I have no clue on how to realize an op amp circuit from a given transfer function. Attached herewith is a problem and its solution. Can someone please help me understand how this problem is solved? The textbook which has this question doesn't have much explanation on this topic. I couldn't find any similar problems on internet also.

 

Is that diagram at the top of the second image supposed to implement the transfer function given in the first image?

 

Hi,

Here is one approach to get from the transfer function to a block diagram, and then you would go from the block diagram to the circuit. So this way you would handle the conversion in two phases.

Starting with the transfer function:
G/(s+A)

which means that is the ratio of output over input:
Vo/Vi=G/(s+A)

We want to be able to quickly visualize the block diagram so we change the form of this equation a little by expanding it and referencing everything to the output Vo. Start by multiplying by Vi, and we get:
Vo=Vi*G/(s+A)

Next multiply out the denominator s+A and we get:
Vo*(s+A)=Vi*G

Next expand that:
Vo*s+Vo*A=Vi*G

Now we divide by s because we want to be able to visualize each integrator, so we get:
Vo+V0*A/s=Vi*G/s

then subtract any extra terms on the left so we get Vo alone, we get:
Vo=Vi*G/s-Vo*A/s

This is the equation we were after, but we can notice that we can factor out (1/s) and get:
Vo=(Vi*G-Vo*A)*(1/s)

This shows us that we have only one integrator as an integrator is represented by (1/s). But back to the previous equation:
Vo=Vi*G/s-Vo*A/s

and with the integrator shown as a factor of each term rather than in the denominator:
Vo=Vi*G*(1/s)-Vo*A*(1/s)

What we do now is look at the equation and note that the contribution to the output Vo from the input Vi is:
Vo=Vi*G*(1/s)

This tells us that the path from Vi to Vo must include a gain of G and a single integration.

Looking at the previous equation we also see that the contribution to the output Vo from the output Vo is:
Vo=-Vo*A*(1/s)

and this tells us that the path from the output back to the output must include a gain -A and a single integration.

From this information we can draw a block diagram that includes two gains and a single integrator.

I will draw a diagram of the block diagram and post it in a minute or two.

Note that the last equation obtained Vo from Vo, so Vo must be the very output. If you intend to insert a gain before the output (as in that block diagram shown in your paper of -10) then you may need to adjust A. The block diagram will change slightly and that will be the second block diagram i'll show in the drawing.

It would be wise to understand how to get from the equation to the block diagram first, then from the block diagram to the circuit next.

 

Thanks for the detailed explanation MrAl.

How do i get the op amp equivalent from the block diagram? That is that part that i don't get..

Hi,

Here is one approach to get from the transfer function to a block diagram, and then you would go from the block diagram to the circuit. So this way you would handle the conversion in two phases.

Starting with the transfer function:
G/(s+A)

which means that is the ratio of output over input:
Vo/Vi=G/(s+A)

We want to be able to quickly visualize the block diagram so we change the form of this equation a little by expanding it and referencing everything to the output Vo. Start by multiplying by Vi, and we get:
Vo=Vi*G/(s+A)

Next multiply out the denominator s+A and we get:
Vo*(s+A)=Vi*G

Next expand that:
Vo*s+Vo*A=Vi*G

Now we divide by s because we want to be able to visualize each integrator, so we get:
Vo+V0*A/s=Vi*G/s

then subtract any extra terms on the left so we get Vo alone, we get:
Vo=Vi*G/s-Vo*A/s

This is the equation we were after, but we can notice that we can factor out (1/s) and get:
Vo=(Vi*G-Vo*A)*(1/s)

This shows us that we have only one integrator as an integrator is represented by (1/s). But back to the previous equation:
Vo=Vi*G/s-Vo*A/s

and with the integrator shown as a factor of each term rather than in the denominator:
Vo=Vi*G*(1/s)-Vo*A*(1/s)

What we do now is look at the equation and note that the contribution to the output Vo from the input Vi is:
Vo=Vi*G*(1/s)

This tells us that the path from Vi to Vo must include a gain of G and a single integration.

Looking at the previous equation we also see that the contribution to the output Vo from the output Vo is:
Vo=-Vo*A*(1/s)

and this tells us that the path from the output back to the output much include a gain -A and a single integration.

From this information we can draw a block diagram that includes two gains and a single integrator.

I will draw a diagram of the block diagram and post it in a minute or two.

Note that the last equation obtained Vo from Vo, so Vo must be the very output. If you intend to insert a gain before the output (as in that block diagram shown in your paper of -10) then you may need to adjust A. The block diagram will change slightly and that will be the second block diagram i'll show in the drawing.

It would be wise to understand how to get from the equation to the block diagram first, then from the block diagram to the circuit next.

 

Yes WBahn, according to the text book, that is the canonical representation of the transfer function.
I don't really think that is correct though....

Is that diagram at the top of the second image supposed to implement the transfer function given in the first image?

 

Thanks for the detailed explanation MrAl.

How do i get the op amp equivalent from the block diagram? That is that part that i don't get..

Hi again,

The first thing you have to understand is how the op amp can be made to function as an integrator. We should look at this because the op amp as a simple gain block is simple, and we need at least one integrator.

First, an ideal op amp can be represented with the following equation:
Vo=(vp-vn)*Aol

where vp is the voltage at the non inverting terminal and vn is the voltage at the inverting terminal.

If we add an impedance in the feedback circuit and another impedance in series with the input and keep vp grounded (vp=0) and allow Aol to go to infinity, this simplifies to:
Vo=-Vi*Zf/Zi

Now if the input Zi is just a resistor Ri, and the feedback is just a resistor Rf, then this simplifies to:
Vo=-Vi*Rf/Ri

so we just have an inverting gain amplifier.

If the input impedance is just a resistor and the feedback impedance is a capacitor, then we get an integrator:
Vo=-Vi*(1/(s*C))/Ri

which simplifies to:
Vo=-Vi/(s*Ri*C)

So what we have here now are two 'gain' blocks:
Pure gain: Vo=-Vi*Rf/Ri
Integrator: Vo=-Vi/(s*Ri*C)

Using these two as 'gain' blocks, we try to arrange them into the transfer function we need, knowing the equation developed in my previous post which makes each gain and integrator more transparent in the equation.

Also interesting however, is if we put a capacitor in parallel with a resistor, we get:
Zp=Rp/(s*C*Rp+1)

If we use that as feedback and a resistor in series with the input, we get:
Vo=-Vi*Zp/Ri

which expands into:
Vo=-Vi*Rp/(Ri*Rp*s*C+Ri)

and with a little division we can turn that into:
Vo=-(Vi/(Ri*C))/(s+1/(Rp*C))

and this is already in the form Vo=G/(s+A) so we might work that as the circuit, or we could just use gain blocks as it looks like your attachment did..

So you see there are a few ways to realize the circuit once we have the block diagram, and you might explore some of these ways so you can become familiar with them. Doing it like this allows us to realize any transfer function, at least in theory.

The main thing to remember is that after you get the drawn out equation and you draw the circuit, you should be able to trace input to output and output back to output (feedback) and get the same equation again just by adding the individual contributions.

Also note that the way we do it here we always ground the non inverting terminal, so vp=0 for all these circuits. That makes the synthesis a little more straightforward. If we wanted to include non grounded non inverting terminal designs then we would start by analyzing a non grounded non inverting terminal gain block and try to incorporate that into the design. This sometimes simplifies the overall design by reducing the parts count (ie one op amp instead of two for example).
The analysis for the non grounded non inverting terminal op amp is not difficult so we could look at that too if you like.