A Supplementary Notes on Block Diagrams

8/8/2019 A Supplementary Notes on Block Diagrams

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ECEN2260 Supplementary notes on block diagrams

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Supplementary Notes on Block Diagrams

R.W. Erickson

IntroductionWe have already viewed a circuit as a box containing an input uin(s) and an output

uout(s). The transfer function H(s) describes the operation which the circuit performs on the

input to obtain the output. The signal processing function performed by the circuit is then

represented as in Fig. 1. In this diagram, the lines represent signals while the boxes represent the

transfer function of a circuit or other funtional block. The details of the inner workings of the

circuit are ignored in this figure, and only the input/output relation is shown.

H(s)uin (s ) uout (s )= H (s ) u in (s )

Fig. 1. Block diagram of a signal processing circuit.

When dealing with a system containing many elements, it is a good idea to break the

circuit down into its basic functional blocks. The transfer functions of each block are described,

and the details of the inner workings of the blocks are ignored. Input and output impedances and

other properties of the blocks are also not represented. This results in a high-level, simpler

functional description of the system, which can be more easily understood. Once this is done, it

then remains to figure out what happens when the blocks are connected together as in the actual

circuit. Doing so is the object of these notes.

Cascaded blocks

u1 (s ) H1(s ) H 2(s )u2(s ) u3(s )

Fig. 2. Cascade connection of two blocks.

Consider next the cascade connection of two blocks, as in Fig. 2. The output of the first

block is connected to the input of the second block. We can write:u2(s) = H1(s) u1(s) (1)u3(s) = H2(s) u2(s) (2)

Hence,u3(s) = H2(s) H1(s) u1(s) (3)


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So the cascaded blocks can be reduced to a single block with transfer function H2(s) H1(s), as

diagrammed in Fig. 3.

u1 (s ) H1(s ) H 2(s )u3(s )

Fig. 3. Reduced block diagram which represents the cascade connection of blocks H1(s) and H2(s).

Some care must be taken when cascading blocks, to ensure that the transfer functions of

the two blocks do not change when the blocks are connected together. If the transfer functions

do change, then they must be modified to properly account for the change.

Example: Effect of cascading simple R-C low-pass filters

Consider the R-C network of Fig. 4. The output V2(s) is

V2(s) = V1(s) 11 + sR1C1 (4)

and hence the transfer function is

H1(s) =V2(s)

V1(s)= 1

1 + sR1C1(5)

So the circuit can be represented by the block diagram given in Fig. 5.

+

R1v1(t)

+

v2(t)

C1R2

+

v3(t)

C2

Fig. 6. Cascade connection of two R-C low-pass filters.

Now let us cascade a second R-C section as in Fig. 6, to obtain a two-pole low pass filter.

We would like to represent this circuit using the block diagram of Fig. 7.

+

R1v1(t)

+

v2(t)

C1

Fig. 4. Simple R-C low-pass filter example.

H1(s)V2(s)V1(s)

H1(s) =1

1 + sR1C1

Fig. 5. Block diagram representation of the R-C low-pass filter.


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V1 (s) H1(s) H 2(s)V2(s) V3(s)

Fig. 7. Block diagram of the cascaded R-C filter circuits.

The transfer function H2(s) can be easily found. Given V2 in Fig. 6, we can solve for V3 to find

H2(s) =V3(s)

V2(s)= 1

1 + sR2C2(6)

But H1(s) is no longer given by Eq. (5). The addition of the second R-C section loads down the

first section, and causes the transfer function H1(s) to change. Now, the impedance (R2 + 1/sC2)

appears effectively in parallel with capacitor C1. The voltage divider relation for the first R-C

section becomes

V2(s) = V1(s)

1sC

1

|| R2 +1

sC2

R1 +1

sC1|| R2 +

1sC2

(7)

So the transfer function H1(s) becomes

H1(s) =V2(s)

V1(s)=

1sC1

|| R2 +1

sC2

R1 +1

sC1|| R2 +

1sC2

(8)

The cascade connection makes H1(s) much more complicated.

