# How is H(jw) = jw correspond to a slope on bode plot?

I cant understand how H(jw) = jw always intersects at "1" , also how the slope of the line is 20dB, Kindly explain, im having a hard time understanding because in book such stuff is not discussed,only a straight foward graph is given for individual components such as $$\H(jw)=jw\$$ and so on.

For a transfer function $$\H(j\omega) = j\omega\$$, we can clearly see that $$\|H| = \omega\$$, and $$\\arg{H} = \frac{\pi}{2}\$$. For a magnitude plot, we care about $$\|H|\$$.
Because the magnitude of $$\H\$$ is simply $$\\omega\$$, multiplying frequency by 10 (a decade) will multiply the magnitude response by a factor of 10 as well (which corresponds to adding 20 dB). Thus it's evident that the slope of this Bode plot will be 20 dB/decade.
The reason you take the intersection at $$\\omega = 1\$$ is simply because $$\|H(j1)| = 1\$$, and unity gain is the same thing as zero decibels of gain.