This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

Loading...

From the course by Georgia Institute of Technology

Introduction to Electronics

372 ratings

This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

From the lesson

Op Amps Part 2

Learning Objectives: 1. Examine additional operational amplifier applications. 2. Examine filter transfer functions.

- Dr. Bonnie H. FerriProfessor

Electrical and Computer Engineering - Dr. Robert Allen Robinson, Jr.Academic Professional

School of Electrical and Computer Engineering

Welcome back to electronics, this is Dr. Robinson.

In this lesson, we're going to look at Cascaded First-Order Filters.

In your previous lesson,

you were introduced to op-amp first-order high pass filters.

And our objectives for this lesson are to introduce cascaded filters and

introduce bandpass filter characteristics.

Now before we begin looking at cascaded first order filters, I want to show you

how we can rewrite the transfer functions that you've seen before for

the first-order low pass and

the first-order high pass filters in a different way.

I want to write them in terms of hertz frequency F rather than omega

as you've seen them before.

Now we know the relationship between omega and

frequency, as omega is equal to 2 Pi F.

And from before you have that omega B is = 1 over Tau is = what I'm

going to call omega nought, or the -3db frequency for this filter.

And then using this relationship, we can write that omega nought = 2piFnot.

Now I want to make these substitutions in to this transfer function

to write it in terms of F.

So we have H of f is equal to KDC times 1 over J,

now instead of omega I'm going to write 2Pif.

In the set of tau, I'm going to write over omega nought,

or I can put a 2Pif naught here plus one.

So we get the H of f = Kdc x 1 over.

jf over f naught + 1.

A completely equivalent way of writing this transfer function for

the low pass filter.

Now let's do the same thing for the high pass filter.

We have again that omega is equal to 2pif.

And from before we have omega c is equal to 1 over tau.

I'm going to call that again I'll make a note, which is now the -3Db for

the high pass-hold.

And we have that omega nought which would be equal to 2 pi

f note to put in terms of f.

From before we have that KPB, the passband gain, is equal to K over tau.

Now I'll make the substitutions.

We now have H(f) is equal to and we bring the K up front,

times j that I write 2 pi f instead of omega

and here I have j2 pi f and

we have just like before, a 2 pi f naught here plus 1.

Now I want the quantity sitting upfront here to be the pass band gain.

So I am going to multiply k here by 2 pi f naught like this.

So, this would be equal to

2 pi fnot K times J 2 pi

f divided by 2 pi f nought

divided by jf over f0 + 1.

So I put a 2 pi f0 out front, and a 2 pi f0 here to cancel.

And I can identify this 2 pi f0 K as being KPB, or the passband gain.

Which means that H of f is equal to Kpb times

times jf over f not

divided by jf over f nought plus 1.

You can see in these two transfer functions we have the same denominators

and both transfer functions have been written in terms of f now.

And this is a common way of expressing these transfer functions when analyzing

filters and I'm going to use these expressions for

the transfer functions in the remainder of the lesson.

On this slide I've drawn two circuit that you're familiar with from your

previous lessons.

This op amp circuit implements a first order low pass filter characteristic.

And this op amp circuit implements a first order high pass filter characteristic.

If we solve for the ratio of the output voltage to the input voltage for

this circuit we end up with a transfer function of this form.

And if we plot the magnitude of this transfer function versus frequency,

we end up with a bode magnitude plot of this shape.

A low pass filter characteristic,

where frequencies in this range are passed unattenuated.

And frequencies in this range are attenuated.

Similarly, if we solve for the ratio of output voltage here,

to input voltage here, we wind up with a high-pass characteristic like this.

We plot its magnitude and we get a characteristic of this shape.

Where higher frequencies are passed unattenuated, but

the lower frequencies are attenuated.

Let's look at how we can form a more complicated filter

by cascading simpler filters.

And a cascaded topology means that we apply the output of one filter

as the input to the next filter.

