hey everyone welcome back to control

system lectures we're continuing our

journey through discrete control and in

this video we're going to expand our

understanding of the bilinear transform

and in doing so we'll learn about how

this transform warps the Z plane what

that means for us and what we can do

about it now this is going to be a

meandering journey but I think the

topics we'll cover will be necessary in

order to set up the proper context

around Z plane warping and this is like

a 25 minute video which is a long time I

know but to entice you to watch the

whole thing at the end there's going to

be this pretty sweet animation so

hopefully you stick around find out what

it is if you've missed any of the

previous videos on discrete control you

can find their links in the description

below let's get to it

there may come a time in your career

where you are asked to design a filter

that will be implemented on a digital

computer how do I go about doing this

you might ask a helpful friend and they

say it's easy just design a continuous

filter with the properties that you want

convert it to a discreet filter with the

bilinear transform and then tada you can

implement it on a digital computer which

you promptly try and find that the

performance is not what you were

expecting the frequency is all wrong why

is that well to answer this question we

first need to talk briefly about filters

you can think of a filter like a sieve

or a strainer that you'd use in cooking

its main purpose is to take a set of

items and remove specific unwanted items

from that set for example after you

finished boiling pasta you have a pot of

pasta and water you then passed the

combined pasta and water through a

strainer or a filter which will allow

the water to pass through and be

discarded so that you're left with only

the pasta

similarly filters can also be used on

signals like the voltage coming from a

sensor over time the voltage from the

sensor is changing and you can see the

frequencies that make up that time

domain signal by viewing it in the

frequency domain now you may not want to

keep the raw unfiltered signal because

it could be a mix of frequencies that

you want the pasta and frequencies that

you don't the water

so we can design a filter that removes

or stops the high-frequency signals from

getting through so that our system only

sees the low frequency sensor

information basically our system is only

getting the pasta in an ideal sense a

signal filter could be specified by

stating exactly which frequencies you

want to keep or pass through the filter

and which frequencies you want to stop

or reject with the filter these ideal

filters have names like low pass and

high pass and band reject or sometimes

called the notch filter and band pass

low pass filters keep frequencies below

the critical frequency and reject or

attenuate frequencies higher than it

high pass filters are the opposite they

keep the frequencies above the critical

frequency and reject those below it the

difference between these two filters in

our cooking example is whether you kept

or threw out the water for pasta you'd

keep the large low frequency noodles and

discard the higher frequency water but

for tea you would keep the water and a

few of the smaller high frequency tea

particles a notch filter is a low pass

and a high pass filter that don't

overlap such that there is a band of

frequencies in between the two that are

rejected and everything above and below

that band are kept you might set up a

notch filter if you're trying to remove

the humming noise and an audio system at

50 Hertz or 60 Hertz the one that's

generated by the powerline a bandpass is

the opposite of a notch filter such that

you only pass through a narrow band of

frequencies and everything above and

below is rejected for example our eye is

a bandpass filter for visible

frequencies in the e/m spectrum we can

only see those specific frequencies and

further still colored objects are

bandpass filters for that specific color

frequency if you shine broad spectrum

light onto a red apple only the red

frequency is reflected in the band that

our eyes can see and sometimes we only

want the metaphorical red frequencies to

get through our control system and in

that case we would design a bandpass

filter but these are ideal

representations of filters and we can't

expect to build a filter that perfectly

rejects or keeps fair

specific frequencies in our pasta

example sometimes a few noodles snake

their way past the strainer and there is

always some water leftover clinging to

the noodles in the pan so not a perfect

separation of the two in reality we're a

filter transitions from passing to

stopping is not a single point but

occurs over a band of frequencies

likewise you don't get perfect unaltered

frequencies in the passband region nor

do you always get perfect attenuation in

the stop band region how you design your

filter is based on the requirements for

the filtering that you are doing and

there are many variations of these types

of filters and their given names like

Butterworth and chebyshev and Bessel and

more and there's tons of them and

they'll probably be even more in the

future but the point I want you to take

away from this section is that almost

every filter will have these key

features in common and those features

are tied to specific frequencies so it's

important that our final digital filter

design maintains these critical

frequencies right where we want them it

does us no good to design a low-pass

analog filter only to have the

transition band move on us when we

convert it to a digital filter so with

that being said let's get back to our

original goal of designing a filter for

a digital computer like our helpful

friend suggested a popular way of

approaching this problem is to design an

