CHAPTER -2
2.4 Electrical Network Transfer Functions
Example 2.6
Transfer Function—Single Loop via the Differential Equation
PROBLEM: Find the transfer function relating the capacitor voltage, VC(s), to the input voltage,
V(s) in Figure 2.3.
Figure 2.3 RLC Network
SOLUTION:
v(t) = vL(t) + vR(t) +vC(t)
variables from current to charge, i(t) = dq(t)/dt
From the voltage-charge relationship for vC(t),
q(t) = C vC(t)
Taking the Laplace transform assuming zero initial conditions,
𝑉𝑐(𝑠)
1
1
=
=
2
2
𝑉(𝑠)
𝐿𝐶𝑠 + 𝑅𝐶𝑠 + 1 𝐿𝐶𝑠[𝑠 + (𝑅/𝐿)𝑠 + 1/𝐿𝐶]
Example 2.7 Transfer Function—Single Loop via Transform Methods
PROBLEM: Repeat Example 2.6 using mesh analysis and transform methods without writing
a differential equation.
SOLUTION:
Using a mesh equation,
V(s) = (Ls + R + 1/Cs). I(s)
------------ eqn: (1)
Page 1 of 14
The voltage across the capacitor, VC(s)
VC(s) = (1/Cs) I(s) ,
I(s) = Cs VC(s)
------------ eqn: (2)
Substitute eqn: (2) into eqn: (1)
V(s) = (Ls + R + 1/Cs). Cs VC(s) =
[𝐿𝐶𝑠2 + 𝑅𝐶𝑠 + 1].VC(s)
𝑉𝑐(𝑠)
1
1
=
=
2
2
𝑉(𝑠)
𝐿𝐶𝑠 + 𝑅𝐶𝑠 + 1 𝐿𝐶𝑠[𝑠 + (𝑅/𝐿)𝑠 + 1/𝐿𝐶]
---------------------------------------------------------------------------------------------------------------------------------------
Example 2.8
Transfer Function—Single Node via Transform Methods
PROBLEM: Repeat Example 2.6 using nodal analysis and without writing a differential
equation.
SOLUTION:
Using nodal analysis,
Figure 2.3 RLC Network
At Node Vc(s) ,
I(s) = Vc(s) / Zc (s) = Vc(s) / [1/Cs]
(1/Cs) . [V(s) _ Vc(s) ] = (Ls + R) . Vc(s)
(1/Cs) . V(s) = (Ls + R + 1/Cs) . Vc(s) = (1/Cs) [LCs2 + R C s+ 1) . Vc(s)
V(s) = [LCs2 + R C s+ 1) . Vc(s)
𝑉𝑐(𝑠)
1
1
=
=
𝑉(𝑠)
𝐿𝐶𝑠 2 + 𝑅𝐶𝑠 + 1 𝐿𝐶𝑠[𝑠 2 + (𝑅/𝐿)𝑠 + 1/𝐿𝐶]
-----------------------------------------------------------------------------------------------------------------Page 2 of 14
Example 2.9
PROBLEM:
Transfer Function—Single Loop via Voltage Division
Repeat Example 2.6 using voltage division and the transformed circuit.