So in this cascade example, I have cascaded a first order low pass filter and

a first order high pass filter to form the overall cascaded filter.

So inside this block could be our op-amp

first order low pass filter from the previous slide and

this could be our op-amp high pass filter from the previous slide.

Let’s label The voltages at various nodes in this block diagram.

This would be the output of the first filter.

This is the input of the second filter and

this is the overall output of the cascaded filter.

Now by definition, the transfer function of the cascade,

"Hc = Vout" the output voltage of the cascade divided by the input voltage Vn.

And we can write that as

Vout1 / Vn times the out,

divided by the In2 because,

due to the cascade topology,

VOut1 in equal to VIn2.

So these two quantities cancel leaving us with VOut over VIn.

Now we can identify this term as the transfer function of the low past filter

and we can identify this as the transfer function of the high past filter.

So that overall the transfer function of the cascade is the product at

the low pass filter transfer function, and the high pass filter transfer function.

So, in general the transfer function of any cascaded filter is equal to

the product of the transfer functions of the individual filters in the cascade.

Let's look at how we can determine the overall bode magnitude plot for

the cascade,

knowing the individual bode magnitude plots of the filters in the cascade.

And remember that the output of a filter, V out, is equal to

the input to the filter, V in, times the transfer function of the filter.

So the magnitude of the output voltage is equal to the magnitude of

the input voltage times the magnitude of the transfer function.

So here I have drawn the magnitude of the transfer function for the low pass filter

and the magnitude of the transfer function for the high pass filter.

And let's assume we applied to the input of the cascade.

A sign wave, a frequency one hertz and amplitude one.

We can represent that sign wave on this bode plot as a spike

located at the frequency one hertz and having amplitude one.

We can see that at that frequency the sin wave is in the pass band of the low pass

filter, so the low pass filter has no effect.

So the output of the low pass filter is applied to the input

of the high pass filter.

This magnitude of one frequency of one hertz sin wave is applied to the high

pass filter.

But it's now in the stop band of the of high pass filter,

where it's attenuated by a factor of 0.1.

So at the output of the high pass filter, we have

a sin wave of amplitude 0.1 and frequency 1.

If we apply to the cascade a sin wave of frequency 10 kHz and amplitude 1.

We can see that at 10 kHz, we are in the stop band of the low-pass filter.

The low-pass filter attenuates the signal before it's applied to the high-pass

filter, By a factor of 0.1.

This signal is applied to the high pass filter

where it's multiplied by this gain of one and passed to the output.

And because of the gain of one,

the high pass filter has no effect on the input signal.

So at the output we have a sin wave of 0.01.

Now if we apply a sin wave of 70 hertz, the 70

hertz sin wave is in the pass band of the low pass filter, so it's unaffected.

The 70 hertz sin wave is in the pass band of the high pass filter,

so it's unaffected.

So the signal passes through the cascade unaffected up to 1.

Now we could do this for lots of frequencies and form the overall booty

magnitude transfer function by connecting the tips of all of these spikes.

We will get a plot that looks something like, passing through this one level.

And then attenuating down to the point one level here.

This is the transfer function of the bandpass filter

where frequencies within this band are passed unattenuated but

frequencies outside the band here and here are attenuated.

So here I have drawn the bode magnitude plot for

the cascaded filter we have been examining.

And I want to look at the plot and identi\fy some of the parameters that

determine the characteristics of the bandpass filter.

Now, just like for a low pass filter and a high pass filter,

their cut off frequencies associating with the bandpass filter.

But because the bandpass filter consists of a combination of a high pass filter and

a low pass filter.

There are actually two cutoff frequencies where the magnitude of the gain is down by

a factor of 0.707.

Or one over root two from the mid-band gain.

For this filter, we have a mid-band gain of one.

We multiply that by one over root two to get 0.707.

I draw a horizontal line at that level and

find the intersection of that line with the bode magnitude plot.

