ua-cam.com/video/sZ1Dg_ufTO4/v-deo.html&ab_channel=ElectricalEngineeringAuthority She shows the table in that video but it would be nice to have it on this one too
@@logozzo85 I know but I was hoping it would help someone else since I was wondering what table she was talking about. I'm barely taking this class now! Glad to have been helpful :)
at 19:29 why is the slope going up? when adding up the slopes 45 deg + 0 deg + (-90 deg) = -45 deg. because it is negative shouldn't the line go down instead of going up?
You're thinking in terms of values, but you should be thinking about slope. Pink and orange have no slope, so the only slope (+45 deg/dec) comes from the green line, so we start at -90 and go up with a slope of +45 deg/dec
I understand that adding a 0db/decade slope to a -20db/decade slope leads to a total of -20db/decade. But why does the slope start at 20 dB at the y-axis? And in the future, how can one know where the slope should start at the y-axis?
5:30 It is mentioned that the magnitude should increase at a rate of 20db per decade... But from 2 to 20, the increase in y should be = (20-2)* (20/10) = 36. However here it is drawn at (20,20)? Am I making a mistake somewhere?
Well, yeah. I feel many sources are bad at explaining it, here looking for the best explanation OR averaging out all the sub-par one if the former's hard to find..It's often the latter sadly.
Very nice video and very clear explanations. I would like know what application you currently use for this course because I want to use such applications in order to improve my work with my students. Thanks
I am a bit confused. why the magnitude graph starts from 0 ? it should be started from +20db as there is a pole 's' at origin? if considered "5" then 20{due to pole at origin}+20log(5) =+34db but why 0db?
@@Abhishek-iq9lo for a zero the slope is +20 dB / dec for a pole the slope is -20 dB / dec the general rule is : slope= n x 20 dB/dec, where "n" is the power on "s"
i dont understand why orange line passes through the 1 frequecny in first plot?how we will find this frequecy plz help me because i have my finals week ago
nah it's 45 degrees/decade , which belongs to the real zero slope, so that means we could not plus (-90) degrees, we use only the +45 degrees to draw the slope, remember "45 degrees/decade", it's different from (-90 or 90). So she's right!
she is right, you should know green line means 45 degree increasing from start point, orange line means stay at -90degree, pink line means stay at o degree, so the total trend means go increasing 45 degree.
When you are drawing the Bode plots, you have to have polynomials in the form of (1+...). For example we do have (10+jw) in the denominator. Since we want to have the from that I mentioned above, we have to divide and multiply the whole polynomial by 10. Which will give use : 10(1+jw/10)
Electrical Engineering Authority thank you , but I already understand that. the part I am asking about is where you divided 2 by 10 to get 0.2 and you multiplied by 10 to get 20.
Because to draw the phase plot, the rule is ... it is 0 degree up to 1 decade below the Corner Freq and then reaches 90 degree at 1 decade above corner freq for a zero and -90 degree for a pole. here corner freq is 2 so up to (2/10)=0.2 rad/sec the phase contribution for the zero is 0 degree and then it becomes +90 degree at (2*10)=20 rad/sec. and reaches 45 degrees at the corner freq of 2 rad/sec in between.
In bode diagrams both poles and zeros are positive regardless of their initial sign 1-s=0, so s=-1, but it doesn't matter, because when you put it into a bode diagram you take it positive, right? So it's +1 regardless (If you're wondering why... well, she didn't write it, but the x axis is actually the logarithm in base 10 of omega, and as you know the logarithm of a negative number doesn't exist That's also the reason why when plotting the pole 0 in the graph you instead write 0.1... an approximation of 0)
s = jω, therefore: (3s + 1) = (3*jω + 1) Finally, rewrite the simple zero as the general form (1 + jω/z) as follows (3*jω + 1) = (1 + jω/(1/3)) Where z = 1/3
I think this qestıon has a little wrong,because "20log(2)" should not be close to 1. 20log(2)=6.02 .logarithmic function should be like this=> 1......................(2 has a maybe here)....3...4...5....6...7..8..9.10
Even after 7 years, your tutorial is the most compact one for bode plot on youtube. Thank you very much, professor :)
Thanks A TON for this video, it's almost a life saver! The example problem is finely explained and covers a lot of questions. Great job!
I'm brasilian. I dont speak english very well and yet that I understood all you explaned. Thanks
seriouslly, I almost gave up on my examn, thank you I did not!!!!
Great 😊
Thank you very much! Whenever I need to remember how to plot Bode Plots, I watch this video!😇👍👍👍
Thank you so much Miss Maryam. You're an amazing teacher! You have no idea how helpful your videos are
ur handwriting is literally beautiful.
