lecture 8: o’ transfer function, o’ transfer function ...ee202/lecture/s18/202 lec 8...

Lec 8 Transfer Functions, Sp 18 1

Lecture 8: O’ Transfer Function, O’ Transfer Function,

What for Art Thou O’ Transfer Function

Transfer Functions are lots of fun.

Remember all, when we’re done.

In 301 they’ll reappear

To cause you nights of sleepless fear.

☺

1. Hey Professor Ray, what’s a transfer function H (s) (of a

circuit or system)?

Input Output


ANSWER 1:

According to the box above,

With input-output hovering like a dove,

The transfer function relates the two,

So computations you can do!

ANSWER 2: The box above has NO INTERNAL STORED

ENERGY—NO INITIAL CONDITIONS—NO SOURCES

INSIDE, then

Definition: H (s) =

L output signal⎡⎣ ⎤⎦L input signal⎡⎣ ⎤⎦

in which case

L output signal⎡⎣ ⎤⎦ = H (s)× L input signal⎡⎣ ⎤⎦

Interpretations: 1. H (s) embodies the s-domain properties

of a 202-like circuit.

2. In general, H (s) embodies the dynamics of a linear time-

invariant finite dimensional physical process which includes

many mechanical and chemical systems.


Hey Professor Ray, can I buy an H (s) at Walmart or Radio

Shack?

ANSWER: No. Transfer functions are figments of our

(mathematical) imagination.

Example 1. Source Transformation, V-division, and transfer

function, H (s) =

Vout (s)Iin(s)

, calculation for the circuit below.

H (s) in this case has units of impedance because it is a ratio

of voltage to current.

Step 1. Execute a source transformation.


Step 2. By V-division

Vout (s) =

Z3(s)Z1(s)+ Z2(s)+ Z3(s)

V (s) =Z3(s)Z1(s)Iin(s)

Z1(s)+ Z2(s)+ Z3(s)

Hence

H (s) =

Vout (s)Iin(s)

=Z3(s)Z1(s)

Z1(s)+ Z2(s)+ Z3(s)

Example 2. Show that H (s) =

Vout (s)Vin(s)

= −Yin(s)Yf (s)

.

Step 1. Iin(s) = − I f (s)

Step 2. By virtual ground property of op amp,

Yin(s)Vin(s) = −Yf (s)Vout (s)


Step 3. Therefore H (s) =

Vout (s)Vin(s)

= −Yin(s)Yf (s)

Example 3. Find R1, R2,C1, and R3, R4,C2 so that

H (s) =

VoutVin

= 30(s+5)(s+ 6)

for the circuit below. The

solution is not unique. There are an infinite number of

solutions.

Step 1. Overview of Design Process. Define the output of

the first op amp to be ⌢

V (s) .


H (s) = 30(s+5)(s+ 6)

=VoutVin

=⌢

VVin

×Vout⌢V

= −5s+5

× −6s+ 6

" H1(s)H2(s)

Step 2. Apply the admittance formula of the previous example to each transfer function.

(a) Stage 1: H1(s) =

⌢VVin

= −5s+5

=−YinYf

=−G1

C1s+G2

Possible Solution: G1 = 5 implies R1 = 0.2 Ω. G2 = 5

implies R2 = 0.2 Ω. Finally, C1 = 1 F.

(b) Stage 2: H2(s) =

Vout⌢V

= −6s+ 6

=−YinYf

=−G3

C2s+G4

Possible Solution: G3 = 6 implies R3 =

16

Ω. G4 = 6

implies R4 = 1

6 Ω. Finally, C2 = 1 F.


Exercise. Find other solutions to the above example by using the formulas

H1(s) = −5s+5

==− 1

R1C1

s+ 1R2C1

and

H2(s) = −6s+ 6

==− 1

R3C2

s+ 1R3C2

2. Impulse Response, h(t) : response of a relaxed (no IC)

circuit to an impulse input, δ (t) :

h(t) ! L−1 H (s)L δ (t)⎡⎣ ⎤⎦⎡⎣ ⎤⎦ = L−1 H (s)⎡⎣ ⎤⎦

3. Step Response: response of a relaxed (no IC) circuit to a

step input, u(t) :

Step Response ! L−1 H (s)L u(t)⎡⎣ ⎤⎦⎡⎣ ⎤⎦ = L−1 H (s)

s⎡

⎣⎢⎤

⎦⎥

4. Fundamental (Practical) RELATIONSHIP:

h(t) ! L−1 H (s)⎡⎣ ⎤⎦ = L−1 s

H (s)s

⎡

⎣⎢⎤

⎦⎥= d

dtStep Response⎡⎣ ⎤⎦


Example 4. H (s) = 30

(s+5)(s+ 6).