analogue filter first get it just right

and then use the bilinear transform to

map it to the Z domain and hopefully

create a digital filter whose critical

frequencies have not been moved the

reason we would approach the design with

an analogue filter rather than directly

designing a digital filter is that we

have a large amount of continuous analog

Theory available to us that will give us

more tools to use during our design

let's see how this works in practice by

trying it out on a standard notch filter

at this point you've understood the

requirements your filter must meet and

you've toiled over the math and

simulations and you're happy with the

following analog filter design where

Omega naught is the critical frequency

that you want to attenuate which in our

case is 0.1 radians per second and Q is

the quality factor of the filter or a

value that corresponds to the steepness

and depth of the notch in this case

we're choosing Q equals 1

this will generate a magnitude response

curve that looks something like this

redlined now this isn't a video about

how to design an analogue filter that

would require its own standalone series

so I'm not going to go into detail about

how we chose this particular notch

filter for now just know that this is

one possible form of a notch filter that

we could have created to meet our

requirements and I'll show you in a bit

in MATLAB that this pink transfer

function does truly create a magnitude

plot that meets our requirements okay so

at this point we have a perfectly fine

analog filter but the catch as you know

is that we want to implement it on a

digital computer and say one running

with a sample time of two seconds which

means we need to map it to the Z domain

using the bilinear transform we did the

calculations by hand in the previous

video so in this one let's just go

straight to MATLAB first things first

let's define our Laplace variable s and

then set our sample time to two seconds

the critical frequency to 0.1 radians

per second and the quality factor to one

with our variables defined we can create

our S domain transfer function using the

equation for a standard notch filter and

no surprise we got the same result that

we did by hand now let's see the bode

plot to you check out the magnitude of

this filter across the frequency

spectrum and that's a good-looking notch

and if we zoom in a bit we can see that

the notch is located precisely at 0.1

radians per second just like we wanted

and we step one out of the way let's use

the bilinear transform or the Tustin

method to convert to the Z domain and we

get a discrete transfer function with a

sample time of two seconds so far so

good we can now plot the bode plots for

both the analog and digital filters on

top of each other and see how we did

that's that's pretty good the red

digital filter lies pretty much on top

of the blue analog filter we did some

fine work here

and all that's left to do is write some

code for this filter and put it on our

digital computer oh wait a second I I'm

just realizing that I got our original

requirement wrong it looked like Omega

naught was written as 0.1 radians per

second but really it was supposed to

read 0.7 Radian

per second okay that's no problem we'll

just update the omega-not variable 20.7

rerun our analog transfer function and

discrete-time transfer function now this

is probably still okay but let's just

check the bode plot once again to be

absolutely sure and they're different

and if we zoom in we can see that the

analog notch is at 0.7 like we want but

the digital notch is much lower at 0.6

one radians per second what happened all

we did was increase the critical

frequency

well this shift in frequency comes down

to how the Z plane is warped by the

bilinear transform differently than e to

the St which is the true definition of Z

so it should be clear that you can't

just blindly apply the bilinear

transform to any function with any

sample time and get a good match in the

Z domain so what's going on here to

answer this we must first talk about how

Z equals e to the St maps from the S

plane to the Z plane I'll just set up a

blank s plane and blank Z plane and

let's try to convert a few points for a

sample time of two seconds we'll start

with the origin s equals zero plus zero

J this gives us Z equals e to the 0 or 1

so the origin in the S plane maps to the

point 1 0 in the Z plane let's try s

equals minus 1 plus 0 J and this maps to

zero point 1 3 5 3 in the Z plane this

is inside the unit circle and fairly

close to the origin if we head up the

imaginary access to s equals 0 plus J we

see that this maps to negative zero

point four two plus zero point 9 1 j

which lies on the unit circle in the z

plane and finally let's try a point at s

equals minus 100 which is practically

negative infinity for this system and it

maps to one point three eight times ten

to the minus 87 basically zero so points

near negative real infinity mapped to

the origin in the Z plane

doing math is fine you know it's fine

but I think it helps to be able to

visualize the mapping here's what it's

doing we start with the Cartesian grid

in the s plane the e to the real part of

s scrunches the left half plane inward

to where the entire left half plane

exists between the real values 0 and 1

and the right half plane extend out past

1 to infinity the e to the imaginary

part rotates this scrunched grid around

the origin and it just keeps rotating it

more and more as Omega increases towards

infinity in this way the J Omega line in

the S plane maps to the unit circle in

the Z plane now to help with this

visualization let's look at something

much more precise than my drawings I

wrote a little program that maps a

location in the S plane on the left to

the Z plane on the right using the