Thanks...neat and precise..keepUp
من لا يشكر الناس لا يشكر الله
Now i'm prepared for my Exam
and Thanks
did u pass ?
@@koala6494 i guess not
Thanks! That helps a lot, I'm now ready for my exam. And your voice is beautiful
Thanks for sharing. Brazilian guy here enjoying your video.
Your video helped me a loooot, thank you very much, from Brazil :)
wouldnt be nice if yu could shown that table into video..
ua-cam.com/video/sZ1Dg_ufTO4/v-deo.html&ab_channel=ElectricalEngineeringAuthority
She shows the table in that video but it would be nice to have it on this one too
@@snakelda bro you actualy replied three years later looool
@@snakelda but the reply helped me so thank you so much
@@logozzo85 lmao
@@logozzo85 I know but I was hoping it would help someone else since I was wondering what table she was talking about. I'm barely taking this class now! Glad to have been helpful :)
Greats from Italy you seem very prepared on what you’re doing 👌🏻
at 19:29 why is the slope going up? when adding up the slopes 45 deg + 0 deg + (-90 deg) = -45 deg. because it is negative shouldn't the line go down instead of going up?
You're thinking in terms of values, but you should be thinking about slope. Pink and orange have no slope, so the only slope (+45 deg/dec) comes from the green line, so we start at -90 and go up with a slope of +45 deg/dec
@@guidos.736 oh I understand now thank you
I understand that adding a 0db/decade slope to a -20db/decade slope leads to a total of -20db/decade. But why does the slope start at 20 dB at the y-axis? And in the future, how can one know where the slope should start at the y-axis?
YOU ARE GOAT..YOU SAVED MY LIFE
You might have just saved my soul. Thank you so much.
Thank you. You are an amazing Professor.
Thank you so much for this example! 😭😭 I appreciate it a lot🙏
Thank you maam . This has saved me.
Great, very clear explaination... it helped me a lot
I wish I would find this video earlier. You are awesome! Thank you so much! ♥
Thank you! Your voice is so beautiful btw! I love it!
I enjoyed this video... From Zambia
Very useful!! I cant understand all of the word.. but its enough to learn how to draw that graph. Thanks!
5:30 It is mentioned that the magnitude should increase at a rate of 20db per decade... But from 2 to 20, the increase in y should be = (20-2)* (20/10) = 36. However here it is drawn at (20,20)? Am I making a mistake somewhere?
This video helped me a lot, thank you
be real here, you searched for this
A lot
Just because shes a girl?
Well, yeah. I feel many sources are bad at explaining it, here looking for the best explanation OR averaging out all the sub-par one if the former's hard to find..It's often the latter sadly.
Tomorrow I have a re-exam of Controll totally saved me !!!
guys if you have your own examples just plug it in matlab to check if you did it correctly but follow the instructions. you cannot go wrong
WOW....Best explanation
Thanks for taking the time to make the video and help us! :)
Good. Better than my teacher.👍
What an explanation! Thanks A lot.😍
madam! which table are you talking about?
Kindly clarify!
Great Explanation, thank you so much
Thank you, ma'am. It really helped me a lot.
most helpful video ever
Thanks! That helps a lot, But in magnitude what the different between drawing of 20 &-20
slope for zero = +20 dB / dec
slope for pole = -20 dB / dec
neat and clean and easy to understand. thank you
Thanks you save my life!!!
Very nice video and very clear explanations. I would like know what application you currently use for this course because I want to use such applications in order to improve my work with my students. Thanks
MS one note
thank you for your clear declaration.
I am a bit confused. why the magnitude graph starts from 0 ? it should be started from +20db as there is a pole 's' at origin? if considered "5" then 20{due to pole at origin}+20log(5) =+34db but why 0db?
wow !!
amazing lecture!!
That was awesome 👌🏻 👏🏻
Clear explanations! thank you so much.
Thank you very much , god bless you .❤
خیلی ممنون
بسیار زحمت کشیدید و لطف کردید
Is she iranian?
Great video 👍🏽👍🏽👍🏽
How do you know that for 200, it reaches 40dB?
The slope is 20 db/dec. from 20 to 200 is one decade so it will go from 20 ----> 40
@@ElectricalEngineeringAuthority how do u find the slipe
@@Abhishek-iq9lo slope =y/x of graph,
@@Abhishek-iq9lo
for a zero the slope is +20 dB / dec
for a pole the slope is -20 dB / dec
the general rule is :
slope= n x 20 dB/dec, where "n" is the power on "s"
i'm so confused, sometimes the values are superpositioned? sometimes the slopes are superpositioned?
This is type 1 system so I think initial slope should be starting from -20db/dec not 0db/dec
Very clear and informative!