It intersects in two places.

At a low frequency to give us the low frequency cut off, "fl", and

the high frequency to give us the higher frequency cut off, "f sub u".

Now the bandwidth of a bandpass filter is to find us the distance and

frequency between the upper cutoff frequency and the lower cutoff frequency.

So bandwidth is the distance between these two frequencies or

I can write it in equation form as the bandwidth Is equal to fu minus fl.

Now another frequency associated with the characteristics of a bandpass

filter is the center frequency.

And by symmetry, you can see that the center frequency for

this particular bandpass filter is 100 hz and can be written as, f nought is equal

to the geometric mean of the upper cut off frequency and the lower cut off frequency.

Another parameter it is defined for

bam test filters is the quality factor or Q.

Where Q can be defined in terms of the resonates frequency or

center frequency and the band width.

Q is equal to the ratio of the center frequency to the bandwidth.

Now for a fixed center frequency,

you can see that by decreasing the bandwidth we have a higher Q.

So the higher the Q indicates a more selective filter, in that the filter

because it has a narrower bandwidth allows fewer frequencies to pass through.

Finally, the last parameter associated with the bandpass filter is the passband

gain KPB where again for this filter has the value of one.

Now, let's look at the actual mathematical expression for the transfer function of

this cascaded filter that we formed from the first order low pass and

first order high pass.

We know the overall transfer function would be the product of the two individual

transfer functions.

And I've written them out here,

here is the transfer function of the first order low pass filter.

And here is the transfer function of the first order high pass filter.

Now I've written f naught the minus 3 DB frequency.

For the low pass filter as F-L-P and I’ve written F nought for

the high pass filter as F-H-P so that we can distinguish between the two.

Now, each one of this filters has associated with it its own parameters

K-D-C and F-L-P for

the low pass filter and K-P-B and F-H-P for the high pass filter.

Now we know that over all associated with the span pass filters

are three parameters, K.

The resonance frequency, or center frequency, f0, and Q, the quality factor.

And with some algebra and a little bit of manipulation, we can solve for

those three parameters in terms of the individual filter parameters.

And I have given you the results here.

In an additional extra lesson in this module,

I derive these from the algebraic expressions.

But you can see that K, for the overall bandpass filter, can be written this way.

The center frequency for the bandpass filter is written this way.

And Q looks like this in terms of the individual filter parameters.

Now there are couple of things to consider about this filter, or to pay attention to.

We have, remember how we have formed this filter.

We have a high pass filter and a low pass filter.

And we've combined these to get an overall band pass filter that looks like this.

With characteristics, it has an F0, it as a K, and it has a quality filter.

It has it's minus 3db frequency called fl.

And it has another minus 3db frequency the upper minus 3db frequency

then we call it f sub u.

Now it turns out that if here is f sub hp

and here is F sub LP the two counter frequencies.

If these frequencies are far apart.

Then we can make the approximation that F HP is approximately equal to FL.

So approximately equal to f sub hp

and f sub lp would be approximately equal to f sub u.

So this would be approximately equal to f sub lp.

Now, these approximations are only true if these are far apart in frequency.

If not, you can't make these approximations.

Another consideration is that when you form a bandpass filter in this way,

there is a limitation on the quality factor.

The maximum quality factor for a filter formed this way is equal to one half.

So for this type of filter, Q must be less than one half.

And Q is equal to one-half

when flp is equal to fhp.

So it is completely possible to form a bandpass filter by cascading

to first-order filters, but you can't create an arbitrary bandpass filter.

There's some limitations that you need consider.

So in summary, during this lesson, we cascaded a first-order lowpass and

a first-order highpass filter to form a bandpass filter.

And we also examined bandpass filter characteristics.

In our next lesson, we'll look at second-order transfer functions.

So thank you, and until next time.

Coursera provides universal access to the world’s best education,
partnering with top universities and organizations to offer courses online.