transformation Z equals e to the st and

in this case T is set to 2 seconds

you'll note that the left half plane is

gray and this maps entirely into the

grey unit circle in the Z plane the

right half plane the white part is

outside of the unit circle the origin of

the S plane the red dot is mapped to the

point 1 0 the black dot now if I move

that red dot to about minus 1 0 you can

see the black dot moves inward towards

the origin and if we keep going more

negative in the S plane it pretty much

Maps right onto the origin of the z

plane and if we start heading up the J

Omega axis the black dot follows the

perimeter of the unit circle that's

because the J Omega axis maps to the

unit circle in the Z plane now here's

the interesting part if we start at a

fairly negative J Omega value and just

start increasing it you'll notice the

black dot rotates around the origin

again and again this is the effect the

imaginary part has on the mapping it

just rotates the mapping around the

origin while the real part dictates how

far you are from the origin now keep

that in mind as we contrast this

behavior with the bilinear transform

we'll start by drawing out a blank s

plane and Z plane again and then convert

the same four points that we did with e

to the S T but this time using the

bilinear transform we'll start with the

origin in the S plane which maps to the

point plus 1 in the Z plane the same as

we got before but this time the point s

equals minus 1 maps exactly to the

origin in the Z plane and the point s

equals 0 plus J maps to J in the Z plane

now that's still on the unit circle I

mean the J Omega lines still maps to the

unit circle using the bilinear transform

but the resulting location around the

circle has changed finally our point at

basically negative infinity maps to

about minus 1 in the Z plane the math is

the math but just like we did before

let's attempt to visualize this mapping

and see if we can use that visualization

to understand how these two transforms

project the S plane onto the Z plane

differently again we start with the

Cartesian grid in the S plane but

instead of scrunching it in from the

negative real line we're going to pinch

it in at three different locations we're

gonna bend the J Omega axis down from

the top and up from the bottom so that

the tips of the line at positive and

negative infinity meet at the point

minus one similarly we're going to pinch

the negative real line so that it's

infinity also meets at minus 1 with this

mapping there are no rotations just a

pinching of the positive and negative

imaginary infinities and negative real

infinity so that they all meet at minus

1 again the pinching is done such that

the J Omega line wraps around of the

unit circle now let's see what this

looks like with a bit more precision

this is the same idea as before but now

we're going to map from the S plane to

the Z plane with the bilinear transform

still with a sample time of 2 seconds

just like with E to the s T the bilinear

transform maps the entire left

half-plane to inside of the unit circle

but now look at the way that this grid

in the Z plane is shaped compared to the

one that we saw with E to the s T the

grid gets more dense towards -1 rather

than the origin and if you follow the

real or imaginary lines in the S plane

they all curve towards this minus 1

point in the Z plane the origin in the S

plane mapped to the point 1 0 as we

expect and as we move along the negative

real line we can watch the black dot

also move along the negative real line

closer and closer to minus 1 but never

quite reaching it and if we move up and

down the J Omega line you can see the

black dot travel along the unit circle

towards minus 1 as well so all

infinities in this left hemisphere in

the S plane mapped to minus 1 in the Z

plane a bit different than e to the St

for sure ok what does this difference in

warping have to do with the frequency

well let's take the bilinear transform

and recognize that s is defined as Sigma

plus J Omega Sigma is the exponential

term which doesn't impact frequency so

we can just look at J Omega for

frequency information we can plug J

Omega into the bilinear transform for s

and we get this equation here I wrote

Omega sub a for analog frequency because

this was the true s domain location

which defined the analog frequency that

we wanted this was our starting point

however to find out what this new

frequency is that we just created we can

convert back to the S plane using the

definition of Z or e to the S T for this

s I'll replace it with Omega sub D for

the discrete frequency that resulted

from the bilinear transform now we can

just solve for Omega D as a function of

Omega a and you can tell from this

equation that Omega D does not equal

Omega a we've moved the frequency by

using the bilinear transform we can

simplify this equation but for now let's

try a few points with this unsimplified

equation and see how the frequencies

move but before we do that let's

recognize that for a sample time of 2

seconds or a sample frequency of 1/2

Hertz the Nyquist frequency is 1/4 Hertz

or half the sample frequency this puts

the Nyquist frequency at PI over 2

radians per second and we're going to

use this shortly for our first point

let's look at an analog frequency of 0.1

radians per second this is about 6% of

the Nyquist frequency which is

relatively slow let's plug this into our

equation and rather than solve it by

hand let's just quickly solve it in

MATLAB first I'll define the Omega a

variable type out the equation for Omega

D and we get a discrete frequency of

zero point zero nine nine seven radians

per second you'll notice that this is

very close to our starting analog

frequency and it's what we saw when we

plotted the discrete and analog notch

filters on top of each other when the