SOLUTION:
Using voltage division,
𝑉𝑐(𝑠) = [
1
𝐶𝑠
1
𝐶𝑠
𝐿𝑆+𝑅+
] V(s)
The transfer function, VC (s) / V(s) :
𝑉𝑐(𝑠)
1/𝐶𝑠
=
𝑉(𝑠)
𝐿𝑆 + 𝑅 + 1/𝐶𝑠
Example 2.10
PROBLEM:
Transfer Function—Multiple Loops
Given the network of Figure 2.6(a), find the transfer function, I2(s) = V(s)
FIGURE 2.6 a. Two-loop electrical network
SOLUTION:
Using KVL:
For Loop 1,
R1 I1(s) + Ls. [ I1(s) – I2(s)] = V(s)
R1 I1(s) + Ls. I1(s) – Ls. I2(s) = V(s)
(R1+ Ls) I1(s) – Ls. I2(s) = V(s)
For Loop 2,
---------------------- eqn: (1)
R2 I2(s) + Ls. [ I2(s) – I1(s)] + (1/Cs) I2(s) = 0
– Ls. I1(s)] + [ R2 + Ls + (1/Cs)] I2(s) = 0
Taking the Laplace Matrix Form,
|
---------------------- eqn: (2)
R1 + Ls
−𝐿𝑠
I1(s)
1
1 | |
| = |V(s)| = | | . V(s)
−𝐿𝑠
[R2 + Ls + (Cs)] I2(s)
0
0
Page 3 of 14
R1 + Ls
−𝐿𝑠
∆=|
| = (R1+ Ls). [ R2 + Ls + (1/Cs)] – (Ls )2
−𝐿𝑠
R2 + Ls + (1/Cs)
R1 + Ls 𝑉(𝑠)
R1 + Ls
∆2 = |
|= |
−𝐿𝑠
−𝐿𝑠
0
I2(s) = ∆2 / ∆ ,
I2(s) =
1
| = 0 − (−𝐿𝑠) = Ls = Ls. V(s)
0
𝐿𝑠 . 𝑉(𝑠)
1
)]−(Ls )2
Cs
(R1+ Ls)[ R2 + Ls + (
I2(s)
𝐿𝑠
=
𝑅1
𝐿
𝑉(𝑠)
𝑅1. 𝑅2 + 𝑅1. 𝐿𝑠 + ( 𝐶𝑠 ) + 𝑅2. 𝐿𝑠 + (𝐿𝑠)2 + 𝐶 − (𝐿𝑠)2
I2(s)
𝐿𝑠
=
𝑅1
𝐿
𝑉(𝑠)
𝑅1. 𝑅2 + 𝑅1. 𝐿𝑠 + ( 𝐶𝑠 ) + 𝑅2. 𝐿𝑠 + (𝐿𝑠)2 + 𝐶 − (𝐿𝑠)2
=
𝐿𝑠
[𝑅1. 𝑅2. 𝐶𝑠 + (𝑅1 + 𝑅2)𝐿. 𝐶𝑠 2 + 𝐿𝑠 + R1]/Cs
=
𝐿𝑠
[(𝑅1 + 𝑅2)𝐿. 𝐶𝑠 2 + (𝑅1. 𝑅2. 𝐶 + 𝐿)𝑠 + R1]/Cs
=
𝐿𝐶 . 𝑠 2
[(𝑅1 + 𝑅2)𝐿𝐶] . 𝑠 2 + (𝑅1. 𝑅2. 𝐶 + 𝐿). 𝑠 + R1]
The Transfer Function,
𝐺(𝑠) =
I2(s)
𝐿𝐶 . 𝑠 2
=
𝑉(𝑠) [(𝑅1 + 𝑅2)𝐿𝐶] . 𝑠 2 + (𝑅1. 𝑅2. 𝐶 + 𝐿). 𝑠 + R1]
Figure 2.7 Block diagram of the circuit
Example 2.11
PROBLEM:
Transfer Function—Multiple Nodes
Find the transfer function, VC(s) = V(s), for the circuit in Figure 2.6(b), Use
nodal analysis.
