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For function;

\$H(jw)=10jw/((1+jw/2)*(1+jw/10)\$

To find starting gain, i did, \$20log(10)\$ which equals 20, but in matlab, it shows -40 ? how ? Bode plot for the function is given below, enter image description here

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  • \$\begingroup\$ why does 0.001 result in -40? Is that your question? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 23 '19 at 3:34
  • \$\begingroup\$ What’s the gain at DC? How about at very low frequency? What does that translate to in dB? \$\endgroup\$ – John D Mar 23 '19 at 3:35
  • \$\begingroup\$ @SunnyskyguyEE75 yes.... \$\endgroup\$ – calculusnoob Mar 23 '19 at 3:37
  • \$\begingroup\$ \$20logH(j0.001) = 20log(10j(0.001)) / (~1)\$ =-40 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 23 '19 at 3:42
  • \$\begingroup\$ @SunnyskyguyEE75 Im trying to plot some transfer functions, how do i decided what value my w should be equal to ? Because if do something like, \$20log(10*1) = 20\$ it gives me 20 and my plots doesnt workout. \$\endgroup\$ – calculusnoob Mar 23 '19 at 3:51
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Your transfer function needs to be rewritten in a low-entropy format as recommended by the fast analytical circuits techniques or FACTs. Factor \$10s\$ in the numerator and \$\frac{s}{2}\$ in the denominator and simplify by \$s\$:

\$H(s)=\frac{10s}{(1+\frac{s}{2})(1+\frac{s}{10})}=\frac{10}{0.5}\frac{1}{(1+\frac{1}{0.5s})(1+\frac{s}{10})}=H_0\frac{1}{(1+\frac{\omega_{p1}}{s})(1+\frac{s}{\omega_{p2}})}\$ with:

\$H_0=20\$ or 26 dB, \$\omega_{p1}=2\;rad/s\$ and \$\omega_{p2}=10\;rad/s\$

This way, you see a plateau gain of 26 dB while the zero being at the origin (for \$s=0\$) is merged into the inverted pole \$\omega_{p1}\$. This is the best way to write the transfer function and see the gain in between the two poles as this might very likely be your design target in a filter design. If you do not rewrite the function in this format, you do not reveal it immediately.

The zero lies in the origin and for \$s=0\$, the magnitude is 0. Therefore, as you approach the origin on the x-axis, you see an attenuation indicated by the negative sign which appears when extracting the log of the transfer function magnitude at this point. The 40-dB attenuation or -40-dB magnitude read at 0.001 rad, simply indicates that the stimulus (your input signal), is divided by 100 at this point.

enter image description here

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By "starting gain" I assume you mean gain at DC, or gain at the left of the Bode plot. If we consider gain at DC, we need to calculate \$|H(0)|\$:

\$|H(0)| = |\frac{0}{1*1}| = 0\$.

But \$20log(0)\$ is negative infinity, which is hard to plot. So we set the far left of the graph at \$\omega = 0.001\$ instead of zero, as in your graph. Now:

\$|H(j0.001)| = |\frac{0.01j}{(1+0.0005j)(1+0.0001j)}| = |\frac{0.01j}{(1+0.0006j-0.0000005)}|\$

Realising the denominator and ignoring insignificantly small numbers, we get:

\$|\frac{0.01j+0.00006}{1+0.0006)}| \approx 0.01\$

Now 20*log(0.01) = -40dB, as your graph shows.

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Well, in the circuit you have we know that:

$$\mathcal{H}\left(\text{s}\right)=\frac{10\cdot\text{s}}{\left(1+\frac{\text{s}}{2}\right)\cdot\left(1+\frac{\text{s}}{10}\right)}\tag1$$

Now, for sinusoidal functions we can write:

$$\text{s}=\text{j}\omega\tag2$$

Where \$\omega=2\pi\cdot\text{f}\$

In order to find some usefull information we need to find the absolute value of the transfer function:

$$\left|\mathcal{H}\left(\text{j}\cdot2\pi\cdot\text{f}\right)\right|=\frac{200\cdot2\pi\cdot\text{f}}{\sqrt{\left(4+\left(2\pi\cdot\text{f}\right)^2\right)\cdot\left(100+\left(2\pi\cdot\text{f}\right)^2\right)}}\tag3$$

Finding some properties:

  • Maximum amplitude: $$\frac{\text{d}}{\text{d}\omega}\left(\frac{200\cdot2\pi\cdot\hat{\text{f}}}{\sqrt{\left(4+\left(2\pi\cdot\hat{\text{f}}\right)^2\right)\cdot\left(100+\left(2\pi\cdot\hat{\text{f}}\right)^2\right)}}\right)=0\space\Longrightarrow\space\hat{\text{f}}=\frac{\sqrt{5}}{\pi}\tag4$$
  • Cut-off frequencies: $$\frac{1}{\sqrt{2}}\cdot\frac{50}{3}=\frac{200\cdot2\pi\cdot\text{f}_\text{c}}{\sqrt{\left(4+\left(2\pi\cdot\text{f}_\text{c}\right)^2\right)\cdot\left(100+\left(2\pi\cdot\text{f}_\text{c}\right)^2\right)}}\space\Longleftrightarrow\space\text{f}_\text{c}=\frac{\sqrt{14}\pm3}{\pi}\tag5$$
  • Roll-off rate (dB/octave): $$\lim_{\text{f}\to\infty}\left(20\log_{10}\left(\left|\mathcal{H}\left(\text{j}\cdot2\pi\cdot2\cdot\text{f}\right)\right|\right)-20\log_{10}\left(\left|\mathcal{H}\left(\text{j}\cdot2\pi\cdot\text{f}\right)\right|\right)\right)=-20\log_{10}\left(2\right)\tag6$$
  • Roll-off rate (dB/decade): $$\lim_{\text{f}\to\infty}\left(20\log_{10}\left(\left|\mathcal{H}\left(\text{j}\cdot2\pi\cdot10\cdot\text{f}\right)\right|\right)-20\log_{10}\left(\left|\mathcal{H}\left(\text{j}\cdot2\pi\cdot\text{f}\right)\right|\right)\right)=-20\tag7$$
  • Bandwidth: $$\mathcal{B}=\left|\frac{\sqrt{14}+3}{\pi}-\frac{\sqrt{14}-3}{\pi}\right|=\frac{6}{\pi}\tag8$$
  • Quality: $$\mathcal{Q}=\frac{1}{\mathcal{B}}=\frac{\pi}{6}\tag9$$
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