Question: Specify the straight -line approximation of the Bode magnitude plot: $$H(j\omega) = 0.04 \cdot \frac{jw+50}{jw+0.2}$$

I don't understand that in the plotting part we add 6dB for two intervals. Why and how we did that?

enter image description here

  • $\begingroup$ "The $6\,\texttt{dB}$ is added to get $20\,\texttt{dB}$ at $0.2\,\texttt{rad/sec}$" $\endgroup$
    – Jdip
    Jan 14, 2023 at 16:33
  • $\begingroup$ @Jdip But why and how? Why we need this? $\endgroup$ Jan 14, 2023 at 17:31

1 Answer 1


The given frequency response is from the following transfer function in the s plane:

$$H(s) = 0.04\frac{s+50}{s+0.2}$$

From which we see this has a real zero at $s=-50$ and a real pole at $s=-0.2$. The frequency response is found by replacing $s$ with $j\omega$, so that we get $H(j\omega)$.

The poles and zeros cause inflection points in the asymptotic lines given in the Bode plot as a log log plot. As shown by the OP's plot, when the frequency reaches $\omega = 0.2$, which is when $s=j\omega$ would be the same magnitude away from the origin on the s-plane as the pole, the asymptote for the magnitude will drop -20 dB/decade (which is $-20Log{10}(\omega)$).

So for purpose of making the plot, we establish the starting magnitude from the DC gain, or when $\omega = 0$ to be:

$H(j\omega =0) = 0.04(50/0.2) = 10$.

In dB quantity this would be $20log_{10}(10) = 20 $ dB.

The asymptote will stay flat at 20 dB, until we reach $\omega = 0.2$, at which point the magnitude will drop at -20 dB /decade or $-20log_{10}(\omega) + C$. Where $C$ is a constant that ensures we also have a magnitude of +20 dB at $\omega = 0.2$. Solving for $C$ results in:

$$ 20 = -20 log_{10}(.2)+ C$$ $$ C = 20-20log_{10}(.2) = 6$$

In general we can use the following key points to quickly create the asymptotes on a Bode plot:

A real pole at $s=-x$ will create an asymptote that drops the magnitude an additional -20 dB/decade starting at $\omega = x$ and will shift the phase $-90$ degrees. (The actual magnitude right at $\omega = x$ will drop -3 dB due to the pole and the phase will have shifted 45°. As we extend the frequency axis towards infinity, the actual response will approach the asymptotes.

Similarly, A real zero at $s=-x$ will create an asymptote that increases the magnitude an additional +20 dB/decade starting at $\omega = x$ and will shift the phase $+90$ degrees (and will be +3 dB and +45 degrees right at $\omega = x$.)

Since the response converges at DC (to a value of 10 in this case), we use the DC gain to set the starting point on the left of the Bode plot: The dB of a magnitude quantity is $20Log_{10}()$ so the starting magnitude in dB would be $20Log_{10}(10)= 20 $ dB.

Similarly by inspection when $s=0$, the phase of H(s) is 0, so the starting phase would be zero.

Using that, we can quickly draw the following Bode plot showing both magnitude and phase:

Bode Plot

We see at $\omega =0.2$ the magnitude starts to drop -20 dB/decade, and the phase changes by 90° (going through -45° right at the pole and then approaching -90° as the frequency increases). Similarly once we reach the zero at $\omega = 50$ the magnitude increases +20 dB/decade, which cancels the prior decrease from the pole and the result is a flat magnitude. The phase similarly advances +90° bringing us back to zero phase.

The reason for the given rules for creating the asymptote lines becomes intuitive once the s-plane is well understood. Consider a single zero in the left half plane on the negative real axis at $s=-50$, consistent with the OP's case, and what the numerator for $H(s)$ would look like as a phasor, as we sweep $s$ along the $j\omega$ axis to create the frequency response plot. For this we are omitting any effect of the pole to understand the effect of the zero alone:

zero on s plane

When $s =0$ the numerator of $H(s)$ as $s-(-50)$ would have a magnitude of $50$ and an angle of zero, as depicted in the first plot on the left. As we increase $s$ along the $j\omega$ axis to the point where the frequency to the point where the frequency is equal to the zero (at $s=50$, the magnitude will have increased by $\sqrt{2}$ (which is +3 dB) and the phase of this numerator phasor given by $s-(-50)$ will be +45°. The numerator is $j50-(-50)$ which has a magnitude of $50\sqrt{2}$ and a phase of +45°. This is depicted in the plot in the middle. As we continue to increase the frequency, as depicted in the plot on the right, the phase will approach 90° (and only reach in the limit as $\omega \rightarrow \infty$). We note that once $\omega$ is large, a doubling of $\omega$ will approximately double the magnitude, which is equivalent to saying +6 dB/octave (an octave is a doubling of the frequency), or +20 dB/ decade as we detailed earlier.

This showed how the location of the zero on the real axis sets the asymptote lines for the magnitude and phase on a Bode plot. Similarly it can be shown how the location of the poles will also result in the opposite slope and phase given that the poles are in the denominator of the transfer function.

  • $\begingroup$ Sir, sorry for late answer but how we found that -28dB graph? $\endgroup$ Jan 17, 2023 at 20:54
  • 1
    $\begingroup$ Solve the same equation—- knowing that the magnitude is going down -20 dB/decade—- plug in the frequency for that point and take the dB of the result $\endgroup$ Jan 17, 2023 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.