FIGURE 2.6 b. transformed two-loop electrical network
SOLUTION:
Using the nodal analysis,
Let, G1 = 1/R1
,
G1 = 1/R2
Page 4 of 14
VL(s)−V(s)
At node VL(s),
𝑅1
1
(R1). [VL(s) −V(s)] +
VL(s)
𝐿𝑠
G1 [VL(s) −V(s)] +
+
VL(s)
𝐿𝑠
+
VL(s)−Vc(s)
𝑅2
=0
1
+ ( ).[VL(s) – VC(s)]= 0
R2
VL(s)
𝐿𝑠
+ G2 [VL(s) – VC(s)] = 0
(G1+ G2) [VL(s) – VC(s)] + (1/Ls) .VL(s) = G1 V(s)
[G1+ G2 +1/Ls] . VL(s) – G2 . VC(s) = G1 V(s)
---------------------- eqn: (1)
1
At node VC(s),
(R2). [VC(s) − VL(s)] + 𝐶𝑠 VC(s)] = 0
− G2 [VC(s) − VL(s)] + 𝐶𝑠 VC(s)] = 0
- G2 VL(s) + (G2+ Cs). VC(s) = 0
---------------------- eqn: (2)
Taking the Laplace Matrix Form,
|
[G1 + G2 + 1/Ls]
– G2
[G1 + G2 + 1/Ls]
∆=|
– G2
VL(s)
−𝐺2
G1
1
||
| = | | V(s) = | | . V(s)
(G2 + Cs) Vc(s)
0
0
−𝐺2
1
| = [G1 + G2 + Ls] . (G2 + Cs) −G2 2
(G2 + Cs)
∆ = G1 C s + G1 G2 + C/L + + G2 C s + G2 2 - G2 2
= [G1 G2 + C/L] + G1 G2 C. s + G2 / Ls
= 1/Ls [ (G1 + G2) LC. s 2 + G1 G2 Ls + Cs + G2]
∆ = 1/Ls [ (G1 + G2) LC. s 2 + (G1 G2 L+C) . s + G2]
1
∆2 = |
G1 + G2 + Ls
−G2
𝐺1
| . 𝑉(𝑠) = 0 - G1 (-G2). V(s) = G1G2. V(s)
0
(Or)
G1 + G2 + 1/Ls 1
∆2 = |
| . 𝑉(𝑠) = 0 - (-G2). V(s) = G2. V(s)
−G2
0
VC(s) = ∆2 / ∆
VC(s) =
G1 .G2 . 𝑉(𝑠)
1/Ls [ (G1 + G2) LC.𝑠2 + (G1 G2 L+C) .s + G2]
Page 5 of 14
Vc(s)
G1 . G2 . 𝐿𝑠
=
𝑠
1
𝑉(𝑠)
LC [ G1 G2 . 𝑠 2 + (G1 G2/C). s + +
G2]
𝐿
𝐿𝐶
G1 . G2 . 𝑠
=
𝑠
1
C [ G1 G2 . 𝑠 2 + (G1 G2 /C) . s + 𝐿 + 𝐿𝐶 G2]
The Transfer Function,
Vc(s)
𝑉 (𝑠 )
=
( G1+ G2).
(G1 G2)
. s
𝐶
(G1 G2 L+C)
G2
𝑠2 +
.s +
𝐿𝐶
𝐿𝐶
Figure 2.7 Block diagram of the circuit
Example 2.12
Transfer Function—Multiple Nodes with Current Sources
PROBLEM:
For the network of Figure 2.6, find the transfer function, VC(s) = V(s), using
nodal analysis and a transformed circuit with current sources.
