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,
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\$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,
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.
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.
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: