <ul> <li>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. </li> <li>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).</li> </ul> https://electronics.stackexchange.com/questions/166143/-/166161#166161 2 Answer by Russell Borogove for How can I read the graph of a low pass filter Russell Borogove https://electronics.stackexchange.com/users/23428 2015-04-22T19:44:35Z 2015-04-22T20:12:41Z <blockquote> <p>An analog signal has only one frequency unless it changes periods.</p> </blockquote> <p>That is absolutely not true, except for the special case of a sine wave signal. See <a href="http://en.wikipedia.org/wiki/Fourier_series" rel="nofollow">Fourier series</a>.</p> <p>Take a square wave, for instance, of period p. Its <em>fundamental</em> frequency (which is what you're thinking of) is f = 1/p, but it also has frequency components called <em>harmonics</em> of decreasing amplitudes at frequencies 3f, 5f, 7f... </p> <p>The lowpass filter graph shows that the <em>gain</em> (V<sub>out</sub> / V<sub>in</sub>) 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.</p> https://electronics.stackexchange.com/questions/166143/-/166164#166164 2 Answer by Magic Smoke for How can I read the graph of a low pass filter Magic Smoke https://electronics.stackexchange.com/users/69910 2015-04-22T19:58:13Z 2015-04-23T13:28:42Z <p><em>Transfer(function)</em> means the relationship between what you put into it to what you get out of it.</p> <p>You for instance eat food and well, poop from time to time.</p> <p>input -> output</p> <p>In mathematical terms this is expressed as a ratio $$\frac{output}{input}$$</p> <p>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 $$\frac{V_{out}}{V_{in}}$$</p> <p><strong>Why is it on the y-axis?</strong></p> <p>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 <strong>function</strong>. Well, things slowly come together now: that's why it's called a transfer<strong>function</strong>, because it is not a constant value but changing depending on some other value.</p> <p>Just like your transfer function depends on several things, like if you suffered from constipation, you would eat as normal, but poop less.</p> <p><strong>What value does it depend on?</strong></p> <p>To answer that question, you have to find the formula of the transfer function. In your case, that's just a voltage divider</p> <p>$$\frac{V_{out}}{V_{in}}=\frac{\frac{1}{j\omega C}}{R + \frac{1}{j \omega C}}$$</p> <p>Recap that the capacitor has a complex resistance (or impedance) \$\frac{1}{j\omega C}\$</p> <p>As you can see, it the depends on \$\omega\$, which is why the frequency is the x axis.</p> <p><strong>Why dB?</strong></p> <p>Basically speaking it stretches the plot of this function into a nicer form. </p> <blockquote> <p>An analog signal has only one frequency unless it changes periods.</p> </blockquote> <p>That's plain wrong. A signal is <code>analog</code> if it is continuous, not just 1 and 0 (<code>digital</code>) but all values in between are possible, to give a <strong>very</strong> simplified example.</p> <blockquote> <p>So if the frequency of \$V_{in}\$ would be 100 Hz, wouldn't there just be a single discrete value when frequency = 100?</p> </blockquote> <p>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.</p> <p>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.</p> <p>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.</p>