FIGURE 2.8 Transformed network ready for nodal analysis
SOLUTION:
Using the nodal analysis,
Let, G1 = 1/R1
At node VL(s),
,
G1 = 1/R2
[G1+1/Ls] . VL(s) – G2 . [VL(s) -VC(s)] = G1 V(s)
[G1+ G2 +1/Ls] . VL(s) – G2 . VC(s) = G1 V(s)
----------- eqn: (1)
Page 6 of 14
At node VC(s),
G2 . [VC(s) -VL(s)] + Cs . VC(s) = 0
(G2 + Cs ).VC(s) – G2 . VL(s) = 0
----------- eqn: (2)
Taking the Laplace Matrix Form,
|
[G1 + G2 + 1/Ls]
– G2
[G1 + G2 + 1/Ls]
∆=|
– G2
VL(s)
−𝐺2
G1
1
||
| = | | V(s) = | | . V(s)
(G2 + Cs) Vc(s)
0
0
−𝐺2
1
| = [G1 + G2 + Ls] . (G2 + Cs) −G2 2
(G2 + Cs)
∆ = G1 C s + G1 G2 + C/L + + G2 C s + G2 2 - G2 2
= [G1 G2 + C/L] + G1 G2 C. s + G2 / Ls
= 1/Ls [ (G1 + G2) LC. s 2 + G1 G2 Ls + Cs + G2]
∆ = 1/Ls [ (G1 + G2) LC. s 2 + (G1 G2 L+C) . s + G2]
1
∆2 = |
G1 + G2 + Ls
−G2
𝐺1
| . 𝑉(𝑠) = 0 - G1 (-G2). V(s) = G1G2. V(s)
0
(Or)
G1 + G2 + 1/Ls 1
∆2 = |
| . 𝑉(𝑠) = 0 - (-G2). V(s) = G2. V(s)
−G2
0
VC(s) = ∆2 / ∆
VC(s) =
G1 .G2 . 𝑉(𝑠)
1/Ls [ (G1 + G2) LC.𝑠2 + (G1 G2 L+C) .s + G2]
Vc(s)
G1 . G2 . 𝐿𝑠
=
𝑠
1
𝑉(𝑠)
LC [ G1 G2 . 𝑠 2 + (G1 G2/C). s + 𝐿 + 𝐿𝐶 G2]
=
G1 . G2 . 𝑠
𝑠
1
C [ G1 G2 . 𝑠 2 + (G1 G2 /C) . s + 𝐿 + 𝐿𝐶 G2]
Page 7 of 14
The Transfer Function,
Vc(s)
𝑉 (𝑠 )
=
( G1+ G2).
(G1 G2)
. s
𝐶
(G1 G2 L+C)
G2
𝑠2 +
.s + 𝐿𝐶
𝐿𝐶
Figure 2.8 Block diagram of the circuit
---------------------------------------------------------------------------------------------------------------------------------------
Example 2.13
Mesh Equations via Inspection
PROBLEM: Write, but do not solve, the mesh equations for the network shown in Figure 2.9.
FIGURE 2.9 Three-loop electrical network
SOLUTION:
Using the mesh analysis,
For Loop 1,
V(s)= (I1 -I3). 1 + (1+ 2s). (I1 - I2)
(2+2s) I1 - (1+2s) I2 - I3 = V(s)
For Loop 2,
0 = (1+2s) (I2 - I1) + 4s (I2 -I3) + 3s I2
-(1+2s) I1 + (1+9s) I2 - I3 = 0
For Loop 3,
----------- eqn: (1)
----------- eqn: (2)
0 = (1/s) I3 + 4s (I3 -I2) + (I3 -I1)
-I1 - 4s I2 + (1+4s+1/s) I3 = 0
----------- eqn: (3)
Page 8 of 14
Taking the Laplace Matrix Form,
(2 + 2s) – (1 + 2s)
|– (1 + 2s) (1 + 9𝑠)
−1
−4𝑠
−1
−4𝑠
𝐼1
1
| |𝐼2 | = |0|V(s)
1
0
(1 + 4s + s ) 𝐼3
-----------------------------------------------------------------------------------------------------------------Example 2.14
Transfer Function—Inverting Operational Amplifier Circuit
PROBLEM:
Find the transfer function, Vo(s) = Vi(s), for the circuit given in Figure 2.11.
FIGURE 2.11 Inverting operational amplifier circuit for Example 2.14
SOLUTION:
For Inverting operational amplifier circuit,
Z1(s) =
1
𝐶1 .𝑠+1/𝑅1
Z2(s) = R2 +
1
𝐶2 . 𝑠
=
1
5.6×10−6 𝑠 +1/360×10−3
= 220 × 103 . 𝑠 +
107
𝑠
=
=
360×103
2.016 𝑠 + 1
220×103 . 𝑠 +107
𝑠
Page 9 of 14
𝑉𝑜 (𝑠)
𝑉𝑖 (𝑠)
𝑉𝑜 (𝑠)
𝑉𝑖 (𝑠)
𝑍2 (𝑠)
=−
=
𝑍1 (𝑠)
220×103 . 𝑠 +107
𝑠
360×103
2.016 𝑠 + 1
= −
=−
220×103 . 𝑠 +10
7
𝑠
443.52×103 . s2 + 2.016×107 . 𝑠 + 220×103 . 𝑠 +107
−[
𝑠
= −[
1.232 s2 +56.611 𝑠 – 27.77
𝑠
𝑉𝑜 (𝑠)
The transfer function,
𝑉𝑖 (𝑠)
Example 2.15
] = −1.232 [
= −1.232 [
.