Ohh god lastly u saved me !! thank you thank you so much
Starts at 4:55
please what is this program you uses?? is it a adobe pdf
why the plotting is done from 0.1
I guess they start at 10^(-1)
Nice explanation
Amazing explanation
i dont understand why orange line passes through the 1 frequecny in first plot?how we will find this frequecy plz help me because i have my finals week ago
Thank you very much, really helpful video!
I wanted to know what to do in -40log(S/100-1)
Can you please give advice?
thank you so much! great job please upload more videos on Electrical engineering.
How is the zeros and poles 2 and 10? How is she getting that
Thank you for the great explanation :)
Thank you for bode diagrams
Whopper video bro sound 🔥
beautiful explanation.. thank u
won't the plot have -45 degrees slope from 0.2 to 1 in the phase plot. because (+45-90=-45)...
nah it's 45 degrees/decade , which belongs to the real zero slope, so that means we could not plus (-90) degrees, we use only the +45 degrees to draw the slope, remember "45 degrees/decade", it's different from (-90 or 90). So she's right!
she is right, you should know green line means 45 degree increasing from start point, orange line means stay at -90degree, pink line means stay at o degree, so the total trend means go increasing 45 degree.
YOU ARE THE MAN!!!!!!!!!!!!!!!!!!!!
may I please know why you are dividing the zeroes and poles by 10 when sketching the phase plot
When you are drawing the Bode plots, you have to have polynomials in the form of (1+...). For example we do have (10+jw) in the denominator. Since we want to have the from that I mentioned above, we have to divide and multiply the whole polynomial by 10. Which will give use : 10(1+jw/10)
Electrical Engineering Authority thank you , but I already understand that. the part I am asking about is where you divided 2 by 10 to get 0.2 and you multiplied by 10 to get 20.
Because to draw the phase plot, the rule is ... it is 0 degree up to 1 decade below the Corner Freq and then reaches 90 degree at 1 decade above corner freq for a zero and -90 degree for a pole. here corner freq is 2 so up to (2/10)=0.2 rad/sec the phase contribution for the zero is 0 degree and then it becomes +90 degree at (2*10)=20 rad/sec. and reaches 45 degrees at the corner freq of 2 rad/sec in between.
Wont the zeroes and poles be negative?
these are always positive
Great video, well done!
Thanks. good explanation!!
thanks miss you're amazing
Tell Book Name Plz?( also edition)
Very helpful video! Thanks!! :)
What in case we have the form (1-s) instead of (1+s)?the formulas aren’t applicable please help
In bode diagrams both poles and zeros are positive regardless of their initial sign
1-s=0, so s=-1, but it doesn't matter, because when you put it into a bode diagram you take it positive, right? So it's +1 regardless
(If you're wondering why... well, she didn't write it, but the x axis is actually the logarithm in base 10 of omega, and as you know the logarithm of a negative number doesn't exist
That's also the reason why when plotting the pole 0 in the graph you instead write 0.1... an approximation of 0)
which book r u talking about ?? where i can see the table
I added the name of the book in the video's caption
this is helpful, thanks !
thank you brader.
why wouldn't you link or show the table you are constantly referring to? "You can see that" @14:17 .. No, I cant fucking see that.
calm down bru
@@johntello8904 You can hear emotion via text? Amazing.
@@brycehazen have you ever read a book before?😂
Why doesnt it go back to zero after 100, because after 100 it would just be 20 db/sec minus 20 db/sec=0
I think she made a mistake
Very useful, thanks.
Nice video thank you!
Very useful video
when you draw phase plot , why decrease at z1=0.2? i think that it must be decrese at z=2 .
plz , expain to me 😭😭
for phase you start one decade before so that at the pole it's 45 degrees
Alp Sezer Orak Thank you!
THAAAAAAAAAAAAAAAAANK U@@flamingoKnight
thanks. very helpfull
Thank you very much.
You sound just like my teacher, are you teaching in Belgium?
that was nice thank you madam
very nice jos kel
How did you know it went up to 40 dB?
The slope is +20dB per decade, so two decades later it's 40dB
where is table?
I think the poles and zeros you calculated are wrong
what if it was (3s+1) on numerator?
jw is actually s here dude
s = jω, therefore:
(3s + 1) = (3*jω + 1)
Finally, rewrite the simple zero as the general form (1 + jω/z) as follows
(3*jω + 1) = (1 + jω/(1/3))
Where z = 1/3
I'll be back on Tuesday!
I think this qestıon has a little wrong,because "20log(2)" should not be close to 1. 20log(2)=6.02 .logarithmic function should be like this=> 1......................(2 has a maybe here)....3...4...5....6...7..8..9.10