Bode Plot

Hi guys ,

I can sovle attached question at step (1) , But i don't know How the combined the curves at the next step (2) ??

How can i find the intersection curves with others and with axes??

No one has any idea of Bode Plot

If you can get the individual curves shown in Step 1, what is causing you problems in adding them together to get the combined curve in Step 2? Show your work and perhaps we can see where you are having problems.

alright , i got the combined curves but i want to know the values intersection with w-axis as shown in the attached fig.

The slope of the line is 20dB/decade, right? (the left skirt is +20db/decade and the right skirt is -20dB/decade).

So, if you need to drop 26dB, how many decades do you need to go out?

The distance between two things on a logarithmic scale is the base-10 log of the ratio.

So, for instance, 1300Hz and 27500Hz would be

log10(27500Hz/1300Hz) = 1.325 decades apart.

Does that help you figure out?

okay ,,

Assumed the intersection points with w-axis X and K

i.e.
X with the left skirt is +26db

26 = 20 log10(10Hz/X Hz)
1.3= log10(10Hz/X Hz) =====> 10Hz/X Hz = 10^1.3
X= 10 /10^1.3 = 0.5 ??? Is it Right??

K with the right skirt is -26db

-26 = 20log10(K Hz/20,000 Hz)
-1.3= log10(K Hz/20,000 Hz) =====> K Hz/20,000 Hz = 10^-1.3
K = 1002.37 HZ It's is wrong because on the fig. 10^5 < K < 10^6

mo2015mo said:

okay ,,

Assumed the intersection points with w-axis X and K

i.e.
X with the left skirt is +26db

26 = 20 log10(10Hz/X Hz)
1.3= log10(10Hz/X Hz) =====> 10Hz/X Hz = 10^1.3
X= 10 /10^1.3 = 0.5 ??? Is it Right??

Click to expand...

You need to track your units:

26dB = 20dB log10(10Hz/X)
1.3 = log10(10Hz/X)
10Hz/X = 10^1.3
X = 10Hz /10^1.3 = 0.5Hz

See how the units work out? Most mistakes you make will screw up the units, so tracking the units will let you catch most (not all) mistakes.

K with the right skirt is -26db

-26 = 20log10(K Hz/20,000 Hz)
-1.3= log10(K Hz/20,000 Hz) =====> K Hz/20,000 Hz = 10^-1.3
K = 1002.37 HZ It's is wrong because on the fig. 10^5 < K < 10^6

Click to expand...

Good! You are asking whether or not the answer makes sense instead of just blindly accepting whatever your calculator spits out.

It might help to drop back and look at the fundamental relationship involved

y(f) = m*log10(f/fo) + b

Don't worry about the fo in the equation. You can take it to be any frequency you want, such as 1Hz. It doesn't matter because it will cancel out eventually. But it needs to be there in the equation as given because the logarithm function is transcendental and can't have an argument that has units.

You know:

y(20kHz) = 26dB
m = -20dB/decade

You want to find x such that:

y(x) = 0dB

So, from the known data point, we have

y(f) = m*log10(f/fo) + b

y(20kHz) = -20dB/decade*log10(20kHz/fo) + b = 26dB

b = 26dB + 20dB/decade*log10(20kHz/fo)

Substituting this into the unknown data point:

y(x) = 0dB = m*log10(x/fo) + b

0dB = -20dB/decade*log10(x/fo) + 26dB + 20dB/decade*log10(20kHz/fo)

20dB/decade*log10(20kHz/fo) - 20dB/decade*log10(x/fo) = -26dB

20dB/decade*[log10(20kHz/fo) - log10(x/fo)] = -26dB

[log10(20kHz/fo) - log10(x/fo)] = -26dB / (20dB/decade) = -1.3decades

log10[(20kHz/fo)/(x/fo)] = -1.3decades

log10(20kHz/x) = -1.3decades

20kHz/x = 10^(-1.3decades) = 0.501 (WRONG - should be 0.0501)

x = 20kHz/0.501 ~= 30kHz (DOUBLE WRONG - should be 400kHz)

Note that the unit of "decade" is very analogous to "radian". It is dimensionless but conveys useful information to help us keep track of the meaning of certain numbers.

WBahn said:

You need to track your units:

26dB = 20dB log10(10Hz/X)
1.3 = log10(10Hz/X)
10Hz/X = 10^1.3
X = 10Hz /10^1.3 = 0.5Hz

See how the units work out? Most mistakes you make will screw up the units, so tracking the units will let you catch most (not all) mistakes.



Good! You are asking whether or not the answer makes sense instead of just blindly accepting whatever your calculator spits out.

It might help to drop back and look at the fundamental relationship involved

y(f) = m*log10(f/fo) + b

Don't worry about the fo in the equation. You can take it to be any frequency you want, such as 1Hz. It doesn't matter because it will cancel out eventually. But it needs to be there in the equation as given because the logarithm function is transcendental and can't have an argument that has units.

You know:

y(20kHz) = 26dB
m = -20dB/decade

You want to find x such that:

y(x) = 0dB

So, from the known data point, we have

y(f) = m*log10(f/fo) + b

y(20kHz) = -20dB/decade*log10(20kHz/fo) + b = 26dB

b = 26dB + 20dB/decade*log10(20kHz/fo)

Substituting this into the unknown data point:

y(x) = 0dB = m*log10(x/fo) + b

0dB = -20dB/decade*log10(x/fo) + 26dB + 20dB/decade*log10(20kHz/fo)

20dB/decade*log10(20kHz/fo) - 20dB/decade*log10(x/fo) = -26dB

20dB/decade*[log10(20kHz/fo) - log10(x/fo)] = -26dB

[log10(20kHz/fo) - log10(x/fo)] = -26dB / (20dB/decade) = -1.3decades

log10[(20kHz/fo)/(x/fo)] = -1.3decades

log10(20kHz/x) = -1.3decades

20kHz/x = 10^(-1.3decades) = 0.501 modify 10^(-1.3decades) = 0.05

x = 20kHz/ 0.0501 ~= 400kHz

Note that the unit of "decade" is very analogous to "radian". It is dimensionless but conveys useful information to help us keep track of the meaning of certain numbers.

Click to expand...

Thanks for catching my mistake. I had to step aside and got rushed at the end and failed to follow my oen cardinal rule -- does the answer make sense. Clearly 10^-1 is 0.1 and so 10^-1.3 has to be smaller than that. I recalled the 0.501 from the earlier problem and didn't bother to recalculate it. As for 20/0.5 being 30 and not 40, that was just a stupid mental math blunder. Another sanity check would have been to simply note that 26dB is more than 20dB and hence the answer would have to be more than one decade above 20kHz, or greater than 200kHz.