2.016 𝑠 + 1
360×103
]
s2 + 45.95 𝑠 − 22.55
𝑠
s2 + 45.95 𝑠 − 22.55
𝑠
]
]
Transfer Function—Noninverting Operational Amplifier Circuit
PROBLEM:
Find the transfer function, Vo(s) = Vi(s), for the circuit given in Figure 2.13.
FIGURE 2.13 Noninverting operational amplifier circuit for Example 2.15
SOLUTION:
For the Noninverting Operational Amplifier Circuit,
Z1(s) = R1 +
Z2(s) = R2 //
1
𝐶2 . 𝑠
Z1(s) + Z2(s) =
=
1
𝐶1 . 𝑠
=
𝑅1𝐶1 . 𝑠 + 1
𝐶1 . 𝑠
𝑅2
𝐶2 . 𝑠
1
𝑅2 +
𝐶2 . 𝑠
𝑅1𝐶1 . 𝑠 + 1
𝐶1 . 𝑠
Z1(s) + Z2(s) =
+
𝑅2
𝐶2 . 𝑠
1
𝑅2 +
𝐶2 . 𝑠
=
𝑅1𝐶1 . 𝑠 + 1
𝐶1 . 𝑠
+
𝑅2
𝑅2𝐶2 . 𝑠 +1
𝑅1𝑅2𝐶1𝐶2 . s2 + (𝑅1𝐶1+𝑅2 𝐶2+𝑅2 𝐶1 ) . 𝑠 + 1
𝐶1 . 𝑠 (𝑅2𝐶2 . 𝑠 + 1)
Page 10 of 14
𝑉𝑜 (𝑠)
𝑍1 (𝑠) + 𝑍2 (𝑠)
=
𝑉𝑖 (𝑠)
𝑍1 (𝑠)
𝑉𝑜 (𝑠)
𝑉𝑖 (𝑠)
=
=
𝑅1𝑅2𝐶1𝐶2 . s2 + (𝑅1𝐶1+𝑅2 𝐶2+𝑅2 𝐶1 ) . 𝑠 + 1
𝐶1 . 𝑠 (𝑅2𝐶2 . 𝑠 + 1)
=
=
The transfer function,
𝑅1𝑅2𝐶1𝐶2 . s 2 + (𝑅1𝐶1 + 𝑅2 𝐶2 + 𝑅2 𝐶1 ) . 𝑠 + 1
𝐶1 . 𝑠 (𝑅2𝐶2 . 𝑠 + 1)
𝑅1𝐶1 . 𝑠 + 1
𝐶1 . 𝑠
.
𝐶1 . 𝑠
𝑅1𝐶1 . 𝑠 + 1
𝑅1𝑅2𝐶1𝐶2 . s2 + (𝑅1𝐶1+𝑅2 𝐶2+𝑅2 𝐶1 ) . 𝑠 + 1
(𝑅2𝐶2 . 𝑠 + 1) (𝑅1𝐶1 . 𝑠 + 1)
𝑅1𝑅2𝐶1𝐶2 . s2 + (𝑅1𝐶1+𝑅2 𝐶2+𝑅2 𝐶1 ) . 𝑠 + 1
𝑅1𝑅2𝐶1𝐶2 . s2 + (𝑅1𝐶1+𝑅2 𝐶2 ) . 𝑠 + 1
𝑉𝑜 (𝑠)
𝑉𝑖 (𝑠)
=
𝑅1𝑅2𝐶1𝐶2 . s2 + (𝑅1𝐶1+𝑅2 𝐶2+𝑅2 𝐶1 ) . 𝑠 + 1
𝑅1𝑅2𝐶1𝐶2 . s2 + (𝑅1𝐶1+𝑅2 𝐶2 ) . 𝑠 + 1
-------------------------------------------------------------------------------------------------------------------
Page 11 of 14
Chapter - 7
7.8
Steady-State Error for Systems in State Space
Example 7.13
Steady-State Error Using the Final Value Theorem
PROBLEM:
Evaluate the steady-state error for the system described by Eqs. (7.90) for
unit step and unit ramp inputs. Use the final value theorem.
SOLUTION:
Using the final value theorem:
Steady-State Error,
e (∞) = lim 𝑠. 𝐸(𝑠) = lim 𝑠. 𝑅(𝑠)[1 − 𝐶(𝑠𝐼 − 𝐴)−1 . 𝐵]
𝑠→0
𝑠→0
0
−5
1
0
A= | 0
−2 1| ; B = |0| ; 𝐶 = |−1
1
20 −10 1
1 0| ;
𝑠
s I= |0
0
0 0
𝑠 0|
0 s
for unit step: R(s) = 1/s ,
e (∞) = lim 𝑠. 𝐸(𝑠) = lim 𝑠. 𝑅(𝑠)[1 − 𝐶(𝑠𝐼 − 𝐴)−1 . 𝐵]
𝑠→0
𝑠→0
1
= lim 𝑠. ( ) [1 − 𝐶(𝑠𝐼 − 𝐴)−1 . 𝐵] = lim [1 − 𝐶(𝑠𝐼 − 𝐴)−1 . 𝐵]
𝑠→0
𝑠→0
𝑠
𝑠 0
(𝑠𝐼 − 𝐴) = |0 𝑠
0 0
0 −5
1
0
𝑠 + 5 −1
0| - | 0
−2 1| = | 0
𝑠+2
s
20 −10 1
−20
10
0
−1 |
s−1
det (sI -A) = (𝑠 + 5)[ (𝑠 + 2)(s − 1) − (−10)] − (−10)[0 − 20] + 0
= (s + 5) [ s2 + s +2]-20
=
s3+s2+8s+5 s2 +5 s + 40-20
= s3+6s2 +13 s + 20
Z11
𝑎𝑑𝑗(𝑠𝐼 − 𝐴) = |Z21
Z31
Z12
Z22
Z32
Z13 𝑇
Z23 |
Z33
Z11 = (𝑠 + 2)(s − 1) − (−10) = 𝑠 2 + s + 18
Z12 = - [0(s − 1)- 20] = 20
Page 12 of 14
Z13 = 0
Z21 = −[−1(𝑠 − 1) − 0] = s-1
Z22 = (𝑠 + 5) (𝑠 − 1) − 0 = 𝑠 2 + 4s − 5
Z23 = −[(𝑠 + 5)10 −20] = −(10𝑠 + 30)
Z31 = 1-0 = 1
Z32= −[−1(𝑠 + 5) − 0] = (𝑠 + 5)
Z33 = (𝑠 + 5) (𝑠 + 2) − 0 = 𝑠 2 + 7s + 10
𝑇
(𝑠 2 + s + 18)
20
0
2
𝑎𝑑𝑗(𝑠𝐼 − 𝐴) = | (s − 1)
(𝑠 + 4s − 5) −(10𝑠 + 30) |
(s + 5)
(𝑠 2 + 7s + 10)
1
(𝑠 2 + s + 18)
(s − 1)
1
(s + 5)
𝑎𝑑𝑗(𝑠𝐼 − 𝐴) = |
|
20
(𝑠 2 + 4s − 5)
2
0
−(10𝑠 + 30) (𝑠 + 7s + 10)
(𝑠𝐼 − 𝐴)−1 =
𝐶(𝑠𝐼 − 𝐴)
−1
𝑎𝑑𝑗(𝑠𝐼−𝐴)
=
det(𝑠𝐼−𝐴)
. 𝐵 = |−1
= |−1
= (𝑠3
(𝑠 2 +s+18)
(s−1)
1
2
|
(s+5) |
20
(𝑠 +4s−5)
0
−(10𝑠+30) (𝑠 2 +7s+10)
(𝑠 3 +6𝑠 2 +13 s + 20)
1 0|
1 0|
(𝑠 2 +s+18)
(s−1)
1
2
|
(s+5)
|
20
(𝑠 +4s−5)
2
0
−(10𝑠+30) (𝑠 +7s+10)
3
(𝑠 +6𝑠 2 +13 s + 20)
1
(s+5)