critical frequency was zero point one

radians per second now let's try an

analogue frequency of zero point seven

radians per second this is a relatively

fast frequency or about 45% of the

Nyquist frequency again we set up our

equation to solve for Omega D and then

we go back to MATLAB change our Omega a

variable and solve for Omega D the

discrete frequency is not zero point

seven radians per second but about zero

point six one which again if you recall

was the difference we saw between our

two notch filters when the critical

frequency was 0.7 radians per second

interesting so at this point you might

be asking what happens if we design a

filter above the Nyquist frequency well

let's try it with an analog frequency of

2.5 radians per second well zip through

the math real quick you can double-check

this on your own but we find that it's

about one point one nine radians per

second which is really different than

our goal

it was mapped to just below the Nyquist

frequency in fact it doesn't matter how

high the analog frequency is when you

map that system using the bilinear

transform it always maps it to just less

than the Nyquist frequency that's the

result of the mapping not rotating

around the origin but rather pinching it

down to minus 1 that minus 1 point is

the Nyquist frequency in the Z plane ok

enough of that the question is what can

we do about this

well remember I said that this equation

can be simplified the math is pretty

straight forward to do the

simplification but it's a bit much to

write out here instead I direct you to

follow along with the Wikipedia page

where they do a pretty good job of

deriving the simplified equation and

it's this equation here where Omega D is

this arctan function of Omega a alright

we are in the homestretch now I'll

rewrite the frequency warping equation

here and if we plot this equation we

find something very telling as we

increase Omega a Omega D continues to

increase as well but asymptotically to

PI over T which is the Nyquist frequency

in our case it's PI over 2 since our

sample time is 2 seconds and what you'll

notice is that this equation has nearly

a one-to-one linear relationship with

Omega a and Omega D at low frequencies

compared to the Nyquist frequency and as

we increase Omega a Omega D differs from

it more and more just like we saw with

our design so ask the question again

what can we do about this well we can

pre-warp Omega a essentially this means

that we are artificially moving our

design frequency before mapping it with

the bilinear transform so that after the

mapping the critical frequency is in the

correct spot to do this we apply the

tangent equation to Omega a first

luckily there is a way to do this pre

warping at the exact same time that we

do the bilinear transform with a

modified version of the transform

instead of having a multiplier of 2 over

T we multiply Z minus 1 over Z plus 1

with Omega naught divided by the tangent

of Omega naught times T over 2

and you know we could do this mapping by

hand but again we could just do it in

MATLAB as well once again we'll set up

our variables and define our S domain

notch filter at zero point seven radians

per second

then we can convert it to the Z domain

with the bilinear transform but this

time we'll apply a pre warping at zero

point seven radians per second during

the conversion remember when we did this

without the pre warping our discrete

knotch filter came back at zero point

six one radians per second so let's see

how this one does and it's perfect look

at this if we zoom in we can see that

with the pre warping we were able to

create a notch filter with the desired

critical frequency okay that's pretty

much what I wanted to cover in this

video just that we can use the bilinear

transform to convert analog filters into

discrete filters but we have to

understand how frequencies are warped

during this process and account for them

during the conversion so thanks for

watching and following along with this

extremely long video but before I leave

you I promised that we would end with a

pretty sweet animation well I think it's

sweet anyway once again I present the Z

plane and it's unit circle I'm going to

use this to show you exactly what is

happening when we apply a pre-warp at a

specific frequency I'm plotting four

poles here the two on the far right are

the two notch filter poles for the

critical frequency of 0.1 radians per

second and the two on the left are the

notch filter poles for the critical

frequency of 0.7 radians per second

these are the true locations of the

poles when converted using Z equals e to

the St the grid pattern you might

recognize as the mapping that occurs

when using e to the S T now I'll add on

top of this the exact same for poles but

this time mapped using the bilinear

transform notice the two on the right

overlay nicely this agrees with what we

saw before which was that relatively low

frequencies compared to the Nyquist

frequency don't get shifted by very much

however look at the two poles for the

zero point seven radians per second

critical frequency not close and also

exactly what we saw before now watch as

I apply a

pre-warp at a larger and larger critical

frequency the whole red grid starts to

move and stretch and moving the poles

along with it and if I stop exactly at

zero point seven radians per second you

can see that all four poles for the zero

point seven Radian per second notch

filter line up with each other

now in doing so the four zero point one

Radian per second poles no longer line

up but that's the penalty we have to pay

with the bilinear transform we have to

choose the frequencies that matter most

to us but all in all the bilinear

transform is pretty awesome okay that is

the end of this video

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