It is usually a good idea to design your circuits so that the blocks do not interact, and so

that loading effects such as the one illustrated above are not significant. The resulting circuits

are much easier to understand, and there is greater likelihood that the analysis will be free of

algebra errors and that the final design will work as intended. Other people can also more easily

understand how your design works, which is a good idea in a team environment.

It is possible to choose the element values in the circuit above so that Eq. (8) and Eq. (5)

are essentially equal. By multiplying out Eq. (8) to express it in factored pole-zero form, one

obtains

H1(s) =1 + sR2C2

1 + s R1C1 + R2C2 + R1C2 + s2 R1C1R2C2 (9)

If R2 >> R1 or C2


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which is the original H1(s) in Eq. (5). So if R2 >> R1 or C2


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block, to push a pick-off point through a block, and to reduce a feedback connection to a single

block.

The last rule, for reducing the feedback connection to a single block, requires some

analysis. We can write

Table 1. Rules for manipulation of block diagrams

Original Equivalent

V1(s)

H1(s)V2(s)

H 2(s)

V1 (s)

H1(s ) H 2(s)

V2(s)

V1 (s)

V2(s)

V3(s)

+H1(s)

V1 (s)

V2(s)

V3(s)

+ H1(s)

1H1(s)

V1 (s)

V2(s)

V3(s)

+ H1(s)

V1 (s)

V2(s)

V3(s)

+H1(s)

H1(s)

V1 (s) V2(s)

V3(s)

H1(s) H1(s)V1 (s) V2(s)

V3(s)1H1(s)

H1(s)V

1(s) V2(s)

V3(s)

V1 (s)V2(s)

V3(s)H1(s)

H1(s)

V1 (s) V2(s)V3(s)

+ G(s)

H(s)

V1(s) V3(s)G(s)

1 + G(s) H(s)


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V2(s) = V1(s) H(s) V3(s) (14)

andV3(s) = G(s) V2(s) (15)

Now eliminate V2(s) by plugging Eq. (14) into Eq. (15):

V3(s) = G(s) V2(s) = G(s) V1(s) H(s) V3(s)(16)

Solve for V3(s):

V3(s) = V1(s)G(s)

1 + G(s) H(s)(17)

which yields the equivalent block shown in the table.

Some examples

Example 1: reduction of a block diagram with two feedback loops

V2(s)

+ G2(s)

H2(s)

V1 (s)

+ G1(s)

H1(s)

Fig. 11. Example 1.

Consider first the block diagram given in Fig. 11. We can reduce this to a single block,and therefore find the transfer function from V1(s) to V2(s), by the following steps. First, we

reduce the feedback connection of the G2(s) and H2(s) blocks, yielding the diagram of Fig. 12.

V2(s)V1 (s)

+ G1(s)

H1(s)

G2(s)

1 + G2(s) H2(s)

Fig. 12. Step 1 in simplification of Fig. 11.

Next, we combine the cascaded blocks to obtain Fig. 13.

V1 (s) V2(s)V3(s)

+ G(s)

H(s)

Fig. 10. Feedback connection of blocks.


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V2(s)V1 (s)

+

H1(s)

G1(s) G2(s)

1 + G2(s) H2(s)

Fig. 13. Step two in the simplification of Fig. 11.

Finally, we reduce the feedback connection, to obtain Fig. 14.

V2(s)V1 (s)G1 G2

1 + G2 H2

1 +G1 G2 H1

1 + G2 H2

Fig. 14. Final step in the simplification of Fig. 11.

The transfer function is therefore

V2(s)

V1(s)=

G1 G21 + G2 H2

1 +G1 G2 H11 + G2 H2

=G1 G2

1 + G2 H2 + G1 G2 H1(18)

Example 2: block diagram of an op-amp summing circuit

++

+

+V1 (s)

V2(s) V3(s)

Z 1(s)

Z 2 (s)

Z 3 (s)V1Z1

V2Z2

V3Z3

Fig. 15. Op amp summing circuit example.