|
|
(𝑠 2 +7s+10)
(𝑠 3 +6𝑠 2 +13 s + 20)
= (𝑠3
1
+6𝑠 2 +13 s + 20)
[-1+(s+5) +0]
(s+4)
+6𝑠 2 +13 s + 20)
1 − 𝐶(𝑠𝐼 − 𝐴)−1 . 𝐵 = 1 − (𝑠3
(s+4)
+6𝑠 2 +13 s + 20)
e (∞) = lim [1 − 𝐶(𝑠𝐼 − 𝐴)−1 . 𝐵] = lim [
𝑠→0
For unit step,
0
|0|
1
𝑠→0
=
𝑠 3 +6𝑠 2 +12 s + 16
𝑠 3 +6𝑠 2 +13 s + 20
𝑠 3 +6𝑠 2 +12 s + 16
𝑠 3 +6𝑠 2 +13 s + 20
] = 4/5
e (∞) = 4/5
Page 13 of 14
R(s) = 1/s2
For unit ramp inputs:
e (∞) = lim 𝑠. 𝐸(𝑠) = lim 𝑠. 𝑅(𝑠)[1 − 𝐶(𝑠𝐼 − 𝐴)−1 . 𝐵]
𝑠→0
𝑠→0
1
e (∞) = lim 𝑠. ( 2) [1 − 𝐶(𝑠𝐼 − 𝐴)−1 . 𝐵]
𝑠
𝑠→0
= lim
𝑠→0
1
[1 − 𝐶(𝑠𝐼 − 𝐴)−1 . 𝐵]
𝑠
1
e (∞) = lim
𝑠→0 𝑠
[
𝑠 3 +6𝑠 2 +12 s + 16
𝑠 3 +6𝑠 2 +13 s + 20
]=
1
0
[
16
20
]= ∞
Example 7.13
Steady-State Error Using Input Substitution
PROBLEM:
Evaluate the steady-state error for the system described by the three
equations. (7.90) for unit step and unit ramp inputs. Use input substitution.
SOLUTION:
Using input substitution:
0
−5
1
0
A= | 0
−2 1| ; B = |0| ; 𝐶 = |−1
1
20 −10 1
−5
−1
1 0| ; 𝐴 =| 0
20
1
0 −1
−2 1| = 1 − 0.2
−10 1
e (∞) = 1 + 𝐶 𝐴−1 𝐵
For unit step ,
e (∞) = 1 + |−1
−5
1 0| | 0
20
For unit ramp inputs ,
1
0 −1 0
−2 1| |0| = 1+(-1/5) = 1-0.2 = 0.8
1
−10 1
e (∞) = lim [(1 + 𝐶 𝐴−1 𝐵). 𝑡 +𝐶(𝐴−1 )2 𝐵]
𝑡→∞
e (∞) = lim [ 0.8𝑡 + 2/25] = lim [ 0.8𝑡 + 0.8] = [ 0.8 × ∞ + 0.8] = ∞ + 0.8 = ∞
𝑡→∞
𝑡→∞
(Or)
e (∞) = lim [ 0.8(𝑡 + 1)] = 0.8(∞ + 1) = ∞
𝑡→∞
------------------------------------------------------------------------------------------------------------------Page 14 of 14