(a) The impulse response is:

h(t) = L−1 H (s)⎡⎣ ⎤⎦ = L−1 30(s+5)(s+ 6)⎡

⎣⎢

⎤

⎦⎥

= L−1 30s+5

⎡

⎣⎢⎤

⎦⎥− L−1 30

s+ 6⎡

⎣⎢⎤

⎦⎥= 30e−5tu(t)− 30e−6tu(t)

(b) Step response is:

L−1 H (s)s

⎡

⎣⎢⎤

⎦⎥= L−1 30

s(s+5)(s+ 6)⎡

⎣⎢

⎤

⎦⎥

= L−1 1s

⎡

⎣⎢⎤

⎦⎥− L−1 6

s+5⎡

⎣⎢⎤

⎦⎥+ L−1 5

s+ 6⎡

⎣⎢⎤

⎦⎥

in which case

Step Response = 1− 6e−5t +5e−6t⎡⎣

⎤⎦u(t)


(c) Check Relationship:

ddt

Step Response⎡⎣ ⎤⎦ =ddt

1− 6e−5t +5e−6t( )⎡

⎣⎢⎤

⎦⎥u(t)

+ 1− 6e−5t +5e−6t⎡⎣

⎤⎦

ddt

u(t)

= 30e−5t − 30e−6t⎡⎣

⎤⎦u(t)+ 1− 6e−5t +5e−6t⎡

⎣⎤⎦δ (t)

= 30e−5t − 30e−6t⎡⎣

⎤⎦u(t)

which is precisely the impulse response. ☺

Example 5. Find H (s) =

Vout (s)Vin(s)

.

Solution. Step 1. iin(t) = i f (t)


Step 2. From the properties of the ideal op amp:

Iin(s) = Yin(s)Vin(s) and I f (s) = Yf (s) Vout (s)−Vin(s)⎡⎣ ⎤⎦

Step 3. Substitute:

Yin(s)Vin(s) = Yf (s) Vout (s)−Vin(s)⎡⎣ ⎤⎦

Step 4. Manipulate:

H (s) =

Vout (s)Vin(s)

=Yin(s)+Yf (s)

Yf (s)= 1+

Yin(s)Yf (s)

Example 6. Consider the circuit below. Find Yin(s) ,

H1(s) =

VLVin

, H2(s) =

VCVin

, H3(s) =

ILVin

, and if IL is taken

as the s-domain output, find the impulse and step responses

in the time world. Suppose R1 = R2 = 2 Ω, L = 1.5 H, and

C = 1

3 F.


Step 1.

Z1 =1

Cs+ 1R1

=

1C

s+ 1R1C

= 3s+1.5

Z2 =R2Ls

R2 + Ls=

R2s

s+R2L

= 2s

s+ 43

Step 2. Yin =

1Z1 + Z2

= 0.5s+ 4

3⎛⎝⎜

⎞⎠⎟

s+1.5( )(s+1)(s+ 2)

Step 3. H1(s) =

VLVin

=Z2

Z1 + Z2= Z2Yin =

s(s+1.5)(s+1)(s+ 2)

Step 4. H2(s) =

VCVin

=Z1

Z1 + Z2= Z1Yin =

1.5 s+ 43

⎛⎝⎜

⎞⎠⎟

(s+1)(s+ 2)


Step 5. IL = 2

3sVL ⇒ H3(s) =

ILVin

= 23× (s+1.5)

(s+1)(s+ 2)

H3(s) = 2

3× (s+1.5)

(s+1)(s+ 2)=

13

s+1+

13

s+ 2

implies that the impulse response is:

h(t) = L−1 H (s)⎡⎣ ⎤⎦ =

13

e−tu(t)+ 13

e−2tu(t)

Finally, the step response is computed as follows:

H3(s)s

= 23× (s+1.5)

s(s+1)(s+ 2)= 0.5

s−

13

s+1−

16

s+ 2

Step Response = 0.5− 1

3e−t − 1

6e−2t⎡

⎣⎢⎤

⎦⎥u(t)

lecture 8: o’ transfer function, o’ transfer function ...ee202/lecture/s18/202 lec 8...

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