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I am doing a Bode diagram plot of the following transfer function:

$$G(s)=\frac{100 \, (s+2)}{s\, (s+10)}$$ that can be put in the form

$$\frac{20 (1+s/2)}{s\, (1+s/10)}$$

At \$\omega =100\$, can the magnitude in dB be calculated as

$$20 \log(20)+20 \log \left (\sqrt{1^2 + 50^2} \right)-20 \log(100) - 20 \log \left (\sqrt{1^2 + 10^2} \right)$$

The above gives \$-0.0415\$ dB as opposed to \$0\$ dB that I get on a Bode magnitde plot. At \$\omega =100\$, \$G(j\omega)\$ has contributions from all terms, so the magnitude calculation should include all terms. Is the above calculation ok?

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2 Answers 2

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Your answer is correct, it's just that you only verified visually the magnitude plot. If you measured it, you would have found out that the results coincide:

ac

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  • \$\begingroup\$ Yes I agree! I am plotting the asymptotic bode plot. The real graph will not totally agree with the asymptotic plot. \$\endgroup\$
    – user11206
    May 29, 2018 at 6:54
  • \$\begingroup\$ May I ask which software output is that? Thanks. \$\endgroup\$
    – user11206
    May 29, 2018 at 6:55
  • \$\begingroup\$ @user11206 LTspice. It's free, but there are plenty other free simulators, this is just one of them. \$\endgroup\$
    – a concerned citizen
    May 29, 2018 at 7:21
  • \$\begingroup\$ @a concerned citizen, you are correct (and my answer was not). I have deleted my answer. \$\endgroup\$
    – LvW
    May 29, 2018 at 11:20
  • \$\begingroup\$ @LvW I'm sorry if it turned out this way, but maybe you shouldn't have deleted it. After all, your answer did describe the usual way of calculating. A minor modification to point this out would have been all there is needed to keep it. \$\endgroup\$
    – a concerned citizen
    May 29, 2018 at 11:26
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Corrected answer:

In case, you do not want to start with log calculations, here is two alternatives:

(1) You can - separately - calculat the magnitude of the numerator and the magnitude of the denominator. The quotient of both gives you the magnitude of the given function G(jw).

(2) Or you have to manipulate the function as a first step - with the aim to arrive at the form G(jw)=Re(w) + jIm(w), which easily can be used to find the magnitude.

For this purpose, both numerator and denominator are to be multiplied by the denominator`s conjugate-complex expression.

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  • \$\begingroup\$ OP wants the magnitude, so ultimately it will be 20*log10(|G(s)|), which G(s) is N(s)/D(s), so log(a/b)=log(a)-log(b). Furthermore, both a and b are products, log(a)=log(x*y)=log(x)+log(y). Some of these are complex, so he writes as hypot(m,n). Mathematically is correct, but a bit unusual, that's all. \$\endgroup\$
    – a concerned citizen
    May 29, 2018 at 10:33
  • \$\begingroup\$ @a concerned citizen If we forget the dB scale and find the magnitude as (mag(20)*mag(1+j100/2))/(mag(j100)*mag(1+j100/10)), we would get about 0.9952 and converting to dB yields -0.04148 which is about the same answer from above. Now the magnitude of a complex term mag(1+j100/10) is computed as sqrt(1^2 + 10^2) = 10.050. In the dB result in my post, I am computing the magnitude in the same manner. Is there another way to compute the magnitude? Thanks. \$\endgroup\$
    – user11206
    May 31, 2018 at 4:55
  • \$\begingroup\$ @LvW Thank you for including your post. Method 1 would be perhaps the easy way to go. Method 2 can become hard to obtain if we have several poles and zeros. Due to my low reputation, I am not unable to give you a vote up. \$\endgroup\$
    – user11206
    May 31, 2018 at 4:57

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