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

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From the course by Georgia Institute of Technology

Introduction to Electronics

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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.

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

Â In your previous lesson,

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

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

Â introduce bandpass filter characteristics.

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

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

Â the first-order low pass and

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

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

Â as you've seen them before.

Â Now we know the relationship between omega and

Â frequency, as omega is equal to 2 Pi F.

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

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

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

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

Â to write it in terms of F.

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

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

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

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

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

Â jf over f naught + 1.

Â A completely equivalent way of writing this transfer function for

Â the low pass filter.

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

Â We have again that omega is equal to 2pif.

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

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

Â the high pass-hold.

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

Â f note to put in terms of f.

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

Â Now I'll make the substitutions.

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

Â times j that I write 2 pi f instead of omega

Â and here I have j2 pi f and

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

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

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

Â So, this would be equal to

Â 2 pi fnot K times J 2 pi

Â f divided by 2 pi f nought

Â divided by jf over f0 + 1.

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

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

Â Which means that H of f is equal to Kpb times

Â times jf over f not

Â divided by jf over f nought plus 1.

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

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

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

Â filters and I'm going to use these expressions for

Â the transfer functions in the remainder of the lesson.

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

Â previous lessons.

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

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

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

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

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

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

Â A low pass filter characteristic,

Â where frequencies in this range are passed unattenuated.

Â And frequencies in this range are attenuated.

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

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

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

Â Where higher frequencies are passed unattenuated, but

Â the lower frequencies are attenuated.

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

Â by cascading simpler filters.

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

Â as the input to the next filter.

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

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

Â So inside this block could be our op-amp

Â first order low pass filter from the previous slide and

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

Â Letâ€™s label The voltages at various nodes in this block diagram.

Â This would be the output of the first filter.

Â This is the input of the second filter and

Â this is the overall output of the cascaded filter.

Â Now by definition, the transfer function of the cascade,

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

Â And we can write that as

Â Vout1 / Vn times the out,

Â divided by the In2 because,

Â due to the cascade topology,

Â VOut1 in equal to VIn2.

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

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

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

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

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

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

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

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

Â the cascade,

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

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

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

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

Â the input voltage times the magnitude of the transfer function.

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

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

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

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

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

Â located at the frequency one hertz and having amplitude one.

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

Â filter, so the low pass filter has no effect.

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

Â of the high pass filter.

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

Â pass filter.

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

Â where it's attenuated by a factor of 0.1.

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

Â a sin wave of amplitude 0.1 and frequency 1.

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

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

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

Â filter, By a factor of 0.1.

Â This signal is applied to the high pass filter

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

Â And because of the gain of one,

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

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

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

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

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

Â so it's unaffected.

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

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

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

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

Â And then attenuating down to the point one level here.

Â This is the transfer function of the bandpass filter

Â where frequencies within this band are passed unattenuated but

Â frequencies outside the band here and here are attenuated.

Â So here I have drawn the bode magnitude plot for

Â the cascaded filter we have been examining.

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

Â determine the characteristics of the bandpass filter.

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

Â their cut off frequencies associating with the bandpass filter.

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

Â a low pass filter.

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

Â a factor of 0.707.

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

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

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

Â I draw a horizontal line at that level and

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

Â It intersects in two places.

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

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

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

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

Â So bandwidth is the distance between these two frequencies or

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

Â Now another frequency associated with the characteristics of a bandpass

Â filter is the center frequency.

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

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

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

Â Another parameter it is defined for

Â bam test filters is the quality factor or Q.

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

Â center frequency and the band width.

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

Â Now for a fixed center frequency,

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

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

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

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

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

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

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

Â first order high pass.

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

Â transfer functions.

Â And I've written them out here,

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

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

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

Â For the low pass filter as F-L-P and Iâ€™ve written F nought for

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

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

Â K-D-C and F-L-P for

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

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

Â are three parameters, K.

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

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

Â those three parameters in terms of the individual filter parameters.

Â And I have given you the results here.

Â In an additional extra lesson in this module,

Â I derive these from the algebraic expressions.

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

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

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

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

Â We have, remember how we have formed this filter.

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

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

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

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

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

Â then we call it f sub u.

Â Now it turns out that if here is f sub hp

Â and here is F sub LP the two counter frequencies.

Â If these frequencies are far apart.

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

Â So approximately equal to f sub hp

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

Â So this would be approximately equal to f sub lp.

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

Â If not, you can't make these approximations.

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

Â there is a limitation on the quality factor.

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

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

Â And Q is equal to one-half

Â when flp is equal to fhp.

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

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

Â There's some limitations that you need consider.

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

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

Â And we also examined bandpass filter characteristics.

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

Â So thank you, and until next time.

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