Consider next the op-amp circuit given in Fig. 15. For this circuit, we can write

V3Z3

=V1Z1

+V2Z2

(19)

and hence the output V3 is given by,

V3 = Z3V1Z1

+V2Z2

(20)


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This has the block diagram representation shown in Fig. 16.

V1 (s)

V2(s)

V3(s)

++

1Z1

1Z2

Z3

Fig. 16. Block diagram representation of the op-amp summing circuit.

Example 3: the biquad filter circuit

This is also called a state variable filter. It can yield a bi-quadratic transfer function,

i.e., a transfer function with up to two poles and two zeroes. The poles and zeroes may be

complex if desired. It is free from a tendency to oscillate, and is relatively easy to tune. The

basic circuit is given in Fig. 17.

+

R1

R2

R3

+

C1R4

+

C2

+

R5

R5

V1 (s)

V2(s) V3(s)

V4(s)

Fig. 17. Basic circuit of the Biquad filter.

A block diagram of this circuit can be developed using a method similar to the previous

example. The equations of each op amp circuit are:

V2 = R1

1 + s R1 C1

V1R4

+V4R3

(21)

V3 = 1

s C2V2R2

(22)

V4 = V3 (23)


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A block diagram which corresponds to these equations is given in Fig. 18.

V2(s)

V1 (s)+

+

1R4

R1

1 + s R1 C1 1

s R2 C2

V3(s)

1V4(s)

1R3

Fig. 18. Block diagram of the basic biquad circuit.

Let us manipulate the block diagram, to solve for the transfer function from V 1(s) to V3(s).Figure

18 can be re-drawn as in Fig. 19.

V1 (s)-

+

1R4

R1

1 + s R1 C1 1

s R2 C2 V3(s)

1R3

Fig. 19. The block diagram of Fig. 18, with V3 treated as the output.

We can next combine the cascaded blocks to obtain Fig. 20.

V1 (s)-

+

1R4 V3(s)

1R3

R1s R2 C2 1 + s R1 C1

Fig. 20. Reduction of cascaded blocks in Fig. 19.

Reduction of the feedback connection, and simplification of the transfer function of the resulting

block into standard normalized form, yields Fig. 21.

V1 (s)

V3(s)

1R4

R3

1 + s C2 R2 R3 / R1 + s2 C1 C2 R2 R3

Fig. 21. Reduction of feedback connection of Fig. 20.


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From Fig. 21, it is apparent that the transfer function is

V3V1

=R3R4

1

1 + s C2 R2 R3 / R1 + s2 C1 C2 R2 R3(24)

Therefore, the transfer function contains two poles. Writing this transfer function in the standard

formV3V1

= H01

1 + sQ 0

+ s0

2(25)

and equating coefficients of s yields the following analytical expressions for the transfer function

salient features:

dc gain H0 =R3R4

(26)

corner frequency 0

= 1

C1 C2 R2 R3(27)

Q-factor Q = R1C1

C2 R2 R3(28)

It can be seen that these quantities can be chosen arbitrarily. For example, the dc gain is given

by R3/R4. R2, R3, C1, and C2 determine the corner frequency 0. The Q is set by R1.

A version of the biquad filter which contains both quadratic poles and quadratic zeroes is

given in Fig. 22. By proper choice of the element values, it is possible to design this circuit to

obtain any desired values of pole and zero corner frequencies and Q-factors.


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+

R1

R2

R3

+

C1R4

+

C2

V1 (s)

+

R5

R5

+

output

Fig. 22. Biquad filter circuit whose transfer function contains quadratic zeroes and quadratic poles.

An invalid manipulationA common mistake made by students is to push a pickoff point through a summing node.

The two block diagrams illustrated in Fig. 23 are not equivalent!

V1 (s)

V2(s)

V3(s)

+

V3(s)

V1 (s)

V2(s)

V3(s)

+

V1 (s)

Fig. 23. It is not valid to push a pickoff point through a summing node! The block diagramsillustrated above are not equivalent.

This suggests that, when attempting to simplify block diagrams, one should usually attempt to

push pickoff points away from summing nodes.

A Supplementary Notes on Block Diagrams

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