Solving Bode Plot Questions: Finding the Gain for Accurate Magnitude Response

[gl0ck]

Hello,

I have the following problem ( which should be uploaded below the text )
I've done part a) correct, I know how to sketch the bode plots but don't know how to find the gain in order to draw correct magnitute response bode plot. This is where I am stucked and if I find out how to find get the magic number of 26dB I think I will be able to solve the rest.

Thanks

 

[berkeman]

Hello,

I have the following problem ( which should be uploaded below the text )
I've done part a) correct, I know how to sketch the bode plots but don't know how to find the gain in order to draw correct magnitute response bode plot. This is where I am stucked and if I find out how to find get the magic number of 26dB I think I will be able to solve the rest.

Thanks

What did you get for part a)?

What is the gain when ω = 0? What does the gain do for ω > 0?

 

[gl0ck]

a) ( 10⁵ x 5 x 10^-2 x 2 x 10^-4 ) x (1+jw0.2)x (1/jw10^-2)x(1/jw2x10^-4)

I think when w = 0 the gain is 0, when w>0 it should be a bandpass kind a bode plot ( from what I can see from the transfer functions) . I think from 5rad/sec the gain should go up since it is inverted low pass transfer function till 100rad/sec and after that stays the same and go down at 5k rad/sec . Not sure if it should go down after 100k or should be 0dB

 

[berkeman]

a) ( 10⁵ x 5 x 10^-2 x 2 x 10^-4 ) x (1+jw0.2)x (1/jw10^-2)x(1/jw2x10^-4)

I think when w = 0 the gain is 0, when w>0 it should be a bandpass kind a bode plot ( from what I can see from the transfer functions) . I think from 5rad/sec the gain should go up since it is inverted low pass transfer function till 100rad/sec and after that stays the same and go down at 5k rad/sec . Not sure if it should go down after 100k or should be 0dB

I don't understand your answer to a). It asked you to express the transfer function as the product of standard forms. A bandpass function is the product of what standard forms?

 

[gl0ck]

for a) the term in the 1st brackets is the constant term, the second bracets is the inverted low pass followed with 2 low pass functions. I just put all the constants in the 1st brackets. Bandpass is made of low pass and high pass filters combined together. I just thought it will look like it because here I have inverted low pass..

 

[NascentOxygen]

Check the steps you followed to change 1/(100+jw) into 10^-2(1/(jw10^-2)).

 

[gl0ck]

oh I see its just a typo.. it should be 1/(1+jw10^-2) same for 1/(1+jw2x10^-4)

 

[donpacino]

it is usually best to have the equations in the form of
V_{DC}*\frac{(1+jw/Z1)*...(1+jw/Zn)}{(1+jw/P1)*...(1+jw/Pn)}
with Z being the system zeros and P being the system poles

you have it in the form of

V_{DC}*\frac{(1+jw*X1)*...(1+jw*Xn)}{(1+jw*Y1)*...(1+jw*Yn)}
with X being the inverse of the zeros and Y being the inverse of the poles


arranging the equation as seen in the first will allow you to easily see the system poles and zeroes, as well as the DC gain.

edit: this system isn't necessary a voltage. I'm not sure why I assumed so.

 

[gl0ck]

Not sure what you mean, but I find it easy to have the equations in the form Vout/Vin = the transfer functions..
As the thread become discussing question a), is it correct what I've written or it is wrong? Because the thing I can't get is part b) and I would appreciate some suggestions on b)
Thanks

 

[donpacino]

while your part A is correct, it can take some effort to draw the bode plot using the answer you gave.

If you get it in the form I gave, part B will be very easy to do. If you don't change it, at the very least simplify your dc response.

also where do you get the "magic number of 26 dB?"

my advice, follow berkman's advice.

 

[gl0ck]

This number is given in the answers as "when w = 6000rad/sec , 20log₁₀(Vout/Vin) = 26dB" I think this is just an approximation. Because when I draw the bode plot and draw line upwards from 6k until I meet the magnitude line, you don't really know the value, I think they just kinda guessed it to be able to find Vout/Vin.. in order after that to find part d) which if at 6k is 26dB becomes 20sin(6000t -Pi/4).
It might be just guess but I think because we have inverting low pass transfer function from 5 rad/sec and it goes +20db per dec until it meet the low pass frequency at 100 rad/sec and they cancel each other. So actually it has been less than 2 decs and the gain is somewhere between 20 and 40 ( or 30 ). That is what makes sense for me but not sure if I am just guessing here..
For c) I find it that the unity gain is before 5rad/sec and at 100k rad/sec.

 

[NascentOxygen]

When you need to determine the gain at some particular ω you substitute that value into your equation (a) once you have it correct. Taking a reading off the roughly sketched graph is no substitute.

 

[donpacino]

When you need to determine the gain at some particular ω you substitute that value into your equation (a) once you have it correct. Taking a reading off the roughly sketched graph is no substitute.

OP is supposed to use a bode plot to find the answer to D

 

[NascentOxygen]

OP is supposed to use a bode plot to find the answer to D

Yes and no. A Bode Plot is not provided, so if a student is to use one he will have to plot it for himself. This seems a lot of trouble when all he needs with accuracy is one point, that at 6000 rad/s. He might as well make do with a rough sketch and just accurately plot that one point he needs. (The alternative that would suffice for the case here is to sketch the plot on log-log paper and read directly off that.)

 

[NascentOxygen]

This number is given in the answers as "when w = 6000rad/sec , 20log₁₀(Vout/Vin) = 26dB" I think this is just an approximation. Because when I draw the bode plot and draw line upwards from 6k until I meet the magnitude line, you don't really know the value, I think they just kinda guessed it

It seems there's a mistake in the question. A suitable correction would be to edit that 6000 rad/s changing it to 600 rad/s.

The Bode Plot gain at 6000 rad/s is around 22 dB and this cannot be read with any accuracy off a freehand sketch.

 

[donpacino]

Yes and no. A Bode Plot is not provided, so if a student is to use one he will have to plot it for himself. This seems a lot of trouble when all he needs with accuracy is one point, that at 6000 rad/s. He might as well make do with a rough sketch and just accurately plot that one point he needs. (The alternative that would suffice for the case here is to sketch the plot on log-log paper and read directly off that.)

true :)

 

[AlephZero]

It seems there's a mistake in the question. A suitable correction would be to edit that 6000 rad/s changing it to 600 rad/s.

The Bode Plot gain at 6000 rad/s is around 22 dB and this cannot be read with any accuracy off a freehand sketch.

I assumed the Bode plot required was an "approximate" plot made from straight line segments (on log-log paper), and the point of the question was about understanding how to make and use the approximate plot, not about substituting one set of numbers into an equation.

Since the approximate plot is defined by frequency breakpoints and slopes, you can "read off" the value at any frequency, to the accuracy of the approximation (which will be pretty good, except close to the break points).

Clearly it makes no pedagogical sense to make an "accurate" Bode plot (most likely using computer software) and then read one number from it.