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Having trouble with bode plots. I understand that the is to be converted to $dB$ but after that I'm stuck. Could someone please show me how this graph gives a frequency response of $10/1+jw10$ enter image description here

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  • $\begingroup$ Its way simpler to solve using Asymptote Piecewise Linear Approximation $\endgroup$ – Ayo Mar 26 '19 at 12:22
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In the most typical sense the Bode plot is actually a, very good, piece-wise linear approximation to a logarithmically plotted (on both x and y axes) nonlinear curve given by the frequency response magnitude of a linear time invariant system (mainly analog) with a rational transfer function which is described in your question by: $$ H(\omega) = \frac{10}{1 + 10j\omega}$$

The Bode plot then displays the base-$10$ logarithm of $|H(\omega)|$ and plots it on a logarithmic frequency scale for positive frequencies, as: $$ Y(\omega) = 20 \log_{10}(|H(\omega)|)$$

Lets compute the magnitude of $H(\omega)$: $$ |H(\omega)| = \frac{10}{ \sqrt{1 + 100 w^2} }$$ and make an analysis of this magnitude wrtt frequency $\omega$ as it moves from $\omega=0$ to infinity.

For example when $\omega=0$, then $|H(\omega)| = 10$ and consequently $Y(\omega) = 20 \log_{10}(10) = 20$ dB. This means that the bode plot asymptotically hits the y-axis at 20 dB point.

Also from an analysis of quotients you can infer that $|H(\omega)| \approx 10$ as long as $100 \omega^2 < 1$ and this will hold as long as $ w < 0.1$. This means that a piecewise linear approximation to the curve of $Y(\omega)$ from $\omega=0$ to approximately $w=0.1$ will be a constant line of 20 dB.

Then for larger frequencies we have the following observation; if $100 \omega^2 > 1 $ then $|H(\omega)| \approx \frac{1}{\omega}$ and consequently $Y(\omega) \approx 20 \log_{10}(1/\omega) = -20 \log_{10}(\omega) $ dB. This means that if we plot $Y(\omega)$ wrt a logarithmic $\omega$ scale, for every multiplication of $\omega$ by $10$ the plot will decrease by $20$ dB. Which is the well known behaviour of a typical Bode plot (for a first order system) defined as $20$ dB per decade rule. In general for an n-th order system it will be $20n$ dB per decade.

So for frequencies $\omega > 0.1$ we shall approximate the nonlinear curve $Y(\omega)$ by a single linear line segment whose slope is $-20$ dB per decade when plotted over a logarithmic frequency axis. Note that this line goes to negative infinity as frequency approaches positive infinity, however for a practical setting the frequency will always be limited by a finite value.

If the maximum frequency on the plot will be $\omega=10$ then since this corresponds to a $100$ times increase (2 decades) of the frequency $\omega$ then the plot will have the y-axis of $20-2 \times 20 = -20 $ dB.

Note that a first order system (as defined in the first paragraph) is either lowpass or highpass (or all pass ?). If it's lowpass then its Bode plot will have a negative slope, and a positive slope if it's highpass (and zero if its allpass).

Going back from the Bode plot to the system transfer function is more difficult (actually impossible) as @FlorentEcochard describes, however if the system whose Bode plot is given is known to have some restrictions then we can obtain partial information about it as applied into your plot:

1- The system is first order (from $-20$ dB per decade slope ) , 2- The system is LowPass (from negative slope) , 3- The systems cutoff frequency is $\omega=0.1$ radians per second (the point where the slope changes), 4- The systems DC gain is $|H(0)|=10$ (from asymptotic y-axis intercept) ,

Assuming (and therefore restricting) a first order LTI lowpass system transfer function of the form; $$H(\omega) = \frac{K}{1 + aj\omega}$$ you can infer that $K=10$ and $a=10$ (from $(a \omega_c)^2 = 1 \rightarrow a \times 0.1 = 1 \rightarrow a=10$)

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  • $\begingroup$ brilliant really cleared it up ! just quick question working backwards from bode plot how would i find starting point ? i know that 20log10 of 10 = 20 . working backwards how would i solve for 20 ? $\endgroup$ – towns Jul 10 '17 at 22:09
  • $\begingroup$ I couldn't understant your question, do you mean this: $$20\log_{10}(X) = Y \rightarrow X = 10^{Y/20} $$ ? $\endgroup$ – Fat32 Jul 11 '17 at 14:14
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If your question is how to use this graph to find the transfer function the answer is simple : you can't.

First of all, this graph give only half of the information: you have a gain plot but not the phase plot, and 2 systems with different TF could have the same gain profile.

Second of all, this is obviously an asymptotic graph, as it presents a discontinuity around the cutoff frequency.

However, you could extract some information from this plot. You can see that it cuts the high frequencies with a slope of -20dB per decade, which means that there is a great chance that it is a 1st order low pass filter. Additionally, you can also read that the cutoff pulsation is about 0.1 rad/s.

Think of it like a normal function, would you be able to tell what a function is if you any saw an asymptotic graph of one half of it? All you can do is make some educated guesses about the form of the TF.

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  • $\begingroup$ my mistake i meant to type frequency response $\endgroup$ – towns Jul 10 '17 at 0:22
  • $\begingroup$ Then what exactly is your question? I think Fat32's answer below should explain everything you need to know on how to plot a Bode diagram. $\endgroup$ – Florent Jul 10 '17 at 0:26

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